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IGCSE Chemistry: Cambridge 0620 tutoring, Malaysia

Balancing Chemical Equations Step by Step

A clear method for balancing chemical equations in IGCSE Chemistry 0620, with worked examples and the errors that lose marks.

Published by IGCSEChemistry.com.my

Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).

Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.

Balancing equations is a skill that appears on every 0620 paper. Paper 2 may give you four equations and ask which is balanced. Paper 4 asks you to write and balance equations from scratch. The method below works for every equation the exam can set, from simple combinations to complex organic combustion. The broader equation-writing process is covered in how to write chemical equations.

The golden rule

You may only change the large number in front of a formula (the coefficient). You must never change the small number within a formula (the subscript). Changing H2O to H2O2 does not balance an equation; it creates a completely different substance.

The method

Step 1: Write the unbalanced equation with correct formulae

Before you can balance, every formula must be correct. This is where most errors start. Check:

  • Diatomic elements: H2, O2, N2, Cl2, Br2, I2, F2 (not H, O, N, etc.).
  • Ionic compound formulae: the charges must cancel. MgCl2 not MgCl; Al2O3 not AlO.
  • Brackets are used correctly: Ca(OH)2 means two OH groups.

Step 2: Count atoms on each side

Make a quick table, either in your head or in the margin:

Example: Fe + O2 -> Fe2O3 (unbalanced)

ElementLeftRight
Fe12
O23

Neither element balances. That tells you coefficients are needed.

Step 3: Balance metals first, then non-metals, then H and O

Balance iron: There are 2 Fe on the right, so put 2 in front of Fe on the left. 2Fe + O2 -> Fe2O3

Balance oxygen: There are 3 O on the right and 2 on the left. You cannot easily make 2 become 3 with a whole number. When oxygen is awkward, try doubling everything first.

Multiply the right side by 2: 2Fe + O2 -> 2Fe2O3. Now the right has 4 Fe and 6 O. Adjust the left: 4Fe + 3O2 -> 2Fe2O3.

Check: 4 Fe on each side, 6 O on each side. Balanced.

Step 4: Final check

Count every element one more time. It takes ten seconds and catches errors that would cost marks.

Worked examples

Example 1: simple neutralisation

NaOH + H2SO4 -> Na2SO4 + H2O

Count: Na: 1 left, 2 right. H: 3 left, 2 right. O: 5 left, 5 right. S: 1 left, 1 right.

Balance Na: put 2 in front of NaOH. 2NaOH + H2SO4 -> Na2SO4 + H2O

Recount: Na: 2, 2. H: 4 left (2 from NaOH + 2 from H2SO4), 2 right. O: 6 left, 5 right.

Balance H and O: put 2 in front of H2O. 2NaOH + H2SO4 -> Na2SO4 + 2H2O

Check: Na: 2, 2. O: 6, 6. H: 4, 4. S: 1, 1. Balanced.

Example 2: combustion of ethanol

C2H5OH + O2 -> CO2 + H2O

Count C: 2 left, 1 right. Put 2 in front of CO2. C2H5OH + O2 -> 2CO2 + H2O

Count H: 6 left (5 + 1 from OH), 2 right. Put 3 in front of H2O. C2H5OH + O2 -> 2CO2 + 3H2O

Count O: 1 (from OH) + 2 (from O2) = 3 left. 4 (from 2CO2) + 3 (from 3H2O) = 7 right. Need 7 total O on left. Already have 1 from ethanol, so need 6 more from O2. Put 3 in front of O2. C2H5OH + 3O2 -> 2CO2 + 3H2O

Check: C: 2, 2. H: 6, 6. O: 1 + 6 = 7, 4 + 3 = 7. Balanced.

Example 3: dealing with fractions

Sometimes balancing produces a fraction. For instance, CH4 + O2 -> CO2 + H2O gives:

C: 1, 1. H: 4, 2. Put 2 in front of H2O. CH4 + O2 -> CO2 + 2H2O

O: 2 left, 2 + 2 = 4 right. Need 4 O on the left, so put 2 in front of O2. CH4 + 2O2 -> CO2 + 2H2O

Check: C: 1, 1. H: 4, 4. O: 4, 4. Balanced. In this case no fractions were needed.

But if the equation were 2C2H6 + 7O2 -> 4CO2 + 6H2O, and you had started with C2H6 + 3.5O2 -> 2CO2 + 3H2O, the fraction (3.5) is removed by multiplying everything by 2. The exam expects whole-number coefficients.

Balancing with brackets

When a formula contains brackets, count the atoms inside by multiplying:

Ca(OH)2 contains: 1 Ca, 2 O, 2 H. Al2(SO4)3 contains: 2 Al, 3 S, 12 O. Mg(NO3)2 contains: 1 Mg, 2 N, 6 O.

Then balance as normal, treating each element individually.

The errors that lose marks

  1. Changing subscripts instead of coefficients. This is the most penalised error because it changes the chemistry.
  2. Leaving the equation unbalanced. An unbalanced equation loses the balancing mark, even if the formulae are correct.
  3. Forgetting diatomic elements. Writing O instead of O2 means your oxygen count is wrong from the start.
  4. Not checking. A five-second recount catches most errors.
  5. Forgetting products. Acid + carbonate reactions produce three products (salt + water + CO2). Missing CO2 means missing atoms that cannot be balanced.

Practice strategy

Take ten unbalanced equations from past papers and balance them without looking at the answers. Time yourself: you should be able to balance a standard equation in under two minutes. If it takes longer, the issue is usually in step 2 (counting atoms in complex formulae) rather than step 3 (adjusting coefficients).

If balancing equations is a persistent difficulty, a trial lesson can identify whether the root cause is formula writing, atom counting, or the balancing logic itself.

Frequently asked questions

Why do I need to balance chemical equations?

Atoms cannot be created or destroyed in a chemical reaction (law of conservation of mass). A balanced equation has the same number of each type of atom on both sides. Unbalanced equations are scientifically wrong and lose marks.

Can I change the subscript numbers to balance an equation?

Never. Changing a subscript changes the substance. H2O is water; H2O2 is hydrogen peroxide. You balance by placing coefficients (large numbers) in front of formulae only.

What order should I balance elements in?

Start with metals, then non-metals, and leave hydrogen and oxygen until last. Oxygen is often the hardest to balance because it appears in many compounds, so balancing it last avoids undoing earlier work.

How do I balance equations with brackets like Ca(OH)2?

Treat the group inside brackets as a unit. Ca(OH)2 contains 1 Ca, 2 O and 2 H. Count each element by multiplying the subscript inside the bracket by the subscript outside. Then balance as normal using coefficients.

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