How do you calculate the coefficients to balance a chemical equation?

Answer 1

There must be the same total number of each element on each side of the equation.

Working through each element in turn:

1. count the total number of each element on each side (don't forget to include any coefficients found so far as multipliers)

2. If the counts are different then:

• divide the counts by their highest common factor if it is greater than 1;
• multiply coefficient of each compound containing the element on the left by the count of the right;

• multiply coefficient of each compound containing the element on the right by the count of the left;

3. repeat the above until all the counts are the same

4. if there is a common factor (greater than 1) of all the coefficients on both sides divide the coefficients by this common factor.

Examples (as I can't do subscripts, they are shown in brackets after the element):

CH(4) + O(2) → CO(2) + H(2)O

C: both sides have 1, leave equation alone

H: left 4, right 2; hcf(4, 2) = 2, so counts become L: 4÷2 = 2, R: 2÷2=1; equation becomes CH(4) + O(2) → CO(2) + 2H(2)O

O: left 2, right 4; hcf(2, 4) = 2, so counts become Left 1, Right 2; equation becomes CH(4) + 2O(2) → CO(2) + 2H(2)O

C: both sides 1, leave the equation alone

H: both sides 4, leave the equation alone

O: both sides 4, leave the equation alone

→ balanced equation is CH(4) + 2O(2) → CO(2) + 2H(2)O

CH(4) + O(2) → CO + H(2)O

C: both sides have 1, leave equation alone

H: left 4, right 2; hcf(4, 2) = 2, so counts become Left 2, Right 1; equation becomes CH(4) + O(2) → CO + 2H(2)O

O: left 2, right 3; hcf(2, 3) = 1, so counts stay the same; equation becomes CH(4) + 3O(2) → 2CO + 4H(2)O

C: left 1, right 2; hcf(1, 2) = 1, so counts stay the same, equation becomes 2CH(4) + 3O(2) → 2CO + 4H(2)O

H: both sides have 8, leave equation alone

O: both sides have 6, leave equation alone

[as only first element had coefficients changed, the equation is now balanced, repeating the check would show both sides of the elements are now equal with C: 2. H: 8, O: 6]

→ balanced equation is 2CH(4) + 3O(2) → 2CO + 4H(2)O