Fundamentals of Pharmacy Calculations
Overview
This text is used as the required textbook for a 1 credit hour Pharmacy Calculations course at SIUE School of Pharmacy. It was written specifically for our course. We have shared it here in case you may find all or parts it useful for your needs.
This textbook is provided as is under the Creative Commons BY license. Anyone may copy, display, and/or distribute the book with appropriate citation of the creators. No warranty, express or implied, is granted.
Disclaimer:
The purpose of this textbook is to develop quantitative competence for pharmacy practice.
Nothing in the textbook is intended, nor should it be inferred, as medical advice or opinion.
Healthcare professionals are solely responsible for their application of the information contained herein.
Introduction to the Text
This text is used as the required textbook for a 1 credit hour Pharmacy Calculations course at SIUE School of Pharmacy. It was written specifically for our course. We have shared it here in case you may find all or parts it useful for your needs.
This textbook is provided as is under the Creative Commons BY license. Anyone may copy, display, and/or distribute the book with appropriate citation of the creators. No warranty, express or implied, is granted.
Disclaimer:
The purpose of this textbook is to develop quantitative competence for pharmacy practice.
Nothing in the textbook is intended, nor should it be inferred, as medical advice or opinion.
Healthcare professionals are solely responsible for their application of the information contained herein.
Module 1: Fundamentals of Calculations
Introduction
Module 1 will introduce and review several fundamental topics: Measurement definitions and conversion between units; Decimal places, significant digits (figures) and rounding, exemplified with syringes; Institutional time notation; Abbreviations common in pharmacy and medicine; some Institute for Safe Medication Practices guidelines; and Ratio and proportion techniques.
Module 1A: Measurement Definitions and Conversions
A. Measurement Definitions and Conversions
Here we review some basic conversion factors you will likely know at this point in your education. Recall the relationships between the Greek and Latin prefixes milli-, micro-, and kilo-. In this course, you will only use these exact conversions. Take some time to verify your familiarity with these units and conversion factors.
Volume
• 1 Liter = 1000 mL (milliliters)
• 1 mL = 1000 mcL (microliters)
• 1 teaspoonful (tsp) = 5 mL
• 1 tablespoonful (Tbsp) = 15 mL
• 1 fluidounce (fl oz) = 30 mL
Mass or Weight
• 1 gram = 1000 mg (milligrams)
• 1 mg = 1000 mcg (micrograms)
• 1 mcg = 1000 ng (nanograms)
• 30 grams = 1 ounce Note: This is not an exact conversion - it is an approximate value used in reference to drug products. Do NOT use this conversion for calculating patient body weights.
• 16 ounces = 1 pound (may be abbreviated 1 lb or 1 #)
• 1000 g = 1 kg (kilogram)
• 1 kg = 2.2 pounds
Length
• 1 inch = 2.54 centimeters
• 12 inches = 1 foot
• 100 centimeters = 1 meter
Time
• 60 seconds = 1 minute
• 1 hour = 60 minutes
• 1 day = 24 hours
• 1 week = 7 days
• 1 month = 30 days
Notational Shorthand
• A height listed as X’ Y” implies X feet and Y inches
5’ 10” means 5 feet and 10 inches.
Example Problems:
1.1 Pounds and ounces to kilograms:
A newborn baby weighs 7 pounds and 5 ounces. How many kilograms does the baby weigh?
\(7\:lbs\:5 oz\equiv 7\:lbs\;+\;\frac{5\;oz}{16\;\frac{oz}{lb}}=\frac{7.3125\;lbs}{2.2\frac{lb}{kg}}=3.32\;kg\)
1.2 Ounces to grams:
A pharmacist dispenses 4 ounces of a steroid cream. How many grams were dispensed?
\(4\;oz\;\times\;\frac{30\;g}{1\;oz}=120\;g\)
1.3 Volume for dispensing:
A patient will receive 1 teaspoonful of antibiotic suspension three times a day for 1 week. How many milliliters should you dispense?
\(1\;\frac{tsp}{dose}\;\times3\;\frac{doses}{day}\;\times7\;\frac{days}{week}\;\times\;5\;\frac{mL}{tsp}\;=\;105\;\frac{mL}{week}\)
1.4 Pounds to kilograms:
\(\frac{121\;lbs}{2.2\;\frac{lbs}{kg}}\;=\;55\;kg\)
1.5 Feet and inches to centimeters and meters:
A patient is 5’ 7” tall. Calculate the patient’s height in centimeters and meters.
\(5\text{'}\;7\text{"}\;=\;67\text{"}\)
\(67\text{"}\;\times\;\frac{2.54\;cm}{1\;inch}\;=\;170.18\;cm\;\times\frac{1\;meter}{100\;cm}\;\simeq \;1.7\;meters\)
1.6 Hours to minutes:
A patient’s urine was collected for 24 hours in order to perform a creatinine clearance evaluation. For how many minutes was the sample collected?
\(24\;hours\;\times\;\frac{60\;min}{1\;hr}\;=\;1440\;minutes\)
1.7 Days and months:
You receive a prescription with directions for the patient to take 1 tablet daily. How many tablets should be dispensed to fill a 3-month supply?
\(1\;\frac{tablet}{day}\;\times\;\frac{30\;days}{1\;month}\;\times\;3\;months\;=\;90\;tablets\)
Module 1B: Decimal Places, Significant Digits (Figures), and Rounding
Decimal numbers represent a whole number and a fractional part of that number. A typical example is 3.125. This notation represents three and one hundred and twenty-five thousandths. We know you are very familiar with this notation, and it is included here as a prelude to discussing significant digits and rounding.
For measurments using a graduated device, calculations should be rounded to match the precision of the device. For example, look at the 5 mL oral syringe. Note that the major markings on the barrel run from one to five milliliters. The minor markings are spaced every 0.2 mL. If a patient’s dosage volume calculation resulted in a value of 3.125 mL, how would you explain to the patient or caregiver how to use this device?
An individual cannot accurately withdraw 3.125 mL with this syringe. A decision must be made about using the appropriate number of significant figures. In this case, you should advise the patient to measure 3.2 mL. While the actual calculation answer may be 3.125 mL, there is no practical way to make that measurement. Using 3.2 mL for the volume results in a relative error of 2.4%. If you selected 3 mL, the relative error would be 4%. Pharmacists are practical people.
The example above demonstrates that selecting the number of milliliters to use is aided by the size and demarcation of the available measuring device. Different parenteral and oral syringes have various markings that range from 0.01 mL to 1 mL. You will see these devices routinely, and it will help you to memorize the table.
| Size (mL) | Type (Oral or Parenteral) | Smallest Division (mL) |
| 1 | Oral | 0.01 |
| 5 | Oral | 0.2 |
| 10 | Oral | 0.2 |
| 1 | Parenteral | 0.01 |
| 3 | Parenteral | 0.1 |
| 5 | Parenteral | 0.2 |
| 10 | Parenteral | 0.2 |
Let’s look at another example where the physical situation helps us to determine a practical volume.
Example 1.8: An 81.4 kg patient requires a drug dose of 5 mg/kg.
\(81.4\;kg\times \frac{5\;mg}{kg}=407\;mg\)
Example 1.9: What volume of drug suspension is required if the drug concentration is 100 mg/mL? The calculator answer is:
\(\frac{407\;mg}{100\;\frac{mg}{mL}}=4.07\;mL\)
We do not have a syringe with that degree of accuracy and precision. Based on the available parenteral syringes, you should recommend a dose of 400 mg and a volume of 4 mL.
Before ending this section, several more issues about significant figures and rounding decimal places will be addressed.
Counting significant digits begins with the farthest digit to the left of the decimal place that is not zero and ends with the digit farthest to the right of the decimal place that is not zero. You may have learned a slightly different definition in physics or chemistry, but this definition will work for pharmacy. Let’s look at some problems.
In the above dosing calculation example, the patient weighs 81.4 kg. We could say that the patient’s weight is accurate to 3 significant figures or 1 decimal place. You will develop an understanding of how many significant figures are needed for particular situations.
Now let’s look at the dose. Based on 5 mg/kg, the patient would receive 407 mg of the drug. How many significant figures are represented in the dose? The concentration of the drug in the vial is 100 mg/mL. Recall that volume = mass/concentration. The actual calculated volume is 4.07 mL. How many significant figures are represented in the volume? How many significant figures are represented in the volume we can accurately measure?
Consider this problem from a recent national exam. The formula for a type of “magic mouthwash” is:
| Diphenhydramine syrup 12.5 mg/5 mL | 80 mL |
| Lidocaine oral solution 2% | 30 mL |
| Maalox antacid suspension | 90 mL |
| Total | 200 mL |
The patient is instructed to orally swish 1 Tbsp (15 mL) of magic mouthwash (MM) three times a day.
Example 1.10: How many milliliters of lidocaine solution will the patient receive with each treatment?
\(\frac{200\;mL\;MM}{30\;mL\;Lidocaine\;soln}=\frac{15\;mL\;MM}{x\;mL\;lidocaine\;soln}\)
\(x\;=\frac{15\;mL\;MM\;\times\;30\;mL\;lidocaine\;soln}{200\;mL\;MM}=2.25\;mL\;lidocaine\;soln\)
What number would you tell a physician if they were curious about the amount of lidocaine the patient received with each dose? This topical therapy is intended to relieve pain in the oral cavity. The amount of lidocaine received is not so critical to the treatment as to require an answer to 2 decimal places. You could tell the physician that the patient receives approximately 2 mL per dose.
Some calculations will require different rounding based, in part, on the needed significant figures and the practical circumstances involved with measurement. Your clinical judgment and decision-making skills will sharpen as you advance through the curriculum. The lectures will contain more guidance on this when specific topics are covered.
Example 1.11: How many significant figures are represented in the numbers?
a. 14.75 – 4 significant figures
b. 2.37 – 3 significant figures
c. 12.3 – 3 significant figures
d. 18.789 – 5 significant figures
e. 0.0205 – 3 significant figures
f. 0.00330 – 2 significant figures
g. 0.09105 – 4 significant figures
Example 1.12: Round the numbers to the indicated decimal places.
a. 8.357 (2 decimal places) = 8.36
b. 8.354 (2 decimal places) = 8.35
c. 12.276 (1 decimal place) = 12.3
d. 12.249 (1 decimal place) = 12.2
Example 1.13: Round the numbers to the indicated significant figures.
a. 12.0041 (3 significant figures) = 12.0, but we do not write trailing zeros after a decimal point, so 12. (See section 1F: ISMP guidelines)
b. 0.693147 (3 significant figures) = 0.693
c. 1.0402 (3 significant figures) = 1.04
Module 1C: Patient Weight and Height
Pharmacists typically use patient weights in kilograms with two or three significant digits depending on the weight and age.
Some healthcare institutions will only use two significant digits for patients weighing over 20 kg. When using metric units for height, either centimeters or meters, continue to use three significant digits. You will be using height in the equation for Body Surface Area. Let’s look at a range of examples.
Example 1.14: Weight and height conversions
a) A premature newborn weighs 1 pound and 4 ounces.
Convert this weight to kg. There are 16 oz in 1 lb, so 1 lb 4 oz = 1.25 lb.
\(1.25\;lb\times\;\frac{1\;kg}{2.2\;lb}=0.5681818\;kg\)
Use 0.568 kg for calculations.
b) An infant weighs 19 pounds and 6 ounces. Use 8.81 kg for calculations.
c) A child weighs 48 pounds. Use 21.8 kg.
d) A teenager weighs 133 pounds. Use 60.5 kg.
e) An adult weighs 189 pounds. Use 85.9 kg.
f) An adult weighs 235 pounds. Use 107 kg.
g) A child is 3’ 6” tall. Use 107 cm, 1.07 m.
h) A teenager is 5’7” tall. Use 170 cm, 1.7 m.
i) An adult is 6’3” tall. Use 191 cm, 1.91 m.
Module 1D: Institutional Time (24 Hour Time)
Institutional or 24-hour time is frequently used in healthcare settings to avoid the common 12-hour AM/PM ambiguity.
- Institutional time is a 4-digit number, with the first two digits indicating the hour and the last two representing the minutes.
- Midnight is 0000
- Morning times are identical in 12- and 24-hour time systems.
- Add 12 to afternoon times to convert 12- to 24-hour time systems.
If you would like more information, there are several tutorials available online.
For example, https://www.militarytime.us/learn-military-time/
Some examples:
| 12 hour time | 24 hour time |
| 12:30 AM | 0030 |
| 8:00 AM | 0800 |
| 12:35 PM | 1235 |
| 2:30 PM | 1430 |
| 6:15 PM | 1815 |
| 11:05 PM | 2305 |
Example 1.15: A patient received an IV bolus drug dose at 1800 on February 3 and another at 0445 on February 4. How much time elapsed between the 2 doses?
1800 -> 0000 = 6 hours + 4 hours and 45 minutes = 10 hours and 45 minutes = 10.75 hours
Example 1.16: A surgeon ordered morphine 2 mg IV every 6 hours if needed for pain relief. The patient received his previous dose at 1730. What is the earliest time he may receive another dose?
1730 + 6 hours = 2330
Example 1.17: A patient is to receive gentamicin 80 mg in 50 mL of normal saline over 30 minutes every 8 hours. If the first infusion was started at 1500, when should the next 2 doses be started?
1500 + 8 hr = 2300 the same day; 2300 + 8 hr = 0700 the next morning
Example 1.18: A patient received 300 mg of a drug by IV infusion starting at 0500. The infusion of 500 mL was completed at 0630. What was the infusion rate in mg/h? What was the solution flow rate in mL/min?
The infusion ran from 0500 to 0630 or 1.5 hours.
300 mg/1.5 hr = 200 mg/hr = 5.6 mL/min
Example 1.19: A patient is scheduled for surgery at 0730 on January 16. She is ordered to receive 2 doses of pre-operative medications 14 hours and 6 hours before surgery. When should the doses be given?
0730 - 6 hours = 0130 on surgery day (Jan 16)
0730 - 14 hours = 1730 the day before surgery (Jan 15)
Module 1E: Common Pharmacy Abbreviations
The electronic transmittal of prescriptions has reduced the use of historical Latin abbreviations. The Institute for Safe Medical Practices (ISMP, www.ismp.org) advises against using any abbreviations. However, the organization acknowledges that there are abbreviations that are so commonly used that restricting their use would lead to hardships. We expect you to memorize the abbreviations in the table.
Table 2.2. Common Abbreviations used in Pharmacy
Abbrev | Meaning |
| Abbrev | Meaning |
Prescription Directions |
| bid | two times a day | |
aa. | of each |
| tid | three times a day |
ad | up to, to make |
| qid | four times a day |
NR | no refills |
| q (t) h | every (t) hours |
q.s. | a sufficient quantity |
| prn | as needed |
q.s. ad | enough to make |
|
|
|
stat | immediately |
| a.c. | before meals |
ut dict | as directed |
| p.c. | after meals |
|
|
| q AM | every morning |
Quantity and Measurement |
| q PM | every evening | |
BSA | body surface area |
| q HS | at bedtime |
m2 | square meters |
|
|
|
|
|
| p. o. | by mouth (orally) |
cc, cm3 | cubic centimeter, mL |
| NPO | nothing by mouth |
tsp | teaspoonful (5 mL) |
|
|
|
Tbsp | tablespoonful (15 mL) |
| a.d. | right ear |
|
|
| a.s. | left ear |
mcg | microgram |
| a.u. | each ear |
ng | nanogram |
|
|
|
mcL | microliter |
| o.d. | right eye |
|
|
| o.s. | left eye |
mEq | milliequivalent |
| o.u | each eye |
mmol | millimole |
|
|
|
mOsm | milliosmole |
| IV | intravenous |
|
|
| IM | intramuscular |
|
|
| ID | intradermal |
|
|
| subQ | subcutaneous |
Module 1F: Some ISMP Guidelines
History has provided many opportunities for learning from our mistakes. Unfortunately, patients have suffered from a lack of exact directions or misinterpreting written instructions. IMSP has developed guidelines, and you must use the following in class.
• Whole numbers should be written without a decimal point and without a terminal zero. For example, write 6 mg, not 6.0 mg. The decimal point might be missed, and the dose is interpreted as 60 mg.
• A number with a value less than one should be written with a leading zero. For example, write 0.4 mg, not .4 mg. The decimal point might be missed, and the dose is interpreted as 4 mg.
• Use whole numbers when possible and not the equivalent decimal fraction. Write 100 mg, not 0.1 g.
• Do not use the abbreviation u or U for units. Spell out the word units. The letter U has been misinterpreted to represent zero (0).
• Be especially careful about using the letter d when representing dose or day. When the situation may be ambiguous, spell out the intended word. What is the meaning of d when written as mg/kg/d? Is this mg/kg/dose or mg/kg/day? Always put the safety of the patient first in your activities.
Consider your safety education to start now.
Module 1G: Ratio and Proportion
Many numerical problems in pharmacy calculations can be solved using ratios and proportions.
A ratio compares two quantities, for example the fraction \(\frac{4}{7}\), which can also be written as 4 : 7, means 4 parts of one component and 7 parts of the other.
In addition to comparing one component to the other, the ratio can also be interpreted as the amount of one component in relation to the sum of all other components.
The proportional relationship is usually written as:
\(\frac{a}{b}=\frac{c}{d}\)
To solve this type of problem, you require three values and solve for the fourth by cross-multiplying and then dividing to isolate the term of interest. Remember to include the units and ensure that like units occupy numerators or denominators.
Example 1.20: An injectable product contains 350 mcg of a drug in each 0.9 mL. How many micrograms are contained in 0.4 mL?
\(\frac{350\;mcg}{0.9\;mL}=\frac{x\;mcg}{0.4\;mL}\\\\\)
\(x\;mL=\frac{350\;mcg\times0.4\;mL}{0.9\;mL}=155.6\; mcg\)
Module 1: Practice Problems
- A patient will receive 5 mL PO bid of a medication x 10 days. What volume is required?
- A patient will receive 3.5 mL PO tid of a medication x 8 days. What volume is required?
- A patient will receive 2.5 mL PO qid of a medication x 7 days. What volume is required?
- A child weighs 35#. What is the weight in kg?
- A newborn weighs 5# 3oz. What is the weight in kg?
- An infant weighs 7# 7oz. What is the weight in kg?
- A patient weighs 186#. What is the weight in kg?
- A patient is 6’4”. What is their height in cm and m?
- A patient is 5’6”. What is their height in cm and m?
- A patient is 3’7”. What is their height in cm and m?
- A patient is 4’3”. What is their height in cm and m?
- If you are paid $56.25 for 1 ½ hours of work, what is your hourly wage?
- If you are paid $1075.20 for 42 hours of work, what is your hourly wage?
- Risankizumab injection contains the following ingredients in each 0.83 mL dose: Risankizumab 75 mg, succinic acid 0.049 mg, sorbitol 34 mg, polysorbate 20 0.17 mg, and sodium succinate anhydrous 0.53 mg. How many milligrams of each ingredient would be present in 10 mL of the solution.
- Prednisone is available in 5 mg tablets. How many tablets would be required to fill an Rx with the directions:
Day 1 - take 40 mg one time
Day 2 - take 30 mg one time,
Day 3 - take 20 mg one time
Day 4 - take 10 mg one time
Days 5 - 10 - take 5 mg one time per day. - A surgeon ordered morphine 2 mg IV every 6 hours if needed for pain relief. The patient received his previous dose at 2230. What is the earliest time he may receive another dose?
- A patient will receive an IV drug over 1 hour every 8 hours. If the first infusion was started at 1000 today, at what times should the subsequent 2 doses be started?
- Levothyroxine sodium tablets are available in 12 different dosages, from 25 mcg to 300 mcg. Express that range in milligrams.
- A patient is prescribed an antibiotic with the directions: 2 tsp qid x 4 days, then 1 tsp bid x 2 days. How many milliliters will the patient take over the course of therapy?
- Each dose from a dry powder inhaler weighs a total of 12.5 mg - see label below. The powder mixture contains fluticasone propionate, salmeteral xinafoate, and lactose. The patient inhales two doses (blisters) a day. How many milligrams of lactose are inhaled in one week?
1.65 grams equals mg.
0.25 grams equals mcg.
0.1 grams equals ng.
0.5 mg = mcg
If the contents of 1 vial of angiotensin II injection is diluted to a total of 500 mL with normal saline, what is the drug concentration in nanograms per mL (ng/mL).
- Read the volume in each syringe
| A | B | C |
| C | D | E |
| G | H | I |
| J | K | L |
Answers:
- 100 mL
- 84 mL
- 70 mL
- 15.9 kg
- 2.36 kg
- 3.38 kg
- 84.5 kg
- 193 cm, 1.93 m
- 168 cm, 1.68 m
- 109 cm, 1.09 m
- 130 cm, 1.3 m
- $37.50/h
- $25.60/h
- Risankizumab 903.6 mg
Succinic acid 0.59 mg
Sorbitol 409.6 mg
Polysorbate 20 2.05 mg
Sodium succinate anh. 6.39 mg - 26 tablets
- 0430
- 1800, 0200 tomorrow
- 0.025 mg – 0.3 mg
- 180 mL
- 166.985 mg (or 167 mg) lactose
- 1650 mg.
250,000 mcg.
100,000,000 ng.
500 mcg.
5000 ng/mL
A. 2.6 mL
B. 0.8 mL
C. 6.8 mL
D. 0.68 mL
E. 3.2 mL
F. 0.62 mL
G. 1.6 mL
H. 0.35 mL
I. 7.6 mL
J. 0.51 mL
K. 8.4 mL
L. 4.8 mL
Module 2: Working with Concentrations
This module will introduce and review calculations involving different concentration units used in Pharmacy.
Concentration
A drug product contains one or more active ingredients mixed with excipients, such as diluents, flavorings, and colors. Different concentration terms are used to express the quantity of any single ingredient in relation to the entire mixture.
- A suspension contains 250 mg of amoxicillin in every 5 mL of liquid, or 250 mg/5 mL
- Tylenol contains 325 mg of acetaminophen in each tablet, or 325 mg/tablet
- A cream contains 1% hydrocortisone
- A cough suppressant elixir contains 5% alcohol as an inactive ingredient
Pharmacists must work with concentrations to calculate the quantity of product to dispense, the dose a patient should take, and the amount of each ingredient required for compounded formulations. Table 2.1 defines the most common concentration terms used in pharmacy. You will need to know these definitions and be able to convert between them.
Table 2.1. Common concentration terms
Concentration term | Definition |
Percent weight in volume (% w/v) | Grams of component in 100 mL of mixture |
Percent weight in weight (% w/w) | Grams of component in 100 g of mixture |
Percent volume in volume (% v/v) | mL of component in 100 mL of mixture |
Molar (M) | Moles of component in 1L of mixture |
Millimolar (mM) | Millimoles of component in 1L of mixture |
Ratio strength (1:X) | 1 gram or mL of component in every X g or mL of mixture |
g/mL, mg/mL, Units/mL, etc. | The amount of component in a specified amount of mixture |
Module 2A: Percent Strength
Percent w/v is commonly used for liquid dosage forms, such as solutions and suspensions. The amount of drug in the mixture is stated as grams (g) per 100 mL of the mixture.
Example 2.1: A minoxidil 5% topical solution contains 5 g of the drug in every 100 mL of the product.
Concentration does not change with the package size, so
\(\frac{5\;g}{100\;mL}=\frac{0.05\;g}{1\;mL}=\frac{2.5\;g}{50\;mL}=\frac{10\;g}{200\;mL}=\frac{50\;g}{L}=5\%\;w/v\)
Example 2.2: How many milligrams of minoxidil are contained in a 60 mL bottle of 5% minoxidil solution?
Solve using dimension analysis:
\(60\;mL \;solution\times\frac{5\;g\;minoxidil}{100\;mL\;solution}=3\;g \;minoxidil\)
Solve using proportionality:
\(\frac{5\;g\;minoxidil}{100\;mL\;solution}=\frac{x\;g\;minoxidil}{60\;mL\;solution}\)
Now solve for x:
\((5\;g\;minoxidil)(60\;mL\;solution)=(x\;g\;minoxidil)(100\;mL\;solution)\)
\(\frac{(5\;g\;minoxidil)(60\;mL\;solution)}{(100\;mL\;solution)}=x\;g\;minoxidil\)
\(x=3\;g\;minoxidil\;in\;every\;60\;mL\;of\;5\%\;solution \)
TIP: Whether you choose to solve problems using the dimension analysis or proportionality method, label all values with a unit (g, mL, etc.) and a name to indicate what the unit refers to. This will help organize your work and reduce errors when performing multi-step calculations.
Example 2.3: What volume of 17% w/v solution can be prepared from 10 g of benzalkonium chloride (BC)?
\(17\%\;BC=\frac{17\;g\;BC}{100\;mL\;solution}\)
There is 10 g of BC to make the solution, so the volume of solution to prepare is given by:
\(\frac{17\;g\;BC}{100\;mL\;}=\frac{10\;g\;BC}{x\;mL},\;x\;=\;58.8\;mL\)
Percent w/w is used for semisolids (ointments, creams) and concentrated acids (for example, concentrated hydrochloric acid). It may also be used for powder mixtures. A 10% salicylic acid ointment contains 10 g of salicylic acid in every 100 g of the ointment. The weight of all ingredients, including any liquid ingredients, must be included in the ointment total.
Example 2.4: An ointment was prepared according to the formula below. Calculate the percent strength (w/w) of triamcinolone acetonide in the ointment.
Triamcinolone acetonide 500 mg
Glycerin 12.5 g
Hydrophilic petrolatum qs 200 g
First, understand what the formula means. Triamcinolone acetonide (TA) is a powder. It is packaged in a jar and weighed out on a balance. This batch of ointment will contain 500 mg of TA. Glycerin is a liquid with a density of 1.25 g/mL. The pharmacist preparing the ointment may either measure out the correct volume of glycerin or weigh it into a tared beaker to obtain the required 12.5 g. Hydrophilic petrolatum (HP) is an ointment base, and the amount required is “qs 200 g”. The abbreviation qs means to add enough HP so the total mixture weighs 200 g. The total weight of the drug, TA, is 500 mg, or 0.5 g, and the total weight of the ointment is 200g. The percent strength of TA, then, can be calculated as:
\(\frac{0.5\;g\;TA}{200\;g\;ointment}=\frac{x\;g\;TA}{100\;g\;ointment}\)
Solving for x, the concentration of the ointment is:
\(\frac{0.25\;g\;TA}{100\;g\;ointment}\;or\;0.25\%\;w/w\)
The concentration of all components in a semi-solid are expressed as w/w, including liquids like glycerin, polysorbate 80, water, etc. The density of a liquid is used to convert between the weight and volume of a material, and it has units of g/mL. Glycerin has a density of 1.25 g/mL, so a 1 mL sample of glycerin weighs 1.25 g. Water has a density of 1 g/mL, so a 1 mL sample of water weighs 1 g. Density values for some common liquids are found in Table 2.2. The symbol for density is d, or the Greek letter rho, 𝜌.
Example 2.5: A pharmacist requires 25 g of glycerin. Calculate the volume this represents.
\(25\;g \;glycerin\times\frac{1\;mL\;glycerin}{1.25\;g\;glycerin}=20\;mL\;glycerin\)
Or solving by proportions,
\(\frac{1\;mL\;glycerin}{1.25\;g\;glycerin}=\frac{x\;mL\;glycerin}{25\;g\;glycerin}\)
\(x=\frac{1\;mL\;glycerin\;\times\;25\;g\;glycerin}{1.25\;g\;glycerin}=20\;mL\;glycerin\)
Example 2.6: Calculate the percent strength of polysorbate 80 (PS80) in the cream formula. PS 80 is a liquid with a density of 1.1 g/mL.
Lidocaine 10 g
Polysorbate 80 10 mL
Cream base 200 g
This formula calls for the pharmacist to weigh out 10 g of lidocaine powder, 10 mL of PS80, and 200 g of cream base. The formula states that 200 g of base are needed. However, the total formula weight is not 200 grams but the sum of the three individual component weights.
The weight of PS80 must be calculated:
\(10\;mL \;PS80\times\frac{1.1\;g\;PS80}{1\;mL\;PS80}=11\;g \;PS80 \)
The formula's total weight is, then, 10 g (lidocaine) + 11 g (PS80) + 200 g (base)= 221 g.
The percent strength of PS80 in the formula is then calculated as:
\(\frac{11\;g\;PS80}{221\;g\;cream}=\frac{x\;g\;PS80}{100\;g\;cream}\)
Solving for x, the PS80 concentration is 4.977 g/100 g of cream, or 4.977% w/w. Applying the rules for significant figures from chapter 1, we should round the answer to 1 decimal place, giving an answer of 5.0. However, we do not write values with trailing zeros after the decimal point, so the proper answer is 5% w/w.
Table 2.2. Densities of some common pharmaceutical liquids at room temperature.
Liquid | Density (g/mL) |
Alcohol USP | 0.81 |
Glycerin | 1.25 |
Mineral oil USP | 0.89 |
Polyethylene glycol (PEG) 400 | 1.03 |
Polysorbate 80 | 1.1 |
Propylene glycol | 1.04 |
Water | 1 |
Values from Merck Index 14th Edition.
Example 2.6. How many grams of petrolatum should be added to 35 g of zinc oxide to prepare a 10% zinc oxide ointment?
10% zinc oxide means every 100 g of ointment contains 10 g of zinc oxide. Therefore, for every 10 g of zinc oxide, there is 90 g of petrolatum in the product. A simple solution to this problem then, is:
\(\frac{10\;g\;Zinc\;oxide}{90\;g\;petrolatum}=\frac{35\;g\;zinc\;oxide}{x\;g\;petrolatum}\)
x = 315 g petrolatum
We can check the answer by using the result to calculate the concentration.
\(\frac{35\;g\;zinc\;oxide}{35\;g\;zinc\;oxide\;+\;315\;g\;petrolatum}\times100=10\%\)
Concentrated acids are traditionally labeled as percent w/w. Concentrated hydrochloric acid, for example, is available as a 37% w/w solution (37 grams of HCl in every 100 g of solution) with a density of 1.2 g/mL. Concentrated phosphoric acid is a viscous solution with a concentration of 85% w/w (85 g of H3PO4 in every 100 g of solution) and a density of 1.84 g/mL.
Example 2.7: How many mL of 85% phosphoric acid solution are required to provide 10 g of phosphoric acid (H3PO4).
\(10\;g\;H_{3}PO_{4}\times\frac{100\;g\;solution}{85\;g\;H_{3}PO_{4}}\times\frac{1\;mL\;solution}{1.84\;g\;solution}=6.4\;mL\;solution\)
6.4 mL of 85% phosphoric acid solution will provide 10 g of H3PO4.
Alternatively, the problem can be solved using 2 proportionalities. First, find how many grams of 85% solution contain 10 g of H3PO4.
\(\frac{x\;g\;solution}{10\;g\;H_{3}PO_{4}}=\frac{100\;g\;solution}{85\;g\;H_{3}PO_{4}}\)
Solving for x, 11.76 g of 85% phosphoric acid solution will provide 10 g of H3PO4.
Next, convert 11.76 grams of 85% phosphoric acid to mL using the density:
\(\frac{11.76\;g\;solution}{x\;mL\;}=\frac{1.84\;g\;solution}{1\;mL}\)
x = 6.4 mL of 85% solution.
Percent v/v is usually used to express the concentration of a liquid solute in another liquid, such as alcohol (ethanol) or glycerin dissolved in water. Pharmacists sometimes use Alcohol USP as the source of ethanol for preparing solutions. Alcohol USP is a solution of ethanol and water with an ethanol concentration of 95% v/v.
Example 2.8: A pharmacist needs to prepare 2 L of 70% v/v alcohol by mixing Alcohol USP and Purified Water. How much Alcohol USP is needed?
\(2\;L\times\frac{70\;mL\;EtOH}{100\;mL\;soln}\times\frac{1000\;mL}{1\;L}\times\frac{100\;mL\;Alc\;USP}{95\;mL\;EtOH}=1473.7\;mL\;Alc\;USP\)
To prepare the solution, the pharmacist should measure out 1,473.7 mL of Alcohol USP and then add enough purified water to make the total volume of 2 L.
What does it mean when a concentration is labeled with “%” without specifying w/w, w/v, or v/v? The meaning can be determined by using the definition of concentration and the conventions associated with specific types of mixtures. If they are not specified otherwise, then:
- Topical products or semisolids are labeled as % w/w
- Concentrated acids are labeled as % w/w
- Alcohol solutions are labeled as % v/v
- Other liquid mixtures are usually labeled as % w/v if the pure solute is a solid or v/v if the pure solute is a liquid. Most pure drugs are solid at room temperature.
- Medicated powders, such as athlete’s foot powder, are labeled as % w/w.
Module 2B: Molar and millimolar
Some clinical chemistry and drug concentrations are expressed in molar units. Recall that 1 mole is the weight in grams equal to the molecular weight of a substance. Sucrose, or table sugar, has a molecular weight of 342, therefore there are 342 g per mole. Drugs are usually used in small amounts, so the more common unit is the millimole, 1/1000th of a mole, or 0.001 mole. A millimole can be defined as the weight in milligrams equal to the molecular weight of a substance. 1 millimole of sucrose weighs 342 mg.
The concentration unit Molar (M) is defined as moles of solute contained in each 1 L of solution. Millimolar (mM) is defined as millimoles of solute contained in each 1 L of solution. These concentration units should be familiar because they are used extensively in chemistry and biology courses.
Example 2.9: The concentration of glucose in blood, i.e. the blood glucose level, is used as a control measure for diabetic patients. The normal blood glucose level in a non-diabetic patient is approximately 5.5 mM or 5.5 millimoles of glucose per liter of blood. Glucose levels are also frequently expressed in mg of glucose per deciliter (dL, or 100 mL) of blood. Convert 5.5 mM glucose into mg/dL. Glucose is a solid with a molecular weight of 180 g/mole.
Solving by dimension analysis:
\(\frac{5.5\;mmoL\;glucose}{1000\;mL\;blood}\times\frac{180\;mg\;glucose}{1\;mmol\;glucose}=\frac{0.99\;mg\;glucose\;USP}{mL}\times100=\frac{99\;mg\;glucose}{100\;mL}=\frac{99\;mg\;glucose}{dL}\)
Alternatively, the problem can be solved by using two proportions to convert mmol of glucose to mg of glucose and 1 L of blood to 1 dL of blood.
First, find the milligrams of glucose represented by 5.5 mmoles.
\(\frac{5.5\;mmol\;glucose}{x\;mg\;glucose}=\frac{1\;mmol\;glucose}{180\;mg\;glucose}\)
x = 990 mg glucose.
\(5.5\;mM=\frac{990\;mg\;glucose}{1\;L\;blood}\;or\;\;\frac{990\;mg\;glucose}{1000\;mL\;blood}\)
Next, convert the concentration to 100 mL (1 dL) in the denominator
\(\frac{990\;mg\;glucose}{1000\;mL\;blood}=\frac{x\;mg\;glucose}{100\;mL\;blood}\)
\(x=99\;mg,\;so\;\frac{99\;mg\;glucose}{100\;mL\;blood}\;or\;\frac{99\;mg\;glucose}{dL\;blood}\)
Phosphate salts are often used for electrolyte replacement, either orally or by intravenous infusion. Intravenous phosphate supplementation is typically ordered in millimoles of phosphate. In water, salts containing phosphates exists as an equilibrium mixture of the dihydrogen phosphate (1– charge) and the monohydrogen phosphate (2– charge), depending on the pH of the solution. According to the Henderson-Hasselbalch equation, the specific concentration of each anion form is determined by the appropriate phosphoric acid pKa and the surrounding solution pH.
\(H_{2}{PO_{4}^\;}^{-}\leftrightharpoons\;{HPO_{4}}^{2-}\)
Both forms have 1 phosphorus (P) atom per anion, so 1 millimole of H2PO4– provides the same amount of phosphorus as 1 millimole of HPO4–2. The “phosphate concentration” is the total of both ion forms present in the solution.
Example 2.10: A solution contains 12.42 g of monobasic sodium phosphate monohydrate (NaH2PO4 · H2O, MW = 138) and 6.39 g of dibasic sodium phosphate anhydrous (Na2HPO4, MW = 142) in a total volume of 45 mL. Calculate the phosphate concentration in mM.
Monobasic (P1):
\(\frac{12.42\;g\;P1}{45\;mL\;soln}\times\frac{1\;mol\;P1}{138\;g\;P1}\times \frac{1000\;mmol\;P1}{mol\;P1}\times\frac{1000\;mL\;soln}{L\;soln}=\frac{2000\;mmol\;P1}{L\;soln}\)
Dibasic (P2):
\(\frac{6.39\;g\;P2}{45\;mL\;soln}\;\times\;\frac{1\;mol\;P2}{142\;g\;P2}\;\times\;\frac{1000\;mmol\;P2}{mol\;P2}\;\times\;\frac{1000\;mL\;soln}{L\;soln}\;=\;\frac{1000\;mmol\;P2}{L\;soln}\)
Total phosphate concentration = 2000 mM P1 + 1000 mM P2 = 3000 mM
Millimoles and millimolar concentrations are also used to calculate solutions' osmolarity, which is an important patient safety consideration in intravenous drug therapy, intramuscular injections, and ophthalmic, otic and nasal solutions.
Module 2C: Ratio Strength
Ratio strength is typically used to express the concentration of dilute solutions, i.e., solutions with very low solute concentrations. It is important for pharmacists to understand ratio strength because the drug epinephrine is frequently expressed in terms of ratio strength. Ratio strength is expressed as 1:X, which means the concentration is 1 part of solute in X parts of solution. 1 part is either 1 g for a solid or 1 mL for a liquid. Epinephrine is a solid, so 1 part of epinephrine is 1 g. An aqueous solution is liquid, so 1 part of the solution is 1 mL. A 1:1000 epinephrine solution, then, contains 1 g of epinephrine in every 1000 mL of solution. The units are rarely written with the concentration.
Most concentration units express how much solute is contained in a fixed amount of the mixture, e.g., millimoles in 1 L, grams in 1 dL, etc. Ratio strength is the exact opposite, because it expresses the amount of the mixture that contains 1 g or 1 mL of the solute. Ratio strength can be converted to other concentration units by expressing the ratio as a fraction and using dimensional analysis or proportions to solve for the desired unit.
Example 2.11: An aqueous solution contains a drug at a concentration of 1:2500. Calculate the percent strength of the solution.
1:2500 means there is 1 g of drug in every 2500 mL of the solution. Percent means grams of drug in every 100 mL of solution. These definitions can be combined as a proportion to solve the problem.
\(\frac{1\;g\;drug}{2500\;mL\;soln}=\frac{x\;g\;drug}{100\;mL\;solution}\)
\(x=\frac{1\;g\;drug\;\times\;100\;mL\;soln}{2500\;mL\;soln}=0.04\;g\;drug\;in\;every\;100\;mL=0.04\%\)
Example 2.12: A patient with a severe allergy to bee venom requires 300 mcg of epinephrine. What volume of a 1:1000 solution is required?
This problem can be solved with proportions, as in the previous example. Alternatively, the dimension analysis approach is:
\(300\;mcg\;epi\times\frac{1\;g\;epi}{1,000,000\;mcg\;epi}\times\frac{1000\;mL\;soln}{1\;g\;epi}=0.3\;mL\;of\;1:1000\;solution\)
Example 2.13: Calculate the concentration of epinephrine 1:1000 in mM units. Epinephrine
MW = 183.
\(\frac{1\;g\;epi}{1000\;mL\;soln}\times\frac{1000\;mL\;soln}{1\;L\;soln}\times \frac{1\;mol\;epi}{183\;g\;epi}\times\frac{1000\;mmoL\;epi}{1\;mol\;epi}=\frac{5.5\;mmol\;epi}{L\;soln}=5.5mM\;epi\)
Example 2.14: A solution is prepared by dissolving 1.5 mg of drug in enough water to make 3 L. Calculate the ratio strength of the finished solution.
The simplest way to solve this problem is to write the concentration as milliliters of solution divided by grams of solute. We need to convert mg to g and L to mL:
\(\frac{3\;L\;\times\frac{1000\;mL}{L}}{1.5\;mg\;\times\frac{1\;g}{1000\;mg}}=2,000,000\)
The ratio strength of the solution is 1:2,000,000.
Module 2D: Units of Activity
Some drugs, especially those derived from natural sources, are labeled in terms of “Units of activity.” The most important drug expressed as Units is insulin. Insulin concentration is expressed as U-100, U-200, or U-300, meaning 100 Units/mL, 200 Units/mL, or 300 Units/mL. Insulin Units are treated like mg, mmol, or any other term describing the weight of drug. Patients measure their daily insulin doses in terms of Units, either with pre-filled injector pens or traditional insulin syringes.
Heparin, a natural anticoagulant, is available in injectable solutions containing 1000 USP Units/mL, 5000 USP Units/mL, 10,000 USP Units/mL, and 20,000 USP Units/mL.
Example 2.15: How many microliters of U-200 insulin is required to provide a dose of 35 Units.
\(\frac{1\;mL}{200\;Units}=\frac{x\;mL}{35\;Units}\)
\(x\;mL=0.175\;mL\\\\\)
\(x=0.175\;mL\times\frac{1000\;mcL}{mL}=175\;mcL\)
Safety note: Always spell out and capitalize Unit to avoid medication errors.
Module 2E: Reducing and enlarging formulas
The amount of a component in a mixture is easily scaled to a larger or smaller amount using proportionalities.
Example 2.15: The formula for Urea Compounded Irrigation USP specifies using 10 g of urea and enough sodium chloride irrigation solution to make 50 mL. If you needed to prepare 240 mL instead of 50 mL, the amount of urea required would be easily calculated using the proportion:
\(\frac{10\;g\;urea}{50\;mL\;irrigation}=\frac{x\;g\;urea}{240\;mL\;irrigation}\)
\(x=48\;g\;urea\)
Example 2.16: Compound Clioquinol Topical Powder USP is a solid mixture containing 4 ingredients. The formula for 1 kg of powder is given.
a) Calculate the percent strength of clioquinol in the formula.
b) Calculate the amount of lactose needed to prepare 250 g of the product.
Ingredient | Amount |
Clioquinol | 250 g |
Lactic acid | 25 g |
Zinc stearate | 200 g |
Lactose | 525 g |
(Total) | 1000 g |
The percent strength should be expressed as % w/w because all of the components are solids.
\(\frac{250\;g\;clioquinol}{1000\;g\;mixture}=\frac{x\;g\;clioquinol}{100\;g\;mixture}\)
x=25, so 25% clioquinol
The amount of lactose required for 250 g of the mixture is calculated as:
\(\frac{525\;g\;lactose}{1000\;g\;mixture}=\frac{x\;g\;lactose}{250\;g\;mixture}\)
\(x=131.25\;g\;lactose\;in\;250\;g\;of\;mixture \)
Example 2.17: A gabapentin cream is compounded according to the formula.
a) Calculate the percent strength of glycerin in the cream.
b) Calculate the grams and milliliters of glycerin required to prepare 60 g of cream.
c) Calculate the ratio strength of methylparaben in the preparation.
Ingredient | Amount |
Gabapentin | 3.5 g |
Glycerin | 7.5 mL |
Methylparaben | 75 mg |
Cream base | qs 100 g |
The product is a cream, so the concentration should be expressed in % w/w. Glycerin is a liquid with a density of 1.25 (Table 2.2). The concentration of glycerin can be calculated as:
\(\frac{7.5\;mL\;glycerin\;\times\;\frac{1.25\;g\;glycerin}{1\;mL\;glycerin}}{100\;g\;cream}=\frac{9.375\;g\;glycerin}{100\;g\;cream}=9.4\%\;w/w\;glycerin \)
The grams of glycerin required for 60 g of cream may be calculated from the percent strength as:
\(\frac{9.4\;g\;glycerin}{100\;g\;cream}=\frac{x\;g\;glycerin}{60\;g\;cream}\)
\(x=5.6\;g\;glycerin\)
The milliliters of glycerin for 60 mL of cream is calculated from the original formula:
\(\frac{7.5\;mL\;glycerin}{100\;g\;cream}=\frac{x\;mL\;glycerin}{60\;g\;cream}\)
\(x=4.5\;mL\)
The formula contains 75 mg or 0.075 g of methylparaben (MP) in 100 g of cream, so the ratio strength of MP is calculated as:
\(\frac{0.075\;g\;MP}{100\;g\;cream}=\frac{1\;g\;MP}{x\;g\;cream},\;x=1333.3\)
\(\text{The ratio strength of MP = 1:1333}\).
Module 2: Practice Problems
1. Calculate the amount of urea in each 60 gram bottle of 40% urea lotion.
2. A solution contains 250 mg of drug, 15 g of sucrose, 250 mg of methylparaben, and 1 mL of grape flavor in every 8 fluidounces. Calculate the percent strength of the drug in the solution.
3. A pharmacist mixed 60 g of cyclodextrin, 20 mL of glycerin (d = 1.25), 1 gram of drug, and 30 mL of water. Calculate the glycerin concentration in % w/w.
4. Convert each concentration to mg/mL.
25% w/v
1:5,000 w/v
15 mM glucose (MW = 180)
45% v/v alcohol (d = 0.81)
5. Convert each concentration to ratio strength.
0.025% w/v
0.05 mg/mL
0.02 mM (MW = 350)
22.5 mL alcohol in 650 mL of solution
6. An ointment contains 1,500 mcg of triamcinolone acetonide in every 60 g tube. Calculate the percent strength of triamcinolone acetonide.
7. How many milliliters of polysorbate 80 (d=1.1) are required to prepare 350 mL of a 3% w/v solution.
8. Convert each concentration to mM.
25% w/v salicylic acid (MW = 138)
1:3300 w/v epinephrine (MW = 183)
50 mg/mL amoxicillin (MW = 365)
45% v/v alcohol (d = 0.81, MW = 46)
9. A pharmacist mixed 800 mg of drug (MW = 159), 20 mL of propylene glycol (d=1.04), and 40 mL of water, resulting in a final solution volume of 62 mL. A: Calculate the drug's millimolar concentration in the solution. B: Calculate the propylene glycol concentration in % w/w.
10. How many mL of 85% phosphoric acid (d=1.84) should be used to prepare 3 L of 2% w/v phosphoric acid solution?
11. Calculate the millimolar concentration of Sodium Bicarbonate Injection USP. Sodium bicarbonate MW = 84.
12. A veterinary product contains neomycin sulfate 2.5 mg and 0.25 mg of triamcinolone acetonide in every mL of solution. Calculate the percent strength of both drugs.
13. A 1% testosterone transdermal gel product delivers 1.25 g of product per actuation of the metered dose pump. How many milligrams of testosterone is contained in each pump?
14. A 0.1% estradiol transdermal gel contains 0.75 g of gel in each packet. How many micrograms of estradiol are contained in each packet?
15. Chlorhexidine gluconate 4% solution is used as an antibacterial skin cleanser. How many grams of chlorhexidine gluconate are contained in each 16-ounce bottle?
16. An allergen extract contains 100 mcg/mL of honey bee venom. Calculate the concentration in percent strength and ratio strength.
17. Sodium phosphates injection contains 276 mg of monobasic sodium phosphate monohydrate (NaH2PO4•H20, MW = 138) and 268 mg of dibasic sodium phosphate heptahydrate (Na2HPO4•7H20, MW = 268) in every milliliter. Calculate the total millimolar concentration of phosphate in the solution. HINT: Calculate the millimolar concentration of each individual salt, then add the numbers together.
18. A 2020 water analysis for Edwardsville IL, reported one sample contained 570 mcg of copper per liter of water. Convert this concentration to ratio strength.
19. A 2020 water analysis for Edwardsville IL, reported one sample contained 2.9 mcg of lead per liter of water. Convert this concentration to ratio strength.
20. A product contains an allergen extract at a concentration of 1:25,000. How many micrograms of allergen does a patient receive with each 0.3 mL dose?
21. Mannitol is available as a 20% solution in water for injection. What volume of mannitol solution is required to provide a dose of 100 g.
22. A pharmacist mixed 35 g of salicylic acid with 20 mL of polyethylene glycol (PEG) 400 (d=1.03) and 15 g of PEG 8000. Calculate the percent strength of salicylic acid, PEG 400, and PEG 8000 in the mixture.
23. Calculate the molarity of 20% mannitol (MW = 182) in water.
24. Sodium acetate injection contains 328 mg of sodium acetate (MW = 82) in every milliliter of solution. Calculate the millimolar concentration.
25. An injectable supplement contains 75.5 mg of manganese sulfate (MW = 151) in every 0.5 mL. Calculate the manganese sulfate concentration in millimolar, percent strength, and ratio strength.
26. Calcitonin salmon injection is a sterile solution containing 400 USP Units in each 2 mL vial. If a patient is ordered 240 USP Units every 6 hours, how many milliliters of the solution should the patient receive daily?
27. Bleomycin for injection is supplied as a vial containing 15 Units of a sterile powder. A hospital pharmacy reconstitutes each vial by adding 3 mL of sterile water for injection. Calculate the drug concentration in the reconstituted solution. Calculate how many milliliters of the reconstituted solution are required to supply a dose of 18 Units. Calculate how many vials are required to provide the dose.
28. How many milligrams of solute are contained in 25 mL of a 1:30,000 solution?
29. Dextrose monohydrate 50% injection is sometimes used to treat severe hypoglycemia. Calculate the solution's molarity. Dextrose·H2O MW = 198.
30. A solution was prepared by dissolving 12 g of monobasic sodium phosphate (NaH2PO4•H20, MW = 138) and 7.1 g of dibasic sodium phosphate (Na2HPO4•7H20, MW = 268) in water to a total volume of 50 mL. Calculate the phosphate concentration in mM.
31. An injectable solution contains aluminum (atomic wt = 27) as a contaminant at approximately 12.5 mcg/L concentration. Calculate the ratio strength, percent strength, and mM concentration of aluminum in the solution.
Answers:
1. 24 g
2. 0.1%
3. 21.6%
4. 250 mg/mL
0.2 mg/mL
2.7 mg/mL
364.5 mg/mL
5. 1:4000
1:20,000
1:142857
1:29
6. 0.0025%
7. 9.5 mL
8. 1,812 mM
1.7 mM
137 mM
7924 mM
9. 81.2 mM
33.8%
10. 38.4 mL
11. 500 mM
12. 0.25% neomycin, 0.025% triamcinolone
13. 12.5 mg testosterone
14. 750 mcg estradiol
15. 19.2 g
16. 0.01%
1:10,000
17. 3000 mM
18. 1:1,754,386
19. 1:344,827,586
20. 12 mcg
21. 500 mL
22. 49.6% salicylic acid, 29.2% PEG400, 21.2% PEG8000
23. 1.1 M
24. 4000 mM
25. 1000 mM, 15.1% w/v, 1:6.6
26. 4.8 mL
27. 5 USP Units/mL
3.6 mL
1.2 vials
28. 0.83 mg
29. 2.5 M
30. 2269 mM
31. 1:80,000,000
0.00000125%
0.00046 mM
Module 3: Dilution, Alligation, and Concentration
This module introduces some mathematical techniques used to alter product strength, either decreasing the active pharmaceutical ingredient (API) concentration or increasing the (API) concentration. The product's physical form could be a solution or a semi-solid, like an ointment or cream. Diluting a product with an appropriate patient-compatible solvent results in lowering the concentration. We can increase the concentration of, or fortify, a product by adding the API in pure form or by using another product with a higher concentration than the one we are working with. Finally, you will learn about an algebraic technique, alligation, which is unique to pharmacy. The method is useful when solving problems that require mixing two different concentrations to make a third concentration between the two starting values.
Brief Review
In the last module, you learned about concentrations and the various expressions pharmacists use to represent these concepts and relate them to products. As a reminder, the strength or concentration of a pharmaceutical product represents the amount of the active pharmaceutical ingredient (API) relative to the total amount of the product.
Example 3.1: If 120 mL of an oral solution contains 6 grams of the drug, then the concentration of the API is 5%.
\(\frac{6\;g}{120\;mL}\times100\;= 5\;\text{%}\)
If you add an additional 50 mL of solvent to the product, the API amount does not change, but the total content volume increases to 170 mL, thus reducing the strength. This is one example of a dilution.
\(\frac{6\;g}{120\;mL\;+\;50\;mL}\times100\;\cong 3.5\;\text{%}\)
This cartoon from Wikipedia depicts a series of dilutions of the most concentrated solute, on the left, to a more dilute solute, on the right.
Source: By Grasso Luigi - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=76044995
A very useful equation commonly used by pharmacists is C1V1 = C2V2. You most likely used this equation in your previous chemistry courses. We will demonstrate this equation with representative problems later in the module.
Module 3A: Dilution Process Involving Liquids
Dilution is a technique where you add a solvent, most commonly water, to a solute already in solution. The amount of the solute is constant, but the final volume is increased, thus decreasing concentration. The Wikipedia picture demonstrates the important formula, C1 · V1 = C2 · V2, and a pictorial cartoon to help visual learners see the outcome.
By Theislikerice - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=69653928
Notice that this equation relates four parameters in an algebraic relationship. You may recall from high school algebra that you need three parameters to uniquely solve an equation that contains four parameters. A key fact to remember is that all units must be consistent. If one volume is in milliliters, the second must be in milliliters, not liters. If one concentration is in percent, the second concentration must be in percent, not mg/mL.
Let’s see how you can use this equation in pharmacy practice.
Example 3.2: What would be the new percent concentration (v/v) if you took 300 mL of a 12% solution and diluted it with water to a new volume of 400 mL? Let’s use C1 · V1 = C2 · V2. Based on the information in the problem, what values do you know? C1 = 12%. V1 = 300 mL. V2 = 400 mL. Now substitute the values into the equation and solve for the unknown value of C2.
\(\left(12\text{%}\right)\times\left(300\;mL\right)=\left(x\text{%} \right)\times\left(400\;mL \right)\)
\(x\text{%}=\frac{12\text{%}\;\times\;300\;mL}{400\;mL}\)
\(x=9\text{%}\)
Example 3.3: How much water would you add to 250 mL of a 15% (v/v) aqueous solution to reduce its concentration to 5%?
\(\left(15\text{%}\right)\times\left(250\;mL\right)=\left(5\text{%} \right)\times\left(x\;mL \right)\)
\(x\;{mL}=\frac{15\text{%}\;\times\;250\;mL}{5\text{%}}=750\;mL\)
** This problem requires further explanation. The answer is easily calculated, but the number needs some further thought. The final volume is 750 mL. You started with 250 mL of solution. Therefore, you need to add 750 – 250 mL = 500 mL of water. Please pay close attention to this type of problem which asks how much solvent must be added to the original solution. **
Example 3.4: You need to prepare 4 L of a solution with a final API concentration of 0.04%. How many milliliters of an 8% stock solution should you use to make the solution?
\(\left(0.04\text{%}\right)\times\left(4000\;mL\right)=\left(8\text{%} \right)\times\left(x\;mL \right)\)
\(x\;{mL}=\frac{0.04\text{%}\;\times\;4000\;mL}{8\text{%}}=20\;mL\)
Remember to verify that all units are the same throughout the problem. The initial volume was given as 4 L. I decided to use 4000 mL because the original problem asks for the answer in milliliters.
Example 3.5: You have 120 mL of a 5% stock solution. What would be the new concentration if you added an additional 50 mL of water to the solution? I will alter the units to demonstrate that technique. Recall that 5% w/v = 50 mg/mL.
\(\frac{50\;mg}{mL}\times0.12\;L=\frac{x\;mg}{mL}\times0.17\;L \)
\( \\\frac{x\;mg}{mL}=\frac{\frac{50\;mg}{mL}\;\times\;0.12\;L}{0.17\;L}\\\\ \)
\(x\cong \frac{35.3\;mg}{mL}\cong 3.5\text{%}\)
Module 3B: Dilution Process Involving Solids and Semisolids
Most people think of solutions when you mention the word dilution, but the same ideas, and equation, C1 · V1 = C2 · V2, apply when you are working with solids and semisolids. When working with powders or ointments, gels, and creams, we usually substitute the letter Q (quantity) for V in the equation, thus C1 · Q1 = C2 · Q2.
Example 3.6: How much petrolatum base would you add to 60 g of a 2.5% (w/v) steroid ointment to reduce its concentration to 1.5%? Notice the similarity of this problem with Example 2 in the previous section.
\(\left(2.5\text{%}\right)\times\left(60\;g\right)=\left(1.5\text{%} \right)\times\left(x\;g \right)\)
\(x\;g=\frac{2.5\text{%}\;\times\;60\;g}{1.5\text{%}}\)
Again, the answer is easily calculated, but the number needs further thought. The final quantity is 100 g. You started with 60 g of ointment. Therefore, you need to add an additional amount of petrolatum equal to 100 – 60 g = 40 g. Recall that this problem asks how much base must be added to the original ointment mass.
Eample 3.7: How much carbomer gel should you mix with 15 g of 0.05% clobetasol propionate gel to reduce the concentration to 0.025%?
\(\left(0.05\text{%}\right)\times\left(15\;g\right)=\left(0.025\text{%} \right)\times\left(x\;g \right)\)
\(x\;g=\frac{0.05\text{%}\;\times\;15\;g}{0.025\text{%}}\)
\(x = 30\;g\)
The final quantity is 30 g. You need to add 30 – 15 g = 15 g of carbomer gel.
Example 3.8: A physician asks you to prepare 8 oz of a cream with a final API concentration of 3%. How many grams of a 10% stock cream should you use to make the product? The problem asks for the answer in grams, so I converted 8 oz to 240 g for the equation.
\(\left(3\text{%}\right)\times\left(240\;g\right)=\left(10\text{%} \right)\times\left(x\;g \right)\)
\(x\;g=\frac{3\text{%}\;\times\;240\;g}{10\text{%}}\)
\(x = 72\;g\)
Consider the process where you are adding the active ingredient as a powder to a dosage form that already contains the API in the ointment.
Example 3.9: What is the final concentration of tapinarof if you added 1 g of the powder to the contents of a 60 g tube of Vtama® (tapinarof) 1%?
In these types of problems, the first task is to determine how much of the API is contained in the starting material. Then you sum the requested amount with the starting mass. Finally, divide by the new total amount of product.
\(\frac{1\;g\;T}{100\;g\;oint}=\frac{x\;g\;T}{60\;g\;oint} \), \(x = 0.6\;g\;tapinarof\;from\;60\;g\;tube\)
\(\frac{1\;g\;+\;0.6\;g\;T}{1\;+\;60\;g\;oint}=\frac{1.6\;g\;T}{61\;g\;oint}=2.6\text{%}\;after\;adding\;1g\;drug\;powder\)
Example 3.10: What is the final concentration of mupiricin if you added 2 g of the powder to the contents of a 22 g tube of Mupiricin ointment 2%?
\(\frac{2\;g\;M}{100\;g\;oint}=\frac{x\;g\;M}{22\;g\;oint}\), \(x = 0.44\;g\;mupiricin\)
\(\frac{2\;g\;+\;0.44\;g\;M}{2\;g\;+\;22\;g\;oint}=\frac{2.44\;g\;M}{24\;g\;oint}=10.2\text{%}\)
Dilution Practice Problems:
1. You have a stock solution of benzalkonium chloride 50% w/v. What is the final concentration if 2 mL of the solution is diluted to a final volume of 85 mL? (2 dp)
(50%) x (2 mL)/(85 mL) = C2= 1.18%
2. How much water would you add to 185 mL of a 17.5% v/v aqueous solution to reduce its concentration to 12.5%?
(17.5%) x (185 mL)/(12.5%) = V2 = 259 – 185 mL = 74 mL water
3. You need to prepare 2 L of a solution with a final API concentration of 25%. How many milliliters of a 70% stock solution should you use to make the solution?
(25%) x (2000 mL/(70%) = V2 = 714 mL
4. You have 150 mL of a 60 mg/mL stock solution. What would be the new concentration if you added an additional 75 mL of water to the solution?
(60 mg/mL) x (150 mL)/(225 mL) = 40 mg/mL and C2 = 4%
5. How much cream base should you mix with 45 g of 1% Vtama cream (tapinarof) to reduce the concentration to 0.3%?
(1%) x (45 g)/(0.3%) = Q2 = 150 – 45 g = 105 g cream base
6. How many milliliters of Lidocaine 4% should be used to prepare 125 mL of an IV solution containing 4 mg/mL?
(4 mg/mL) x (125 mL)/(40 mg/mL) = V2 = 12.5 mL
7. If you add 5 g of azelaic acid powder to 50 grams of an ointment containing 10% azelaic acid, what would be the final concentration in the ointment?
5 g + 5 g/55 g total ointment = 18.2%
8. What would be the final concentration of benzocaine in an ointment if you mixed 3 g of the API powder with 4 ounces of an ointment containing 10% benzocaine?
3 g + 12 g/123 g total ointment = 12.2%
Module 3C: Alligation
Alligation is an arithmetical technique for solving problems that require mixing two solutions to make a third solution. Words fail to describe the technique. It is much easier to demonstrate and work problems than to explain. The lecture will make the problem easier to understand. Alligation works very well when we must make a dilution that involves two solutions, one at a higher concentration and one at a lower concentration. The target concentration is between the upper and lower concentrations. Consider the common problem in pediatric Parenteral Nutrition solutions. The pharmacy prepared a PN solution with a dextrose concentration of 15%. The physicians want to increase the dextrose concentration to 20%. Do you need to prepare a new solution? No. We can add more dextrose solution to the bag to make the desired concentration. We can use Dextrose 70% (D70) or Dextrose 50% (D50), both base compounding solutions are readily available.
Note, you cannot solve this problem using C1 · V1 = C2 · V2 , because there are three concentrations involved. Let’s look at the mechanics of the process using D 70.
Example 3.11:
Draw two vertical lines. The desired concentration stated in the problem goes in the middle space. The order of the concentrations on the LHS does not matter, but I put the larger concentration of the solution I am going to use (the juice) on the top left and the smaller starting concentration on the lower left. Now perform two diagonal subtractions, always resulting in a positive number.
In this example D70 – D20 = 50. D20 – D15 = 5. Note that the difference between D70 and the desired strength, D20, represents the number of parts of D15 to be used.
The actual volume of the D15 solution in the bag represents 50 parts. You need 5 parts of D70, so that would be equal to 1/10 the volume in the bag. After the addition, the PN volume will be larger.
Example 3.12: How many milliliters of Dextrose 50% must be added to a bag of PN solution containing 150 mL of Dextrose 20% such that after the addition, the new dextrose concentration will equal 25%?
There are a total of 30 parts. We know the PN bag that needs to be altered contains 150 mL of D20, and that represents the 25 parts in the calculation. Now we can solve for the 5 parts using ratios.
\(\frac{25\;parts}{150\;mL}=\frac{5\;parts}{x\;mL},\;x=30\;mL\;D50\;required\)
New PN volume = 180 mL.
Example 3.13: You need to prepare an ointment with a concentration of 2.5%. You have a 20% ointment that can be diluted with an ointment base that contains no drug. How many grams of each component do you need to make 16 ounces (480 g)?
There is a total of 20 parts. The 20 parts represent the prescribed mass of ointment. Now we can solve for either part using ratios.
\(\frac{20\;parts}{480\;g}=\frac{17.5\;parts}{x\;g},\;x=420\;g\;ointment\;base\)
\(\frac{20\;parts}{480\;g}=\frac{2.5\;parts}{x\;g},\;x=60\;g\;20\%\;ointment\;base\)
Total ointment weight = 420 + 60 = 480 g
Example 3.14: How many grams of salicylic acid powder should be added to 60 g of polyethylene glycol ointment base to prepare a product containing 6% w/w of salicylic acid?
Note that you are starting with 60 g of base, which contains no drug. That 60 g mass represents the 94 parts obtained from the alligation. Now you can solve for the amounts using ratios.
\(\frac{94\;parts}{60\;g}=\frac{6\;parts}{x\;g},\;x=3.83\;g\text{ salicylic acid powder}\)
You can check your work, 3.83 g/63.83 g x 100 = 6%.
Module 3: Practice Problems
- What would be the new percent concentration (v/v) if you took 45 mL of a 15% solution and diluted it with water to a new volume of 390 mL?
- How much water should you add to 75 mL of a 9% (v/v) aqueous solution to reduce its concentration to 5%?
- You need to prepare 2.4 L of a solution with a final API concentration of 0.07%. How many milliliters of an 11% stock solution should you use to make the solution?
- You have 85 mL of a 4% stock solution. What would be the new concentration, in mg/mL, if you added an additional 35 mL of water to the solution?
- You have a stock solution of benzalkonium chloride 85% w/v. What is the final concentration if 12 mL of the solution is diluted to a final volume of 2250 mL?
- How much petrolatum base should you add to 45 g of a 5% (w/v) steroid ointment to reduce its concentration to 2%?
- How much ointment base should you mix with 45 g of 0.05% clobetasol propionate gel to reduce the concentration to 0.03%?
- A physician asks you to prepare 6 oz of a 4% cream. How many grams of a 10% stock cream should you use to provide the correct amount of the API?
- How much water would you add to 165 mL of an 18% v/v aqueous solution to reduce its concentration to 15%?
- You need to prepare 1.5 L of a solution with a final API concentration of 0.9%. How many milliliters of a 23.4% stock solution should you use to make the solution?
- You have 150 mL of a stock solution with a 40 mg/mL concentration. What would be the new concentration (%w/v) if you added an additional 75 mL of water to the solution?
- How much cream base should you mix with 60 g of 1% Vtama cream (tapinarof) to reduce the concentration to 0.6%?
- How many milliliters of Dobutamine 1.25% should be used to prepare 100 mL of an IV solution containing 4 mg/mL?
- If you add 10 g of salicylic acid powder to 4 ounces of an ointment containing 100 mg/g of the API, what would be the final concentration in the ointment?
- What would be the final concentration of benzocaine in an ointment if you added an additional 6 g of the API powder with 8 ounces of an ointment containing 10% benzocaine?
- How many milliliters of Dextrose 70% must be added to a 300 mL bag of PN solution containing Dextrose 15% so that after the addition, the new dextrose concentration will be equal to 20%? (whole number)
- How many milliliters of Dextrose 50% must be added to a 300 mL bag of PN solution containing Dextrose 15% so that after the addition, the new dextrose concentration will be equal to 20%? (whole number)
- How many milliliters of Dextrose 70% must be added to a 180 mL bag of PN solution containing Dextrose 12.5% so that after the addition, the new dextrose concentration will be equal to 17.5%? (whole number)
- How many milliliters of Dextrose 50% must be added to a 180 mL bag of PN solution containing Dextrose 12.5% so that after the addition, the new dextrose concentration will be equal to 17.5%? (whole number)
- How many milliliters of Dextrose 70% must be added to a 200 mL bag of PN solution containing Dextrose 12.5% so that after the addition, the new dextrose concentration will be equal to 15%? (whole number)
- How many milliliters of Dextrose 50% must be added to a 200 mL bag of PN solution containing Dextrose 12.5% so that after the addition, the new dextrose concentration will be equal to 15%? (whole number)
- How many milliliters of Dextrose 70% must be added to a 450 mL bag of PN solution containing Dextrose 15% so that after the addition, the new dextrose concentration will be equal to 22.5%? (whole number)
- How many milliliters of Dextrose 50% must be added to a 450 mL bag of PN solution containing Dextrose 15% so that after the addition, the new dextrose concentration will be equal to 22.5%? (whole number)
- How many milliliters of Dextrose 70% must be added to a 150 mL bag of PN solution containing Dextrose 20% so that after the addition, the new dextrose concentration will be equal to 25%? (whole number)
- How many milliliters of Dextrose 50% must be added to a 150 mL bag of PN solution containing Dextrose 20% so that after the addition, the new dextrose concentration will be equal to 25%? (whole number)
- How many grams of 1% bexarotene gel and a gel base should be mixed in order to prepare 60 g of bexarotene 0.75%?
- In what proportion should 5% and 1% ointments be mixed in order to prepare a 2.5% ointment
- You have a 20% solution and an 8% solution. How many milliliters of each should be mixed in order to prepare 4 ounces of a 10% solution?
- You have a 90% solution and a 50% solution. How many milliliters of each should be mixed in order to prepare 6 ounces of a 70% solution?
Answers:
- 1.7%
- 60 mL
- 15.3 mL
- 28.3 mL
- 0.45%
- 67.5 g
- 30 g
- 72 g
- 33 mL
- 57.7 mL
- 2.7%
- 40 g
- 32 mL
- 16.9%
- 12.2%
- 30 mL
- 50 mL
- 17 mL
- 28 mL
- 9 mL
- 14 mL
- 71 mL
- 123 mL
- 17 mL
- 30 mL
- 45 g bexarotene gel and 15 g of base
- 1.5 parts of 5% and 2.5 parts of 1%
- 20% solution = 20 mL, and 8% solution = 100 mL
- 90 mL of each solution
Module 4: Milliequivalents and Milliosmoles
This module will introduce and review calculations involving milliequivalents and milliosmoles and the concentration units mEq/L and mOsm/L.
Module 4A: Equivalents and Milliequivalents
Equivalents and milliequivalents express the concentration of ionic compounds or salts such as sodium chloride, potassium chloride, or sodium bicarbonate. An equivalent is a unit of mass related to a mole and has the symbol Eq. A milliequivalent is simply one thousandth (1/1000) of an equivalent and has the symbol mEq.
The definition of an equivalent is based on the charge of the ions produced when the electrolyte dissolves. Recall that an ionic compound has an equal number of positive and negative charges, so the overall salt charge is zero. Positively charged ions are called cations, and negatively charged ions are called anions. Table 4.1 lists the common ions frequently used in pharmacy and medicine and their charges. The dose of an electrolyte may be ordered in terms of mEq (e.g. potassium chloride 40 mEq by IV infusion over 4 hours) and the concentration of an electrolyte solution may be expressed in mEq/mL or mEq/L (e.g. sodium bicarbonate 50 mEq/L infusion to run at 75 mL/hour).
Table 4.1 Common pharmaceutical ions
Cations | Anions |
Sodium (Na+) | Chloride (Cl–) |
Potassium (K+) | Sulfate (SO42–) |
Calcium (Ca2+) | Dihydrogen phosphate ion (H2PO4–) |
Magnesium (Mg2+) | Acetate (C2H3O2–) |
Lithium (Li+) | Gluconate (C6H11O7–) |
Iron (Fe2+) | Bicarbonate (HCO3–) |
| Carbonate (CO32–) |
An equivalent is the weight, measured in grams, equal to the molecular weight of a substance divided by the total cation or total anion charge in the chemical formula. Ionic charges are always whole numbers of 1 or higher, so the equivalent weight of an electrolyte is less than or equal to its molecular weight.
\(Equivalent\;weight\left( \frac{g}{Eq}\right)=\frac{Molecular\;weight\;(g)}{Total\;cation\;charges}\)
\(Milliequivalent\;wt\left( \frac{mg}{mEq} \right)=\frac{Molecular\;weight\;\left( mg \right)}{Total\;cation\;charges}\)
Sodium chloride (NaCl) has a molecular weight of 58 g/mole. NaCl contains one mole of Na+ and one mole of Cl– ions. The total cation charge of NaCl is equal to 1, and the total anion charge is equal to 1, so each mole of NaCl represents 1 equivalent of NaCl. The equivalent weight of NaCl, therefore, is (58 g/mole)/(1 Eq/mole) = 58 g/Eq. One millimole of NaCl weighs 58 mg, so the milliequivalent weight of NaCl is 58 mg/mEq.
Magnesium sulfate (MgSO4) has molecular weight of 120 g/mole. MgSO4 contains 1 mole of Mg2+ and 1 mole of SO42–. The total cation and anion charges are each 2, so one mole of MgSO4 contains 2 equivalents. The equivalent weight of MgSO4 is therefore (120 g/mole)/(2 Eq/mole) = 60 g/Eq. One millimole of MgSO4 weighs 120 mg, so the milliequivalent weight of MgSO4 is 60 mg/mEq.
Lithium carbonate (Li2CO3) has a molecular weight of 74 g/mole. Li2CO3 contains 2 moles of Li+ and 1 mole of CO32–. The total cation and anion charges are each 2, so one mole of Li2CO3 contains 2 equivalents. The equivalent weight of Li2CO3 is therefore (74 g/mole)/(2 Eq/mole) = 37 g/Eq. One milliequivalent of Li2CO3 weighs 74 mg, so the milliequivalent weight is 37 mg.
Example 4.1: Lithium carbonate is available as capsules containing 600 mg of Li2CO3 per capsule, so the number of milliequivalents per capsule is:
\(600\;mg\;Li_{2}CO_{3}\times\frac{1\;mEq\;Li_{2}CO_{3}}{37\;mg\;Li_{2}CO_{3}}=16.2\;mEq\;LiCO_{3}\;per\;capsule\)
Clinicians sometimes refer to specific ions individually rather than as neutral salt. For example, Li+ is the pharmacologically active component of Li2CO3. The carbonate ion does not contribute to the pharmacologic effect; it is only a convenient salt-forming ion to deliver Li+. The target concentration of Li+ in blood is approximately 1 mEq/L, stated only in terms of the Li+ cation concentration, without reference to any anion. We have calculated that each 600 mg capsule of Li2CO3 represents 16.2 mEq of Li2CO3, but how many mEq of Li+ are present in each capsule? The mEq number for a salt refers to both the cation and the anion, so 16.2 mEq of Li2CO3 contains 16.2 mEq of Li+ and 16.2 mEq of CO32–.
Example 4.2: Human plasma contains approximately 140 mEq of sodium ion per liter. How many grams of NaCl contain the same amount of Na+ as 5 liters of plasma.
\(5\;L\;plasma\times\frac{140\;mEq\;Na^{+}}{L\;plasma}=700\;mEq\;Na^{+}\;in\;5\;L\;of\;plasma\)
Since 1 mEq of NaCl contains 1 mEq of Na+ ion,
\(700\;mEq\;Na^{+}\times\frac{1\;mEq\;NaCl}{1\;mEq\;Na^{+}}\times\frac{1\;mmol\;NaCl}{1\;mEq\;NaCl}\times\frac{58\;mg\;NaCl}{1\;mmol\;NaCl}\times\frac{1\;g}{1000\;mg}=40.6\;g\;NaCl\)
Example 4.3: A solution contains 3.36 g of monobasic potassium phosphate (KH2PO4, MW = 136) and 3.54 g of dibasic potassium phosphate (K2HPO4, MW = 174) in every 15 mL vial. Calculate the potassium ion concentration in mEq/mL.
Both phosphate salts contribute potassium ions to the solution, so the problem is solved by calculating the mEq/mL concentration of each salt individually and then adding the numbers together. The chemical formulas show that KH2PO4 has 1 Eq/mol, while K2HPO4 has 2 Eq/mol.
\(Monobasic:\frac{3.36\;g}{15\;mL}\times\frac{1\;mol}{136\;g}\times\frac{1\;Eq}{mol}\times\frac{1000\;mEq}{Eq}=\frac{1.65\;mEq}{mL}\)
\(Dibasic:\frac{3.54\;g}{15\;mL}\times\frac{1\;mol}{174\;g}\times\frac{2\;Eq}{mol}\times\frac{1000\;mEq}{Eq}=\frac{2.71\;mEq}{mL}\)
\(Total:\;1.65+2.71=4.36=4.4\;\frac{mEq\;K^{+}}{mL}\)
Example 4.4: A pharmacist added 8 mL of sodium phosphates injection to a 250 mL bag of D5W. What is the phosphate concentration in the bag in mmol/mL and mmol/L? What is the sodium concentration in the bag in mEq/mL and mEq/L?
This label shows that sodium phosphates injection contains “3 mM P per mL” and “4 mEq Na+ per mL.”
Please note: “3 mM P” is an unusual and confusing abbreviation for 3 mmol/mL of phosphate. In every other context, “mM” means millimoles/L. DO NOT use “mM” as an abbreviation for mmole in this course.
D5W contains no sodium or phosphate, so the concentrations are 0 mmol/mL phosphate and 0 mEq/mL Na+. The simplest solution to this problem is the C1 · V1 = C2 · V2 method.
Solve for the phosphate concentration:
\(\left( \frac{3\;mmol}{mL} \right)\left( 8\;mL \right)=\left( \frac{x\;mmol}{mL} \right)\left( 8\;mL\;+\;250\;mL \right)\)
\(x\;=\;\frac{0.093\;mmol}{mL}\;\times\;\frac{1000\;mL}{1\;L}\;=\;93\;\frac{mmol}{L}\)
Solve for the sodium concentration:
\(\left( \frac{4\;mEq}{mL} \right)\left( 8\;mL \right)=\left( \frac{x\;mEq}{mL} \right)\left( 8\;mL\;+\;250\;mL \right)\)
\(x\;=\;\frac{0.124\;mEq}{mL}\;\times\;\frac{1000\;mL}{1\;L}\;=\;124\;\frac{mEq}{L}\)
After mixing, the bag contains 0.093 mmol/mL or 93 mmol/L of phosphate and 0.124 mEq/mL or 124 mEq/L of sodium ion.
Example 4.5: A physician orders potassium chloride 30 mEq/L in 250 mL of D5W. How many mL of potassium chloride injection (2 mEq/mL) must be added to the 250 mL bag of D5W.
Notice that the order is written in mEq/L, while the drug vial is labeled as mEq/mL. To use the alligation method, the concentrations must all be expressed in the same units. We will solve this problem in terms of mEq/L.
250 mL of D5W represents 1970 parts, so the proportion to solve the problem is:
\(\frac{250\;mL}{1970\;parts}=\frac{x\;mL}{30\;parts}\)
\(\\\\x=\text{3.8 mL of KCl injection should be added to the 250 mL bag of D5W.}\)
Module 4B: Osmoles and Milliosmoles
Osmolarity is a measure of the osmotic pressure or tonicity of a solution. Recall from biology and physiology courses that water molecules freely diffuse through semi-permeable cell membranes in response to osmotic pressure differences between the intracellular and extracellular solutions. You probably did a lab exercise where you placed a drop of blood on a microscope slide and then watched what happened to the red blood cells when you added different solutions to the slide. A ‘hypotonic’ solution has lower osmolarity than the interior of the blood cells. The solute concentration difference causes water to diffuse into the cells. The cells swell to a larger size and will eventually burst due to the influx of excess water attempting to reduce the solute concentration. This process is called lysis. A ‘hypertonic’ solution has a higher osmolarity than the interior of the cells. Here, the solute concentration difference causes water to diffuse out of the cells, causing their size to decrease. This process is called crenation. Finally, an ‘isotonic’ solution has equal osmolarity to the interior of the cells and does not cause any net change in the cells. Pharmacists must be aware of the osmolarity of the products that are injected or infused into the body, placed into the eye or ear, or inserted into the rectum or vagina to minimize the pain or tissue necrosis effects of hypotonic and hypertonic solutions.
One osmole (Osm) is one mole of dissolved ions or molecules. Every dissolved ion or molecule has the same effect on osmotic pressure, irrespective of molecular weight. One milliosmole (mOsm) is one thousandth (1/1000) of an Osmole. To calculate milliosmoles, you must determine how many millimoles of dissolved ions or molecules result when the solid completely dissolves. For non-electrolytes, i.e., those compounds that do not dissociate into ions in solution, one millimole is equal to one milliosmole. Electrolytes dissociate into cations and anions when they dissolve, so the number of milliosmoles (mOsm) is greater than the number of millimoles (mmol) for the same mass of material. The chemical formula of the electrolyte tells you how many ions can be produced when complete dissociation occurs. If your background includes physical chemistry you are aware that complete dissociation does not occur, but that is beyond the scope of this course. However, this concept is often shown on parenteral product labels where the term osmolarity has calculated (calc) appended to the value. See the label below.
For the purposes of calculations in this course, we define the normal osmolarity of body fluids as 308 mOsm/L. The terms hypotonic (or hypo-osmolar), hypertonic (hyper-osmolar), and isotonic (iso-osmolar) can be defined quantitatively with respect to the normal physiologic osmolarity of 308 mOsm/L. Hypo-osmolar refers to a solution with total solute concentration less than 308 mOsm/L. Iso-osmolar refers to a solution with total solute concentration equal to 308 mOsm/L. Hyper-osmolar refers to a solution with total solute concentration greater than 308 mOsm/L. The maximum osmolarity that should be infused via a peripheral vein is 600 mOsm/L. Any higher osmolarity must be administered via a central venous catheter which terminates into the inferior vena cava, where the blood flow rate is high enough to quickly dilute the solution to physiologic osmolarity.
The formula for sodium chloride is NaCl. One millimole of NaCl dissociates into 1 mmol of Na+ and 1 mmole of Cl– in solution, so one mmol of NaCl produces 2 mOsm of dissolved ions. NaCl, therefore, contains 2 mOsm/mmol.
Potassium chloride (KCl) dissociates into 1 mmol of K+ and 1 mmole of Cl– in solution, so KCL also contains 2 mOsm/mmol.
Magnesium chloride (MgCl2) dissociates into 1 Mg2+ and 2 Cl–, so MgCl2 has 3 mOsm/mmol.
Please note: the “polyatomic ions,” including acetate, sulfate, gluconate, and carbonate, remain intact when they dissolve:
Sodium acetate:
\(NaC_{2}H_{3}O_{2}\;\to \;Na^{+}\;+\;C_{2}H_{3}O_{2}^{\;–}\)
Sodium acetate contains 1 mEq/mmol and 2 mOsm/mmol.
Potassium sulfate:
\(K_{2}SO_{4}\;\to \;2K^{+}\;+\;SO_{4}^{\;2–}\)
Potassium acetate contains 2 mEq/mmol and 3 mOsm/mmol.
Calcium gluconate:
\(Ca(C_{6}H_{11}O_{7})_{2}\;\to \;Ca^{2+}\;+\;2C_{6}H_{11}O_{7}^{\;–}\)
Calcium gluconate contains 2 mEq/mmol and 3 mOsm/mmol.
Sodium carbonate:
\(Na_{2}CO_{3}\;\to \;2Na^{+}\;+\;CO_{3}^{\;2–}\)
Sodium carbonate contains 2 mEq/mmol and 3 mOsm/mmol.
Please note: “waters of hydration” in a chemical formula are ignored when counting milliosmoles per millimole. Magnesium sulfate is produced in several crystalline forms with different waters of hydration and different molecular weights. However, they all have one Mg2+ and one SO42– per mole, so they all have 2 mOsm/mmol.
Table 4.2. Hydrated forms of magnesium sulfate
Crystal form | Formula weight (g/mol) | mOsm/mmol |
MgSO4 (anhydrous) | 120 | 2 |
MgSO4 · H2O | 138 | 2 |
MgSO4 · 3 H2O | 174 | 2 |
MgSO4 · 5 H2O | 210 | 2 |
MgSO4 · 6 H2O | 228 | 2 |
MgSO4 · 7 H2O | 246 | 2 |
Example 4.6: A solution was prepared by dissolving 15 g of magnesium sulfate heptahydrate (MgSO4 · 7 H2O, MW = 246), 10 g of dextrose monohydrate (C6H12O6 · H2O MW = 198), and 1.8 g of sodium chloride (NaCl, MW = 58) in enough water to make 200 mL. Calculate the osmolarity (mOsm/L) of the solution.
There are 3 solutes with different molecular weights and mOsm/mmol values. Calculate the mOsm/L concentration for each solute and add the numbers together.
\(Mag\;sulf:\;\frac{15\;g\;MS}{200\:mL}\times\frac{1000\;mL}{L}\times\frac{1\;mol\;MS}{246\;g\;MS}\times\frac{2\;Osm}{1\;mol\;MS}\times\frac{1000\;mOsm}{1\;Osm}=610\frac{mOsm}{L}\)
\(NaCl:\;\frac{1.8\;g\;NaCl}{200\:mL}\times\frac{1000\;mL}{L}\times\frac{1\;mol\;NaCl}{58\;g\;NaCl}\times\frac{2\;Osm\;NaCl}{1\;mol\;NaCl}\times\frac{1000\;mOsm}{1\;Osm}=310\frac{mOsm}{L}\)
\(Dextrose: \frac{5\;g}{100\;mL}\times \frac{1000\;mL}{L}\times \frac{1\;mol}{198\;g}\times \frac{1\;Osm}{mol}\times \frac{1000\;mOsm}{Osm}=253\frac{mOsm}{L}\)
The total solution osmolarity is: 610 + 310 + 253 = 1173 mOsm/L.
This solution is very hyperosmolar.
The osmolarity of injectable solutions is usually printed on the label. A 10% calcium chloride injection has an osmolarity of 2.04 mOsm/mL. A 0.45% sodium chloride injection, also referred to as half normal saline or ½ NS, has an osmolarity of 154 mOsm/L or 0.154 mOsm/mL.
Example 4.7: A patient requires a 2 g dose of calcium chloride in ½ NS by IV infusion. The physician wants the solution to be as close to iso-osmolar as possible. ½ NS is available in bags containing 100, 250, 500, or 1000 mL. Which bag size should be used?
To solve the problem, determine the volume of calcium chloride injection needed, then calculate the volume of ½ NS that would produce an iso-osmolar solution. Finally, choose the bag size closest to the calculated volume of ½ NS.
\(CaCl_{2}\;volume:\:2\;g\;CaCl_{2}\times\frac{100\;mL\;injection}{10\;g\;CaCl_{2}}=20\;mL\;CaCl_{2}\;injection\)
Volume of ½ NS to make an iso-osmolar solution (osmolarities expressed as mOsm/mL)
The volume of CaCl2 injection is 20 mL, so the proportion to solve for ½ NS volume is:
\(\frac{0.154\;parts}{20\:mL}=\frac{1.732\;parts}{x\:mL},\:x=224.9\;mL\)
Adding 20 mL CaCl2 injection to 225 mL of ½ NS would produce an iso-osmolar solution. The closest available bag size to 225 mL is 250 mL, so the infusion should be prepared with 250 mL of ½ NS.
Milliequivalents and milliosmoles are related to each other through millimoles, while millimoles and milligrams are related through molecular weight. These relationships allow conversion between milligrams, millimoles, milliequivalents, and milliosmoles for any substance.
Example 4.8: Calculate the osmolarity of a solution containing 30 mEq/L of KCl in water.
KCl has 1 mEq/mmol and 2 mOsm/mmol. These conversions can be used individually or combined into the single conversion 1 mEq/2 mOsm to solve the problem.
\(\frac{30\;mEq\;KCl}{L}\times\frac{1\;mmol\;KCl}{1\:mEq\;KCl}\times\frac{2\;mOsm\;KCl}{1\;mmol\;KCl}=60\frac{mOsm\;KCl}{L}\)
\(Or,\;\frac{30\;mEq\;KCl}{L}\times\frac{2\;mOsm\;KCl}{1\:mEq\;KCl}=60\frac{mOsm\;KCl}{L}\)
Example 4.9: A patient is ordered 60 mmol of potassium phosphates in 250 mL of 0.9% sodium chloride injection, also called normal saline or NS. Would this solution be safe to administer in the peripheral vein?
The osmolarity limit for peripheral infusion is 600 mOsm/L. Calculate the solution's osmolarity as ordered and determine whether it is greater than or less than 600 mOsm/L.
KPhos injection contains 3 mmol/mL of phosphate and 7.4 mOsm/mL osmolarity. NS contains 308 mOsm/L or 0.308 mOsm/mL osmolarity. Calculate the milliosmoles contributed to the total from each solution, then divide by the total volume in liters.
\(60\;mmol\;KPhos\times\frac{1\;mL}{3\;mmol\;KPhos}=20\;mL\;of\;injection\times\frac{7.4\;mOsm}{mL}=148\;mOsm\)
\(\\\\\text{The volume of KPhos injection required is 20 mL, and it contributes 148 mOsm to the mixture.}\)
\(NS\;mOsm:\:250\;mL\times\frac{0.308\;mOsm}{mL\;NS}=77\;mOsm\)
NS 250 mL contributes 250 mL of volume and 77 mOsm.
The mixture, therefore, contains 148 + 77 = 225 mOsm in a total volume of 20 + 250 = 270 mL. The osmolarity is then calculated as:
\(\frac{225\;mOsm}{270\;mL}\times\frac{1000\;mL}{L}=833.3\frac{mOsm}{L}\)
The maximum osmolarity for peripheral infusion is 600 mOsm/L, so this order with an osmolarity of 833 mOsm/L should be administered through a central venous catheter.
Module 4: Practice Problems
- Calculate the mEq/L concentration of a 15% potassium chloride (MW 74.6) solution.
- A solution is prepared by dissolving 410 grams of sodium acetate (MW 82) in enough water to make 2.5 liters. Calculate the concentration in mEq/L.
- Sodium bicarbonate injection is a sterile solution containing 8.4% sodium bicarbonate (MW 84) in water. A pharmacist added 150 mL of sodium bicarbonate injection to a 1 L bag of sterile water for injection. Calculate the solution concentration in mEq/L.
- Calculate the osmolarity of a solution prepared by adding 100 mL of sodium bicarbonate injection (50 mEq/50 mL) to 1000 mL of sterile water for injection.
- One liter of solution contains 6 g of sodium chloride (MW 58), 300 mg of potassium chloride (MW 74.6), 200 mg of calcium chloride dihydrate (MW 146), and 3.1 g of sodium lactate (NaC3H5O3, MW 112). Calculate the sodium concentration in mEq/L. Calculate the chloride concentration in mEq/L.
- A supplement product contains 1.25 g of calcium gluconate (MW 430) per tablet. How many milliequivalents of calcium does the patient receive if the dose is 2 tablets?
- A product contains 17.5 g of sodium sulfate (Na2SO4 MW 142), 3.1 g of potassium sulfate (K2SO4 MW 174), and 1.6 g of magnesium sulfate (MgSO4 MW 120) in 180 mL of solution. Calculate the sulfate concentration in mmol/L and mEq/L.
- A product contains 17.5 g of sodium sulfate (Na2SO4 MW 142), 3.1 g of potassium sulfate (K2SO4 MW 174), and 1.6 g of magnesium sulfate (MgSO4 MW 120) in 180 mL of solution. Calculate the solution's osmolarity in mOsm/L.
- Calculate the osmolarity of a 3% sodium chloride ophthalmic solution. Is the solution hypo-, iso-, or hyperosmolar?
- Calculate the osmolarity of a sterile 10% sodium sulfacetamide (NaC8H10N2O3S; MW 236) ophthalmic solution. Is the solution hypo-, iso-, or hyperosmolar?
- Calculate the osmolarity of 25% mannitol injection (non-electrolyte; mw 182). Can this product be safely infused into a peripheral vein?
- Calculate the osmolarity of a solution prepared by adding 2 g of calcium chloride dihydrate (MW 147) using 10% calcium chloride dihydrate injection, USP. The injection volume is added to 250 mL of normal saline. Can this product be infused into a peripheral vein?
- One liter of a total parenteral nutrition (TPN) solution for newborns has 10% dextrose monohydrate (198 g/mol) and 2.5 % amino acids (along with several additives that do not significantly affect the osmolarity). The source of amino acids is 7% amino acids in water with an osmolarity of 561 mOsm/L. Can the TPN solution be infused into a peripheral vein?
- Calculate the osmolarity of an oral colon gavage solution containing:
Sodium sulfate 21.5 g (142 g/mole)
Sodium chloride 5.53 g (58.5 g/mole)
Potassium chloride 2.82 g (74.5 g/mole)
Sodium bicarbonate 6.36 g (84 g/mole)
*PEG 3350 227.1 g (3350 g/mole; non-electrolyte
Water qs 4L - Calculate the osmolarity of 20 mmol of sodium phosphates IV added to 250 mL D5W. Is the solution safe to administer via a peripheral vein?
- Calculate the osmolarity of 20 mmol of sodium phosphate IV added to 100 mL ½ NS. Is the solution safe to administer via a peripheral vein?
- Calculate the osmolarity of calcium gluconate (mw 430) 2 g in 500 mL of sterile water. Is the solution safe to administer via a peripheral vein?
- Calculate the volume of ½NS that should be added to 30 mEq of KCl injection solution to produce an iso-osmolar solution.
- Calculate the volume of sterile water for injection that should be added to 30 mmol of sodium phosphates to produce an iso-osmolar solution.
- Calculate the volume of 50 mM NaCl solution that should be added to 1 gram of calcium chloride dihydrate solution to produce an iso-osmolar solution.
Answers:
- 2010 mEq/L
- 2000 mEq/L
- 130 mEq/L
- 182 mOsm/L
- 131 mEq/L Na+ and 110 mEq/L Cl–
- 11.6 mEq Ca2+
- 857.7 mM; 1715 mEq/L
- 2500 mOsm/L
- 1027 mOsm/L
- 848 mOsm/L
- 1374 mOsm/L; osmolarity too high for peripheral vein
- 437 mOsm/L; okay to give by peripheral vein
- 704 mOsm/L; osmolarity too high for peripheral vein
- 235 mOsm/L
- 427 mOsm/L; okay to give by peripheral vein
- 582 mOsm/L; okay to give by peripheral vein
- 28 mOsm/L; osmolarity may be too low for peripheral vein
- 360 mL
- 217 mL
- 83 mL
Module 5: Dose Calculation and Dose Checking
We believe pharmacists should be responsible for verifying the right drug and appropriate dose of their patient’s prescribed medication. As you advance through the curriculum, you will learn how to apply the knowledge gained from your coursework to the correct selection of the pharmacologic agent. And it is probably clear that one dose cannot be appropriate for all patients. You will learn to use guidelines and dosing recommendation references. Those references often suggest dosing based on body weight, body surface area, estimated renal function using creatinine clearance, or liver function. Ethically, we are bound to consult with physicians when experience informs us that the drug or dose may be incorrect. This module will provide practice in calculating correct amounts based on different dosing recommendations.
Module 5A: Patient-Specific Dosing (mg/kg)
Many drugs used in pediatrics are dosed according to body weight (kilograms) or body surface area (m2). In this section, we will focus on mg/kg dosing. Typical calculations involve checking a prescribed dose.
Example 5.1: A patient weighs 44 pounds. Recall that to convert patient weight to kilograms, divide the weight in pounds by 2.2. The physician orders Amoxicillin 250 mg PO tid, targeting 30 – 45 mg/kg/day. What is the mg/kg value per day and dose basis? Is the dose consistent with the guideline?
\(44\;lb\times \frac{1\;kg}{2.2\;lb}=20\;kg\)
\(\frac{\frac{750\;mg}{day}}{20\;kg}=37.5\;\text{mg/kg/day}\)
\(\frac{\frac{250\;mg}{dose}}{20\;kg}=12.5\;\text{mg/kg/dose}\)
You can check the medication order and see that the physician’s prescribed dose is consistent with the recommended guidelines. When presented with the drug and dosing schedule, divide the dose by the patient’s weight and compare the answer to the recommendation.
Practice problems.
1. A physician orders an IV loading dose of Remdesivir of 75 mg for a 33-pound patient. The recommended dose is 5 mg/kg. Is the dose correct?
75 mg/15 kg = 5 mg/kg. The dose is correct.
2. A physician orders Erythromycin ethylsuccinate 300 mg PO tid for a 50-pound patient. The recommended dose is 40 – 50 mg/kg/day. Is the dose correct?
900 mg/day/22.7 kg = 40 mg/kg/day. The dose is correct.
3. A physician orders Gentamicin sulfate 50 mg IV q 8° for a 22-pound patient. The recommended dose is 7.5 mg/kg/day. Is the dose correct?
150 mg/day/10 kg = 15 mg/kg/day. The dose is too high. Contact the physician.
4. A physician orders Gentamicin sulfate 50 mg IV for a 44-pound patient. The recommended dose is 2.5 mg/kg/dose. Is the dose correct?
50 mg/dose/20 kg = 2.5 mg/kg/dose. The dose is correct.
5. The recommended dose for Bactrim (Sulfamethoxazole 200 mg and Trimethoprim 40 mg per 5 mL) based on the TMP component, is 6 – 12 mg/kg/day. Physician orders 2-teaspoonsful
q 12° for a 22-pound child. Is the dose correct?
160 mg TMP/day/10 kg = 16 mg/kg/day. The dose is too high. Contact the physician.
Some of you will have the opportunity to “write” the drug orders for some patients. Let’s see how this works. I will also give you the product concentrations so you may practice writing realistic volumes along with the calculated doses.
Example 5.2: A full-term newborn starts on Digoxin using the TDD 30 mcg/kg protocol. (TDD = total digitalizing dose). The recommended dosing schedule is ½ the TDD initially, followed by ¼ of the TDD for each subsequent dose at 6- to 8-hour intervals for 2 doses. The newborn weighs 6 pounds and 9 ounces. Digoxin oral solution is available as a 50 mcg/mL product. What dose and volume should you calculate for the two different doses?
\(6\;lb+\left( 9\;oz\times \frac{1\;lb}{16\;oz} \right)=6.5625\;lb\)
\(6.5625\;lb\times \frac{1\;kg}{2.2\;lb}=2.98\;kg\)
The doses are divided into 15 mcg/kg for the first dose, then 7.5 mcg/kg for two additional doses. The TDD = 30 mcg/kg.
\(2.98\;kg\times \frac{15\;mg}{kg}=44.7\;mg,\;\text{round to 45 mg}\)
\(45\;mcg\times \frac{1\;mL}{50\;mcg}=0.9\;mL\)
This is the initial dose. The next two doses are administered anywhere from 6 – 12 hours later, depending on the cardiologist's protocol.
\(2.98\;kg\times \frac{7.5\;mg}{kg}=22.4\;mg,\;\text{round to 22.5 mg}\)
\(22.5\;mcg\times \frac{1\;mL}{50\;mcg}=0.45\;mL\)
The volume of the first dose will be 0.9 mL, and the two subsequent dose volumes will be 0.45 mL.
Practice problems.
6. Acetaminophen solution contains 160 mg/5 mL. The recommended children's dose is 10 – 15 mg/kg/dose. What reasonable volume of solution should you recommend for a child who weighs 48 pounds?
21.8 kg x 10 – 15 mg/kg = dosage range 220 mg – 330 mg.
220 mg/160 mg/5 mL = 6.9 mL. 330 mg/160 mg/5 mL = 10.3 mL.
A reasonable dose is 240 mg (1.5 teaspoonsful or 7.5 mL)
7. Amoxicillin suspension contains 250 mg/5 mL. The recommended children's dose is 15 mg/kg/dose. What reasonable volume of solution should you recommend for a child who weighs 23 pounds?
10.5 kg x 15 mg/kg/dose = 157.5 mg.
157.5 mg/50 mg/mL = 3.15 mL.
A practical, measurable dose is 150 mg or 3 mL.
8. Fer-in-sol solution contains 15 mg/mL of elemental iron (75 mg/mL of ferrous sulfate). A recommended dose for iron deficiency anemia is 2 mg/kg/dose. What reasonable volume of solution should you recommend for a child who weighs 13 pounds?
5.91 kg x 2 mg/kg/dose = 11.8 mg.
11.8 mg/15 mg/mL = 0.79 mL.
A reasonable volume with a 1 mL oral syringe is 0.8 mL or 12 mg.
9. Amikacin sulfate injection contains 250 mg/mL. A recommended pediatric dose is 7 mg/kg/dose. What reasonable volume of solution should you recommend for a child who weighs 23 pounds?
10.5 kg x 7 mg/kg/dose = 73.5 mg.
73.5 mg/250 mg/mL = 0.29 mL.
A reasonable volume with a 1 mL syringe is 0.3 mL or 75 mg.
Module 5B: Body Surface Area
Some drugs are dosed based on the patient’s body surface area value, calculated in square meters (m2). Dubois and Dubois published some of the earliest works on BSA determination in 1916. Mosteller published a simplified equation in 1987, which is routinely used today. The equation uses the patient's weight in kg and their height in centimeters. You learned how to do these conversions in Module 1. Now, you finally have the opportunity to use the square root key on your calculator! 😊. Calculate all BSA values to 2 decimal places, for example, 1.62 m2.
To calculate the patient’s BSA, use this formula, using the patient’s actual body weight:
\(BSA\;(m^{2})=\sqrt{\frac{wt\;(kg)\;\times\;ht\;(cm)}{3600}}\)
Example 5.3: A patient weighs 21 kg and is 3’ 10” in height. Calculate the BSA. 3’ 10” = 46” = 117 cm.
\(BSA=\sqrt{\frac{21\;kg\;\times\;117\;cm}{3600}}=\sqrt{0.6825}=0.83\;m^{2}\)
These problems require you to calculate the BSA, then multiply the dose given in mg/m2 to obtain the dose.
Example 5.4: Calculate the dose of Allopurinol for a patient who weighs 25 kg and is 4’ 2”. The recommended dose is 200 mg/m2.
\(BSA(m^{2})=\sqrt{\frac{25\;kg\;\times\;127\;cm}{3600}}=\sqrt{0.8819}=0.94\;m^{2}\)
\(0.94\;m^{2}\times \frac{200\;mg}{m^{2}}=188\;mg\)
Example 5.5: Calculate the dose of Methotrexate for a patient who weighs 48 kg and is 5’ 3”. The recommended dose is 12 g/m2.
5’ 3” = 63” = 160 cm
\(BSA(m^{2})=\sqrt{\frac{48\;kg\;\times\;160\;cm}{3600}}=\sqrt{2.1333}=1.46\;m^{2}\)
\(1.46\;m^{2}\times \frac{12\;g}{m^{2}}=17.5\;g\)
Example 5.6: A physician prescribes hydrocortisone 20 mg for a patient. The recommended dose is 20 – 25 mg/m2/day. Is the dose correct? The patient weighs 22 kg and is 118 cm tall.
\(BSA(m^{2})=\sqrt{\frac{22\;kg\;\times\;118\;cm}{3600}}=\sqrt{0.7211}=0.85\;m^{2}\)
\(\frac{20\;mg}{0.85\;m^{2}}=\frac{23.5\;mg}{m^{2}}\)
The dose is within the recommended range.
Practice problems:
10. Caspofungin is an anti-fungal drug administered IV (over 1 hour) as a 70 mg/m2 loading dose the first day, followed by 50 mg/m2 daily thereafter. Calculate the loading dose and maintenance doses for a patient who weighs 24 kg and is 118 cm tall.
11. Dexamethasone is used in treating many conditions. A typical dosing range is 0.6 – 9 mg/m2/day in 3 or 4 divided doses. It is available as an oral solution containing 0.5 mg/5 mL.
Calculate the amount and solution volume for a patient who weighs 18 kg and is 108 cm tall, based on 1.5 mg/m2/day of dexamethasone divided into three doses. What volume of the oral solution (0.5 mg/5 mL) should be taken at each dose?
12. Dronabinol is used for nausea prophylaxis in patients receiving chemotherapy. The National Cancer Institute (NCI) guideline for pediatric patients is 5 mg/m2 PO every 6 – 8 hours before beginning chemotherapy treatment, then 5 mg/m2 PO every 4 – 6 hours for up to 12 hours after the treatment. Calculate the dose for a patient who weighs 23 kg and is 114 cm tall.
13. Acyclovir is used to treat viral infections in children. The dosing guideline can range up to 600 mg/m2 every 6 hours for 10 days. The drug may be administered orally as a 40 mg/mL suspension or by an intermittent IV infusion. Each sterile vial (10 mL) contains acyclovir sodium equivalent to 500 mg of acyclovir. Calculate the dose for a patient who weighs 26 kg and is 125 cm tall.
Module 5C: Estimating Body Surface Area Anatomically
Several medical researchers devised an estimate of the body surface area based on anatomical regions from 1944 to 1960. These include Berkow, Boyd, Brewer, Chu, Lund, Pulaski, Tennison, and Wallace. These individuals were searching for a rapid way to estimate the areas involved with severe burns. The method is often referred to as the Rule of Nines. While the method has been criticized as selective, it has clinically proven its worth. Further information can be found on Wikipedia pages. One other note, many physicians use the rule of thumb that involves the area of the patient's palm. The patient's palm approximates 1% of the body surface area.
The method is called the Rule of Nines because the body can be divided into 11 approximate areas with a common factor of 9% (head (1), arms (2), chest/abdoman (2), back (2), legs (4)) . The table explains the concept better than words. Note that children differ in some key areas. We’ll restrict our problems to adults. Note that in the color cartoon, a distinction is made between front and back. The entire torso represents approximately 36% of the BSA value.
Body Part | Estimated BSA Percentage | |
Adults | Children | |
Left arm | 9% | 9% |
Right arm | 9% | 9% |
Head & neck | 9% | 18% |
Chest | 9% | 9% |
Abdomen | 9% | 9% |
Back | 18% | 18% |
Left leg | 18% | 14% |
Right leg | 18% | 14% |
Example 5.7: An adult oncology patient had their left leg amputated at the hip because of an osteogenic sarcoma tumor. Chemotherapy treatment includes methotrexate 15 g/m2. The patient’s pre-surgical BSA was 1.7 m2. What dose of methotrexate should be considered? Preoperatively, the patient would have received:
\(1.7\;m^{2}\times \frac{15\;g}{m^{2}}=25.5\;g\)
The loss of the left leg results in an 18% reduction in body surface area, so the chemo dose will also need to be reduced by 18%:
25.5 g × 82% = 20.9 g
Example 5.8: A treatment protocol requires the administration of IV fluids at a rate of 2 L per m2 per day. The patient had their right arm removed because of an accident. Before the accident, the patient’s BSA was 1.9 m2. What fluid rate (mL/h) should the patient receive?
2 L/ m2/day × 1.9 m2 = 3.8 L/day
The loss of the right arm results in a 9% surface area reduction, so:
3.8 L/day × 0.91 = 3.458 L/day
3.458 L/day ≡ 3458 mL/24 h = 144 mL/h
Module 5D: Ideal Body Weight
The impact of height and weight on medical conditions has been studied since the early 1900s. Studies from the 1950s used the Metropolitan Life Insurance Company height and weight tables to understand these two variables' influence on health. One outcome of that statistical research was the development of the “ideal” body weight (IBW) for adult women and men. This concept is intimately associated with dosing for certain drugs. Devine (1974), Robinson (1983), and Miller (1983) have developed equations to estimate the IBW. A Google search will give you a good starting point for further reading if you are interested. A good review of the topic is the paper by Pai and Paloucek in the Annals of Pharmacotherapy (2000). As you progress through the curriculum, you will use the concept of IBW for calculating doses and estimating a patient’s creatinine clearance.
We will use the Devine equation in this course.
For Females: IBW = 45.5 kg + 2.3 kg for each inch in height over 60 inches.
For Males: IBW = 50 kg + 2.3 kg for each inch in height over 60 inches.
Example 5.9: Calculate the IBW for a female patient who is 5’ 5” tall.
45.5 kg + 2.3 kg × (65 – 60”) = 57 kg
Example 5.10: Calculate the IBW for a male patient who is 5’ 11” tall.
50 kg + 2.3 kg × (71 – 60”) = 75.3 kg
Module 5E: Adjusted Body Weight
Certain drugs, when dosed on the patient’s total body weight, may lead to under or overdoses for patients defined as obese. While somewhat controversial, obesity is usually defined as the patient weighing more than 130% of their ideal body weight.
In some therapeutic situations, it may be more appropriate to use the adjusted body weight for patients who are obese. There is a continuing controversy over the use of these formulas in patients. We recommend you follow the guidelines in place at your institution.
The adjusted body weight formula can be confusing because it contains 2 words starting with the letter A. The adjusted body weight and the actual body weight are used in the formula, so pay close attention to the calculation. The same equation is used for women and men. Also, note that the adjustment factor can vary from one institution to another. The factor is usually within the range of 0.25 to 0.4. While this factor can vary among institutions and practitioners, for all calculations in this book we will use 0.4 as the adjustment factor.
This equation should only be used when the product labeling or institutional policy specifies its use.
Adj BW = IBW + 0.4 x (Actual BW – IBW)
Example 5.11: Calculate the Adj BW (0.4) for a female patient weighing 194 pounds and 5 feet 4 inches tall.
IBW = 45.5 kg + 2.3 kg × (64 – 60”) = 54.7 kg
194 lb = 88.2 kg
Adj BW = 54.7 kg + 0.4 × (88.2 – 54.7) = 68.1 kg
Module 5F: Creatinine Clearance
Many drugs are cleared from the body by renal excretion. Depending on the drug, its clearance may be significant or minor. Reduced kidney function can lead to potentially toxic drug levels if renal excretion is a major clearance route.
The kidneys receive about 20% of the cardiac output, which equates to a normal male glomerular filtration rate (GFR) of 90 – 120 mL/minute. Serum creatinine has been used as a marker of renal function for many years. Creatinine is a breakdown product of creatine phosphate resulting from the metabolism of muscle and protein. In 1976, D. W. Cockcroft and M. H. Gault published an equation that attempted to estimate an adult patient’s GFR using serum creatinine values. While the equation has been criticized over the years, (for example, see the National Kidney Foundation website https://www.kidney.org/professionals/KDOQI/gfr_calculatorCoc), the C-G CrCL equation remains useful for many drugs. For most patients and most drugs, it has been demonstrated that the Cockcroft and Gault equation estimates the GFR as well as the MDRD equation. An added benefit to using the C-G equation is related to its long historical use in various drug studies. During your career, it is possible that more accurate estimates of renal function may replace the C-G equation. For completeness, we note that the C-G CrCL overestimates the actual glomerular filtration rate by 10 – 20% due to the active secretion of creatinine by the peritubular capillaries.
Now, let’s see the equations.
\(CrCL_{male}=\frac{(140-age)\;\times \;wt\;(kg)}{72\;\times \;SCr}\)
where wt is the lower of ideal or actual body weight and SCr is the patient’s serum creatinine level in mg/dL.
\(CrCL_{female}=0.85\times \frac{(140-age)\;\times \;wt\;(kg)}{72\;\times \;SCr}\)
The units for the CrCL are mL/min. Round the value to the closest whole (integer) number.
Please note: The most common question students ask about this equation is which body weight should be used. Calculate ideal body weight and actual body weight in kilograms. Then use the lower of these values in the equation.
Please note: The most common error students make when using this equation is forgetting to multiply by 0.85 for female patients.
Example 5.12: A male patient is 60 years old, 5’ 6” tall, weighs 145#, with an ideal body weight of 140#. His serum creatinine is 1.2 mg/dL. Calculate his CrCL.
\(CrCL_{male}=\frac{(140\;-\;60)\;\times\;63.8\;kg}{72\;\times\;1.2}=59\;mL/min\)
Example 5.13: A female patient is 45 years old, 5’3” tall, weighs 112#, with an ideal body weight of 115#. Her serum creatinine is 1.6 mg/dL. Calculate her CrCL.
\(CrCL_{female}=0.85\times \frac{(140\;-\;45)\;\times\;50.9\;kg}{72\;\times\;1.6}=36\;mL/min\)
BSA-Adjusted Creatinine Clearance
Certain drugs are dosed based on Body Surface Area adjusted Creatinine Clearance. The creatinine clearance calculation is dependent on the patient's muscle mass, so smaller or less muscular patients will have a lower CrCL value than larger more muscular patients, even if both patients have similar kidney function. BSA adjustment is done in an attempt to account for this effect.
The formula used to adjust the patient's CrCL for BSA is:
\(BSA\;adjusted\;CrCL\;(mL/min/1.73\;m^{2})=CrCL (mL/min)\times \frac{1.73}{Patient's\;BSA}\;m^2\)
The value 1.73 m2 is used because it has traditionally represented the average adult male BSA value. You may question the validity of using 1.73 m2 in light of the changes to the size of the average man in the U.S. since the factor was introduced. Like the C-G CrCL equation, the argument is made that the equation has proved its worth in clinical situations thereby justifying its continued use. Who knows, you may see a change in the value of that factor during your professional career. In summary, the BSA-adjusted CrCL is used to allow easier comparison of an individual patient's CrCL to the average normal adult.
Consider an example. Patient A is 40 year old female patient, 5' tall and 45.5 kg (ideal body weight). Patient B is a 40 year old female, 6' tall and 73.1 kg (ideal body weight). Both patients have serum creatinine of 1.1 mg/dL.
Table 5.1. Creatinine Clearance for Example Patients A and B.
| Statistic | Patient A | Patient B |
| Age | 40 | 40 |
| Height (feet) | 5 | 6 |
| Weight (kg) | 45.5 | 73.1 |
| Creatinine Clearance (mL/min) | 50 | 79 |
| BSA (m2) | 1.39 | 1.93 |
| BSA-adjusted CrCL (mL/min/1.73 m2) | 62 | 71 |
Patient A has CrCL of 50 mL/min, which would indicate possible minor decrease in kidney function. However, the BSA-adjusted CrCL calculation indicates that the patient's kidney function is actually better than CrCL would suggest.
Similarly, Patient B has CrCL of 79 mL/min, which would suggest good kidney function. However, the BSA-adjusted CrCL calculation indicates that the patient's kidney function is not quite as good as CrCL would suggest.
Example 5.14: A male patient is 60 years old, 5’ 6” tall, weighs 145#, with an ideal body weight of 140#. His serum creatinine is 1.2 mg/dL. Calculate his BSA-adjusted CrCL.
There are a few steps to the calculation:
i. Calculate CrCL using the lower of ideal or actual body weight.
Actual body weight = 145 lb/2.2 = 65.9 kg
Ideal body weight = 50 kg + 2.3 kg (66" - 60") = 63.8 kg. Use ideal body weight for CrCL.
\(CrCL_{male}=\frac{(140\;-\;60)\;\times\;63.8\;kg}{72\;\times\;1.2}=59\;mL/min\)
ii. Calculate BSA using actual body weight
\(BSA\;(m^{2})=\sqrt{\frac{65.9\;kg\;\times\;168\;cm}{3600}}=1.75\;m^{2}\)
This patient's BSA is very close to the normal value of 1.73 m2, so BSA-adjusted CrCL will be very similar to the non-adjusted CrCL value.
iii. Calculate the BSA-adjusted CrCL.
\(59\;mL/min\;\times\;\frac{1.73}{1.75}\;m^2=58\;mL/min/1.73\;m^{2}\)
Example 5.15: Calculate the BSA-adjusted CrCL for a 34 year old male, 5'1" tall and 51 kg, with SCr of 0.9 mg/dL.
\(CrCL=\frac{(140\;-\;34)\;\times\;51\;kg}{72\;\times\;0.9\;mg/dL}=83\;mL/min \)
\(BSA\;(m^{2})=\sqrt{\frac{51\;kg\;\times\;152\;cm}{3600}}=1.47\;m^{2} \)
\(BSA-adjusted\;CrCL=83\;mL/min\times \frac{1.73}{1.47}\;m^2=98\;mL/min/1.73\;m^{2}\)
In this case, the patient's BSA is smaller than 1.73 m2, so the BSA-adjusted CrCl is larger than the previous example.
Example 5.16: Calculate the BSA-adjusted CrCL for the female patient in example 5.13. Recall that the C-G equation uses the lesser value for actual body weight or ideal body weight. The Mosteller equation for BSA uses the patient's actual body weight.
\(BSA\;(m^{2})=\sqrt{\frac{50.9\;kg\;\times\;160\;cm}{3600}}\;=\;\sqrt{2.26}=1.5\;m^{2} \)
\(CrCL_{BSA-adj}\;=\;36\;mL/min\;\times\;\frac{1.73}{1.5}\;m^{2}\;=\;42\;mL/min\)
Please note: BSA-adjusted CrCL has units of mL/min/1.73 m2.
Please note: Only use BSA-adjusted CrCL for patient dose calculations when the guidelines for a particular drug specify that it should be used. As an example, the guideline for a particular drug states that the standard dose for adults with BSA-adjusted CrCL > 70 mL/min/1.73 m2 is 250 mg every 6 hours. For a patient of the same weight, but with BSA-adjusted CrCL of 30 mL/min/1.73 m2 should be administered 250 mg every 12 hours, or 1/2 the daily dose of the patient with normal renal function.
Pediatric Estimation of Creatinine Clearance
The Cockcroft & Gault equation was developed using data from male patients over 30.
In 1976, George Schwartz and a team of researchers published a new equation for estimating the glomerular filtration rate (eGFR) in pediatric patients from one week old to 17 years old. The CG equation was used for patients 18 years and older. Schwartz’s equation was updated in 2009 and is currently considered the best estimate of the GFR in that age range. Similar to the CG equation, the primary assumption is that the renal function is not acutely changing, and the serum creatinine value is constant. The original equation used six terms with multiple exponents. Applying some judicious statistical assumptions resulted in a more concise equation colloquially called the “bedside Schwartz.” The bedside Schwartz equation predicts the estimated GFR with the units mL/min/1.73 m2.
\(eGFR \left( mL/min/1.73^{2} \right) = 0.413\:\times\:\left( \frac{ht\;\left( cm \right)}{SCr\;\left( mg/dL \right)} \right)\)
Example 5.17: Calculate the eGFR for a 6-yo male patient who is 45” tall and has a SCr = 1.5.
\(eGFR = 0.413\: \times \:\left( \frac{114}{1.5} \right)\:=\: 31\: mL/min/1.73^{2}\)
Example 5.18: Calculate the eGFR for a 9-yo female patient who is 52” tall and has a SCr = 1.8.
\(eGFR = 0.413\: \times \:\left( \frac{132}{1.8} \right)\:=\: 30\: mL/min/1.73^{2}\)
Example 5.19: Calculate the eGFR for a 14-yo female patient who is 5 ' 3” tall with a SCr = 0.8.
\(eGFR = 0.413\: \times \:\left( \frac{160}{0.8} \right)\:=\: 83\: mL/min/1.73^{2}\)
Example 5.20: Calculate the eGFR for a 5-yo male patient who is 3' 4” tall with a SCr = 0.7.
\(eGFR = 0.413\: \times \:\left( \frac{102}{0.7} \right)\:=\: 60\: mL/min/1.73^{2}\)
Module 5G: Body Mass Index (BMI)
The body mass index was first used in a paper from 1972 in the Journal of Chronic Diseases. The calculation was first proposed in about 1840 to represent a scale useful in comparing different sized people. The application and interpretation of the BMI continues to invoke some controversy in medicine. We will not take a position on the socio-political ramifications of the index. Our task is to focus on the calculation and application of the index in appropriate clinical situations. The calculation is intended to be used in people over 20 years old. Please note, as with BSA-adjusted CrCl, only use BMI in dose calculations when a specific drug guideline tells you to.
Please note: BMI can be confusing because it is not used directly to calculate a drug dose. Rather, it is usually used to decide whether to use ideal body weight, actual body weight, or adjusted body weight in the dose calculation.
The calculation of the BMI is straightforward.
\(BMI=\frac{Weight\;(kg)}{(Height\;(m))^{2}}\)
This formula is an application of earlier conversion factors. Recall the conversion process for height in feet and inches to meters.
\(6'4^{"}=76^{"}\times \frac{2.54\;cm}{inch}\times \frac{1\;m}{100\;cm}=1.93\;m\)
Example 5.17: Calculate the BMI for a 35 yo patient who weighs 180 pounds and is 5’ 7” tall.
\(BMI=\frac{180\;lb\;\times\;\frac{1\;kg}{2.2\;lb}}{(67\;inches\;\times\;\frac{2.54\;cm}{inch}\;\times \;\frac{1\;m}{100\;cm})^{2}}\;=\;\frac{81.82\;kg}{\left( 1.7\;m \right)^2}\;=\;28.3\;kg/m^{2}\)
Example 5.18: Calculate the BMI for a 22 yo patient who weighs 215 pounds and is 5’ 8” tall.
\(BMI=\frac{215\;lb\;\times\;\frac{1\;kg}{2.2\;lb}}{(68\;inches\;\times\;\frac{2.54\;cm}{inch}\;\times\; \frac{1\;m}{100\;cm})^{2}}\;=\;\frac{97.7\;kg}{\left( 1.73\;m \right)^2}\;=\;32.6\;kg/m^{2}\)
Example 5.18: Calculate the BMI for a 61 yo patient who weighs 165 pounds and is 5’ 4” tall.
\(BMI\;=\;\frac{\frac{165}{2.2}}{\left( \frac{64\;\times\;2.54}{100} \right)^{2}}\;=\;\frac{75}{\left( 1.63 \right)^{2}}\;=\;28.2\)
Module 5: Practice Problems
1. A physician wants to start a 7.3 kg pediatric patient on phenobarbital, 2.5 mg/kg/dose given twice daily. Phenobarbital liquid has a concentration of 20 mg/5 mL. Calculate the volume of solution needed to supply each dose. Where should you place the black line dosage marker on the syringe?
2. A physician wants to start a 4.4 kg pediatric patient on phenobarbital, 2 mg/kg/dose given tid. Phenobarbital liquid has a concentration of 20 mg/5 mL. What volume should you instruct the parents to administer?
3. A patient weighs 13 lbs 14 oz. Calculate the IV dose of sodium ampicillin at 100 mg/kg. Round the dose to the nearest 10 mg.
4. A physician orders 325 mg of acetaminophen PO every 4 – 6 hours as needed for pain relief for a patient weighing 35 pounds. The recommended dosage range is 10 – 15 mg/kg/dose. Is the order correct? If not, what dose should you recommend to the doctor?
5. An anesthesiologist orders fentanyl 0.75 mcg/kg/dose for a 30-pound patient. What amount should be prepared? Fentanyl is available as a 50 mcg/mL solution. What volume of fentanyl should be in the syringe?
6. A physician orders Tobramycin 50 mg IV for a 28-pound patient. The recommended dose is 2.5 mg/kg/dose. Is the dose correct? If not, what dose should you recommend to the doctor? The drug concentration is 40 mg/mL. What volume should be prepared?
7. A 25-pound patient needs ondansetron to treat their nausea and vomiting. The recommended dose is 0.15 mg/kg. The drug concentration is 2 mg/mL. How many milligrams of the drug should you order? What volume should be prepared?
8. Bactrim is a suspension containing Sulfamethoxazole 200 mg and Trimethoprim 40 mg per 5 mL. The recommended daily dose of the TMP component is 6 – 12 mg/kg/day. The product is usually administered twice daily. What dose should be prescribed at 8 mg/kg/day for a 43-pound patient? What volume of suspension is required for each dose?
9. What volume of Augmentin suspension (Amoxicillin/Clavulanate - 600 mg/42.9 mg/5 mL) should be administered per dose based on a 90 mg/kg/day amoxicillin component divided twice daily? The patient weighs 24 pounds.
10. Fer-in-sol solution contains 15 mg/mL of elemental iron (75 mg/mL of ferrous sulfate). A recommended dose for iron deficiency anemia is 2 mg/kg/dose administered three times a day. What reasonable volume of solution per dose should you recommend for a child who weighs 20 pounds?
11. A pediatric patient weighs 20 pounds. A physician orders:
Amoxicillin suspension: 250 mg/5 mL
Sig: 4 mL PO tid x 7 days
The recommended dose for the patient’s condition is 45 mg/kg/day in divided doses, every 8 h.
Is the dose correct? If not, what should the amount be? What volume should be administered?
12. A pediatric patient weighs 36 pounds. A physician orders:
Azithromycin suspension: 200 mg/5 mL
Sig: Day 1: 1 tsp PO, then
Days 2 – 5: 1 tsp PO qd x 4 doses
The recommended loading dose for the patient’s condition is 12 mg/kg on day 1. Subsequent daily doses are 6 mg/kg.
Is the dose correct? If not, what should the amount be? What volume should be administered?
13. A pediatric patient weighs 42 pounds. A physician orders:
Ibuprofen suspension: 100 mg/5 mL
Sig: 1 tsp PO q 6 – 8 hours up to 5 days for pain relief
The recommended dose for the patient’s condition is up to 40 mg/kg/day divided into 3 or 4 doses. Is the dosage correct according to the guidelines? If not, what should the amount be? Which size oral syringe should be provided to the parents?
14. The recommended IV dose of Amikacin is 15 mg/kg administered via an infusion over 1 hour. What dose should a 59-pound patient receive? The amikacin vial contains 250 mg/mL. What volume of the injection should be added to the IV bag?
15. A full-term newborn starts on Digoxin using the TDD 30 mcg/kg protocol. (TDD = total digitalizing dose). The recommended dosing schedule is ½ the TDD initially, followed by ¼ of the TDD for each subsequent dose at 8-hour intervals for 2 doses. The newborn weighs 7 pounds and 5 ounces. Digoxin oral solution is available as a 50 mcg/mL product. What dose and volume should you calculate for the two different doses?
16. A patient is 5’ 11” and weighs 206 pounds. Calculate their BSA.
17. A patient is 5’ 3” and weighs 122 pounds. Calculate their BSA.
BSA Dosing Problems
Use the drug dosing guidelines below to answer the questions about the patients whose height and weight is provided in the table.
Patient | Sex | Age | Ht (cm) | Wt (kg) |
AS | M | 7 | 115 | 21 |
CY | F | 5 | 118 | 22 |
ES | M | 4 | 118 | 24 |
HT | F | 4 | 108 | 18 |
IM | F | 7 | 114 | 23 |
KE | M | 3 | 104 | 17 |
OS | M | 5 | 101 | 17 |
PC | M | 6 | 109 | 21 |
RX | F | 6 | 95 | 14 |
WL | F | 3 | 91 | 14 |
Refer to the drug labels for concentration information.
Acyclovir can be used to treat viral infections in children. The guideline recommends doses as high as 600 mg/m2 every 6 hours for 10 days. The product is a 40 mg/mL oral suspension and sterile vials containing acyclovir sodium solution equivalent to 50 mg/mL.
Allopurinol is used to prevent tumor lysis syndrome associated with certain cancer treatments. It is given as a short IV infusion of 200 mg/m2. Allopurinol is marketed as a sterile vial containing the equivalent of 500 mg of allopurinol (as the sodium salt) that is reconstituted to 30 mL before use, 16.7 mg/mL.
Caspofungin is an anti-fungal drug given IV (over 1 hour) as a 70 mg/m2 loading dose the first day, followed by 50 mg/m2 daily thereafter.
Dexamethasone is used in treating many conditions. The typical dose is 0.6 – 9 mg/m2/day in
3 or 4 divided doses. It is available as an oral solution containing 0.5 mg/5 mL.
Dronabinol is used for nausea prophylaxis in chemotherapy patients. The National Cancer Institute (NCI) guideline for pediatric patients is 5 mg/m2 PO every 6 – 8 hours before beginning chemotherapy treatment, then 5 mg/m2 PO every 4 – 6 hours until 12 hours after the treatment.
Hydrocortisone may be used to treat many conditions. For normal replacement therapy, the dosing guideline is 20 – 25 mg/m2/day. For congenital adrenal hyperplasia, the dosing guideline is 30 – 36 mg/m2/day, with 1/3 of the dose administered in the morning and 2/3 in the evening.
18. A physician orders acyclovir 600 mg/m2 for AS. What volume of the oral suspension is required for each dose?
19. CY requires 32 mg/m2/day of hydrocortisone for congenital adrenal hyperplasia. How should you instruct her mother to give the daily dose?
20. WL requires hydrocortisone for the treatment of congenital adrenal hyperplasia. The physician ordered hydrocortisone 12 mg in the morning and 24 mg in the evening daily. Is this dose within the recommended guidelines?
21. According to the dosing guidelines for immunocompromised patients, what is the total dose RX should receive for a 10-day course of acyclovir therapy?
22. PC requires allopurinol 200 mg/m2. The pharmacy prepared an infusion solution by transferring 12.5 mL of the sterile allopurinol solution into a 1L bag of normal saline. Is the correct amount of drug in the bag? What volume should have been added?
23. KE requires allopurinol before his chemotherapy. What volume of the reconstituted allopurinol solution should be added to KE’s infusion solution?
24. OS requires acyclovir 450 mg/m2 by IV infusion. What volume of acyclovir solution is required for each dose?
25. PC requires hydrocortisone replacement therapy. What dose should he receive each day?
26. RX requires caspofungin therapy. Calculate the appropriate dosing regimen.
27. WL is ordered dexamethasone 8.4 mg/m2/day in four divided doses. What volume of the oral solution (0.5 mg/5 mL) should she receive at each dose?
28. AS requires dronabinol per the NCI protocol. What dose should he receive? What volume?
29. CY requires allopurinol 200 mg/m2. What volume of the drug solution should be used to prepare the infusion?
30. A physician orders acyclovir 600 mg/m2 for ES. What volume of the drug solution should be used to prepare the bag for IV infusion?
31. HT requires normal hydrocortisone replacement therapy. What is the normal milligram range for this indication?
32. Calculate the caspofungin dosage regimen for IM.
33. KE has a severe allergic reaction, and his physician orders dexamethasone 2 mg every 6 hours. Is the prescribed amount correct? What is the normal range according to the guidelines?
34. Calculate the dose of dronabinol for OS. What volume should be given?
35. What percent loss in BSA would a patient experience whose right arm was amputated at the elbow.
36. A patient enrolled in an NCI protocol should receive an IV solution at 1.8 L/ m2 over 24 hours. Before their surgery, the patient’s BSA = 1.9 m2. What IV fluid rate should be used for the patient if their left arm has been amputated at the shoulder?
37. A patient with psoriasis has skin involvement over the upper chest area. What percentage of their skin surface area has psoriasis?
38. A patient has received severe 2nd and 3rd degree burns over the front of their two legs in a BBQ grilling accident. What percentage of their BSA has been burned?
39. Calculate the Ideal Body Weight for a female patient who is 5’ 6” tall.
40. Calculate the Adjusted Body Weight for a female patient who is 5’ 6” tall and weighs 210 pounds. Factor = 0.4.
41. A drug is dosed based on the adjusted body weight for obese patients using the factor 0.4. For these patients, the dose is 15 mg/kg. Calculate the dose for a female patient who is 5’ 3” and weighs 155 pounds.
42. A drug is dosed based on the adjusted body weight for obese patients using the factor 0.4. For these patients, the dose is 8 mg/kg. Calculate the dose for a female patient who is 5’ 7” and weighs 183 pounds.
43. Calculate the Ideal Body Weight for a male patient who is 5’ 8” tall.
44. Calculate the Adjusted Body Weight for a male patient who is 5’ 8” tall and weighs 215 pounds. Factor = 0.4.
45. A drug is dosed based on the adjusted body weight for obese patients using the factor 0.4. For these patients, the dose is 6 mg/kg. Calculate the dose for a male patient who is 6’ 5” and weighs 329 pounds.
46. A drug is dosed based on the adjusted body weight for obese patients using the factor 0.4. For these patients, the dose is 5 mg/kg. Calculate the dose for a male patient who is 5’ 10” and weighs 215 pounds.
47. Is this patient considered obese? Female, 5’ 4”, weighs 148 pounds.
48. Is this patient considered obese? Female, 5’ 5”, weighs 165 pounds.
49. Is this patient considered obese? Male, 5’ 9”, weighs 200 pounds.
50. Is this patient considered obese? Male, 5’ 11”, weighs 218 pounds.
Problems 51 - 56 use the table from the Famvir® prescribing information,
"Penciclovir dose adjustments in patients with renal insufficiency."
51. Male, 29 yo, SCr = 1.1 mg/dL, 6’ 1”, 185 pounds, suppressing genital herpes.
52. Male, 64 yo, SCr = 1.5 mg/dL, 5’ 2”, 120 pounds, treatment of herpes zoster.
53. Male, 75 yo, SCr = 1.3 mg/dL, 5’ 11”, 190 pounds, treatment of herpes zoster.
54. Female, 30 yo, SCr = 2.3 mg/dL, 5’ 0”, 210 pounds, suppressing genital herpes.
55. Female, 69 yo, SCr = 2.2 mg/dL, 5’ 2”, 105 pounds, treatment of herpes zoster.
56. Female, 64 yo, SCr = 2.7 mg/dL, 4’ 10”, 165 lb, suppressing genital herpes.
Use Table 4 for problems #57 - 62.
The dose calculation for this drug is based on the BSA-corrected CrCL. Calculate the patient's CrCL and BSA as usual, then use the following equation to calculate the BSA-adjusted value.
\(BSA\;adjusted\;CrCl\;(mL/min/1.73\;m^{2})=CrCl (mL/min)\times \frac{1.73}{Patient's\;BSA}\;m^2\)
In the left column, choose the weight closest to the patient’s actual body weight. For example, if the patient weighs 52 kg, use the 50 kg dosing line. For a patient weighing 56 kg, use the 60 kg dosing line.
Table 4: Reduced intravenous dosage of Primaxin I.V. in adult patients with
impaired renal function and/or body weight < 70 kg
57. Male 75 yo, 4’ 11”, 119 pounds, SCr 1.1 mg/dL, imipenem 1 g/day.
58. Male 52 yo, 5’ 8”, 125 pounds, SCr 0.9 mg/dL, imipenem 1.5 g/day.
59. Male 31 yo, 6’ 4”, 202 pounds, SCr 1.5 mg/dL, imipenem 2 g/day.
60. Female 29 yo, 5’ 11”, 190 pounds, SCr 1 mg/dL, imipenem 1 g/day.
61. Female 63 yo, 5’ 2”, 103 pounds, SCr 2.4 mg/dL, imipenem 1.5 g/day.
62. Female 52 yo, 5’ 4”, 137 lb, SCr 1.6 mg/dL, imipenem 2 g/day.
Practice calculating BMI
63. Calculate the BMI for a 47 yo patient who weighs 155 pounds and is 5’ 7” tall.
64. Calculate the BMI for a 28 yo patient who weighs 188 pounds and is 5’ 9” tall.
65. Calculate the BMI for a 22 yo patient who weighs 126 pounds and is 5’ 5” tall.
66. Calculate the BMI for a 36 yo patient who weighs 223 pounds and is 6’ 4” tall.
Practice estimating affected BSA
67. A patient (BSA = 1.6 m2) is diagnosed with an osteogenic sarcoma involving the left fibula (lower leg). The oncologists recommended treatment with methotrexate 12 g/m2. Before chemotherapy starts, the patient will have a below-knee amputation. What methotrexate dose should be administered after the surgery?
68. An army veteran (pre-injury BSA = 1.8 m2) has been diagnosed with testicular cancer. The oncologists recommended treatment with dactinomycin 1000 mcg/m2. The patient lost their right arm during his tour of duty. What dactinomycin dose should be administered? What volume of solution is needed? The dactinomycin solution concentration is 500 mcg/mL.
69. A patient (pre-injury BSA = 1.5 m2) assigned to an NCI protocol should receive Lactated Ringer’s solution, 1800 mL/m2/day. The patient has had their left arm and left leg removed. Calculate the hourly fluid rate.
70. A woman comes to your store with a severe rash caused by exposure to poison ivy. Both legs are affected, from just above the knees to her ankles. She was walking on an overgrown woodland path. Estimate the percentage of her BSA affected by the rash.
Practice using the "bedside" Schwartz equation
Schwartz (bedside) eGFR (mL/min/1.73m2 ) | Lean Body Weight (kg) q 12 h Dose (mcg) | |
5 kg | 10 kg | |
30 | 10 | 20 |
50 | 12 | 24 |
70 | 14 | 28 |
90 | 16 | 32 |
The q 12 h drug dose is based on the eGFR and the closest lean body weight. For example, if the patient weighs 7 kg, select the dose from the 5 kg column. If the patient weighs 8 kg, select the dose from the 10 kg column.
71. Calculate the drug dose for a 4.8 kg and 23 inches long baby. The SCr = 0.35 mg/dL.
72. Calculate the drug dose for an 11 kg and 33 inches long baby. The SCr = 1.2 mg/dL.
73. Calculate the drug dose for a 6.3 kg and 24 inches long baby. The SCr = 0.8 mg/dL.
74. Calculate the drug dose for a 9.5 kg and 30 inches long baby. The SCr = 0.35 mg/dL.
Answers:
1. 4.6 mL
2. 2.2 mL
3. 630 mg
4. No! 160 – 240 mg
5. 10 mcg (0.2 mL)
6. No! 32 mg (0.8 mL)
7. 1.7 mg (0.8 – 0.85 mL))
8. 160 mg (10 mL)
9. 4 mL
10. 1.2 mL
11. No! 135 mg (2.7 mL)
12. LD – OK. 100 mg (2.5 mL)
13. 190 mg (9.5 – 10 mL)
14. 400 mg (1.6 mL)
15. TDD = 100 mcg; 50 mcg (1 mL), 25 mcg (0.5 mL)
16. 2.2 m2
17. 1.6 m2
18. 12.3 mL
19. 9 mg AM, 18 mg PM
20. No! The range is 17.7 – 21.2 mg/day.
21. 14.64 g
22. No! ~ 9.6 mL
23. 8.4 mL
24. 6.2 mL
25. 16 – 20 mg/day
26. LD = 42.7 mg, then 30.5 mg/day
27. 12.5 mL
28. 4.1 mg (0.82 mL)
29. 10.2 mL
30. 10.7 mL
31. 14.6 – 18.3 mg
32. LD = 59.5 mg, then 42.5 mg/day
33. No, the dose is high. 0.4 – 6.3 mg/day.
34. 3.5 mg (0.7 mL)
35. ~ 4.5%
36. ~ 130 mL/h
37. ~ 9%
38. ~ 18%
39. 59.3 kg
40. 73.8 kg
41. 59.6 kg, 894 mg
42. 70.2 kg, 562 mg
43. 68.4 kg
44. 80.1 kg
45. 113.3 kg, 680 mg
46. 82.9 kg, 415 mg
47. No. < 130% of IBW
48. Yes. > 130% of IBW
49. No. < 130% of IBW
50. Yes. > 130% of IBW
51. 112 mL/min, 250 mg q12°
52. 38 mL/min, 500 mg q24°
53. 52 mL/min, 500 mg q12°
54. 26 mL/min, 125 mg q12°
55. 18 mL/min, 250 mg q24°
56. 14 mL/min, 125 mg q24°
57. 45 mL/min/1.73 m2, 125 mg q6°
58. 81 mL/min/1.73 m2, 250 mg q6°
59. 68 mL/min/1.73 m2, 500 mg q8°
60. 77 mL/min/1.73 m2, 250 mg q6°
61. 21 mL/min/1.73 m2, 250 mg q12°
62. 37 mL/min/1.73 m2, 250 mg q8°
63. 24.3
64. 27.8
65. 21
66. 27.1
67. 10.9 g. (Approx 9% loss of BSA)
68. 1650 mcg; 3.3 mL. (Approx 9% loss of BSA)
69. 82 mL/hr (Approx 27% loss of BSA)
70. Approx 18%
71. Dose = 14 mcg - q 12 h
72. Dose = 20 mcg - q 12 h
73. Dose = 10 mcg - q 12 h
74. Dose = 32 mcg - q 12 h
Module 6: Intravenous Fluids and Drug Therapy
This module will introduce and review the types of calculations associated with intravenous fluid therapy and intravenous drug administration.
Module 6A: Intravenous Therapy
A major component of hospital pharmacy practice involves preparing and assisting patient care staff with the appropriate use of intravenous drug therapy. Intravenous therapy can be given in different ways:
- IV bolus injection, where a small volume of drug solution (e.g. up to 10 mL) is quickly introduced into a vein with a syringe.
- Intermittent IV infusion, where a moderate volume of drug solution (e.g. 50 – 250 mL) is infused into a vein over a short period (e.g. 30 minutes – 2 hours) and repeated on a regular schedule. Most IV drug therapy in the hospital is administered via intermittent infusion. For example, a patient may receive a 30-minute infusion of ampicillin in 100 mL of normal saline every 8 hours for 5 days.
- Continuous infusion, where an LVP solution (e.g. 500 or 1000 mL bag) is infused into a vein at a constant rate for several hours or longer. When one bag of solution empties, it is replaced with another bag. Therapy continues until the physician cancels or modifies the order.
Hospitalized patients unable to drink enough are often administered IV maintenance fluids to help maintain circulatory volume. Intravenous fluid administration also plays an important role for patients being treated for dehydration, unusual fluid shifts within the body (third-spacing), or patients with heart and lung dysfunction. A commonly used weight-based dosing guideline is the Holliday-Segar method, which will be explained in the next section.
Module 6B: Maintenance Fluid Calculations
Everyone needs water and electrolytes to survive. Patients unable to eat and drink require fluids and salts to be administered via the vascular system (parenteral administration). While fluid management can be complex depending on the patient’s condition, this section will only introduce the basics. We will only deal with simple maintenance volume calculations.
A reasonable question is how much water and salt is normally needed. The Holliday-Segar method was introduced for pediatric patients, but it has found wide applicability for all ages. The method uses the patient’s weight to calculate the daily fluid volume.
Table 6.1. Holliday-Segar maintenance fluid calculation
Weight | Daily Requirement |
1 – 10 kg | 100 mL/kg |
11 – 20 kg | 1000 mL + 50 mL/kg for each kg > 10 kg |
> 20 kg | 1500 mL + 20 mL/kg for each kg > 20 kg |
Example 6.1: What is the maintenance fluid requirement for a 2-month-old baby who weighs 5.3 kg? Calculate the hourly IV flow rate.
\(5.3\;kg\;\times\;100\;\frac{mL}{kg·day}\;=\;530\;\frac{mL}{day}\)
\(530\;\frac{mL}{day}\;\times\;\frac{1\;day}{24\;hours}\;=\;\frac{530\;mL}{24\;hours}\;=\;22.1\;mL/h\;or\;22\;mL/h\)
Example 6.2: What is the maintenance fluid requirement for a 16-month-old toddler who weighs 16.5 kg? Calculate the hourly IV flow rate.
\(16.5\;kg\;=\;1000\;\frac{mL}{day}\;+\;\left( 50\;\frac{mL}{kg·day}\;\times\;6.5\;kg\right)=\;1325\;\frac{mL}{day}\)
\(1325\;\frac{mL}{day}\;\times\;\frac{1\;day}{24\;hours}\;=\;\frac{1325\;mL}{24\;hours}\;=\;55\;mL/h\)
Example 6.3: What is the maintenance fluid requirement for a 10-year-old child who weighs 33 kg? Calculate the hourly IV flow rate.
\(33\;kg\;=\;1500\;\frac{mL}{day}\;+\;\left(20\;\frac{mL}{kg·day}\;\times\;13\;kg\right)=\;1760\;\frac{mL}{day}\)
\(1760\;\frac{mL}{day}\;\times\;\frac{1\;day}{24\;hours}\;=\;\frac{1760\;mL}{24\;hours}\;=\;73\;mL/h\)
Depending on a patient’s fluid status or organ condition, a reduction or an increase in the maintenance fluid may be necessary. You may see an order to “run the patient a little dry at ¾ maintenance (75%).” Or, “Let’s run the IV at 1 ¼ maintenance (125%).” In these cases, you calculate the IV flow rate and multiply the hourly rate by the appropriate fraction.
Example 6.4: For the 10-year-old child in example 3, calculate the IV flow rate if the physician decides to run the IV rate at 1 ¼ maintenance (125%).
\(1325\;\frac{mL}{day}\;\times\;\frac{1\;day}{24\;hours}\;=\;\frac{1325\;mL}{24\;hours}\;=\;55\;mL/h\;\times\;1.25\;=\;69\;mL/h\)
Example 6.5: For the 16-month-old toddler in example 2, calculate the IV flow rate if the physician decides to run the IV rate at ¾ maintenance (75%).
\(1760\;\frac{mL}{day}\;\times\;\frac{1\;day}{24\;hours}\;=\;\frac{1760\;mL}{24\;hours}\;=\;73\;mL/h\;\times\;0.75\;=\;55\;mL/h\)
Module 6C: Product Selection and Consideration
Many drugs are manufactured in ready-to-use IV bags. The only preparation required is for the caregiver to connect the bag to the IV tubing and program the infusion pump to run at the correct flow rate and duration of time. Other drugs are supplied in vials as crystals or lyophilized powders that must be reconstituted before administration. The pharmacist must calculate the correct volume of drug to remove from the vial and inject the drug into the correct IV ‘base fluid.’ The most common base fluids are 0.9% sodium chloride injection (also called normal saline or NS) and dextrose 5% in water (also called D5W). Base fluids are available in IV bags containing 50, 100, 150, 250, 500, or 1000 mL to allow flexibility in drug compounding.
It is important for patient safety and, sometimes, drug effectiveness to administer IV drugs at the correct infusion rate. Intermittent infusions are usually ordered to run over a convenient time appropriate to the drug and the patient size, e.g. gentamicin 40 mg in 50 mL NS over 1 hour. Some continuous infusions, especially those used in critical care, are administered at a rate depending on the patient body weight or body surface area. Dobutamine, for example, is used to increase a patient's cardiac output in cases of shock. The prescribing information document states, “The rate of infusion needed to increase cardiac output usually ranged from 2.5 to 15 mcg/kg/min ... On rare occasions, infusion rates up to 40 mcg/kg/min have been required to obtain the desired effect.” See: https://dailymed.nlm.nih.gov/dailymed/drugInfo.cfm?setid=89becb0c-da60-4f43-0a98-29ff7a9eca58.
Note that the infusion rate is expressed in terms of drug mass per time. The solution flow rate in mL/hour to achieve a specified mcg/kg/min rate depends on the dobutamine concentration in the solution (mcg/mL) and the patient’s body weight (kg).
As mentioned earlier, many critical care drugs are commercially available in standardized concentrations in NS or D5W. However, there are patients where the use of these standardized concentration products may not be useful because of total daily fluid volume restrictions. In those cases, pharmacists frequently prepare small volumes of a base solution with a concentration different from those available using vials of the drug.
Module 6D: Overfill in IV bags
Manufacturers overfill bags of IV fluids to ensure patients can get the full labeled amount out of the bag. Overfill also helps decrease the effect of water evaporation from the bag during storage. The amount of overfill varies by manufacturer and may vary from batch to batch, so it is impossible to state how much overfill is in any particular bag. For example, a customer service representative for an IV fluid manufacturer stated that their bags of IV fluid labeled as 100 mL contained between 105 and 113 mL at the time of manufacture. Since the volume of the IV bag is unknown, it is impossible to accurately state the drug concentration of a compounded IV solution. The standard of practice is to calculate the drug concentration assuming the bag contains the labeled amount of fluid. If the drug concentration in a solution is critical, the base fluid and drug can be individually added to an empty IV bag. This method will usually result in a more accurate and precise concentration.
Example 6.6: A physician orders a patient to receive norepinephrine by IV infusion. The hospital policy is to prepare norepinephrine infusions by adding 16 mcg of norepinephrine injection for every 1 mL of fluid in the bag. Norepinephrine is provided in vials containing 4 mg in every 4 mL. How many mL of norepinephrine injection must be added to a 250 mL bag of NS to prepare the infusion?
\( 250\;mL\times \frac{16\;mcg\;Nor}{mL\;infusion}\times \frac{1\;mg\;Nor}{1000\;mcg}\times \frac{4\;mL\;inj}{4\;mg\;Nor}=4\;mL\; Nor\;injection\)
Example 6.7: Most adult patients achieve blood pressure control with norepinephrine infusion of 2 – 4 mcg/min. How long will 250 mL of a 16 mcg/mL solution last if the drug is infused at the average rate of 3 mcg/min?
\(250\;mL\;infusion\times \frac{16\;mcg\;Nor}{mL\;infusion}\times \frac{1\;min}{3\;mcg\;Nor}\times \frac{1\;hr}{60\;min}=22.2\;hr\)
NOTE: The norepinephrine concentration is slightly less than 16 mcg/mL (due to bag overfill and not including the added drug volume) but this difference is ignored in clinical settings.
Example 6.8: A 95 lb patient was ordered a dobutamine infusion to start at 7.5 mcg/kg/min. The pharmacy sent a bag containing 250 mg of dobutamine in 250 mL of D5W. The nurse set the infusion pump to run at 5.5 mL per hour. Did the nurse enter the correct flow rate?
\(95\;lb\times \frac{1\;kg}{2.2\;lb}\times \frac{7.5\;mcg\;Dob}{kg\;\times \;min}\times \frac{1\;mg}{1000\;mcg}\times \frac{250\;ml\;soln}{250\;mg\;Dob}\times \frac{60\;min}{hr}=19.4\;mL/hr\)
The starting infusion rate of 7.5 mcg/kg/min requires a solution flow rate of 19.4 mL/hr. The nurse did not set the correct flow rate on the pump.
Example 6.9: A pharmacist added 50 mL of a 4% drug solution to 250 mL of D5W. What is the flow rate (mL/hr) required to deliver 0.02 mg/kg/min for a 175 lb patient?
First, calculate the drug concentration in the infusion solution.
\(Drug\;concentration=\frac{50\;mL\;soln\;\times\; \frac{4000\;mg\;drug}{100\;mL\;soln}}{250\;mL\;+\;50\;mL}=6.67\;mg/mL\;infusion\)
Now, find the flow rate based on the patient's body weight.
\(175\;lb\times \frac{1\;kg}{2.2\;lb}\times \frac{0.02\;mg}{kg\;\times \;min}\times \frac{1\;mL\;infusion}{6.67\;mg}\times \frac{60\;min}{hr}=14.3\;mL/hr\)
Example 6.10: A pharmacist prepared an IV solution by adding 4 mL of a 1 Unit/mL solution to 100 mL of normal saline. The nurse set the infusion pump to run at 2 mL/hr. Calculate the infusion rate in Units/min.
\(Drug\;concentration=\frac{4\;mL\;\times \;\frac{1\;Unit}{4\;mL}}{100\;mL\;+\;4\;mL}=0.038\;Units/mL\\\\\)
\(\frac{2\;mL}{hr}\times \frac{0.038\;Units}{mL}\times \frac{1\;hr}{60\;min}=0.0013\;Units/min\)
Module 6E: Bag Overfill and Added Drug Volume Considerations
We know the pre-filled IV bags must contain an overfill, so we cannot know the true volume in the bag at the start of our compounding procedure. A reasonable question is if we are required to compound a solution with a prescribed drug concentration, thus adding more volume to the bag, how do we know the final concentration?
We can consider two scenarios. In one case, the entire contents of the bag are infused into the patient. As long as the entire volume is infused, we can conclude that the patient received the correct dose. For example, the pharmacy prepared a 50 mL bag of NS containing 1 g of ampicillin (total volume injected = 4 mL). What volume is in the bag?
50 mL + 4 mL + overfill > 54 mL
If all of the solution is infused, the patient receives the prescribed 1-gram dose.
The other case involves setting the infusion pump to deliver 50 mL over ½ or 1 hour. In this scenario, the patient’s dose will be decreased by at least 10%, depending on the overfill volume, since the bag contains over 55 mL.
As mentioned in the introduction section, there are times when the patient’s fluid status is critical because of heart or kidney dysfunction, pulmonary gas exchange concerns, edema, etc. In these cases, the medical team carefully keeps track of the daily total fluid volume the patient receives. In these situations, you may decide to use a syringe instead of an IV bag to hold the contents.
We administer many drugs as IV infusions in critical care settings, like dopamine, dobutamine, nitroglycerin, adenosine, lidocaine, etc. If the bags are pre-mixed by the manufacturers, we can safely conclude the concentrations are correct since they are required to assay the product before release. If the pharmacy prepares the bag, the contents are rarely, if ever, analyzed. How should you handle that situation? While there are recommended starting doses for these drugs, the IV flow rate is frequently tweaked to target a specific response, like blood pressure or heart rate. The actual starting concentration, while unknown, will be close to that prescribed.
You can consider a rule of thumb regarding the amount of the added drug in these critical care infusions. You can ignore the added amount if the volume added to the bag is less than 2% of the labeled volume. For example, adding 1 mL or less to a 50 mL bag or 2 mL or less to a 100 mL bag. If the volume to be added is greater than 2% of the stated bag volume, you might consider initially withdrawing that volume of base solution from the bag before injecting the drug.
Let’s consider this case. A patient is prescribed 400 mg of infliximab in NS 500 mL to be infused over 2 hours. Each 100 mg vial of the lyophilized product is reconstituted with 10 mL of sterile water for injection. The pharmacy adds 40 mL of the solution to the bag, which is now very full. The infusion is started and set to run at 250 mL/h. After two hours, the nurse was ready to disconnect the IV and send the patient home. However, there was still over 50 mL in the bag, which contained the valuable life-altering drug. This happened to me (BK). I had to request to allow all of the drug to be infused.
Remember, depending on the drug and patient condition, the added volume may be significant or insignificant.
Module 6: Practice Problems
Note: (IVR = IV Flow Rate)
1. Calculate the daily fluid requirement for a newborn weighing 3.1 kg. Calculate the hourly IVR.
2. Calculate the daily fluid requirement for a 6-month-old baby weighing 8.5 kg. Calculate the hourly IVR.
3. Calculate the daily fluid requirement for a 1-year-old baby weighing 9.7 kg. Calculate the hourly IVR.
4. Calculate the daily fluid requirement for a 1-year-old baby weighing 10.5 kg. Calculate the hourly IVR.
5. Calculate the daily fluid requirement for a 3-year-old child weighing 15.6 kg. Calculate the hourly IVR.
6. Calculate the daily fluid requirement for a 5-year-old child weighing 18.3 kg. Calculate the hourly IVR.
7. Calculate the daily fluid requirement for a 6-year-old child weighing 20.2 kg. Calculate the hourly IVR.
8. Calculate the daily fluid requirement for an 8-year-old child weighing 25.4 kg. Calculate the hourly IVR.
9. Calculate the daily fluid requirement for a 16-year-old teen weighing 60.8 kg. Calculate the hourly IVR.
10. Calculate the daily fluid requirement for a 19-year-old weighing 73 kg. Calculate the hourly IVR.
11. A patient weighs 87 kg. Calculate the hourly flow rate for ¾ maintenance.
12. A patient weighs 61 kg. Calculate the hourly flow rate for 1 ¼ maintenance.
13. A patient weighs 7.6 kg. Calculate the hourly flow rate for ¾ maintenance
14. A patient weighs 31 kg. Calculate the hourly flow rate for 1 ¼ maintenance.
15. What IVR (mL/h) will deliver 5 mcg/kg/min for a 35 lb child?
D5W 250 mL
Dopamine 600 mg/L
16. What IVR (mL/h) will deliver 3 mcg/kg/min for a 40 lb child?
D5W 100 mL
Dobutamine 400 mg/L
17. What IVR (mL/h) will deliver 20 mcg/kg/min for a 10 kg child?
D5W 250 mL
Lidocaine 900 mg/L
18. What IVR (mL/h) will deliver 7.5 mcg/kg/min for a 52 lb child?
D5W 500 mL
Dobutamine 500 mg/L
19. What IVR (mL/h) will deliver 0.3 mcg/min for a 25 lb child?
D5W 50 mL
Norepinephrine 100 mg/L
20. For problem #15, how many milliliters of Dopamine HCl injection 40 mg/mL do you need to add to a 250 mL bag to make the requested solution?
21. For problem #16, how many milliliters of Dobutamine HCl injection 12.5 mg/mL do you need to add to a 100 mL bag to make the requested solution?
22. For problem #17, how many milliliters of Lidocaine HCl injection 4% do you need to add to a 250 mL bag to make the requested solution?
23. For problem #18, how many milliliters of Dobutamine HCl injection 12.5 mg/mL do you need to add to a 500 mL bag to make the requested solution?
24. For problem #19, how many milliliters of Norepinephrine bitartrate injection 0.1% do you need to add to a 50 mL bag to make the requested solution?
25. You have three commercially available pre-mixed infusions of Dopamine HCl with concentrations of a) 0.8 mg/mL, b) 1.6 mg/mL, and c) 3.2 mg/mL. What IV infusion rates must the pump be set to deliver the dose in problem #1?
26. You have two commercially available pre-mixed infusions of Lidocaine HCl with concentrations of a) 4 mg/mL and b) 8 mg/mL. What IV infusion rates must the pump be set to deliver the dose in problem #17?
27. You have three commercially available pre-mixed infusions of Dobutamine HCl with concentrations of a) 1 mg/mL, b) 2 mg/mL, and c) 4 mg/mL. What IV infusion rates must the pump be set to deliver the dose in problem #18?
28. What IVR (mL/h) will deliver 0.25 mcg/kg/min to a 6lbs 6oz infant? D5W 250 mL
Nitroglycerin 50 mg
29. Nitroglycerin is also available as a pre-mixed solution in D5W: a) 25 mg/250 mL, and b) 100 mg/250 mL. What would be the IVR for those solutions to deliver the same dose in problem #28?
30. Labetalol injection contains 100 mg of labetalol in every 20 mL. If 30 mL of labetalol injection is added to 100 mL of NS, what flow rate (mL/hr) is required to infuse 2 mg of labetalol per minute?
31. Vancomycin is available as 500 mg of solid that is reconstituted with 10 mL of sterile water for injection before use to produce a 50 mg/mL solution. How many mL of the reconstituted solution is required to prepare a vancomycin dose of 1250 mg?
32. Vancomycin must be diluted for infusion to a concentration of 5 mg/mL or less. A patient is ordered a dose of 1750 mg of vancomycin in NS. The pharmacy has bags of 50, 100, 250, and 500 mL NS. Which NS bag(s) should be used for this order?
33. Gentamicin injection is a sterile solution containing 80 mg in every 2 mL. The normal dose of gentamicin for certain infections is 5 mg/kg/day in 3 equal doses. How much gentamicin injection is required for each individual dose for a 46 kg child?
34. If the normal dose of gentamicin for certain infections is 5 mg/kg/day in 3 equal doses, what is the recommended daily dose for a 185 lb patient?
35. A patient is ordered 250 mg of paclitaxel by IV infusion. How much paclitaxel injection (30 mg/5 mL) is required to prepare the infusion?
36. Mesna is administered after ifosfamide to protect the bladder from the excreted metabolite acrolein, which causes hemorrhagic cystitis. The recommended mesna dose is 240 mg/m2 to be administered immediately (0 hours) after, then 4 and 8 hours after each dose of ifosfamide. Mesna injection is a solution containing 1 g in 10 mL. How many total mL of mesna injection are required for 3 doses (0, 4, and 8 hours) for a patient with a BSA of 2.2 m2?
37. Argatroban is a ready-to-use injection containing 50 mg in 50 mL. A patient is ordered argatroban infusion at 180 mcg/minute. Calculate the flow rate in mL/hr.
38. A patient receives 50 mg/50 mL of argatroban at 12.5 mL/hr. Calculate the infusion rate in mcg/min.
39. Aggrastat® injection is a ready-to-use solution containing 12.5 mg in every 250 mL. It is administered as a loading dose by IV bolus injection of 25 mcg/kg, followed immediately by a maintenance infusion of 0.15 mcg/kg/min. Calculate the volume of Aggrastat required for the loading dose and the infusion flow rate for a 145 lb patient.
40. Dexmedetomidine is an anesthetic drug and is supplied as a 200 mcg/2 mL injection. It is recommended to be diluted to 4 mcg/mL for infusion by adding the correct volume of drug solution and normal saline to an empty sterile IV bag. A patient is ordered a 0.6 mcg/kg/hr infusion of dexmedetomidine. How many mL of dexmedetomidine injection and how many mL of NS should be mixed to provide enough solution for 28 hours of therapy for a 196 lb patient?
41. Propofol is marketed as an injectable emulsion containing 200 mg of propofol in every 20 mL. The emulsion is administered without further dilution. What flow rate (mL/hr) should be used to provide 125 mcg/kg/min for a 130 lb patient?
42. Micafungin is an antifungal drug. It is supplied as a sterile solid reconstituted with 5 mL of sterile water for injection to produce a 10 mg/mL solution. The normal dose is 2.5 mg/kg once daily for patients ≥ 30 kg or 3 mg/kg for patients < 30 kg. It should be diluted with D5W to a concentration less than 1.5 mg/mL for infusion. Calculate the volume of reconstituted micafungin to provide the dose for a 55 lb patient. The pharmacy has D5W bags in 50, 100, 250, and 500 mL. Which bag size should be used?
43. Lidocaine infusions over 1 hour can be used to treat pain in certain conditions. The dose is based on the smaller actual or ideal body weight value. Calculate the infusion rate needed to deliver 2 mg/kg for: a) male, 5’ 11”, 174#, and b) female, 5’ 3”, 113#. The IV bag contains 2000 mg Lidocaine HCl in 250 mL of D5W.
44. The manufacturer recommends diluting angiotensin II (Giapreza®, 2.5 mg/1 mL) to a total volume of 500 mL with normal saline. What is the solution concentration in ng/mL?
45. Angiotensin II (Giapreza®, 2.5 mg/1 mL) increases blood pressure in patients with septic shock. A typical infusion conentration is 2.5 mg in a total volume of 500 mL normal saline. What IV infusion rate should be used to deliver 20 ng/kg/min for a 55 kg patient?
Answers:
1. 310 mL/day - 13 mL/hr
2. 850 mL/day - 35 mL/hr
3. 970 mL/day - 40 mL/hr
4. 1025 mL/day - 43 mL/hr
5. 1280 mL/day - 53 mL/hr
6. 1415 mL/day - 59 mL/hr
7. 1504 mL/day - 63 mL/hr
8. 1608 mL/day - 57 mL/hr
9. 2316 mL/day - 97 mL/hr
10. 2560 mL/day - 107 mL/hr
11. 89 mL/hr
12. 121 mL/hr
13. 24 mL/hr
14. 90 mL/hr
15. 8 mL/hr
16. 8.2 mL/hr
17. 13.3 mL/hr
18. 21.3 mL/hr
19. 2 mL/hr
20. 3.8 mL
21. 3.2 mL
22. 5.6 mL
23. 20 mL
24. 5 mL
25. a) 6 mL/hr, b) 3 mL/hr, c) 1.5 mL/hr
26. a) 3 mL/hr, b) 1.5 mL/hr
27. a) 10.6 mL/hr, b) 5.3 mL/hr, c) 2.7 mL/hr
28. 0.2 mL/hr
29. a) 0.4 or 0.5 mL/hr, b) 0.1 mL/hr
30. 104 mL/hr
31. 25 mL
32. 500 mL bag
33. 1.9 mL/dose
34. 420 mg/day
35. 41.7 mL
36. 15.8 mL
37. 10.8 mL/hr
38. 208 mcg/min
39. Load - 33 mL; maintenance flow rate = 11.9 mL/hr
40. 15 mL of drug; 360 mL of NS; empty 500 mL bag
41. 44.3 mL/hr
42. 25 kg, 75 mg dose = 7.5 mL, 50 mL bag D5W
43. a) 18.8 mL over 1 h, b) 12.8 mL over 1 h.
44. 5000 ng/mL
45. 13.2 mL/hr
Module 7: Applications of Linear Regression in Pharmacy
Introduction
This module is about linear regression (LR). This will be an introduction to the topic for some. For others, it may be a review of material previously learned. In either case, we will concentrate on the mechanics of LR using the TI-84 Plus CE line of calculators. If you don’t own one, you can check one out from the Dean’s office for use during the semester. You will use a TI-84 Plus CE in PHPS 720 Pharmacokinetics next semester to run some PK programs that I have written to make your life a little easier in that course.
In this section, you will use your calculator to find the equation of the best-fitting line to a data set. After the data is entered, the calculator will return values for the slope, y-intercept, and r2. R-squared (r2) measures how well a linear regression model predicts the data set. We will not be concerned with the mathematics of the process. If you are interested, you will find plenty of material online, in YouTube videos, or in a statistics course.
Module 7A: What is Linear Regression?
What is LR? LR is a method for applying a straight-line model between the explanatory variable (x), and the response variable (y). Recall that the equation for a straight-line is:
\(y\;=\;mx\;+\;b\)
where the response (y) can be predicted by multiplying the variable (x) by the slope of the line and then adding the intercept. Too many words! Let’s look at a picture.
The red diamonds are scattered, but the y-value seems to increase as the x-value increases. We could ask, “If we know the value of x can we predict the value of y?” Can we mathematically model that relationship? At the most basic level, this is what LR is about. If we know something about the x-value, can we predict the value of y?
Let’s look at another case you will frequently be asked to solve.
In this case, a straight line does not predict the plasma concentration values very well. The data points are below the line in the middle of the plot. The values are above the prediction line at either end of the data range. Some of you will recall that when dealing with drug plasma concentrations, the relationship is not linear but exponential.
\(C_{p}=C_{p,0}\;\times\;\,e^{-kt}\)
We can linearize the equation by taking the natural log of both sides.
\(ln\;C_{p}=ln\;C_{p,0}-kt\)
Let’s plot the above pharmacokinetic data on a logarithmic scale for the y-axis Cp values and see what we get.
Now, this graph appears to show a linear relationship between the ln Cp value and time.
What is a model? A model is a mathematical equation used to predict the value of y if you know x, or predict the value of x if you know y. For instance, is there a relationship between a patient’s height and their weight? How about between a patient’s serum creatinine and their creatinine clearance? Is there a relationship between the size of the drug dose and the patient’s peak plasma level? Is there a relationship between the patient’s plasma level and the time elapsed after the dose? You know these relationships exist based on your classes, but you may not be able to predict the y-value given an x-value because you are unaware of the particular model.
Here are the two models you will apply most often during your time in the SoP.
1. Logarithmic, 1st – order model (the rate depends on the value of y):
\(C_{p}=C_{p,0}\;\times\;\,e^{-kt}\)
a) Drug degradation from Solution dosage forms,
b) Patient drug plasma levels, and
c) Radioisotope counts and concentrations.
2. Linear, 0 – order model (the rate is constant):
\(C_{t}=C_{0}-kt\)
d) Drug degradation from Suspension dosage forms,
e) In this course, other examples that are not 1st order.
What do you need to memorize? If the problem involves radioactivity, drug plasma levels, or drug degradation in solution, the model is 1st – order. If the problem does not involve a, b, or c, but only drug degradation from a suspension, then the model is 0 – order. Four representative examples are provided.
Module 7B: Drug Degradation from Suspension Dosage Forms, 0-order, Linear Model
A suspension dosage form degrades according to the 0-order model. The equation that predicts the concentration is the starting concentration, C0, minus the product of the 0-order rate constant and time:
\(C_{t}=C_{0}-kt\)
Here is the data for the amount of drug remaining in a suspension over time. Note that time is the x-variable. The y-variable is the measured amount of drug, expressed as the concentration remaining in the bottle.
Time (days since prepared) | Concentration (mg/mL) |
0 | 49.8 |
20 | 46.6 |
30 | 46.1 |
45 | 44.8 |
60 | 41.7 |
90 | 39.9 |
Let’s take a look at the graph.
The graph looks linear, although the data is not perfectly straight. We also know that suspensions degrade via a 0 – order process, so based on that scientific reasoning, we are justified in fitting the data to a straight line. We always see some degree of scatter in the data. This is expected since no man-made measurements are absolute.
Now, let’s look at the mechanics of entering the data into your calculator. This module has a recorded lecture for the 707 course. If you use this text in an environment that cannot access the recorded lecture, you will find several videos on YouTube demonstrating how to perform linear regression with the TI family of calculators.
Setting up your TI-84 Plus CE calculator (You only need to do this once.)
1. Goal: Verify that the stat diagnostics is turned ON.
- Turn on the calculator.
- Select mode.
- Arrow down to stat diagnostics.
- Arrow over to ON
- Hit enter.
2. Goal: Clear the data columns of all data and equations.
- Hit 2nd mem 3 (Clear all entries)
- Hit 2nd mem 4 (Clear all lists)
3. Goal: Enter the data
- Select stat
- Select 1: Edit…
- Enter all the x data (usually time)
- Arrow over to L2
- Enter all the y data (usually concentration)
4. Goal: Running the Regression calculation
- Select stat
- Arrow over to CALC
- Tab down to 4, or just enter 4. (LinReg(ax+b)
5. Goal: Verify the correct columns for the regression analysis.
- You will see this screen.
CRITICAL POINTS
- You must specify which columns contain the x and y-values
- You must verify that the Xlist is L1 and the Ylist is L2.
- In this screenshot my Ylist shows L3 from another problem.
- To change the Ylist, arrow down and type 2nd 2 (above the keypad for 2 is L2).
Now, arrow down to Calculate and select enter.
Let me start off by reminding everyone that we are focusing on the mechanics of performing a linear regression with a calculator. This is NOT a statistics course. We will not be asking questions about the interpretation of r or r2.
6. Goal: Interpreting the results of the regression calculation.
• The heading reminds you of the regression type, in this case, linear regression.
• Line 1 again reminds you of the model, y = ax + b.
• Line 2 is the slope, - 0.111. The minus sign indicates the value of the slope is decreasing with increasing x (time) values. Another fine point to note. While the slope is – 0.111, the value of the rate constant is + 0.111. (The negative sign is accounted for in the model equation.) This is represented in the equation by m. The units for m are y units/x units, in this case, mg/mL/day.
• Line 3 is the intercept, 49.339. That is the best statistical estimate of the time 0 concentration value. This is represented in the equation by b. The units for b are the units for y, in this case, mg/mL. The regression intercept is rarely, if ever, the same as the zero time value in the table.
• Line 4 is r2, the coefficient of determination. It has a formal statistical definition, but we will not concern ourselves with the meaning. The value is not used in any calculations. It does provide interested researchers with information about the “goodness” of the model fit to the data.
•Line 5 is r, the square root of r2 although here it is negative because the regression slope is negative.
7. Goal: Summary and writing the answer to the problem.
You need to write down the linear regression equation and be able to use it to solve problems. For this problem, the equation is:
\(C_{t}=49.34\;mg/mL\;-0.111\;\frac{mg}{mL\;\times\; day}\times t \;(days) \)
Here are some typical calculations you will be expected to perform.
Example 7.1: Find the time required for the concentration to decrease to 45 mg/mL.
\(45\;\frac{mg}{mL}=49.34\;\frac{mg}{mL}-0.111\frac{mg}{mL\;\times\;day}\times t\)
\(\frac{(45\;-\;49.34)\;\frac{mg}{mL}}{-0.111\;\frac{mg}{mL\times day}}=39.1\;days\)
Example 7.2: Find the concentration at 25 days.
\(C_{t=25d}=49.34\;\frac{mg}{mL}\;-0.111\;\frac{mg}{mL\;\times\;day}\times\;25\;days=46.6\;mg/mL \)
Module 7C: Drug Degradation from Solution Dosage Forms, 1st-Order, Exponential Model
Here is the data for the amount of drug remaining in a solution over time. Note that time is the x-variable. The y-variable is the measured amount of drug, expressed as the concentration remaining in the bottle.
Time (days since prepared) | Concentration (mg/mL) |
2 | 15.8 |
5 | 7.9 |
8 | 4.2 |
12 | 1.3 |
Let’s take a look at two graphs. In (a), we observe the data points do not fall on a straight line. Recall that drug degradation in solution occurs via a 1st -order process. For a 1st-order process, a plot of ln C versus time is linear. And that is what is observed in (b).
We can start to perform the linear regression analysis, but first, let’s clear the data entry table.
- Hit 2nd mem 3 (Clear all entries)
- Hit 2nd mem 4 (Clear all lists)
Enter the data, as described in the previous section (select stat, then Edit… . Your screen should look like this.
Now, as shown here, use the arrow (toggle) keys to move the black insertion bar to the cell containing L3, at the top of the column. Note that the text under the table shows L3 = .
We know the relationship between x and y should approximate a straight line when the y data is transformed to its natural log value. The command you use to transform all of the numbers at once is:
ln ( then 2nd then L2 then enter
The calculator inserts the values of ln(L2) in L3. [Note: you can select the number of digits the calculator displays by using the mode key, and tabbing down to float, then tabbing over to the number of digits you prefer. 2nd quit returns you to your previous screen.] Your screen should look like this:
Now that the data is entered into the calculator, the next step is to run the regression, as previously described.
Select stat, arrow to CALC, select 4: LinReg(ax+b).
CRITICAL POINT
In this case, the x-values are in L1 and the y-values are in L3. Tab down to Ylist, type 2nd L3 (L3 is above the key labeled 3, how convenient).
Tab down to CALCULATE, then select enter.
Here are your linear regression results for the ln transformed data.
Results
Here is how to interpret your results.
• The screen heading reminds you that you performed a linear regression.
• The model is y = ax + b. Of course, we think of it as y = mx + b.
• The slope = - 0.248. This is the 1st-order degradation rate constant. The sign indicates the slope is negative and the concentration decreases with time. Another fine point to note. While the slope is – 0.248, the value of the rate constant is + 0.248. The negative sign is accounted for in the model equation. The units of k for the 1st order model are reciprocal time. In this case, the time units are days. The value of k is interpreted as 0.248/day or 0.248 d-1.
• The y-intercept is 3.302. This is the best-fit value of the solution concentration at time = 0. Note the intercept is a transformed value. Recall that the concentration value at 2 hours was approximately 15 mg/mL. The intercept at time = 0 will be bigger than 15 mg/mL. Since the data was transformed by taking the ln (C), to return the real value, you perform the inverse of ln, raise e to the power b. That is eb, or e3.302 = 27.2 mg/mL. A common error is made when students forget to transform the intercept.
• r2 = 0.994. (This represents a “good” fit.)
• You can ignore the value of r in this course.
Summary
You need to correctly write and use the linear regression equation to solve problems. In this case, the equation is:
\(ln\;C_{t}=ln\;C_{0}-kt\)
\(ln\;C_{t}=3.302-0.248/day\times t\)
Here are some typical calculations you will solve in the course.
Example 7.3: Find when the solution concentration was 22.5 mg/mL.
ln 22.5 = 3.302 – 0.248 × t
3.114 = 3.302 – 0.248 × t
3.302 – 3.114 = 0.248 × t
\(t=\frac{3.302\;-\;3.114}{0.248}=0.76\;days\cong 18\;hours\)
Example 7.4: What is the expected concentration on day 4?
\(ln\;C(t=4\;days)=3.302-0.248/day\times 4\; days\)
\(ln\;C(4\;t=days)=2.31 \)
\(C(t=4\;d)=e^{2.31}=10.1\;mg/mL\)
Here is another way to manipulate this equation that you may find quicker to perform with a calculator. This technique will be emphasized in the Pharmacokinetics course.
\(C_{smaller}=C_{larger}\;\times\;e^{-kt}\)
Example 7.5: Let me redo 7.3,
\(22.5\;mg/mL=27.2\;mg/mL\times e^{-0.248t} \)
\(\frac{22.5\;mg/mL}{27.2\;mg/mL}=e^{-0.248t}\)
\(ln\;(\frac{22.5\;mg/mL}{27.2\;mg/mL})=ln(e^{-0.248t})\)
\(ln(0.827)\;=\;-0.248\;\times\;t\)
\(\frac{-0.19}{-0.248}=t=0.77\;days\;=\;18\;hours\)
The difference between 7.3, (0.76 days) and 7.5, (0.77 days) is due to using three decimal places and general rounding and truncation errors. This is not significant difference as both answers round to 18 hours.
Example 7.6: Let me redo 7.4,
\(C_{smaller}=27.2\;mg/mL\times e^{-0.248/day\;\times\;4\;days}=10.1\;mg/mL\)
I am accustomed to using the exponential form of the equation (7.5 and 7.6). You should choose the form with which you feel most comfortable.
Module 7D: Plasma Concentrations after an IV Bolus Dose, 1st-Order, Exponential Model
The 1st-order model is used for drug degradation in solution dosage forms, plasma concentrations following an IV bolus dose, and in radioactive decay processes. The mathematics and the regression procedures are the same, so we will go a little quicker through these calculations. Please make sure to solve enough problems so you will feel confident when performing these calculations in practice.
Here is data for a patient who received an IV bolus dose of a drug. Note that time is the x-variable. The y-variable is the measured plasma drug concentration. This problem is analogous to those in the previous section.
Time (h post dose) | Concentration (mcg/mL) |
1 | 28.1 |
4 | 16.2 |
8 | 6.8 |
15 | 1.9 |
Verify that the columns in your calculator are clear of data.
- Hit 2nd mem 3 (Clear all entries)
- Hit 2nd mem 4 (Clear all lists)
Enter the data, and transform L2 to ln(L2) in column L3.
Your screen should look like this.
Run the regression, verifying that the Xlist is L1 and the Ylist is L3.
Summary
You need to correctly write the linear regression equation and use it to solve problems. In this case, the equation is:
\(ln\;C_{t}=ln\;C_0-kt\)
\(ln\;C_{t}=3.5265-0.1938/hr\times t\;(hr)\)
Example 7.7: a) List the values for k, intercept (ln b) and eb, and r2.
0.1938/h, 3.5265, 34 mcg/mL, and 0.9988.
b) At what time post-dose was the Cp = 4 mcg/mL? (Solve for t)
\(ln\;C_{t}=3.5265-0.1938/hr\times t\;(hr)\)
\(ln\;(4)=3.5265-0.1938/hr\times t\;(hr) \)
\(1.3863=3.5265-0.1938/hr\times t\;(hr) \)
\(\frac{1.3863\;-\;3.5265}{-\;0.1938\;\frac{1}{h}}\;=\;11\;h\)
c) When will the Cp = 0.7 mcg/mL? (Solve for t)
\(ln\;(0.7)=3.5265-0.1938/hr\times t\;(hr)\)
\(-\;0.3567=3.5265-0.1938/hr\times t\;(hr) \)
\(\frac{-\;0.3567\;-\;3.5265}{-\;0.1938\;\frac{1}{h}}\;=\;20\;h\)
d) What is the expected Cp 24 hours after administering the dose? (Solve for Ct)
\(ln\;C_{t}=3.5265-0.1938/hr\times 24\;hr\)
\(ln\;C_{t}=3.5265-4.6512\)
\(ln\;C_{t}=-\;1.1247\)
\(e^{ln\;C_t}=e^{-1.1247} \)
\(C_t\;=\;0.3\;\frac{mcg}{mL} \)
Module 7E: Radioactive Isotope Decay, 1st-Order, Exponential Model
Radioactive decay occurs via a 1st-order process. The decay constant is usually denoted by the Greek letter, 𝜆. Lambda (𝜆) has units of reciprocal time, time-1, for example: h-1, day-1, week-1, month-1, or year-1. Most clinically useful medical isotopes have a half-life measured in hours to days. Recall half-life = 0.693/𝜆. As is usually the case in pharmacy, time is the x-variable. The y-variable is the radioactivity, often measured in milliCurie or microCurie (mCi or 𝜇Ci). (The unit is named in honor of Marie and Pierre Curie.)
Please note: Radioactivity remaining in a sample is often denoted by the letter A (activity). Therefore the general first-order equation is:
\(A_{t}=A_{0}e^{-\lambda t}\) or
\(Ln\;A_{t}=Ln\;A_{0}-\lambda t\)
The mathematical approach to this example problem is analogous to the problems in sections C and D. Let's perform another regression.
A radiopharmaceutical was prepared on day 0.
Time (days) | Activity (mCi) |
2 | 800 |
5 | 565 |
10 | 316 |
24 | 62 |
Before starting the regression procedure, verify that the columns are clear of data.
- Hit 2nd mem 3 (Clear all entries)
- Hit 2nd mem 4 (Clear all lists)
Enter the data, and transform L2 to ln(L2) in L3. Your screen should look like this.
Run the regression, verifying that the Xlist is L1 and Ylist is L3.
\(Ln\;A_{t}=Ln\;A_{0}-\lambda t\)
\(ln\;A_{t}=6.9718-0.1163/day\times t\)
Example 7.8: a) List the values for 𝜆, intercept (ln b), eb, and r2.
0.1163/d, 6.9178, 1010 mCi, 0.999.
b) At what time post-prep was the Activity = 400 mCi? (Solve for t)
\(ln\;A_{t}=6.9178-0.1163/day\times t\;(days)\)
\(ln\;(400)=6.9178-0.1163/day\times t\;(days) \)
\(5.9915=6.9178-0.1163/day\times t\;(days) \)
\(\frac{5.9915\;-\;6.9178}{-\;0.1163\;\frac{1}{day}}\;=\;8\;days\)
c) What is the expected Activity on day 30? (Solve for Ct)
\(ln\;A_{t}=6.9178-0.1163/day\times 30\;days\)
\(ln\;A_{t}=6.9178-3.489\)
\(ln\;A_{t}=3.4288\)
\(e^{ln\;A_t}=e^{3.4288} \)
\(A_t\;=\;30.8\;mCi\)
Module 7F: Working with Regression Results, the 0-Order Model
The motivation for the next 4 sections is based on our experience helping students use the linear regression results to make valid conclusions and predictions.
As mentioned earlier in the Module, the equation that describes the degradation of a drug in a suspension dosage form is:
\(C_{t}=C_0-kt\)
Here Ct is the drug concentration at some time t after the suspension was prepared. C0 is the regression drug concentration at time = 0. You can use this equation to find the time when a sample will be a particular concentration. You can also use the equation to predict the sample concentration at a particular time. The important linear regression parameters are b, the y-intercept (C0 value), and k, the 0-order rate constant (the slope). The units for the zero-order rate constant are Conc × t-1.
There are four parameters in the equation. LR returns C0 and k. If you want to predict the concentration at a particular time, past or future, solve the equation for Ct by inserting the desired value of t. If you want to predict when a suspension will be a particular concentration, use that value and solve for t.
Below is a screen capture of a regression result for a oral suspension. Let’s examine how to manipulate the equation to extract our desired information. In this case, the rate constant units are mg/mL/days, and the concentration is mg/mL.
Example 7.9: a) Write out the correct form of the linear regression equation:
\(C_{t}=C_0-kt\)
b) Enter the calculated regression parameters in their correct places:
\(C_{t}=99.9\;mg/mL\;-0.61\;mg/mL/day\times t\;(days)\)
c) What is the concentration of the suspension after 12 days?
\(C_{t}=99.9\;mg/mL\;-0.61\;mg/mL/day\times 12\;(days)=92.6\;mg/mL\)
d) How long will it take for the concentration to decrease to 90 mg/mL?
\(\frac{(99.9\;-\;90)\;mg/mL}{0.61\;\frac{mg}{mL\;\times\;days}}=16.2\;days\)
So, in summary, the equation has four parameters (or values). You obtain two of the values from the regression procedure. Thus, there are only 2 questions that can be asked. Given a time interval, how much drug is remaining? How long will it take to reach a certain concentration?
Let’s look at the other three types.
Module 7G: Working with Regression Results, the 1st-Order Model, Solution Degradation
Below is a screen capture of a regression result from a solution stability study. Let’s review how to manipulate the equation to extract our desired information. In this case, the rate constant units are days-1, and the concentration is mg/mL.
Example 7.10: What is the best estimate of the solution concentration at time = 0?
\(e^{4.3294}=75.9\;mg/mL\)
When will the concentration equal 70 mg/mL? (ln 70 = 4.2485)
\(ln\;C_{t}=ln\;C_{0}-kt \)
4.2485 = 4.3294 - 0.0067 × t
\(t=\frac{4.3294\;-\;4.2485}{0.0067/day}=13.3\;days\)
What is the expected solution concentration thirty days after preparation?
\(ln\;C_{t}=4.3294\;-\;0.0067\;1/days\;\times\;30\;days\)
\(ln\;C_{t}=4.3294\;-\;0.201\)
\(ln\;C_{t}=4.1284\)
\(C_{t=30\;days}=e^{4.1284}=62.1\;mg/mL\)
Module 7H: Working with Regression Results, the 1st-Order Model, Drug Plasma Concentrations
Here is another example of the 1st-order model applied to the interpretation of drug plasma concentrations following an IV bolus dose of a medication. This is a typical problem you will solve in the pharmacokinetics course.
Below is a screen capture of a regression result from a patient pharmacokinetic study. In this case, the rate constant units are hours-1, and the concentration is mcg/mL.
Example 7.11: What is the best estimate of the solution concentration at time = 0?
\(e^{2.915}=18.4\;mg/mL\)
The next dose should be given when Cp = 3 mcg/mL. (Ln 3 = 1.0986) What time will that occur?
\(ln\;C_{t}=ln\;C_{0}-kt\)
\(1.0986=2.9150\;-\;0.3048\;\times\;t\)
\(t=\frac{2.915-1.0986}{0.3048/hr}=6\;hr \;post\;dose\)
What is the expected Cp value three hours after the dose was administered?
\(ln\;C_{t}=2.9150\;-\;0.3048/h\;\times\;3h\)
\(ln\;C_{t}=2.9150\;-\;0.9144\)
\(ln\;C_{t}=2.0006\)
\(C_{t=3h}=e^{2.0006}=7.4\;mcg/mL\)
Module 7I: Working with Regression Results, the 1st-Order Model, Radioactive Isotope Decay
Recall that radioactive decay also occurs via a 1st-order process. The decay constant is usually denoted by the Greek letter 𝜆. Lambda (𝜆) has units of reciprocal time, time-1, for example, h-1, day-1, week-1, month-1, or year-1. Most clinically useful medical isotopes have a half-life measured in hours or days. Recall half-life = 0.693/𝜆. The units of radioactivity are often measured in milliCurie or microCurie (mCi or mCi). If you like to arise before dawn, nuclear pharmacy might be your specialty!
A calibration study was done for 169Yb. The regression results are shown. Units for 𝜆 are in days-1.
Example 7.12: What is the activity at t = 0?
\(A_{0}=e^{8.962}=7800\;mCi\)
When will the activity = 3000 mCi? (Ln 3000 = 8.0064)
\(Ln\;A_{t}=Ln\;A_{0}-\lambda t\)
\(8.0064=8.9620\;-\;0.0216\;\times\;t\)
\(t=\frac{8.9620\;-\;8.0064}{0.0216/day}=44.2\;days\;after\;preparation\)
Module 7: Practice Problems
1. Data for a drug suspension stability study are shown.
Time (months) | Conc (mg/mL) |
0 | 100.0 |
12 | 94.1 |
24 | 86 |
36 | 81.4 |
48 | 69.5 |
Calculate the equation of the regression line and r2.
Find the time required for the drug concentration to decrease to 40 mg/mL.
Find the concentration at 18 months.
2. The stability of an extemporaneously prepared drug suspension was studied.
Time (months) | Conc (mg/mL) |
3 | 46.8 |
6 | 44.4 |
12 | 40.5 |
24 | 24.5 |
|
|
Calculate the equation of the regression line and r2.
Find the time required for the drug concentration to decrease to 30 mg/mL.
Find the initial concentration and the concentration at 4 months.
3. The stability of an experimental drug suspension was studied.
Time (months) | Conc (mg/mL) |
15 | 13 |
40 | 7.8 |
60 | 5.0 |
85 | 0.3 |
Calculate the equation of the regression line and r2.
Find the time required for the drug concentration to decrease to 5 mg/mL.
Find the initial concentration and the concentration at 48 months.
4. The stability of a drug suspension was studied.
Time (months) | Conc (mg/mL) |
1 | 247.6 |
4 | 240.4 |
12 | 221.9 |
30 | 196.6 |
Calculate the equation of the regression line and r2.
Find the time required for the drug concentration to decrease to 225 mg/mL.
Find the initial concentration and the concentration at 36 months.
5. A small-scale clinical study evaluated the utility of predicting a diabetic patient’s average serum glucose (y) based on their Hgb A1c values (x). These data are expected to follow a linear model.
A1c (%) | Ave Glucose (mg/dL) |
4.7 | 85 |
5.2 | 96 |
5.7 | 126 |
6.4 | 125 |
6.8 | 140 |
6.9 | 161 |
7.5 | 175 |
7.7 | 163 |
7.9 | 174 |
8.2 | 193 |
Calculate the equation of the regression line and r2.
If a patient had an A1c value = 7, what would their average serum glucose have been?
If a patient had an average serum glucose value = 150, what would be their % A1c?
6. The stability of a drug suspension was studied.
Time (months) | Conc (mg/mL) |
3 | 0.982 |
6 | 0.973 |
12 | 0.951 |
24 | 0.858 |
Calculate the equation of the regression line and r2.
Find the time required for the drug concentration to decrease to 0.8 mg/mL.
Find the initial concentration and the concentration at 18 months.
7. A group of pediatric specialists conjecture they can predict the 50th percentile weight for patients between 3 and 30 months. These data are expected to follow a linear model.
Age (months) | 50th percentile Weight (kg) |
3 | 6 |
12 | 10.2 |
30 | 13.7 |
Calculate the equation of the regression line and r2.
A patient is 18 months old. Predict their weight.
A patient weighs 12 kg. Predict their age.
8. The stability of a commercial drug suspension was studied.
Time (months) | Conc (mg/mL) |
12 | 48 |
24 | 42 |
36 | 40 |
48 | 33 |
|
|
Calculate the equation of the regression line and r2.
Find the time required for the drug concentration to decrease to 25 mg/mL.
Find the initial concentration and the concentration at 6 months.
9. A small-scale clinical study evaluated the relationship between diabetic patients' BMI (y) and Hgb A1c values (x). These data are expected to follow a linear model.
A1c (%) | Ave BMI (kg/m2) |
5.7 | 25.3 |
5.8 | 26.2 |
5.9 | 27.1 |
6.0 | 28.4 |
6.1 | 29.6 |
6.2 | 30.7 |
6.3 | 27 |
6.4 | 35.1 |
Calculate the equation of the regression line and r2.
If a patient had an A1c value = 6, what is the best estimate of their BMI?
If a patient had a BMI = 32, what is the best estimate of their A1c value?
10. A small-scale clinical study evaluated the relationship between diabetic patients' Waist-to-Hip ratio (y) (WHR) and Hgb A1c values (x). These data are expected to follow a linear model.
A1c (%) | Waist/Hip |
5.7 | 0.84 |
5.8 | 0.85 |
5.9 | 0.84 |
6.0 | 0.88 |
6.1 | 0.98 |
6.2 | 0.93 |
6.3 | 0.88 |
6.4 | 0.89 |
Calculate the equation of the regression line and r2.
If a patient had an A1c value = 6.1, what is the best estimate of their WHR?
If a patient had a WHR = 0.9, what is the best estimate of their A1c value?
11. Data from a drug solution stability study are shown.
Time (months) | Conc (mg/mL) |
3 | 10.39 |
12 | 9.78 |
24 | 9.65 |
36 | 9.34 |
Calculate the equation of the regression line and r2.
Find t90, the time required for the conc to decrease to 90% of the starting conc.
Find the time required for the drug concentration to decrease to 8 mg/mL.
Find the concentration at 15 months.
12. Data from a drug solution stability study are shown.
Time (days) | Conc (mg/mL) |
2 | 50 |
14 | 49 |
20 | 43 |
30 | 41.3 |
Calculate the equation of the regression line and r2.
Find t90, the time required for the conc to decrease to 90% of the starting conc.
Find the time required for the drug concentration to decrease to 40 mg/mL.
Find the concentration at 25 days.
13. Data from a drug solution stability study are shown.
Time (days) | Conc (mg/mL) |
7 | 24.3 |
14 | 23.4 |
21 | 20.5 |
28 | 19.1 |
Calculate the equation of the regression line and r2.
Find t90, the time required for the conc to decrease to 90% of the starting conc.
Find the time required for the drug concentration to decrease to 15 mg/mL.
Find the concentration at 35 days.
14. Data from a drug solution stability study are shown.
Time (days) | Conc (mg/mL) |
10 | 59.1 |
20 | 57.1 |
25 | 56.1 |
30 | 55.5 |
Calculate the equation of the regression line and r2.
Find the time = 0 concentration.
Find the time required for the drug concentration to decrease to 55 mg/mL.
Find the predicted concentration at 25 days.
15. Data from a drug solution stability study are shown.
Time (days) | Conc (mg/mL) |
3 | 11.9 |
7 | 11.5 |
12 | 10.1 |
15 | 9.9 |
Calculate the equation of the regression line and r2.
Find the time = 0 concentration.
Find the time required for the drug concentration to decrease to 9 mg/mL.
Find the predicted concentration at 10 days.
16. Data from a drug solution stability study are shown.
Time (days) | Conc (mg/mL) |
2 | 4.7 |
5 | 4.1 |
10 | 3.4 |
15 | 2.9 |
Calculate the equation of the regression line and r2.
Find the time = 0 concentration.
Find the time required for the drug concentration to decrease to 4.5 mg/mL.
Find the predicted concentration at 6 days.
17. A hospital pharmacist formulated a pediatric extemporaneous solution. Samples were sent to SIUE for analysis. The data from the stability study are shown.
Time (days) | Conc (mg/mL) |
2 | 28.3 |
5 | 17.1 |
9 | 7.8 |
12 | 5.0 |
Calculate the equation of the regression line and r2.
Find the time = 0 concentration.
Find the time required for the drug concentration to decrease to 10 mg/mL.
Find the predicted concentration at 10 days.
18. A drug solution stability study was conducted at an elevated temperature, 32 °C. The results are shown.
Time (days) | Conc (mg/mL) |
2 | 96.5 |
4 | 92.2 |
10 | 81.4 |
16 | 75.5 |
Calculate the equation of the regression line and r2.
Find the time = 0 concentration.
Find the time needed for the drug concentration to decrease to 90 mg/mL.
Find the predicted concentration at 15 days.
19. An oral drug solution product has the stability data below.
Time (months) | Conc (mg/mL) |
6 | 44.6 |
12 | 39.7 |
24 | 32.5 |
48 | 21.3 |
Calculate the equation of the regression line and r2.
Find the time = 0 concentration.
Find the time needed for the drug concentration to decrease to 40 mg/mL.
Find the predicted concentration at 3 months.
20. Once opened, an injectable solution has a short half-life. The results of a stability study are shown.
Time (hours) | Conc (mcg/mL) |
0.5 | 7.1 |
1.5 | 6.6 |
5 | 4.6 |
10 | 2.8 |
Calculate the equation of the regression line and r2.
Find the time = 0 concentration.
Find t90, the time required for the conc to decrease to 90% of the starting conc.
21. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mcg/mL) |
1 | 29.5 |
5 | 13.6 |
11 | 5.1 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0 h.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
When will the Cp = 10 mcg/mL?
22. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mcg/mL) |
2 | 8.6 |
8 | 2.9 |
12 | 1.4 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0 h.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
When will the Cp = 1 mcg/mL?
23. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mg/L) |
2 | 74 |
6 | 67.2 |
20 | 16.3 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0 h.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
When will the Cp = 0.7 mg/L?
24. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mg/L) |
2 | 16.8 |
12 | 11 |
20 | 5.6 |
36 | 4.4 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0 h.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
Calculate the Cp @ 48 h.
25. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mg/L) |
0.5 | 211.8 |
6 | 94.1 |
10 | 31.9 |
16 | 16.6 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0 h.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
Calculate the Cp @ 12 h.
26. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mcg/mL) |
1 | 33 |
6 | 12.6 |
10.5 | 8.8 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
When will the Cp = 4 mcg/mL?
27. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mcg/mL) |
3 | 9 |
8 | 7.2 |
16 | 4.4 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0 h.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
When will the Cp = 2 mcg/mL?
28. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mcg/mL) |
2 | 12.6 |
7.5 | 3.1 |
10 | 1.7 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0 h.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
When will the Cp = 1 mcg/mL?
29. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mcg/mL) |
4 | 25.9 |
9 | 7.9 |
12.5 | 2.9 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0 h.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
When will the Cp = 1.5 mcg/mL?
30. Patient plasma concentration data is shown.
Time (h post dose) | Conc (mcg/mL) |
4 | 15.6 |
8.5 | 4.4 |
12 | 0.9 |
Calculate the equation of the regression line and r2.
Find the concentration at time = 0 h.
Calculate the drug’s half-life for this patient. (t0.5 = 0.693/ke)
When will the Cp = 2 mcg/mL?
31. A solution contains 60 mCi of a radiopharmaceutical five hours after preparation. At 10 hours, the sample contains 25 mCi. Calculate 𝜆 and t½.
32. The data from an analysis of a radioisotope is shown.
Time (days) | Activity (mCi) |
2 | 1000 |
6 | 358 |
12 | 77 |
Calculate 𝜆 and t ½. Find r2.
Determine the activity on Day 0, when the isotope was prepared.
33. The data from an analysis of a radioisotope is shown. The isotope was prepared on Day 0.
Time (days) | Activity (mCi) |
2 | 143 |
6 | 129 |
12 | 111 |
Calculate 𝜆 and t ½.
Determine the activity on Day 0, when the isotope was prepared.
34. The data from an analysis of a radioisotope is shown. The isotope was prepared on Day 0.
Time (days) | Activity (mCi) |
3 | 101 |
9 | 26 |
15 | 6.6 |
Calculate 𝜆 and t ½.
Determine the activity on Day 0, when the isotope was prepared.
35. The data from an analysis of a radioisotope is shown. The isotope was prepared on Day 0.
Time (h) | Activity (mCi) |
1.5 | 55 |
4 | 48 |
8 | 39 |
Calculate 𝜆 and t ½.
Determine the activity on Day 0, when the isotope was prepared.
36. The data from an analysis of a radioisotope is shown. The isotope was prepared on Day 0.
Time (h) | Activity (mCi) |
2 | 79 |
5 | 56 |
10 | 32 |
Calculate 𝜆 and t ½.
Determine the activity on Day 0, when the isotope was prepared.
37. The data from an analysis of a radioisotope is shown. The isotope was prepared on Day 0.
Time (h) | Activity (mCi) |
1 | 342 |
4 | 110 |
16 | 1.2 |
Calculate 𝜆 and t ½.
Determine the activity on Day 0, when the isotope was prepared.
38. The data from an analysis of a radioisotope is shown. The isotope was prepared on Day 0.
Time (days) | Activity (mCi) |
1 | 9 |
7 | 5 |
14 | 2 |
Calculate 𝜆 and t ½.
Determine the activity on Day 0, when the isotope was prepared.
39. The data from an analysis of a radioisotope is shown. The isotope was prepared on Day 0.
Time (days) | Activity (mCi) |
5 | 31 |
17 | 24 |
40 | 15 |
Calculate 𝜆 and t ½.
Determine the activity on Day 0, when the isotope was prepared.
40. The data from an analysis of a radioisotope is shown. The isotope was prepared on Day 0.
Time (days) | Activity (mCi) |
1 | 71 |
20 | 28 |
35 | 14 |
Calculate 𝜆 and t ½.
Determine the activity on Day 0, when the isotope was prepared.
41. The regression parameters for a drug suspension stability study are shown. The rate constant units are mg/mL/day.
a) What is the calculated concentration at time = 0?
b) When will the concentration = 40 mg/mL?
c) What is the expected concentration after 15 days?
42. The regression parameters for a drug suspension stability study are shown. The rate constant units are mg/mL/month.
a) What is the calculated concentration at time = 0?
b) When will the concentration = 110 mg/mL?
c) What is the expected concentration after 105 days?
43. The regression parameters for a drug solution stability study are shown. The rate constant units are 1/month. The concentration units are mg/mL.
a) What is the calculated concentration at time = 0?
b) When will the concentration = 11 mg/mL?
c) What is the expected concentration after 180 days?
44. The regression parameters for a drug solution stability study are shown. The rate constant units are 1/day. The concentration units are mg/mL.
a) What is the calculated concentration at time = 0?
b) When will the concentration = 36 mg/mL?
c) What is the expected concentration after 40 days?
45. A patient received an IV bolus drug dose. The regression parameters are shown. The rate constant units are 1/hour. The concentration units are mcg/mL.
a) What is the calculated concentration at time = 0?
b) When will the concentration = 9 mcg/mL?
c) What is the expected concentration after 18 hours?
46. A patient received an IV bolus drug dose. The regression parameters are shown. The rate constant units are 1/hour. The concentration units are mg/L.
a) What is the calculated concentration at time = 0?
b) When will the concentration = 6 mg/L?
c) What is the expected concentration after 7 hours?
47. The regression parameters for a radiopharmaceutical product are shown. The rate constant units are 1/hour.
a) What is the calculated activity (mCi) at time = 0?
b) How many hours after preparation will the activity (mCi) = 50?
c) What is the expected activity 12 hours after preparation?
48. The regression parameters for a radiopharmaceutical product are shown. The rate constant units are 1/days.
a) What is the calculated activity (mCi) at time = 0?
b) How many hours after preparation will the activity (mCi) = 10?
c) What is the expected activity 30 hours after preparation?
Answers
1. C = 100.94 mg/mL – 0.614 mg/mL/month · t; r2 = 0.979; 99.251 months to reach 40 mg/mL; 89.888 mg/mL at 18 months.
2. C = 51.07 mg/mL – 1.0684 mg/mL/month · t; r2 = 0.976; 19.721 months to reach 30 mg/mL; C0 = 51.07 mg/mL; 46.796 mg/mL at 4 months.
3. C = 15.44 mg/mL – 0.178 mg/mL/month · t; r2 = 0.996; 58.652 months to reach 5 mg/mL; C0 = 15.44 mg/mL; 6.896 mg/mL at 48 months.
4. C = 246.986 mg/mL – 1.732 mg/mL/month · t; r2 = 0.983; 12.694 months to reach 225 mg/mL; C0 = 246.986 mg/mL; 184.634 mg/mL at 36 months.
5. Ave Glu = - 51.09 mg/dL + 29.09% · A1c; r2 = 0.9443; %A1c = 6.9.
6. C = 1.0087 mg/mL – 0.00602 mg/mL/month · t; r2 = 0.966; 34.668 months to reach 0.8 mg/mL; C0 = 1.0087 mg/mL; 0.9 mg/mL at 18 months.
7. Wt @ 50th % = 0.272 kg/month × age (months) + 5.9 kg; 18 months old = 10.8 kg; 12 kg = 22.4 months of age.
8. C = 52.5 mg/mL – 0.392 mg/mL/month · t; r2 = 0.9625; 70.15 months to reach 25 mg/mL; C0 = 52.5 mg/mL; 50.1 mg/mL at 6 months.
9. Ave BMI (kg/m2) = - 32.26 (kg/ m2) + 10.07 kg/m2/% · A1c %; r2 = 0.6164; BMI = 28.16 kg/m2; A1c = 6.4%.
10. WHR = 0.1036 1/% · A1c % + 0.2596; r2 = 0.2596; WHR = 0.89; A1c = 6.2%.
11. Ln C = 2.336 – 0.0029 · t, r2 = 0.8957; (90% of C0 = 9.31 mg/mL) t90 = 36.2 months; 88.5 months; 9.9 mg/mL.
12. Ln C = 3.9442 – 0.0074 · t, r2 = 0.8457; (90% of C0 = 46.47 mg/mL) t90 = 14.2 days; 34.4 days; 42.9 mg/mL.
13. Ln C = 3.2920 – 0.0122 · t, r2 = 0.9608; (90% of C0 = 24.21 mg/mL) t90 = 8.6 days; 47.9 days; 17.6 mg/mL.
14. Ln C = 4.1101 – 0.0032 · t, r2 = 0.9926; C0 = 60.9 mg/mL; 31.8 days; 56.2 mg/mL.
15. Ln C = 2.5373 – 0.0169 · t, r2 = 0.9526; C0 = 12.6 mg/mL; 19.9 days; 10.6 mg/mL.
16. Ln C = 1.6071 – 0.0369 · t, r2 = 0.9950; C0 = 5 mg/mL; 2.9 days; 4 mg/mL.
17. Ln C = 3.6969 – 0.1765 · t, r2 = 0.9975; C0 = 40.3 mg/mL; 7.9 days; 6.9 mg/mL.
18. Ln C = 4.5957 – 0.0177 · t, r2 = 0.9844; C0 = 99.1 mg/mL; 5.4 days; 76 mg/mL.
19. Ln C = 3.8982 – 0.0175 · t, r2 = 0.9998; C0 = 49.3 mg/mL; 11.9 months; 46.8 mg/mL.
20. Ln C = 2.0219 – 0.0991 · t, r2 = 0.9994; C0 = 7.6 mcg/mL; (90% of C0 = 6.84 mcg/mL) t90 = 1.1 hours.
21. Ln C = 3.5304 – 0.1746 · t, r2 = 0.9978; C0 = 34.1 mcg/mL; t ½ = 4 h; C = 10 @ 7 h post-dose.
22. Ln C = 2.5153 – 0.1815 · t, r2 = 0.9999; C0 = 12.4 mcg/mL; t ½ = 3.8 h; C = 1 @ 13.9 h post-dose.
23. Ln C = 4.5939 – 0.0885 · t, r2 = 0.9758; C0 = 98.9 mg/L; t ½ = 7.8 h;
C = 0.7 @ 55.9 h post-dose.
24. Ln C = 2.8184 – 0.0407 · t, r2 = 0.9051; C0 = 16.8 mg/L; t ½ = 17 h;
C @ 48 h post-dose = 2.4 mg/L.
25. Ln C = 5.4297 – 0.1707 · t, r2 = 0.9741; C0 = 228 mg/L; t ½ = 4.1 h;
C @ 12 h post-dose = 29.4 mg/L.
26. Ln C = 3.5523 – 0.1401 · t, r2 = 0.9491; C0 = 34.9 mg/L; t ½ = 4.9 h;
C = 4 @ 15.5 h post-dose.
27. Ln C = 2.3852 – 0.0557 · t, r2 = 0.9934; C0 = 10.9 mg/L; t ½ = 12.4 h;
C = 2 @ 30.4 h post-dose.
28. Ln C = 3.031 – 0.2511 · t, r2 = 0.9998; C0 = 20.7 mg/L; t ½ = 2.8 h;
C = 1 @ 12.1 h post-dose.
29. Ln C = 4.3064 – 0.2562 · t, r2 = 0.9972; C0 = 74.2 mg/L; t ½ = 2.7 h;
C = 1.5 @ 15.2 h post-dose.
30. Ln C = 4.2579 – 0.3531 · t, r2 = 0.9814; C0 = 70.7 mg/L; t ½ = 2 h;
C = 2 @ 10.1 h post-dose.
31. 𝜆 = 0.175 h-1, t ½ = 4 h.
32. 𝜆 = 0.2564 h-1, t ½ = 2.7 days. r2 = 0.9999. C0 = 1668 mCi.
33. 𝜆 = 0.025 d-1, t ½ = 27.4 days. C0 = 150 mCi.
34. 𝜆 = 0.2273 d-1, t ½ = 3.05 days. C0 = 200 mCi.
35. 𝜆 = 0.0528 h-1, t ½ = 13.1 h. C0 = 59 mCi.
36. 𝜆 = 0.1129 h-1, t ½ = 6.1 h. C0 = 99 mCi.
37. 𝜆 = 0.3767 h-1, t ½ = 1.8 h. C0 = 498 mCi.
38. 𝜆 = 0.1161 d-1, t ½ = 6 d. C0 = 11 mCi.
39. 𝜆 = 0.0207 d-1, t ½ = 33.5 d. C0 = 34 mCi.
40. 𝜆 = 0.0478 d-1, t ½ = 14.5 d. C0 = 74 mCi.
41. 49.7 mg/mL, 59.7 days, 47.3 mg/mL.
42. 118.6 mg/mL, 10.2 months, 115.7 mg/mL.
43. 12.2 mg/mL, 23.9 months, 11.9 mg/mL.
44. 40 mg/mL, 27.7 days, 34.4 mg/mL.
45. 20.1 mcg/mL, 7.9 hours post-dose, 3.3 mcg/mL.
46. 45.1 mg/L, 5.8 hours post-dose, 4 mg/L.
47. 98 mCi, 6 hours, 26 mCi.
48. 19.4 mCi, 14.7 hours, 5 mCi.
Module 8: Standardized Dosing Protocols
This module will demonstrate some examples for calculating patient-specific doses according to institutional protocols and patient lab data.
Standard protocols are established by medical societies or hospital committees as a guide for initiating and/or adjusting treatment with a particular drug based on the patient’s weight, kidney function, or other lab test results. They are intended to ensure appropriate personalized drug therapy for patients. It is probably easiest to understand with an example, so let’s jump into the first protocol.
Note: Every hospital or institution is free to set their own drug therapy guidelines. These examples are intended to give experience and confidence in implementing treatment protocols. They are not intended as instructions for patient care. You must follow the treatment protocols in use at the institution where you practice.
Module 8A: Amikacin
Amikacin is an aminoglycoside antibiotic that is administered by IV infusion to treat infections with sensitive strains of Gram-negative bacteria. The dose and frequency of amikacin administration must be chosen appropriately to achieve effective therapy without causing excessive toxicity. If the dose is too low or not given frequently enough, then the amikacin concentration in the patient’s blood and tissues will be too low to cure the infection. If the dose is too high or given too frequently, then the drug concentration will build up in the patient’s blood and tissues to a potentially toxic level. An example protocol is shown below.
Table 8.1 Amikacin Protocol | ||
CrCl > 60 mL/min | CrCl 40 – 60 mL/min | CrCl 20 – 40 mL/min |
5 – 7.5 mg/kg* q8h | 5 – 7.5 mg/kg* q12h | 5 – 7.5 mg/kg* q24h |
Use IBW if BMI < 40 kg/m2 |
The normal dose is 5 to 7.5 mg of amikacin per kg of body weight, and the dose is to be administered either every 8 hours, 12 hours, or 24 hours depending on the patient’s creatinine clearance. Body Mass Index (BMI) is used to determine which weight to use in the calculation. For patients with BMI < 40, the dose is calculated using the lower of actual body weight or ideal body weight (IBW). For patients with BMI ≥ 40, the dose is calculated using adjusted body weight (ABW), ABW = IBW + 0.4(Actual wt - IBW)
Recall the equation for calculating BMI is the patient’s actual body weight in kg divided by the square of their height in meters. BMI therefore has units of kg/m2. Do not forget to square the patient height.
\(BMI=\frac{Pt\;actual\;wt\;(kg))}{(Pt\;height\;(m))^{2}}\)
IBW and ABW are calculated as you learned in Module 5.
Once the correct dose has been determined, the frequency of administration is based on the patient’s creatinine clearance, as you learned in Module 5. Remember to use the lower of ideal or actual body weight when calculating creatinine clearance. Let’s walk through an example.
Example 8.1: A physician orders amikacin 7 mg/kg for a 35 year old female patient (165 lb and 5’7”). The patient’s serum creatinine is 1.5 mg/dL. Calculate the appropriate amikacin dose in mg and the correct frequency. Amikacin is supplied in sterile vials containing 500 mg in every 2 mL of solution. Calculate how many milliliters of the drug solution are required to prepare each dose of the drug for this patient.
First calculate BMI to determine which weight to use in the dose calculation.
\(BMI=\frac{165\;lb\times \frac{1\;kg}{2.2\;lb}}{(67\;in\times \frac{2.54\;cm}{in}\times \frac{1\;m}{100\;cm})^{2}}=\frac{25.9\;kg}{m^{2}}\)
BMI < 40, so calculate dose using the lower of actual body weight or ideal body weight.
\(Actual\;weight=165\;lb\times \frac{1\;kg}{2.2\;lb}=75\;kg\)
\(IBW=45.5\;kg +2.3\;kg(67"-60")=61.6\;kg\)
Use IBW to calculate the dose.
\(Dose=61.6\;kg\times \frac{7\;mg}{kg}=431.2\;mg\)
The calculated dose is 431.2 mg, so round this to a reasonable practical dose of 430 mg.
Next calculate creatinine clearance to determine the frequency of administration. Remember to use the 0.85 factor for a female and IBW because it is lower than actual body weight.
\(CrCl=\frac{0.85\;\times \;(140-35)\;\times \;61.6\;kg}{72\;\times \;1.5}=50.9\;mL/min,\text{round to 51}\; mL/min\)
51 mL/min falls between 40 and 60 mL/min, so the appropriate frequency of administration according to Table 8.1 is every 12 hours. An appropriate dosage regimen for this patient, according to the prescriber’s order of 7 mg/kg, is 430 mg every 12 hours.
The volume of amikacin solution required for each infusion is calculated as:
\(430\;mg\times \frac{2\;mL}{500\;mg}=1.7\;mL\)
Module 8B: Vancomycin AUC:MIC Ratio
Vancomycin is another antibiotic that is administered by IV infusion. Similar to amikacin, the dose and frequency must be chosen to ensure effective therapy while minimizing toxicity. The AUC:mic ratio, or ratio of vancomycin area under the curve (AUC) to minimum inhibitory concentration of the drug for the infecting bacterium can be used to calculate the optimal vancomycin dose. MIC is reported by the microbiology lab, while AUC is estimated from the patient’s creatinine clearance. Vancomycin is primarily eliminated from the body by urinary excretion, so the patient’s kidney function is a major factor in determining the dose. The vancomycin dose is calculated to target AUC as a chosen multiple of MIC, typically 400 or 450.
The vancomycin dose is calculated using the equation
\(\text{Vancomycin daily dose} (\frac{mg}{24\;hr})=0.045\times CrCl\times MIC\times target\;multiple\)
Calculate CrCl using the lower of IBW or actual body weight as usual. The factor 0.045 is a unit conversion factor so that the result gives an answer in mg of vancomycin. We note here that some physicians advocate for using the actual body weight when estimating the CrCl for use in this vancomycin calculation. As we have mentioned previously, this text is not offering medical or therapeutic advice. When in practice adhere to the standards in use at your facility.
NOTE 1: The daily dose is rounded up to the nearest 250 mg (250, 500, 750, 1000, 1250, 1500, 1750, 2000, etc.)) If the calculated value is 2050 mg per day, then the daily dose should be rounded up to 2250 mg per day. (You will learn more about vancomycin dosing in Pharmacokinetics and Integrated Therapeutics courses.)
NOTE 2: The equation calculates the required daily dose of vancomycin, and vancomycin is typically administered every 12 hours. Each individual dose is ½ of the daily dose.
Example 8.2: A 60 yo female patient (5’5”, 145 lb, SCr = 0.9 mg/dL) has an infection the physician would like to treat with vancomycin. The MIC for the organism is 1.2 mcg/mL. Calculate the daily dose to target 400 × MIC.
The patients actual body weight is 145 lb or 65.9 kg.
\(IBW=45.5\;kg +2.3\;kg\times\;(65"-60")=57\;kg\)
Calculate CrCl using 57 kg.
\(CrCl=\frac{0.85\;\times \;(140-60)\;\times \;57\;kg}{72\;\times \;0.9}=59.8\;mL/min,\text{round to 60}\; mL/min\)
\(\text{Daily dose (mg per 24 hours)}=0.045\times 60\times 1.2\times \;400=1296\;mg/24\;hr\)
The calculated daily dose of 1296 mg is rounded up to the nearest multiple of 250 mg, or 1500 mg day.
Since vancomycin in given every 12 hours, each individual dose is 750 mg.
This patient should receive 750 mg every 12 hours.
Module 8C: Intravenous Phosphate Supplementation
Phosphate is an important plasma electrolyte. Patients with low phosphate levels may require supplements to prevent the negative health effects. Phosphate supplements should be given orally if the patient is able to take it. Many institutions have treatment guidelines to ensure safe intravenous phosphate supplementation when the oral route is not sufficient.
Intravenous phosphate supplement products include Sodium Phosphates Injection and Potassium Phosphates Injection. Both products are highly concentrated, containing 3 mmoles of phosphate per mL or 3 moles per liter. Sodium phosphates contains 4 mEq/mL of Na+ and potassium phosphates contains 4.4 mEq/mL of K+.
Sodium and potassium are also important electrolytes, and their concentrations must be maintained at a healthy level when administering phosphate supplements. The normal sodium level in plasma is approximately 140 mEq/L, and excess sodium is excreted in the urine. A patient with normal kidney function will excrete any excess sodium that is administered as sodium phosphate supplementation.
The normal potassium level in plasma is approximately 3.5 – 5 mEq/L. A patient can experience serious adverse effects if their potassium level falls too low below 3.5 mEq/L or is elevated above 5 mEq/L. For an adult patient, an intravenous dose of 40 mEq of K+ is expected to raise the plasma potassium level by approximately 0.5 mEq/L. Pharmacists must be aware of the patient’s potassium status and the amount of potassium in an order for potassium phosphates infusion to avoid raising the patient’s potassium level above 5 mEq/L.
If a patient receives an order for phosphates infusion and their phosphate level is normal, the best choice for phosphate supplementation would be sodium phosphates. If the patient requires both potassium and phosphate supplementation, then potassium phosphates is the convenient choice.
Finally, both sodium and potassium phosphates are highly concentrated solutions with osmolarity of 7 or 7.4 mOsm/mL, or 7000 mOsm/L or 7400 mOsm/L. These products must be diluted in a large enough volume to bring the infusion to an osmolarity less than 600 mOsm/L for peripheral infusion.
It is useful to memorize the osmolarity of the common base fluids so that you do not need to calculate them for every problem:
- 0.9% Sodium Chloride Injectin (Normal Saline, NS), = 308 mOsm/L
- 0.45% Sodium Chloride Injection (1/2 Normal Saline, 1/2 NS) = 154 mOsm/L
- 5% Dextrose Injection (D5W) = 252 mOsm/L
- Sterile Water for Injection (SWI) has no solutes, so 0 mOsm/L
Example 8.3: A patient with Na+ of 138 mEq/L and K+ 4.9 mEq/L is ordered 50 mmoles of potassium phosphates in 250 mL of D5W over 6 hours to supplement their low serum phosphate level. How many mEq of potassium would the patient receive from this solution? How much would this raise the patient’s potassium level? Is this the best choice to treat the patient? If not, what should you recommend.
Potassium phosphates contains 3 mmoles phosphate and 4.4 mEq potassium per mL. The potassium dose represented by 50 mmoles of potassium phosphate and the resulting expected increase in blood potassium level is:
\(50\;mmol\;P\times \frac{4.4\;mEq\;K^{+}}{3\;mmol\;P}=73.3\;mEq\;K^{+}\times \frac{0.5\;mEq/L}{40\;mEq\;K^{+}}=0.9\;mEq/L\;increase\)
50 mmoles of potassium phosphate carries a K+ dose of 73.3 mEq, which would raise the K+ level to approximately 4.9 + 0.9 = 5.8 mEq/L. This is higher than the upper limit of normal K+, so it should be avoided. Recommend for patient safety that the supplement order be changed to sodium phosphates 50 mmoles.
Example 8.4: An intravenous phosphate supplement is prepared by adding 20 mL of sodium phosphates injection to 250 mL of D5W. Calculate the osmolarity of the solution. Should this be given via peripheral IV line?
The solution is prepared by mixing 20 mL of a 7000 mOsm/L solution with 250 mL of a 252 mOsm/L solution. The mixture osmolarity is calculated (with volumes in liters):
\(\frac{(0.02L\;\times \;\frac{7000\;mOsm}{L})\;+\;(0.25\;L\;\times \;\frac{252\;mOsm}{L})}{0.02\;+\;0.25\;L}=\frac{752\;mOsm}{L}\)
The infusion osmolarity is greater than the recommended limit of 600 mOsm/L for peripheral infusion. The sodium phosphate supplement should be diluted in a larger volume. For example, if 20 mL of sodium phosphates injection was added to 500 mL of D5W the osmolarity would be 511 mOsm/L, which is acceptable for peripheral infusion.
If the prescriber does not want to use 500 mL of D5W, the phosphate dose could be prepared in other ways. For example, it could be combined with 435 mL of sterile water for injection in an empty IV bag to give an isoosmolar solution (308 mOsm/L).
Module 8D: Cisplatin-Etoposide for Non-Small Cell Lung Cancer
Cancer chemotherapy is generally administered according to a standard protocol based on the type of cancer being treated and patient-specific variables.
As an example, a protocol for treating non-small cell lung cancer (NSCLC) uses two drugs on a defined schedule:
- Cisplatin: 80 mg/m2 by IV infusion on day 1 of cycle,
- Etoposide 100 mg/m2/day by IV infusion on days 1, 2, and 3 of cycle. Repeat the cycle every 21 days for 4 cycles.
One cycle is 3 days of treatment, with both cisplatin and etoposide on day 1 and only etoposide on days 2 and 3. The patient then waits 18 days to start the next cycle. Cisplatin and etoposide doses are based on the patient’s body surface area.
Cisplatin is diluted in 2 L of fluid containing 5% dextrose, ⅓ to ½ normal saline, and 37.5 g of mannitol. Etoposide must be diluted to a concentration of 0.2 to 0.4 mg/mL in D5W or NS and should be infused over 60 minutes.
Example 8.5: A 47 year old male patient (5’11” and 198 lb) is ordered cisplatin and etoposide according to the protocol above. Calculate the total amount of cisplatin and etoposide the patient should receive in the first cycle.
The cycle represents one dose cisplatin and 3 doses of etoposide. Calculate the dose of each.
\(BSA=\sqrt{\frac{71\;in\;\times \;\frac{2.54\;cm}{in}\;\times \;198\;lb\;\times \;\frac{1\;kg}{2.2\;lb}}{3600}}=2.12m^{2}\)
- Cisplatin dose: 2.12 m2 x 80 mg/m2 = 169.6 mg; round this to 170 mg.
- Etoposide dose: 2.12 m2 x 100 mg/m2 = 212 mg; round this to 210 mg.
- The total cisplatin for 1 cycle is 170 mg.
- The total etoposide for 1 cycle is 3 x 210 mg = 630 mg.
Module 8E: Dose Rounding for Biologic Drugs
Many institutions have implemented a dose rounding policy for biologic drugs, especially those used to treat cancer. These products tend to be expensive and any leftover partial vials represent significant loss of money. Dose rounding protocols are used to manage the cost of cancer therapy. There are several papers in the pharmacy literature describing how hospital or health system pharmacies have saved millions of dollars per year using this approach.
The protocol is generally very simple: if the ordered dose for a patient is within ± 10% of a whole number of the drug vials, then the dose is automatically rounded to the nearest whole vial size. Doses may be rounded up or down within the accepted range.
NOTE: Some hospitals may use a different percent of the dose when rounding, e.g., ± 5%. As always, follow the specific instructions at your institution. For the purpose of this course, we will use ± 10%.
The procedure for determining the appropriate dose for a patient under a dose rounding protocol is:
- Calculate dose according to treatment guideline.
- Calculate (dose – 10%) and (dose + 10%).
- Determine if whole vial sizes fall within the acceptable dose range. If yes – round dose to the whole vial size. If no – use the dose calculated from treatment guideline.
NOTE: There may be more than 1 acceptable answer for a particular problem, depending on the vial sizes for a particular drug. Any combination of full vials to provide the correct dose is acceptable for this course.
Example 8.6: Bevacizumab is used treat some cancers and is available in 100 mg and 400 mg vials. A 56 kg patient is ordered bevacizumab 5 mg/kg. Calculate the appropriate dose for this patient.
\(56\;kg\;\times\;5\;\frac{mg}{kg}=280\;mg\)
280 mg ± 10% spans the range 252 – 308 mg. 300 mg represents 3 full 100 mg vials. The acceptable dose range (252 – 308 mg) contains 300 mg, which represents 3 full 100 mg vials. The dose should be rounded up to 300 mg.
The figure shows the available full vial doses from combining 100 mg and/or 400 mg vials. The blue box represents the acceptable dose range of 252-308 mg. The mark for 300 mg (3 x 100 mg vials) is within the acceptable range, so the dose should be rounded to 300 mg.
Example 8.7: A 70 kg patient is ordered 5 mg/kg of bevacizumab = 350 mg.
350 mg ± 10% is 315 – 385 mg. No combination of full vials falls within this range. Leave dose at 350 mg. Some drug waste is unavoidable in this case.
Module 8: Practice Problems
- MM is a 53 y.o. female, 5’3” tall and 130 lb, with serum creatinine of 2.1 mg/dL.
MM’s physician orders amikacin 7.5 mg/kg.
a. Calculate the correct dose and frequency for MM. Round it to the nearest 10 mg.
b. Calculate the volume of drug solution for each dose. Round it to a reasonable number. - RP is a 64 y.o. male, 5’8” tall and 265 lb, with serum creatinine of 1.7 mg/dL.
RP’s physician orders amikacin 5 mg/kg.
a. Calculate the correct dose and frequency for RP. Round it to the nearest 10 mg.
b. Calculate the volume of drug solution for each dose. Round it to a reasonable number. - DH is a 39 y.o. male, 6’ tall and 210 lb, with serum creatinine of 1.4 mg/dL.
DH’s physician orders amikacin 6 mg/kg.
a. Calculate the correct dose and frequency for DH. Round it to the nearest 10 mg.
b. Calculate the volume of drug solution for each dose. Round it to a reasonable number. - A 34 yo male patient (5’8”, 155 lb, SCr = 0.8 mg/dL) has an infection the physician would like to treat with vancomycin. The MIC for the organism is 1.2 mcg/mL. Calculate the daily dose to target 400*mic and the volume of drug solution required for each individual dose.
- A 45 yo female patient (5’11”, 165 lb, SCr = 1.1 mg/dL) has an infection the physician would like to treat with vancomycin. The MIC for the organism is 1.3 mcg/mL. Calculate the daily dose to target 500*mic and the volume of drug solution required for each individual dose.
- A 62 yo male patient (5’10”, 150 lb, SCr = 1.4 mg/dL) has an infection the physician would like to treat with vancomycin. The MIC for the organism is 1.4 mcg/mL. Calculate the daily dose to target 450*mic and the volume of drug solution required for each individual dose.
- Calculate the volume of vancomycin injection required per dose for each patient according to the protocol.
a. CrCl = 25 mL/min, AUC/MIC multiple = 400, MIC = 1.1 mcg/mL.
b. CrCl = 35 mL/min, AUC/MIC multiple = 500, MIC = 1.3 mcg/mL.
c. CrCl = 45 mL/min, AUC/MIC multiple = 600, MIC = 1.5 mcg/mL.
d. CrCl = 55 mL/min, AUC/MIC multiple = 500, MIC = 1.3 mcg/mL.
e. CrCl = 75 mL/min, AUC/MIC multiple = 400, MIC = 1.1 mcg/mL.
f. CrCl = 90 mL/min, AUC/MIC multiple = 500, MIC = 1.3 mcg/mL. - A patient with Na+ of 138 mEq/L and K+ 4.9 mEq/L is ordered 40 mmoles of potassium phosphates in 250 mL of D5W over 6 hours to supplement their low serum phosphate level. How many mEq of potassium would the patient receive from this solution? Is this the best choice to treat the patient? If not, what would you recommend.
- A patient with Na+ of 142 mEq/L and K+ 3.7 mEq/L is ordered 35 mmoles of potassium phosphates in 250 mL of D5W over 6 hours to supplement their low serum phosphate level.
How many mL of drug solution is required to fill the order.
What is the osmolarity of the solution as ordered. Should it be given by peripheral line? - A patient with Na+ of 148 mEq/L and K+ 4.1 mEq/L is ordered 25 mmoles of potassium phosphates in 250 mL of D5W over 6 hours to supplement their low serum phosphate level.
How many mL of drug solution is required to fill the order.
How many mEq of potassium would the patient receive from this solution? - A 63 year old female patient, 4’11” and 150 lb, is ordered cisplatin-etoposide.
Calculate the cisplatin and etoposide dose for day 1 of the treatment cycle. - Calculate the cisplatin and etoposide dose according to the protocol for a 32 year old male, 6’2” and 195 lb.
- A patient is ordered cisplatin 150 mg to be added to 2L of fluid containing 5% dextrose, ½ normal saline, and 37.5 g of mannitol. The 2L of fluid diluent is prepared by mixing the appropriate volumes of sterile water for injection, dextrose 70% injection, sodium chloride 4 mEq/mL injection, and mannitol 20% injection. Recall that normal saline is 140 mEq/L of NaCl. Calculate the volume of each component to prepare the 2L of base fluid.
Dose Rounding Problems
Bevacizumab is supplied in 100 mg and 400 mg vials and is administered at 5, 10, or 15 mg/kg of body weight. - A physician orders a dose of 10 mg/kg for a 75 kg patient. What dose should the patient receive.
- A physician orders a dose of 5 mg/kg for a 52 kg patient. What dose should the patient receive.
- A physician orders a dose of 15 mg/kg for a 91 kg patient. What dose should the patient receive.
- A physician orders a dose of 10 mg/kg for a 147 kg patient. What dose should the patient receive.
Daratumumab is supplied in 100 mg and 400 mg vials. It is administered at 16 mg/kg of body weight. - A physician orders a dose of 16 mg/kg for a 75 kg patient. What dose should the patient receive.
- A physician orders a dose of 16 mg/kg for a 52 kg patient. What dose should the patient receive.
- A physician orders a dose of 16 mg/kg for a 91 kg patient. What dose should the patient receive.
- A physician orders a dose of 16 mg/kg for a 63 kg patient. What dose should the patient receive.
Trastuzumab is supplied in 150 mg and 420 mg vials. It is administered at 2, 4, 6, or 8 mg/kg of body weight. - A physician orders a dose of 2 mg/kg for a 75 kg patient. What dose should the patient receive.
- A physician orders a dose of 4 mg/kg for a 52 kg patient. What dose should the patient receive.
- A physician orders a dose of 6 mg/kg for a 91 kg patient. What dose should the patient receive.
- A physician orders a dose of 8 mg/kg for a 82 kg patient. What dose should the patient receive.
Answers
- 390 mg q24h, 1.6 mL
- 450 mg q12h, 1.8 mL
- 470 mg q8h, 1.9 mL
- 2750 mg/day, 27.5 mL per dose
- 2250 mg/day, 22.5 mL per dose
- 1500 mg/day, 15 mL per dose
- a. 5 mL; b. 12.5 mL; c. 20 mL; d. 17.5 mL; e. 15 mL; f. 27.5 mL.
- 58.7 mEq K+ would raise patient level 0.7 mEq/L to 5.6 mEq/L. Sodium phosphate would be a better choice.
- 11.7 mL of solution required, infusion osmolarity 567 mOsm/L. Okay for peripheral infusion.
- 8.3 mL, 36.7 mEq K+.
- (134.4 mg) 135 mg cisplatin and (168 mg) 170 mg etoposide.
- (172 mg) 170 mg cisplatin and 215 mg etoposide.
- 142.9 mL 70% dextrose, 35 mL NaCl 4 mEq/mL, 187.5 mL mannitol 20%, and 1634.6 mL sterile water for injection.
Ordered dose 10 mg/kg x 75 kg = 750 mg ± 10% = 675 – 825 mg.
2 x 400 mg vials = 800 mg or 1 x 400 + 3 x 100 = 700 mgOrdered dose 5 mg/kg x 52 kg = 260 mg ± 10% = 234 – 286 mg.
This range does not allow for full vial use. The dose of 260 mg should be used.Ordered dose 15 mg/kg x 91 kg = 1365 mg ± 10% = 1228 – 1502 mg.
3 x 400 mg + 1 x 100 mg = 1300 mg or 3 x 400 mg + 2 x 100 mg = 1400 mg
Ordered dose 10 mg/kg x 147 kg = 1470 mg ± 10% = 1323 – 1617 mg.
4 x 400 mg = 1600 mg (fewest vials), or 3 x 400 mg + 2 x 100 mg = 1400 mg,
or 3 x 400 mg + 3 x 100 mg = 1500 mg (closest to ordered dose)Ordered dose 16 mg/kg x 75 kg = 1200 mg ± 10% = 1080 – 1320 mg.
3 x 400 mg = 1200 mgOrdered dose 16 mg/kg x 52 kg = 832 mg ± 10% = 748 – 915 mg.
2 x 400 mg = 800 mg (fewest vials and closest to ordered dose)Ordered dose 16 mg/kg x 91 kg = 1456 mg ± 10% = 1310 – 1601 mg.
4 x 400 mg = 1600 mg (fewest vials) or 3 x 400 mg + 3 x 100 mg = 1500 mg (closest to ordered dose)Ordered dose 16 mg/kg x 63 kg = 1008 mg ± 10% = 907 – 1109 mg.
2 x 400 mg + 2 x 100 mg = 1000 mg. (closest to ordered dose and fewest vials)Ordered dose 2 mg/kg x 75 kg = 150 mg ± 10% = 135 – 165 mg.
1 x 150 mg = 150 mgOrdered dose 4 mg/kg x 52 kg = 208 mg ± 10% = 187 – 229 mg.
This range does not allow for full vial use. The dose of 208 mg should be used.Ordered dose 6 mg/kg x 91 kg = 546 mg ± 10% = 491 – 601 mg.
1 x 150 mg + 1 x 420 mg = 570 mgOrdered dose 8 mg/kg x 82 kg = 656 mg ± 10% = 590 – 722 mg.
1 x 420 mg + 2 x 150 mg = 720 mg
Module 9: Case Based Problems
This modele applies concepts from the prior 8 modules to case-based problems with more than one calculated answer. The solution to a particular problem may require calculations obtained from earlier steps.
Module 9A: Compound an Oral Suspension for Marshmallow the Cat
Example 9.1: Marshmallow, a 3 year old cat that weighs 9 lb 4 oz, is diagnosed with a bacterial infection in her intestines. A veterinarian has prescribed metronidazole at a dose of 8 mg/kg twice a day for 7 days. A compounding pharmacy prepared a suspension containing 50 mg/mL of metronidazole.
- How many milligrams of metronidazole does Marshmallow require per dose?
- How many milliliters of the suspension are required per dose?
\(9\;lb\;4\;oz=9.25\;lb\times \frac{1\;kg}{2.2\;lb}\times \frac{8\;mg}{kg}=\frac{33.6\;mg}{dose}\)
\(\frac{33.6\;mg}{dose}\times \frac{1\;mL}{50\;mg\;MTZ}=0.67\;mL/dose,\;round\;to\;0.7\;mL\)
Marshmallow refuses to take the metronidazole suspension, probably due to the bad taste of the drug. The vet then orders metronidazole benzoate suspension because cats tend to find the taste less offensive. Metronidazole benzoate is an ester prodrug of metronidazole with a molecular weight of 275.3 g/mole. The prodrug is hydrolyzed in the body to generate metronidazole, which has molecular weight of 171.2 g/mole. The vet wants Marshmallow to receive the equivalent of 8 mg/kg of metronidazole twice daily for 7 days.
Metronidazole benzoate suspension may be compounded using the formula:1
Metronidazole benzoate 8 g
Glycerin 10 mL
Flavoring agent (optional) qs
Simple syrup qs 100 mL
- How many milligrams of metronidazole benzoate are required to provide the equivalent of 8 mg/kg of metronidazole?
- How many milliliters of metronidazole benzoate suspension are required to provide this dose?
We already determined that 8 mg/kg of metronidazole is 33.6 mg. We need to adjust the dose using the molecular weights to account for the drug being delivered as the benzoate ester.
\(33.6\;mg\;MTZ\times \frac{\frac{275.3\;g\;MTZ Benz}{mole}}{\frac{171.2\;g\;MTZ}{mole}}=54\;mg\;MTZ\;Benz\;per\;dose\)
The suspension formula specifies 8 g (8000 mg) of metronidazole benzoate per 100 mL of the suspension, so the volume of suspension required per dose is:
\(54\;mg\;MTZ\;Benz\times \frac{100\;mL}{8000\;mg\;MTZ\;Benz}=0.68\;mL,\;round\;to\;0.7\;mL\)
- How much of the metronidazole benzoate suspension should be dispensed to provide 7 days of therapy?
\(\frac{0.7\;mL}{dose}\times \frac{2\;doses}{day}\times7\;days =9.8\;mL\;required\)
You should dispense more than 10 mL to account for some waste, slight measurement errors, etc. Dispensing 15 mL (approximately 5 mL more than the required volume) should allow 7 full days of therapy at 0.7 mL per dose.
Marshmallow’s owner is concerned about giving her metronidazole benzoate because they read that benzoates are toxic to cats. A veterinary medical reference states that benzoate is safe in cats if the exposure is less than 200 mg/kg/day.
- How many mg/kg of benzoate will Marshmallow ingest per day at a dose of 0.7 mL of metronidazole benzoate suspension twice a day?
Metronidazole molecular weight is 171.2 g/mole while metronidazole benzoate molecular weight is 275.3 g/mole. Therefore, metronidazole benzoate is approximately equivalent to 171.2 g of metronidazole and 104.1 g (275.3 – 171.2) of benzoate per mole.
\(\frac{2\;doses}{day}\times \frac{54\;mg\;MTZ\;Benz}{dose}\times \frac{104.1\;g\;Benzoate}{275.3\;g\;MTZ\;Benz}=40.8\;mg\;benzoate\;per\;day\)
Marshmallow weighs 9¼ lb or 4.2 kg, so benzoate intake per day is
\(\frac{40.8\;mg}{day}=9.7\;mg/kg/day\)
This is much lower than the 200 mg/kg/day limit. According to the guidelines, this dose should be safe.
Marshmallow takes the metronidazole benzoate suspension without struggling and, after a followup visit, the vet orders another 7 days supply at the same dose. Marshmallow’s owner finds it inconvenient to measure 0.7 mL per dose and would like the suspension to be compounded so that each dose is given in a larger volume.
- What volume of suspension should be dispensed if each dose is 2.5 mL?
- How much metronidazole benzoate is required to compound this volume of suspension?
- How much glycerin is required if the same proportion is used, i.e. 10 mL of glycerin per 100 mL of suspension?
\(\frac{2.5\;mL}{dose}\times \frac{2\;doses}{day}\times 7\;days=35\;mL\;for\;7\;days.\;Dispense\;40\;mL\)
The amount of metronidazole per dose is 54 mg and it will be delivered in 2.5 mL per dose. The total amount of metronidazole benzoate required to compound 40 mL of suspension is:
\(40\;mL\times \frac{54\;mg\;MTZ\;Benz}{2.5\;mL}=864\;MTZ\;Benz\;needed\)
Glycerin is included in the original published suspension formulation at 10 mL per 100 mL of suspension, so for 40 mL of suspension:
\(\frac{10\;mL\;glycerin}{100\;mL\;suspension}=\frac{x\;mL\;glycerin}{40\;mL\;suspension}=4\;mL\;glycerin\;needed\)
1U.S. Pharmacist 2023;48(4):71-72.
Module 9B: Phenytoin Products for Treating Seizure Disorder
Phenytoin is an anticonvulsant drug used to treat some types of seizure disorders. It is available in 5 dosage forms as shown in table 9.1. The different dosage forms have different absorption rates and different potencies, so switching from one dosage form to another must be done carefully.
Phenytoin chewable tablets and phenytoin suspension contain phenytoin as the free acid. Extended phenytoin sodium capsules and phenytoin injection contain phenytoin sodium. Each milligram of phenytoin sodium has the equivalent of 0.92 mg of phenytoin free acid, and this difference must be taken into account when switching from phenytoin sodium and phenytoin free acid products. Fosphenytoin sodium is an injectable prodrug of phenytoin and each 1.5 mg of fosphenytoin sodium contains the equivalent of 1 mg of phenytoin sodium. In an effort to reduce errors in dose calculation, fosphenytoin sodium is labeled in terms of phenytoin sodium equivalent (PE) dose, where 1.5 mg of fosphenytoin sodium represents 1 mg PE (1 mg of phenytoin sodium). Phenytoin product equivalencies are summarized as:
1 mg PE = 1 mg phenytoin sodium = 0.92 mg phenytoin
Table 9.1 Some Phenytoin Dosage Forms
Example 9.2: CJ (10 years old, 45 kg, 4’9” tall) has been healthy, taking no regular medications or over-the-counter products until experiencing a serious seizure. CJ is taken to the hospital and the physician orders a fosphenytoin loading dose of 15 mg PE/kg to be infused at a rate of 2 mg PE/kg/min or 150 mg PE/min, whichever is slower. This hospital dilutes fosphenytoin sodium in the smallest volume of D5W that gives a drug concentration less than 25 mg PE/mL.
- Calculate the fosphenytoin loading dose in mg PE for CJ.
\(45\;kg\times \frac{15\;mg\;PE}{kg}=675\;mg\;PE\)
- Calculate the volume of fosphenytoin injection required for the dose (see label in Table 9.1).
The hospital pharmacy stocks D5W in 50 mL, 100 mL, 250 mL and 500 mL bags. Calculate which bag should the dose be prepared in.
\(675\;mg\;PE\times \frac{1\;mL}{50\;mg\;PE}=13.5\;mL\;fosphenytoin\;injection\)
\(\frac{675\;mg\;PE}{x\;mL}=\frac{25\;mg\;PE}{1\;mL},\;x=\;27\;mL\)
If the dose was diluted to a total of 27 mL the concentration would be 25 mg PE/mL. The smallest bag is 50 mL, so 50 mL D5W + 13.5 mL fosphenytoin inj = 63.5 mL. Use the 50 mL bag.
- Calculate the required solution flow rate in mL/min.
From the calculations above, the drug concentration is 675 mg PE/63.5 mL. The preferred flow rate is 2 mg PE/kg/min.
\(45\;kg\times \frac{2\;mg\;PE}{kg\;\times\;min}\times\frac{63.5\;mL}{675\;mg\;PE} =8.5\;mL/min\)
- How many minutes will it take to infuse the loading dose?
\(63.5\;mL\times \frac{1\;min}{8.5\;mL}=7.5\;min\)
After the loading dose has been administered, the physician orders a maintenance dose of fosphenytoin 4 mg PE/kg by IV infusion every 12 hours at a rate of 2 mg PE/kg/min or 150 mg PE/min, whichever is slower.
- Calculate the fosphenytoin maintenance dose in mg PE for CJ.
\(45\;kg\times \frac{4\;mg\;PE}{kg}=180\;mg\;PE\)
- Calculate the volume of fosphenytoin injection required for each maintenance dose.
\(180\;mg\;PE\times \frac{1\;mL}{50\;mg\;PE}=3.6\;mL\;fosphenytoin\;injection\)
- What D5W bag size should be used?
\(\frac{180\;mg\;PE}{x\;mL}=\frac{25\;mg\;PE}{mL},\;x=7.2\;mL\)
If the dose was diluted to a total of 7.2 mL the concentration would be 25 mg PE/mL. The smallest bag is 50 mL, so 50 mL D5W + 3.6 mL fosphenytoin inj = 53.6 mL. Use the 50 mL bag.
- Calculate the required solution flow rate in mL/min.
\(45\;kg\times \frac{2\;mg\;PE}{kg\;\times\;min}\times \frac{53.6\;mL}{180\;mg\;PE}=26.8\;mL/min\)
- How many minutes will it take to infuse each maintenance dose?
\(53.6\;mL\times \frac{1\;min}{26.8\;mL}=2\;min\)
When CJ is stabilized and ready for discharge, the physician would like to switch from fosphenytoin to phenytoin chewable tablets for CJ’s outpatient maintenance therapy.
- Read the phenytoin product labels. Calculate the dose of phenytoin free acid that is equivalent to each fosphenytoin maintenance dose.
\(180\;mg\;PE\times \frac{0.92\;mg\;phenytoin}{1\;mg\;PE}=165.6\;mg\;phenytoin\)
- The physician orders the phenytoin chewable tablets to be administered twice daily. Calculate how many tablets should CJ take per dose.
\(165.6\;mg\;phenytoin\times \frac{1\;tab}{50\;mg\;phenytoin}=3.3\;tablets\)
It is not practicle to break the chewable tablets to obtain a dose of 3.3 tablets. Suggest to the physician to round the dose up to 3.5 tablets.
- Calculate how many tablets should be dispensed for a 90 days’ supply.
\(90\;days\times \frac{2\;doses}{day}\times \frac{3.5\;tabs}{dose}=630\;tablets\)
Module 9C: Gentamicin Extended Interval Dosing Nomogram
Gentamicin is an aminoglycoside antibiotic administered by IV infusion for treating infections caused by susceptible strains of Gram-negative bacteria. The dose and frequency of gentamicin administration must be carefully chosen to provide effective therapy while minimizing the risk of toxicity to the kidneys. Gentamicin has traditionally been administered in 3 doses per day. Extended interval dosing is an alternative approach to treat patients with normal renal function and uncomplicated infections. These protocols involve giving a single dose of drug every 1 to 2 days, guided by the plasma concentration of gentamicin after the first dose. Extended interval dosing uses a standard weight-based dose and a nomogram to determine the frequency of administration.
SIUE Gentamicin protocol2
- The standard dose of gentamicin is 7 mg/kg. Use actual body weight. For patients whose actual body weight is higher than 1.2 x IBW, use adjusted body weight (factor 0.4): (IBW + 0.4 × (Actual weight - IBW)).
- Gentamicin is diluted in 100 mL of D5W and infused over 60 minutes.
- Obtain the gentamicin plasma level 6 – 12 hours after the first dose is started.
If the dose begins at 0800, order a gentamicin level to be drawn between 1400 and 2000 the same day. - Plot the gentamicin concentration and the blood draw time on the nomogram (Figure 9.1) to determine the appropriate interval for subsequent doses. If the concentration is above the borderline for the Q48h interval, then the patient is not a candidate for extended interval dosing and should be treated according to the traditional approach (not covered here).
- Check the gentamicin trough plasma level one time per week. Draw the trough blood sample six hours prior to a dose. The concentration should be less than 1 mcg/mL. If the trough value is higher than 1 mcg/mL, refer to the infectious disease team for dose adjustment (not covered here).
Examples (see nomogram below):
- A - the blood sample was drawn10 hours after the first dose was started and the gentamicin concentration was 4 mcg/mL. The patient should continue to receive 7 mg/kg at an interval of every 24 hours.
- B - the blood sample was drawn 8 hours after the first dose was started and the gentamicin concentration was 8 mcg/mL. The patient should continue to receive 7 mg/kg at an interval of every 36 hours.
- C - the blood sample was drawn 11 hours after the first dose was started and the gentamicin concentration was 7 mcg/mL. The patient should continue to receive 7 mg/kg at an interval of every 48 hours.
- D - the blood sample was drawn 9.5 hours after the first dose was started and the gentamicin concentration was 12 mcg/mL. The patient should not be treated according to the extended dosing protocol.
Figure 9.1 - Gentamicin nomogram
Example 9.3: Patient BZ (70 yo male, 210 lb, 5’9”) is prescribed gentamicin 7 mg/kg per the protocol.
- What weight should you use to calculate the dose?
\(Actual\;weight=210\;lb\times \frac{1\;kg}{2.2\;lb}=95.5\;kg\)
\(IBW=50\;kg\;+\;2.3\;kg(69-60^{"})=70.7\;kg\)
\(\frac{Actual}{IBW}=\frac{95.5\;kg}{70.7\;kg}=1.35\gt 1.2,\;use\;adjusted\;body\;weight\)
\(ABW=70.7\;kg\;+\;0.4(95.5-70.7\;kg)=80.6\;kg\)
- Calculate the dose in milligrams for BZ.
\(Dose=80.6\;kg\times\frac{7\;mg}{kg}=564\;mg,\;round\;to\;565\;mg \)
- Calculate how many milliliters of the gentamicin injection are required for each dose and the flow rate (mL/min) that should be used to infuse the drug.
\(565\;mg\;gent\times \frac{1\;mL\;inj}{40\;mg\;gent}=14.1\;mL\;injection \;per\;dose \)
The drug is added to 100 mL D5W, so concentration = 565 mg/114.1 mL. The drug is infused over 60 minutes, so flow rate = 114.1 mL/60 min = 1.9 mL/min.
- BZ’s first infusion started at 0800 on 1/15/24. A blood sample was drawn at 1830 the same day and the gentamicin level was 5.6 mcg/mL. When (dates and times) should the next 3 doses be administered?
The gentamicin level was 5.6 mcg/mL at 10.5 hours after the infusion was started. This point falls into the Q36h section of the nomogram. The next 3 doses should be given at:
- Dose 2: 0800 1/15/24 + 36 hours = 2000 1/16/24
- Dose 3: 2000 1/16/24 + 36 hours = 0800 1/18/24
- Dose 4: 0800 1/18/24 + 36 hours = 2000 1/19/24
- The physician wants to check the gentamicin trough level before the 4th dose. When (date and time) should the blood sample be drawn.
Troughs are drawn 6 hours prior to a dose. Dose 4 scheduled for 2000 on 1/19/24, so trough should be drawn at 1400 on 1/19/24.
2Based on Stanford Health Care Aminoglycosides Dosing Guidelines (https://med.stanford.edu/content/dam/sm/bugsanddrugs/documents/antimicrobial-dosing-protocols/SHC-Aminoglycoside-Dosing-Guide.pdf)
Module 9D: Intravenous Phosphate Supplementation
Begin by reviewing the phosphate supplementation examples in Module 8
- Sodium phosphates injection contains 3 mmoles of phosphate and 4 mEq sodium per mL.
- Potassium phosphates injection contains 3 mmoles of phosphate and 4.4 mEq potassium per mL.
- The plasma potassium level is expected to increase by 0.5 mEq/L for every 40 mEq of potassium administered.
- The normal potassium plasma level for this course is 3.5 – 5 mEq/L.
(Some institutions or testing labs may have slightly different normal values.)
The osmolarity of phosphates infusions must be checked to determine if peripheral infusion is appropriate. The maximum osmolarity for peripheral infusion is 600 mOsm/L.
Example 9.4: Patient HP has K+ level of 4.1 mEq/L and requires 45 mmoles of phosphate. The expected K+ level after the dose is less than 5, so this KPhos dose is saf
- Calculate the expected potassium level after administering 45 mmoles of potassium phosphates.
- Is it safe to administer the dose as potassium phosphates?
\(45\;mmol\;Phos\times \frac{4.4\;mEq\;K^{+}}{3\;mmol\;Phos}=66\;mEq\;K^{+}\\\\ \)
\(\frac{4.1\;mEq\;K^{+}}{L}\;+\;66\;mEq\times \frac{0.5\;mEq/L}{40\;mEq\;K^{+}\;dose}=4.9\;mEq\;K^{+}/L\;after\;dose \)
The expected K+ level after the dose is less than 5, so this KPhos dose is safe.
- What is the minimum volume of D5W that the dose be diluted in (50, 100, 250, 500, or 1000 mL) for peripheral administration, i.e. what volume of D5W will give osmolarity < 600 mOsm/L.
\(45\;mmol\;KPhos\times \frac{1\;mL}{3\;mmol\;Phos}=15\;mL\;KPhos\;injection\)
Find the volume of D5W that will give a final 600 mOsm/L using alligation.
The KPhos label shows the injection osmolarity is 7.4 mOsm/mL. Recall D5W osmolarity is 252 mOsm/L or 0.252 mOsm/mL. 600 mOsm/L or 0.6 mOsm/mL is the target value.
\(\frac{15\;mL}{0.348\;parts}=\frac{x\;mL}{6.4\;parts},\;x=276\text{ mL or larger D5W required. Use 500 mL bag}\)
- If HP receives 45 mmoles of potassium phosphates in the smallest appropriate bag for peripheral infusion, what flow rate (mL/hr) should be used to administer the dose over 6 hours.
The drug volume is 15 mL and the bag volume is 500 mL = 515 mL total.
515 mL/6 hr = 85.8 or 86 mL/hour
Example 9.5: RG has K+ level of 4.8 mEq/L. A physician ordered potassium phosphates 25 mmoles in 250 mL of ½ NS. Is this order safe and appropriate? Determine if any changes are needed, i.e. is the appropriate phosphates salt ordered (final K+ level 5 mEq/L or less) and the osmolarity of the solution 600 mOsm/L or less? If the K+ dose is too high, recommend using the Na Phosphates injection. If the osmolarity is too high, use a larger volume of fluid.
\(25\;mmol\;Phos\;\times\;\frac{4.4\;mEq\;K^+}{3\;mmol\;Phos}=\;36.7\;mEq\;of\;K^+\;ordered\)
\(\frac{4.8\;mEq\;K^{+}}{L}\;+\;36.7\;mEq\;\times \frac{0.5\;mEq/L}{40\;mEq\;K^{+}\;dose}=5.3\;mEq\;K^{+}/L\;after\;dose\)
This dose is not safe. Recommend sodium phosphates injection.
\(25\;mmol\;Phos\times \frac{1\;mL}{3\;mmol\;Phos}=8.3\;mL\;NaPhos\;injection\)
\(Osmolarity=\frac{\left(8.3\;mL\times \frac{7\;mOsm}{mL} \right)+\left(250\;mL\times \frac{0.154\;mOsm}{mL} \right)}{8.3\;+\;250\;mL}\times \frac{1000\;mL}{L}=373\;mOsm/L. \)
NaPhos 25 mmol in 250 mL of 1/2 NS is appropriate for administration via a peripheral infusion.
Module 9E: Parenteral Nutrition Calculations
Stanley Dudrick’s development and introduction of parenteral nutrition (PN) solutions in the 1960s profoundly influenced patient care. From the beginning, Dr. Dudrick insisted that pharmacists be members of the Nutrition Support Team.
The American Society for Parenteral and Enteral Nutrition (ASPEN) is a professional organization dedicated to advancing the science and practice of clinical nutrition and metabolism. It publishes detailed information on feeding guidelines for patients receiving parenteral nutrition. If you are interested in clinical nutrition, consider joining this organization.
A PN solution typically consists of amino acids, dextrose, fat emulsions, and water. Electrolytes, vitamins, and micronutrients are also used to support normal physiology. The pharmacist typically prepares a PN solution by calculating the component volumes and caloric contributions.
You may recall the caloric contributions from biochemistry for dextrose (a monosaccharide carbohydrate), amino acids, and fat. Carbohydrates provide 4 kcal/g, but parenteral solutions use dextrose monohydrate with a caloric value of 3.4 kcal/g. In general, we do not count the calories provided by amino acids since these molecules are used for tissue synthesis and are not intended as an energy source for the nourished patient. If fully metabolized, they would contribute 4 kcal/g. Finally, fat is an essential source of calories and essential fatty acids and has a caloric contribution of 9 kcal/g.
Component | Calories |
Amino Acids | Not counted, but 4 kcal/g |
Dextrose · H2O | 3.4 kcal/g |
Fat | 9 kcal/g |
Several manufacturers offer Amino Acid Solutions (AA). Amino acid product solutions are available in several w/v concentrations. Some brand names include Aminosyn, Plenamine, Prosol, Travasol, and Troph-Amine. There are others. We will not discuss the individual amino acid amounts as they do not directly influence calculation issues. The nutrition team usually refers to amino acids as protein.
Dextrose USP injection is available in solutions at 50% and 70% w/v concentrations.
Fat emulsions (ILE – intravenous lipid emulsions) are available with several different oil components, and the three commercial concentrations are 10%, 20%, and 30% w/v.
Sterile Water for injection (SWFI) makes the final volume after adding electrolytes, vitamins, and minerals.
These calculations are identical to the w/v calculations seen in Module 2A.
When writing PN orders use 1 decimal place in the percentages for protein, dextrose, and fat.
Consider the PN solution order for a hospitalized 180-pound, medically stable patient unable to eat following GI tract surgery. The electrolytes, vitamins, and micronutrients add 65 mL to the total volume. Total daily volume = 2000 mL. The nutrition team wants to provide 20 - 30 kcal/kg/day of energy, not counting the contribution from protein.
Amino Acids 3 %
Dextrose 19 %
Lipids 3 %
Your PN compounding supplies include Dextrose 70% injection, an Amino Acid 15% solution, a 20% IV fat emulsion, and sterile water for injection. What volume (mL) of each component is needed to prepare the 2L solution?
Dextrose
How many grams of Dextrose are in 2000 mL of the PN solution?
\(\frac{19\;g\;Dextrose}{100\; mL}\;=\frac{X\;g\;Dextrose}{2000\;mL};\;X\;=\;380\;g\)
What volume of D70 provides 380 g?
\(\frac{70\;g\;Dextrose}{100\; mL}\;=\frac{380\;g\;Dextrose}{X\;mL};\;X\;=\;543\;mL\)
Amino Acids
How many grams of Protein are in 2000 mL?
\(\frac{3\;g\;Plenamine}{100\; mL}\;=\frac{X\;g\;Plenamine}{2000\;mL};\;X\;=\;60\;g\)
What volume of Plenamine 15% provides 60 g?
\(\frac{15\;g\;Plenamine}{100\; mL}\;=\frac{60\;g\;Plenamine}{X\;mL};\;X\;=\;400\;mL\)
Fat
How many grams of Fat are in 2000 mL?
\(\frac{3\;g\;Fat}{100\; mL}\;=\frac{X\;g\;Fat}{2000\;mL};\;X\;=\;60\;g\)
What volume of SMOFLipid 20% provides 60 g?
\(\frac{20\;g\;Fat}{100\; mL}\;=\frac{60\;g\;Fat}{X\;mL};\;X\;=\;300\;mL\)
SWFI
How much SWFI is needed to qs to 2000 mL?
\(2000\;mL\;-\;543\;mL\:-\;400\;mL\;-\;300\;mL\;-\;65\;mL\;=\;692\;mL\;water \)
Dextrose calories
How many kcal are provided by Dextrose when the patient receives the 2000 mL?
\(380\; g\;\times \;3.4\;\frac{kcal}{g}\;=\;1292\;kcal\)
Fat calories
How many fat calories are provided by the 2000 mL PN solution per day?
Intravenous lipid emulsion (ILE) products use emulsifiers and glycerin in the formulation. The egg phospholipid emulsifiers and glycerin increase the caloric value beyond what you might calculate based on 9 kcal/g of fat. The caloric values per milliliter for several products are summarized in the table.
Product | kcal/mL |
Intralipid 10 % | 1.1 |
Intralipid 20 % | 2 |
Intralipid 30 % | 3 |
Nutrilipid 20 % | 2 |
Omegaven 10 % | 1.1 |
SMOFLipid 20 % | 2 |
\(300\; mL\;\times \;2\;\frac{kcal}{mL}\;=\;600\;kcal\)
Total Nonprotein Calories
An ASPEN guideline suggests providing a total energy (carbohydrate and fat) component of 20 – 30 kcal/kg/day. Does the PN solution attain that target?
\(1292\;+\;600\;=\;1892\;kcal/day\)
\(\frac{1892\;kcal/day}{82\;kg}\;=\;23\;kcal/kg/day\)
Yes, the formulation meets the target goal.
Module 9: Practice Problems
1. Chewbacca needs metronidazole too.
Chewbacca is a 17 lb male Maine Coon cat. The veterinarian ordered metronidazole 50 mg/mL suspension, 2 mL PO BID x 10 days for an infection.
- The usual dose for this indication in cats is 8 mg/kg BID. Check the dose – if it is not the usual dose, what should you report as the usual dose and volume of 50 mg/mL suspension per dose for Chewbacca when checking with the vet?
- Chewbacca’s vet thanks you and accepts your recommendation for the dose. What volume of the drug suspension should be dispensed according to the dosing guideline? (Add about 5 mL extra to allow for some loss during the week of therapy.)
- What size of oral syringe should you dispense with the suspension.
Your pharmacy compounds metronidazole suspension according to the formula:
Metronidazole 5 g
Ora Plus® suspending agent 40 mL
Flavor concentrate 2.5 mL
Purified water qs 100 mL
- How many 250 metronidazole tablets should be used to compound the required amount of suspension for Chewbacca?
- The flavor concentrate contains fish flavor and 35% v/v propylene glycol in water.
- Calculate the percent strength of propylene glycol in the metronidazole suspension.
2. BK (9 years old, 53 kg) is hospitalized for new onset seizures. The physician orders fosphenytoin loading dose and maintenance dose according to the guidelines.
- Calculate the recommended loading dose of fosphenytoin for BK
- Calculate the recommended flow rate and time required to admininster the loading dose.
- Calculate the recommended maintenance dose of fosphenytoin.
- Calculate the recommended flow rate and time required to administer the maintenance dose.
- Calculate the recommended oral phenytoin dose and the number of infatabs required per dose.
- Calculate the number of infatabs that should be dispensed for one month of therapy.
Gentamicin Practice Cases
3. Patient IB (38 yo male, 6’1”, 320 lb) is prescribed gentamicin 7 mg/kg per protocol.
- Calculate the dose in milligrams for IB.
- Calculate how many milliliters of the gentamicin injection are required to prepare each dose and the appropriate flow rate in mL/min.
- IB’s first infusion started at 1900 on 2/12/24. A blood sample was drawn at 0630 the next morning and the gentamicin level was 8.8 mcg/mL. When (dates and times) should the next doses be administered?
4. Patient HD (52 yo female, 5’10”, 155 lb) is prescribed gentamicin 7 mg/kg per protocol.
- Calculate the dose in milligrams for HD.
- Calculate the volume of gentamicin injection required to prepare the dose and the appropriate flow rate in mL/hr.
- The first dose infusion was started at 1300 on 2/3/24. What is the window of time (earliest to latest) for the first blood sample to be drawn to determine the gentamicin plasma level according to the protocol.
- The first gentamicin plasma level was 3.2 mcg/mL, taken 11 hours after starting the infusion. When should the next dose be administered.
Phosphate Practice Cases
5. A 45 year old patient has K+ level of 4.3 mEq/L. The patient is ordered 55 mmol of KPhos to be administered as an isotonic solution by IV infusion over 6 hours. Determine if the order is safe as written. If it is not safe, what should you recommend. The pharmacy will prepare the dose by adding the correct amounts of phosphates injection and sterile water for injection to an empty IV bag. Calculate the volume of sterile water required to make the infusion isotonic.
6. A patient with potassium level of 4.1 mEq/L is ordered 40 mmol of K phos in 250 mL of D5W. Is the order safe and appropriate? If not, recommend an appropriate order.
7. A patient with potassium level of 4.8 mEq/L is ordered 50 mmol of K phos in 250 mL of ½ NS. Is the order safe and appropriate? If not, recommend an appropriate order.
8. A patient with potassium level of 4.6 mEq/L is ordered 30 mmol of K phos to be administered as an isotonic solution in sterile water for injection. Is the order safe and appropriate? If not recommend an appropriate order. Calculate the volume of sterile water required for the final order.
9. A PN solution order is written for a 150-pound, medically stable patient who will use the solution at home after discharge from the hospital. The electrolytes, vitamins, and micronutrients add 90 mL to the total volume. Total daily volume = 1800 mL.
Amino Acids 4.1 %
Dextrose 20 %
Lipids 2.6 %
Your PN compounding supplies include Dextrose 50% for injection, an Amino Acids 15%, a 20% IV Fat emulsion, and Sterile Water For Injection.
What volume (mL) of each component is needed to prepare 2 L of the solution?
How many g/kg/day of amino acids does the patient receive?
How many carbohydrate calories are provided by the formula?
How many fat calories are provided by the ILE product?
How many non-protein kcal/kg/day does the patient receive?
How much SWFI is required to make the 1800 mL total volume?
10. Write a PN solution order for a 70 kg patient that provides 1 g/kg/day of amino acids, 0.8 g/kg/day of fat, and 1540 kcal of carbohydrate. Your starting materials are a 10% amino acid solution, a 20% ILE product, Dextrose 70%, and SWFI. The electrolytes, vitamins, and micronutrients add 110 mL to the total volume. The total PN daily volume is 2100 mL. How much SWFI is needed to bring the total volume to 2100 mL?
To start, calculate the number of grams needed for each component. Calculate the final concentrations, as in the above problem. You will need to calculate the component volumes to find the required volume of SWFI.
11. A PN solution order is written for a 66-pound, medically stable patient who will use the solution at home after discharge from the hospital. The electrolytes, vitamins, and micronutrients add 50 mL to the total volume. Total daily volume = 1360 mL.
Amino Acids 4.4 %
Dextrose 19.5 %
Lipids 4.4 %
Your PN compounding supplies include Dextrose 50% injection, an Amino acid 15% solution, a 20% IV Fat emulsion, and Sterile water for injection.
What volume (mL) of each component is needed to prepare the 2 L solution?
How many g/kg/day of amino acids does the patient receive?
How many carbohydrate calories are provided by the formula?
How many fat calories are provided by the ILE product?
How many non-protein kcal/kg/day does the patient receive?
How much SWFI is required to make the 1360 mL total volume?
12. Write a PN solution order for a 3 kg patient that provides 3 g/kg/day of amino acids, 2.5 g/kg/day of fat, and 176 kcal of carbohydrate. Your starting materials are a 15% amino acid solution, a 30% ILE product, Dextrose 70%, and SWFI. The electrolytes, vitamins, and micronutrients add 41 mL to the total volume. The total PN daily volume is 240 mL. How much SWFI is needed to bring the total volume to 240 mL?
To start, calculate the number of grams needed for each component. Calculate the final concentrations, as in the above problem. You will need to calculate the component volumes to find the required volume of SWFI.
Follow the advice provided in problem 10.
13. A PN solution order is written for a 55-pound, medically stable patient who is hospitalized and unable to eat. The electrolytes, vitamins, and micronutrients add 60 mL to the total volume. Total daily PN volume = 1200 mL.
Amino Acids 4.6 %
Dextrose 27 %
Lipids 4.4 %
Your PN compounding supplies include a 70% dextrose injection, a 15% amino acid solution, a 20% IV Fat emulsion, and Sterile Water for Injection.
What volume (mL) of each component is needed to prepare the 2 L solution?
How many g/kg/day of amino acids does the patient receive?
How many carbohydrate calories are provided by the formula?
How many fat calories are provided by the ILE product?
How many non-protein kcal/kg/day does the patient receive?
How much SWFI is required to make the 1200 mL total volume?
14. Write a PN solution order for a 7 kg patient that provides 2.5 g/kg/day of amino acids, 3 g/kg/day of fat, and 515 kcal of carbohydrate. Your starting materials are a 15% amino acid solution, a 30% ILE product, Dextrose 70%, and SWFI. The electrolytes, vitamins, and micronutrients add 50 mL to the total volume. The total PN daily volume is 630 mL. How much SWFI is needed to bring the total volume to 630 mL?
To start, calculate the number of grams needed for each component. Calculate the final concentrations, as in the above problem. You will need to calculate the component volumes to find the required volume of SWFI.
Follow the advice provided in problem 10.
Answers:
Practice Case 1 – Chewbacca the cat
- The ordered dose represents 12.9 mg/kg; the dose guideline for Chewbacca is 61.8 mg or 1.2 – 1.3 mL BID.
- 1.3 mL BID x 10 days is 26 mL; dispense 30 mL
- Dispense a 3 mL (preferred) or 5 mL oral syringe
- Six (6) tablets are needed
- The propylene glycol concentration is 0.875 or 0.9% v/v
Practice Case 2 – Phenytoin
- Loading dose 780 mg PE
- 15.6 mL fosphenytoin injection required; dose should be diluted in a 50 mL bag of D5W, and infused at 8.7 mL/min, infusion will run for 7.5 minutes
- Maintenance dose 210 mg PE (rounded up from 208)
- MD should be diluted in 50 mL D5W and infused at 27.1 mL/min, for 2 minutes.
- Oral dose 200 mg (rounded up from 191 mg – with MD approval), requires 4 infatabs per dose; dispense 240 tablets for 30 days supply.
Practice Cases 3 and 4 - Gentamicin
- Calculate 743 mg, round to 745 mg
- 18. 6 mL of injection, flow rate 2 mL/min for 118.6 mL
- 1900 0630 is 11.5 hours. 8.8 mcg/mL @ 11.5 hours is above the 48 hour line
- Protocol is not appropriate for this patient – refer to the infectious disease team
- Calculate 494 mg, round to 495 mg
- 12.4 mL of injection, 1.9 mL/min to infuse 112.4 mL
- Draw 1st level between 6 and 12 hours after starting first dose – so between 1900 2/3 and 0100 2/4
- 3.2 mcg/mL @ 11 hours indicates 24 hr interval, so next dose at 0800 2/4
Practice Cases 5, 6, 7, and 8 - Phosphate Supplementation
5. The K+ level would be 5.3 mEq/L after the dose – recommend 55 mmol Na Phos instead. To produce an isotonic solution, mix 398 mL sterile water + 18.3 mL of NaPhos injection.
6. The K+ level would be 4.8 mEq/L after the dose, so K Phos is appropriate. The osmolarity will be too high (614 mOsm/L) in 250 mL of D5W – recommend 500 mL D5W (438 mOsm/L).
7. The K+ level would be 5.7 mEq/L after the dose – recommend Na Phos. The osmolarity would be 582 mOsm/L in 250 mL of ½ NS, so this volume is appropriate.
8. The K+ level would be 5.2 mEq/L after the dose – recommend Na Phos. The required amount of sterile water is 217 mL.
Parenteral Nutrition
9. A PN solution order for a 150-pound, medically stable patient who will use the solution at home after discharge from the hospital.
- AA 15% 492 mL
D 50% 720 mL
ILE 20% 238 mL
(Other 90 mL)
SEFI 264 mL
- Protein ~ 1.1 g/kg/day
- Carbs 1224 kcal/day
- Fat 476 kcal/day
- Calories ~ 25 kcal/kg/day
10. A PN solution for a 70 kg patient.
- Amino acids 3.3 %
Dextrose 21.6 %
Lipids 2.7 %
SWFI = 369 mL
11. A PN solution for a 66-pound patient.
- AA 15% 400 mL
D 50% 529 mL
ILE 20% 300 mL
(Other 50 mL)
SWFI 81 mL
- Protein ~ 2 g/kg/day
- Carbs 902 kcal/day
- Fat 600 kcal/day
- Calories ~ 50 kcal/kg/day
12. A PN solution for a 3 kg patient.
- Amino acids 3.75 %
Dextrose 21.6 %
Lipids 3.1 %
SWFI = 40 mL
13. A PN solution for a 55-pound patient.
- AA 15% 368 mL
D 70% 463 mL
ILE 20% 264 mL
(Other 60 mL)
SWFI 45 mL
- Protein 2.2 g/kg/day
- Carbs 1102 kcal/day
- Fat 526 kcal/day
- Calories ~ 65 kcal/kg/day
14. A PN solution for a 7 kg patient.
- Amino acids 2.8 %
Dextrose 24 %
Lipids 3.3 %
SWFI = 177 mL