Balancing chemical equations for IGCSE: 4 steps that work on every equation, with state symbols and ionic equations
Balancing an equation means making the number of atoms of each element the same on both sides of the arrow. It sits under almost every IGCSE Chemistry calculation. This guide gives a 4-step method that works on every equation, worked examples up to combustion, state symbols and the Extended-tier ionic equations, and lets you balance some yourself at the end.
The short answer
Atoms are not created or destroyed in a reaction (conservation of mass), so each element must have the same number of atoms on both sides.
Change only the numbers in front of formulae (coefficients), never the subscripts: changing H₂O to H₂O₂ makes hydrogen peroxide, not water.
This is Cambridge 0620 topic 3 (Stoichiometry) and Edexcel 4CH1 section 1, on every tier. Ionic equations are Extended-tier content in 0620.
Balancing in 4 steps
Exam questions often give a partly balanced equation and ask for the missing coefficient, or ask you to write one from a word equation. The same 4 steps handle both.
| Step | What to do | Watch out |
|---|---|---|
| 1. Correct formulae | Write the formula of every reactant and product | Seven gases are diatomic: H₂ N₂ O₂ F₂ Cl₂ Br₂ I₂ |
| 2. Count atoms | Tally each element on the left and the right | Brackets multiply everything inside: Ca(OH)₂ has 2 O and 2 H |
| 3. Adjust coefficients | Start with the element in the fewest formulae; leave H and O to last | Treat unchanged groups such as NO₃⁻ and SO₄²⁻ as one block |
| 4. Check, then add state symbols | Recount every element; use the lowest whole numbers | 4H₂ + 2O₂ → 4H₂O balances but is not the answer: 2H₂ + O₂ → 2H₂O |
Example 1: aluminium burning in oxygen
Unbalanced: Al + O₂ → Al₂O₃
Count: Al 1 left, 2 right · O 2 left, 3 right. The lowest common multiple of 2 and 3 is 6, so write 3O₂ and 2Al₂O₃ to get 6 O on each side.
The right now has 4 Al, so write 4Al: 4Al(s) + 3O₂(g) → 2Al₂O₃(s). Recount: Al 4 = 4 · O 6 = 6. Balanced.
Example 2: acid and base (count ion groups as blocks)
Unbalanced: Ca(OH)₂ + HNO₃ → Ca(NO₃)₂ + H₂O
There are 2 NO₃ blocks on the right, so write 2HNO₃. The 2 OH from Ca(OH)₂ and the 2 H from the acid make 2 water molecules.
Ca(OH)₂(aq) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + 2H₂O(l). Count: Ca 1 = 1 · NO₃ 2 = 2 · H 4 = 4 · O in OH 2 = 2.
Example 3: combustion (oxygen last)
Unbalanced: C₃H₈ + O₂ → CO₂ + H₂O
3 C on the left, so 3CO₂ · 8 H on the left, so 4H₂O · the right now has 3 × 2 + 4 × 1 = 10 O, so 5O₂.
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l). Every hydrocarbon combustion goes C → H → O like this (more organic chemistry).
State symbols
When a question says "include state symbols", there is a mark for them alone. A common slip is writing water as (aq): water formed in a reaction is (l).
| Symbol | Meaning | Examples |
|---|---|---|
| (s) | solid | metals, precipitates, crystalline salts |
| (l) | liquid | pure water, molten substances |
| (g) | gas | O₂, CO₂, H₂ |
| (aq) | aqueous, dissolved in water | solutions of acids, alkalis and soluble salts |
Ionic equations (Extended)
An ionic equation shows only the ions that change. Ions that stay the same in solution before and after (spectator ions) are left out.
Neutralisation: H⁺(aq) + OH⁻(aq) → H₂O(l) · Precipitation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Ionic equations must balance both atoms and total charge, e.g. Cu²⁺ + 2e⁻ → Cu has a total charge of 0 on each side (half-equations in electrolysis).
Balance some yourself
Use + / − on the coefficients. Each element's atom count turns green when both sides match. The subscript button is locked; tap it to see why.
Pick an equation, then press + / − on the numbers in front
| Atom | Left | Right |
|---|---|---|
| H | 2 | 2 |
| O | 2 | 1 |
Not balanced yet: look at the red rows
Balanced, now what
A balanced equation gives the mole ratio of the reaction, which is step 2 of every reacting-mass calculation (the mole formulas · molar mass, limiting reactants and % yield).
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