How to balance chemical equations

The one rule people break

You may change the coefficient in front of a formula. You may never change a subscript inside it.

Changing H₂O to H₂O₂ does not balance an equation — it replaces water with hydrogen peroxide, which is a different substance with different chemistry. This is the error that turns a balancing mistake into a chemistry mistake, and markers treat it as the latter.

The order to work in

Balancing by inspection is systematic, not guesswork, if you follow the order.

• Start with an element that appears in exactly one reactant and one product.
• Then metals, then non-metals other than oxygen and hydrogen.
• Leave oxygen and hydrogen until last — they usually appear in several species, so balancing them early forces you to redo everything.
• Treat a polyatomic ion that survives the reaction intact, such as SO₄²⁻, as a single unit rather than as separate atoms.
• Count every element on both sides before you stop.

Worked example: combustion of propane

C₃H₈ + O₂ → CO₂ + H₂O

Carbon appears in one reactant and one product, so start there: 3 carbons on the left means 3CO₂ on the right.

Hydrogen next: 8 hydrogens on the left means 4H₂O on the right.

Oxygen last, because it now appears in both products: 3CO₂ gives 6 oxygens, 4H₂O gives 4, so 10 in total, which needs 5O₂.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Check: C 3=3, H 8=8, O 10=10.

The fraction trick

Sometimes the tidiest route produces a fraction. That is fine — use it, then clear it.

C₂H₆ + O₂ → CO₂ + H₂O. Balancing carbon and hydrogen gives 2CO₂ and 3H₂O, which needs 7 oxygen atoms, so 3½O₂.

Multiply everything by 2: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.

Students often abandon a correct line of reasoning at the fraction, thinking they have gone wrong. They have not — clearing it is a step, not a failure.

When inspection stops working

Complex redox equations resist inspection. KMnO₄ + HCl → KCl + MnCl₂ + H₂O + Cl₂ balances to 2, 16, 2, 2, 8, 5 — findable by trial, but slowly.

For these, use the half-equation method: split the reaction into oxidation and reduction halves, balance each for atoms then for charge with electrons, multiply so the electrons cancel, and recombine. If your course has taught half-equations, an exam question of this shape is asking for them.

Our balancer solves the underlying linear system directly, which is why it handles these instantly — but the marks in your paper are for the method, so use it to check your working rather than to replace it.

Do not forget state symbols

(s), (l), (g) and (aq) are usually worth a mark and are usually the mark students drop.

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l). If the question gives conditions, use them — water is (l) at room temperature and (g) in a combustion analysis at high temperature.