How to balance chemical equations for SPM

Balancing chemical equations is one of the most useful skills in SPM Chemistry, because it appears everywhere, in the mole concept, chemical formula and equation, in stoichiometry calculations, and across almost every reaction you will write. The good news is that balancing is a mechanical skill: once you know the method, you can balance almost any SPM equation reliably. This guide teaches that method step by step, with worked examples.

The one rule behind balancing

Balancing exists because of the law of conservation of mass: atoms are never created or destroyed in a reaction, only rearranged. So the number of atoms of each element must be the same on both sides of the equation. To make them match, you adjust the big numbers in front of formulae, the coefficients. You must never change the small numbers inside a formula, the subscripts, because that would change the substance itself. Water is H₂O; if you “balance” by writing H₂O₂ you have written hydrogen peroxide, a completely different compound. Remember: change coefficients, never subscripts.

The step-by-step method

1. Write correct formulae for every reactant and product, and leave them fixed. This step relies on knowing your formulae, sulfuric acid is H₂SO₄, not HSO₄. If the formulae are wrong, no amount of balancing will fix the equation.
2. Count the atoms of each element on both sides.
3. Balance one element at a time by adjusting coefficients, usually starting with the element that appears in only one compound on each side. Leave oxygen and hydrogen, which often appear in several places, until near the end.
4. Save uncombined gases like O₂ and H₂ for last, because a single coefficient there fixes many atoms at once.
5. Re⁻count everything to confirm every element balances, then simplify the coefficients to the smallest whole-number ratio.

Worked example 1: metal and acid

Balance the reaction of magnesium with hydrochloric acid.

• Correct formulae: Mg + HCl → MgCl₂ + H₂
• Count: chlorine is 1 on the left but 2 on the right; hydrogen is 1 on the left but 2 on the right.
• Fix chlorine by putting a 2 in front of HCl: Mg + 2HCl → MgCl₂ + H₂
• Re⁻count: Mg 1 = 1, H 2 = 2, Cl 2 = 2. Balanced.

Worked example 2: combustion of propane

Combustion questions look intimidating but follow the method cleanly. Balance the complete combustion of propane, C₃H₈.

• Correct formulae: C₃H₈ + O₂ → CO₂ + H₂O
• Balance carbon first: 3 carbons on the left, so 3CO₂.
• Balance hydrogen next: 8 hydrogens on the left, so 4H₂O (4 × 2 = 8).
• Now count oxygen on the right: 3 × 2 (from CO₂) + 4 × 1 (from H₂O) = 10 oxygen atoms. So put 5O₂ on the left (5 × 2 = 10).
• Final: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Re⁻count and it all matches.

Notice the strategy: carbon and hydrogen first, oxygen last, because oxygen is in two products and is easiest to fix once the others are set.

Worked example 3: metal carbonate and acid

Balance calcium carbonate reacting with hydrochloric acid.

• Correct formulae and products: CaCO₃ + HCl → CaCl₂ + H₂O + CO₂
• Calcium and carbon are already balanced (1 each side).
• Chlorine is 1 on the left, 2 on the right, put a 2 in front of HCl: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
• Re⁻count: Ca 1, C 1, Cl 2 = 2, H 2 = 2, O 3 = 3 (1 from H₂O + 2 from CO₂). Balanced.

The polyatomic-ion trick

When a group of atoms, like sulfate SO₄, nitrate NO₃, or hydroxide OH, stays intact on both sides of the equation, balance it as a single unit instead of counting its atoms separately. It is faster and less error-prone.

Balance aluminium displacing copper from copper(II) sulfate: Al + CuSO₄ → Al₂(SO₄)₃ + Cu. Treat SO₄ as one block. Aluminium needs 2 on the left; sulfate needs 3 units, so 3CuSO₄ and 3Cu: 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu. Check: Al 2 = 2, SO₄ 3 = 3, Cu 3 = 3. Balanced, without ever splitting the sulfate into sulfur and oxygen.

Handling the fraction case

Sometimes balancing an odd number of atoms forces a fraction. Balance the combustion of ethane, C₂H₆: carbon gives 2CO₂, hydrogen gives 3H₂O, and oxygen on the right is 2 × 2 + 3 = 7, which needs 3½O₂. Since equations should use whole numbers, multiply everything by 2: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. Balance with the fraction first, then double to clear it, a reliable trick for combustion of hydrocarbons.

Do not forget state symbols

In Paper 2, equations are usually expected to include state symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous. For example: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). A balanced equation without required state symbols can still lose a mark, so make adding them the last step of every equation you write. You can see many reactions written correctly, with states, in the reactions reference.

Practise until it is automatic

Balancing rewards drilling more than almost any other Chemistry skill, the method never changes, so with practice it becomes seconds of work. While you build the skill you can check your answers with an equation balancer, but the goal is always to do it yourself in the exam, since no tool is allowed there. If balancing keeps tripping you up, usually because a formula underneath is wrong, or the reaction type is unclear, that is a quick thing for a teacher to fix. Our online one-to-one lessons run in English from RM50 an hour, with a paid one-hour trial; see how it works if you would like the method taught directly against your own weak reactions.