IGCSE Mole Calculations: Worked Examples for Chemistry 0620

IGCSE mole calculations nearly all rest on one relationship: moles = mass ÷ molar mass. Convert what you are given into moles, use the balanced equation to compare moles, then convert back into the mass, gas volume or concentration the question asks for. In Cambridge IGCSE Chemistry 0620, the mole calculations themselves are Supplement content, so they are tested on the Extended papers. You also need one mole of gas = 24 dm³ at r.t.p., and concentration = moles ÷ volume in dm³.

Key takeaways

• One mole contains 6.02 × 10²³ particles. This number is the Avogadro constant.

• Moles = mass (g) ÷ molar mass (g/mol).

• Moles of gas = volume (dm³) ÷ 24 at room temperature and pressure.

• Moles of solute = concentration (mol/dm³) × volume (dm³).

• The ratio in a balanced equation applies to moles, never directly to grams.

• Calculators are allowed in all parts of the exam, and only your final answer should be rounded.

What do IGCSE mole calculations cover in Chemistry 0620?

Moles sit in section 3 of the 0620 syllabus for 2026, 2027 and 2028, under "Stoichiometry". The split between Core and Supplement matters when you plan your revision.

Syllabus level	What you need	Papers
Core	Relative atomic and molecular mass; reacting masses in simple proportions without the mole concept; knowing concentration can be measured in g/dm³ or mol/dm³	Papers 1 and 3
Supplement (Extended)	The mole and Avogadro constant; molar gas volume; reacting masses; limiting reactants; concentrations; titration calculations; empirical and molecular formulae; percentage yield, composition and purity	Papers 2 and 4

If you are taking Core, you don't need the mole calculations in this guide. If you are taking Extended, these calculations can come up in your multiple-choice paper and your theory paper, so they deserve regular practice. Paper 2 gives you 45 minutes for 40 questions, and Paper 4 gives you 1 hour 15 minutes for 80 marks. Speed matters on both.

What is a mole in chemistry?

A mole is the unit for amount of substance. Think of it as a counting word, like a dozen, only far bigger. One mole of anything contains 6.02 × 10²³ particles, whether those particles are atoms, ions or molecules.

The useful part is what this means for mass. The mass of one mole of a substance, in grams, is its relative atomic mass or relative formula mass. So one mole of carbon atoms has a mass of 12 g, and one mole of water, H₂O, has a mass of (2 × 1) + 16 = 18 g. Chemists can't count atoms directly, but they can weigh things, and moles connect the two.

You will find relative atomic masses on the Periodic Table in your exam paper. The Periodic Table in the syllabus also carries the note that one mole of any gas has a volume of 24 dm³ at r.t.p. Even so, learn the value, because you will work faster if you don't have to look for it.

Which mole formulas do you need for IGCSE Chemistry?

What you know	Formula	Units to use
Mass	moles = mass ÷ molar mass	mass in g, molar mass in g/mol
Number of particles	particles = moles × 6.02 × 10²³	no unit, but say which particle
Gas volume at r.t.p.	moles = volume ÷ 24	volume in dm³
Solution	moles = concentration × volume	mol/dm³ and dm³
Concentration in g/dm³	g/dm³ = mass ÷ volume	mass in g, volume in dm³

Two conversions come up all the time. To change cm³ into dm³, divide by 1000, so 250 cm³ = 0.25 dm³. To change mol/dm³ into g/dm³, multiply by the molar mass.

How do you calculate moles from mass?

All the worked examples in this guide are original practice questions, not taken from past papers. They use these relative atomic masses: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, Cl = 35.5, Ca = 40.

Example 1: moles and particles

Question: How many moles of carbon dioxide are in 8.8 g of CO₂, and how many molecules is that?

1. Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol.

2. Moles = 8.8 ÷ 44 = 0.20 mol.

3. Molecules = 0.20 × 6.02 × 10²³ = 1.2 × 10²³ molecules (to 2 significant figures).

Example 2: finding the molar mass

Question: 0.15 mol of an unknown compound has a mass of 9.0 g. What is its relative formula mass?

1. Rearrange the formula: molar mass = mass ÷ moles.

2. Molar mass = 9.0 ÷ 0.15 = 60 g/mol, so Mr = 60.

Notice that Mr has no units, while molar mass is in g/mol.

How do you calculate reacting masses using moles?

Use the same three steps every time. They also work for gases and solutions:

1. Change to moles: work out the moles of the substance you have full information about.

2. Use the ratio: use the balanced equation to find the moles of the substance you want.

3. Change back: convert those moles into what the question asks for.

Example 3: reacting masses

Question: Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. What mass of magnesium oxide forms from 6.0 g of magnesium?

1. Moles of Mg = 6.0 ÷ 24 = 0.25 mol.

2. The ratio of Mg to MgO is 2 : 2, which is 1 : 1, so moles of MgO = 0.25 mol.

3. Molar mass of MgO = 24 + 16 = 40 g/mol. Mass = 0.25 × 40 = 10 g.

How do you calculate gas volumes at r.t.p.?

Treat a gas just like any other substance, but use 24 dm³ per mole for the conversion. This value only applies at room temperature and pressure, and questions will normally say so.

Example 4: volume of gas produced

Question: Calcium carbonate reacts with excess hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. What volume of carbon dioxide forms at r.t.p. from 5.0 g of calcium carbonate?

1. Molar mass of CaCO₃ = 40 + 12 + (3 × 16) = 100 g/mol. Moles = 5.0 ÷ 100 = 0.050 mol.

2. The ratio of CaCO₃ to CO₂ is 1 : 1, so moles of CO₂ = 0.050 mol.

3. Volume = 0.050 × 24 = 1.2 dm³, which is also 1200 cm³.

How do you calculate concentration in mol/dm³ and g/dm³?

Concentration tells you how much solute is dissolved in each dm³ of solution. Most lost marks here come from forgetting to convert cm³ into dm³ first.

Example 5: concentration of a solution

Question: 4.0 g of sodium hydroxide, NaOH, is dissolved to make 250 cm³ of solution. Find the concentration in mol/dm³ and in g/dm³.

1. Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles = 4.0 ÷ 40 = 0.10 mol.

2. Volume = 250 ÷ 1000 = 0.25 dm³.

3. Concentration = 0.10 ÷ 0.25 = 0.40 mol/dm³.

4. In g/dm³: 4.0 ÷ 0.25 = 16 g/dm³. As a check, 0.40 × 40 = 16 g/dm³.

How do you do titration calculations for IGCSE?

The syllabus expects you to use titration data to find the moles of solute, or the concentration or volume of a solution. The method is the same three steps as before: start with the solution where you know both the concentration and the volume.

Example 6: finding an unknown concentration

Question: 25.0 cm³ of sodium hydroxide solution is exactly neutralised by 20.0 cm³ of 0.10 mol/dm³ sulfuric acid. The equation is 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. What is the concentration of the sodium hydroxide?

1. Moles of H₂SO₄ = 0.10 × (20.0 ÷ 1000) = 0.0020 mol.

2. The ratio of NaOH to H₂SO₄ is 2 : 1, so moles of NaOH = 0.0020 × 2 = 0.0040 mol.

3. Concentration of NaOH = 0.0040 ÷ (25.0 ÷ 1000) = 0.16 mol/dm³.

The most common slip is ignoring the ratio and treating every titration as 1 : 1. Always look at the equation first.

How do you find the limiting reactant?

The limiting reactant is the one that runs out first, and it decides how much product you get. Find the moles of each reactant, then use the equation to see which one falls short.

Example 7: limiting reactant

Question: 1.2 g of magnesium is added to 50 cm³ of 1.0 mol/dm³ hydrochloric acid. The equation is Mg + 2HCl → MgCl₂ + H₂. What volume of hydrogen forms at r.t.p.?

1. Moles of Mg = 1.2 ÷ 24 = 0.050 mol.

2. Moles of HCl = 1.0 × (50 ÷ 1000) = 0.050 mol.

3. 0.050 mol of Mg would need 0.10 mol of HCl, but only 0.050 mol is there. So HCl is the limiting reactant and magnesium is in excess.

4. The ratio of HCl to H₂ is 2 : 1, so moles of H₂ = 0.050 ÷ 2 = 0.025 mol.

5. Volume of H₂ = 0.025 × 24 = 0.60 dm³.

The two reactants have equal moles here, which is a good trap to practise on. Equal moles don't mean that neither is limiting, because the equation needs twice as much acid as magnesium.

How do you calculate empirical and molecular formulae?

An empirical formula is the simplest whole-number ratio of atoms. A molecular formula gives the actual number of atoms in one molecule. The table layout below keeps the working tidy.

Example 8: empirical and molecular formula

Question: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its Mr is 60. Find its empirical and molecular formulae.

Step	C	H	O
Mass in 100 g	40.0	6.7	53.3
Divide by Ar	40.0 ÷ 12 = 3.33	6.7 ÷ 1 = 6.7	53.3 ÷ 16 = 3.33
Divide by the smallest	1	2	1

The empirical formula is CH₂O, which has a formula mass of 12 + 2 + 16 = 30. Then 60 ÷ 30 = 2, so the molecular formula is C₂H₄O₂.

How do you calculate percentage yield and percentage purity?

These three percentage formulae are easy marks once you have the correct numbers to put in.

• Percentage yield = (actual mass ÷ theoretical mass) × 100

• Percentage composition by mass = (mass of the element in one formula unit ÷ Mr) × 100

• Percentage purity = (mass of pure substance ÷ total mass of the sample) × 100

Example 9: yield, composition and purity

Yield: In Example 3, the theoretical mass of MgO was 10 g. If a student collects 8.5 g, the percentage yield is (8.5 ÷ 10) × 100 = 85%.

Composition: Ammonium nitrate, NH₄NO₃, has Mr = 80 and contains two nitrogen atoms. The percentage of nitrogen by mass is (28 ÷ 80) × 100 = 35%.

Purity: Heating 12.5 g of impure calcium carbonate gives 4.4 g of CO₂ in the reaction CaCO₃ → CaO + CO₂. Moles of CO₂ = 4.4 ÷ 44 = 0.10 mol, so moles of CaCO₃ = 0.10 mol and its mass = 0.10 × 100 = 10 g. Percentage purity = (10 ÷ 12.5) × 100 = 80%. This assumes that all of the carbon dioxide came from the calcium carbonate.

What are the most common mistakes in mole calculations?

• Using the ratio on grams. A 2 : 1 ratio in an equation compares moles. Convert to moles first.

• Skipping the cm³ to dm³ conversion. Check every volume before you use it in a concentration or gas calculation.

• Using 24 for gases in cm³. One mole of gas is 24 dm³, or 24 000 cm³.

• Forgetting the coefficients. The big numbers in the equation change the ratio. The small subscript numbers change the molar mass.

• Getting molar mass wrong. In Ca(OH)₂, the bracket means two O atoms and two H atoms, so Mr = 40 + (2 × 17) = 74.

• Rounding halfway through. Keep the full value on your calculator and round only the final answer.

• Leaving out units. Write dm³, g or mol/dm³ next to your answer so the examiner can see what you found.

How should you revise moles for the 0620 Extended papers?

Moles improve with short, regular practice rather than one long session. Try this order, ticking off a section once you can get two questions in a row right without your notes:

1. Moles from mass, and formula mass.

2. Reacting masses from an equation.

3. Gas volumes and concentrations.

4. Titrations and limiting reactants.

5. Formulae, yield and purity.

Then practise by topic with past-paper questions, marking them against the mark scheme. Our guide to the benefits of IGCSE past papers explains how to learn from your errors, and our guide to IGCSE study resources can help you choose where to practise. If mocks are coming up, add moles to the plan in our 6-week IGCSE mock exam revision plan.

Try these yourself

Use the same relative atomic masses as above. Answers are at the end of this post.

1. What is the mass of 0.40 mol of sodium chloride, NaCl?

2. What volume does 0.30 mol of oxygen occupy at r.t.p.?

3. What is the concentration, in mol/dm³, of a solution with 0.050 mol of solute in 200 cm³?

Frequently asked questions

Are mole calculations only for Extended in IGCSE Chemistry?

Mostly, yes. In the 2026 to 2028 syllabus, the mole, molar gas volume and calculations using them are Supplement content. Core students still need relative masses, reacting masses in simple proportions and the units of concentration.

Is the Avogadro constant given in the exam?

Don't rely on it being given. The syllabus requires you to state that one mole contains 6.02 × 10²³ particles, so learn the value.

What is the molar gas volume at r.t.p.?

One mole of any gas has a volume of 24 dm³ at room temperature and pressure. For the same number of moles at r.t.p., hydrogen and carbon dioxide take up the same volume.

How do I know which formula to use?

Look at what the question gives you. If it gives a mass, use mass ÷ molar mass. For a gas volume at r.t.p., use volume ÷ 24. For a solution with its concentration and volume, multiply them. Each route leads to moles, and that is your starting point for the ratio.

Answers to the practice questions

1. Mr of NaCl = 23 + 35.5 = 58.5, so mass = 0.40 × 58.5 = 23.4 g. 2. Volume = 0.30 × 24 = 7.2 dm³. 3. Concentration = 0.050 ÷ 0.200 = 0.25 mol/dm³.

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