A Mathematical Method of Balancing Chemical Equations

We can use mathematical methods to balance chemical equations when we realize that a balanced chemical equation is just a way of representing that there are equal number of each atom on both sides of the equation. If we have the balanced chemical equation:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

we can see that we have equal number of carbon atoms on both side: 1 carbon on the left = 1 carbon on the right. The same is true for the hydrogen and oxygen atoms.

How can we use this to balance a chemical equation? First of all, we assign variables to the stoichiometric coefficients (the numbers in front of the chemical formulas in the equation). Then we can set up a series of equations, one for each element, which we can solve to determine the relative coefficients.

For example, let’s say we want to balance this equation:

NaN₃ + KNO₃ + SiO₂ → Na₄SiO₄ + K₄SiO₄ + N₂

We would assign a variable to each of the coefficients:

a NaN₃ + b KNO₃ + c SiO₂ → d Na₄SiO₄ + e K₄SiO₄ + f N₂

Then we can start writing mathematical equations for each element. For the first term we have a NaN₃ so we can see that we have a atoms of Na and 3a atoms of N. Our series of equations for each element would look like:

Na: a = 4 d

N: 3a + b =2f

K: b = 4e

O: 3b + 2c = 4d + 4e

Si: c = d + e

Notice that we have 1 fewer equations than we have variables. This means that there is no unique solution to the problem. However, that is OK because there is no unique way of balancing a chemical equation. If we have a balanced chemical equation, we can multiply all of the coefficients by any number and it’s still balanced.

We now need to solve this system of equations for which there are an infinite number of solutions. Because there are an infinite number of solutions, we just need to pick one. If we choose to have d = 1 that help us to find the other values.

For d = 1:

a = 4

we can then do some substitutions to find the other values.

3b + 2(d + e) = 4d + 4e substituting the equation for Si in to the one for O

3b = 2d + 2e simplifying

3(4e) = 2d + 2e substituting the equation for K in to the previous equation

10 e = 2d simplifying

so,

5e = d and e = 1/5

b = 4/5

c = 6/5

and

f = (3a + b) / 2 = (3(4) + 4/5)/2 = 64/10

Because we generally want whole number coefficients in the balanced chemical equation we need to clear the fractions by multiplying everything by 10 which gives the coefficients as:

a = 40

b = 8

c = 12

d = 10

e = 2

f = 64

40 NaN₃ + 8 KNO₃ + 12 SiO₂ → 10 Na₄SiO₄ + 2 K₄SiO₄ + 64 N₂

which can be divided by 2 to give the lowest whole number ratios:

20 NaN₃ + 4 KNO₃ + 6 SiO₂ → 5 Na₄SiO₄ + 1 K₄SiO₄ + 32 N₂

In hindsight, we could have made this easier to solve by setting e = 1 (which it is in the balanced equation) and then getting b = 4 from the equation for K, d = 5 via the above substitutions, c = 6 (from c = d + e), a = 20 (from a = 4d), and then f = 32 from the equation for N. There is generally more than one way to solve these systems.