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Revision guide · IGCSE

IGCSE Chemistry: the mole and chemical calculations

A mole is the amount of substance containing 6.02 x 10^23 particles (the Avogadro constant). Almost every IGCSE calculation uses three relationships: moles = mass ÷ molar mass, moles = concentration x volume (in dm^3), and, for gases at room temperature and pressure, moles = volume ÷ 24 dm^3. Then the balanced equation gives the mole ratio between substances. In Cambridge 0620 the mole is Supplement content (Core students calculate reacting masses by simple proportion only); in Edexcel 4CH1 moles, reacting masses, percentage yield and empirical formulae are for everyone, while solution concentrations (1.34C) and gas volumes (1.35C) are examined in Paper 2.

Facts checked:

At a glance

Cambridge section
0620 topic 3 Stoichiometry (3.1 to 3.3)
Edexcel section
4CH1 statements 1.25 to 1.36
Molar gas volume
24 dm^3 (24 000 cm^3) at r.t.p.
Avogadro constant
6.02 x 10^23 per mole
Unit change
1 dm^3 = 1000 cm^3

The formulas you need

To findFormulaUnits
Moles from massmoles = mass ÷ molar mass (Mr in g)mol, g, g/mol
Moles in a solutionmoles = concentration x volumemol, mol/dm^3, dm^3
Moles of a gas at r.t.p.moles = volume ÷ 24mol, dm^3 (or volume in cm^3 ÷ 24 000)
Number of particlesparticles = moles x 6.02 x 10^23no units
Percentage yieldactual yield ÷ theoretical yield x 100%
Percentage by mass of an element(Ar x number of atoms) ÷ Mr x 100%

Concentration can also be given in g/dm^3: multiply mol/dm^3 by Mr to convert.

What each board expects

Cambridge 0620 Core students calculate Mr, use relative atomic masses, and calculate reacting masses in simple proportions without the mole concept: if 24 g of magnesium makes 40 g of magnesium oxide, then 6 g makes 10 g. They also state that concentration can be measured in g/dm^3 or mol/dm^3. Everything about the mole (3.3.2 to 3.3.8) is Supplement: the Avogadro constant, moles from mass, molar gas volume, reacting masses, limiting reactants, solution concentrations including cm^3 to dm^3 conversion, titration calculations, empirical and molecular formulae, and percentage yield, composition and purity.

Edexcel 4CH1 requires all students to calculate Mr, use moles with masses, calculate reacting masses and percentage yield, and find empirical and molecular formulae from experimental data (1.25 to 1.33), plus the practical determining the formula of a metal oxide (1.36). Calculations with solution concentration in mol/dm^3 (1.34C) and gas volumes using 24 dm^3 or 24 000 cm^3 at rtp (1.35C) are examined only in Paper 2.

Worked example 1: reacting mass, gas volume and yield

Question: 25.0 g of calcium carbonate is heated. CaCO3 -> CaO + CO2. Calculate (a) the maximum mass of calcium oxide, (b) the volume of carbon dioxide at r.t.p., (c) the percentage yield if 12.6 g of CaO is obtained. (Ar: Ca = 40, C = 12, O = 16.)

Mr of CaCO3 = 40 + 12 + (3 x 16) = 100. Moles of CaCO3 = 25.0 ÷ 100 = 0.250 mol. The equation is 1 : 1 : 1, so 0.250 mol of CaO and 0.250 mol of CO2 form.

(a) Mr of CaO = 56, mass = 0.250 x 56 = 14.0 g. (b) Volume of CO2 = 0.250 x 24 = 6.00 dm^3. (c) Percentage yield = 12.6 ÷ 14.0 x 100 = 90.0%.

Worked example 2: a titration

Question: 25.0 cm^3 of sodium hydroxide solution is neutralised by 20.0 cm^3 of 0.100 mol/dm^3 sulfuric acid. H2SO4 + 2NaOH -> Na2SO4 + 2H2O. Find the concentration of the sodium hydroxide.

Moles of H2SO4 = 0.100 x (20.0 ÷ 1000) = 0.00200 mol. The ratio H2SO4 : NaOH is 1 : 2, so moles of NaOH = 0.00400 mol. Concentration = 0.00400 ÷ (25.0 ÷ 1000) = 0.160 mol/dm^3. In g/dm^3: 0.160 x 40 = 6.40 g/dm^3 (Mr of NaOH = 23 + 16 + 1 = 40).

Worked example 3: empirical and molecular formula

Question: a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its Mr is 180. Find the empirical and molecular formulae.

Divide each by Ar: C 40.0 ÷ 12 = 3.33; H 6.7 ÷ 1 = 6.7; O 53.3 ÷ 16 = 3.33. Divide by the smallest (3.33): C 1, H 2.0, O 1. Empirical formula CH2O, with mass 12 + 2 + 16 = 30. 180 ÷ 30 = 6, so the molecular formula is C6H12O6.

Common mistakes

  • Forgetting to convert cm^3 to dm^3 (divide by 1000) before using concentration x volume.
  • Ignoring the mole ratio from the balanced equation, especially 1 : 2 in titrations with sulfuric acid.
  • Using the Mr of O instead of O2, or of Cl instead of Cl2, when a gas is involved.
  • Rounding too early. Keep full calculator values and round only the final answer, usually to 3 significant figures.
  • Rounding 1.5 or 1.33 ratios to whole numbers in empirical formulae instead of multiplying (by 2 or by 3).
  • Using 24 dm^3 for a liquid or a solid. Molar gas volume applies only to gases at r.t.p.

Exam technique

Write every step with a label: "moles of CaCO3 = 25.0 ÷ 100 = 0.250". Mark schemes award marks for each correct step and carry errors forward, so a student who slips early but sets out the method clearly can still earn most of the marks. A bare final answer that is wrong earns nothing.

For limiting reactant questions (Cambridge Supplement), work out the moles of each reactant, divide each by its number in the equation, and the smaller result is the limiting reactant. The product is calculated from the limiting reactant only. Give units and significant figures matching the data, usually 3.

How one-to-one lessons help

Mole calculations are the single biggest source of lost marks in IGCSE Chemistry because they combine maths, equations and units. A tutor watches the student work through problems on the shared whiteboard and spots the exact step that breaks, whether it is unit conversion, the mole ratio or rearranging. Practice then targets that step with graded past-paper questions until the routine is reliable.

Self-check

  1. Calculate the Mr of CaCO3, H2SO4 and Ca(OH)2.
  2. Convert between mass, moles and number of particles.
  3. Use 24 dm^3 to find a gas volume from moles and back.
  4. Calculate a concentration from titration results using the mole ratio.
  5. Find the limiting reactant from two masses.
  6. Calculate an empirical formula and then a molecular formula from Mr.
  7. Calculate percentage yield and percentage purity.

Common questions

Do Cambridge Core students need moles?

No. In the 2026 to 2028 Cambridge 0620 syllabus the mole and Avogadro constant (3.3) are Supplement. Core students calculate reacting masses by simple proportion and should know the units of concentration.

Is the molar gas volume given in the exam?

The syllabuses state the value: Cambridge uses 24 dm^3 at r.t.p. and Edexcel 24 dm^3 or 24 000 cm^3 at rtp. Exam papers often state it, but learn it in case a question does not.

How many significant figures should I give?

Match the least precise data in the question, usually 3 significant figures. If the question specifies, follow it.

How much do LiveTutor chemistry lessons cost?

$15 a lesson for every subject and level, on a weekly plan of 1 to 5 lessons billed monthly. Lessons are one to one, 60 minutes and online, and the first is a free trial.

Is percentage yield ever more than 100%?

Not for a pure, dry product. A result above 100% usually means the product was still wet or impure, which is a common evaluation point.

Sources

Dates and figures on this page come from these official and published sources. Always confirm deadlines on the official page before acting on them.