Mole Calculations MCQs — 40 Practice Questions with Answers & Explanations

Mole calculations are the most tested calculation topic in school chemistry, so this page gives you 40 free practice questions with fully worked answers. They are ideal revision for GCSE, IGCSE, A-Level and AP Chemistry students, and they get harder as you work through each section. Try each question first, then open the answer to check the working step by step.

Quick formula recap: moles = mass / Mr • mass = moles × Mr • particles = moles × 6.02 × 1023 • moles = concentration × volume (dm3) • gas volume = moles × 24 dm3 at RTP.

Keep the full formula sheet for your exam board open while you practise: GCSE Chemistry Formula Sheet • IGCSE Chemistry Formula Sheet • A-Level Chemistry Formula Sheet • AP Chemistry Formula Sheet

Section 1: The Mole Triangle — Mass, Moles and Mr

Q1. How many moles are there in 8 g of sodium hydroxide, NaOH? (Mr of NaOH = 40)

  • (A) 0.02 mol
  • (B) 0.2 mol
  • (C) 2.0 mol
  • (D) 5.0 mol
Show Answer

Answer: B — moles = mass / Mr = 8 / 40 = 0.2 mol.

Q2. What is the mass of 0.25 mol of calcium carbonate, CaCO3? (Mr of CaCO3 = 100)

  • (A) 2.5 g
  • (B) 40 g
  • (C) 25 g
  • (D) 250 g
Show Answer

Answer: C — mass = moles × Mr = 0.25 × 100 = 25 g.

Q3. What is the relative formula mass (Mr) of sulfuric acid, H2SO4? (Ar: H = 1, S = 32, O = 16)

  • (A) 98
  • (B) 82
  • (C) 49
  • (D) 96
Show Answer

Answer: A — Mr = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98.

Q4. How many molecules are there in 0.5 mol of carbon dioxide? (Avogadro constant = 6.02 × 1023 per mol)

  • (A) 6.02 × 1023
  • (B) 1.20 × 1024
  • (C) 1.51 × 1023
  • (D) 3.01 × 1023
Show Answer

Answer: D — particles = moles × Avogadro constant = 0.5 × 6.02 × 1023 = 3.01 × 1023 molecules.

Q5. How many moles are there in 11 g of carbon dioxide, CO2? (Mr of CO2 = 44)

  • (A) 0.25 mol
  • (B) 0.44 mol
  • (C) 2.5 mol
  • (D) 4.0 mol
Show Answer

Answer: A — moles = mass / Mr = 11 / 44 = 0.25 mol. (Dividing the wrong way round, 44 / 11, wrongly gives 4.0 mol.)

Q6. What is the mass of 2 mol of ammonia, NH3? (Mr of NH3 = 17)

  • (A) 17 g
  • (B) 34 g
  • (C) 8.5 g
  • (D) 51 g
Show Answer

Answer: B — mass = moles × Mr = 2 × 17 = 34 g.

Q7. A sample contains 1.204 × 1024 atoms of zinc. How many moles of zinc atoms is this? (Avogadro constant = 6.02 × 1023 per mol)

  • (A) 0.5 mol
  • (B) 1 mol
  • (C) 2 mol
  • (D) 4 mol
Show Answer

Answer: C — moles = particles / Avogadro constant = 1.204 × 1024 / 6.02 × 1023 = 2 mol.

Q8. What is the relative formula mass (Mr) of aluminium sulfate, Al2(SO4)3? (Ar: Al = 27, S = 32, O = 16)

  • (A) 123
  • (B) 150
  • (C) 278
  • (D) 342
Show Answer

Answer: D — Mr = (2 × 27) + 3 × (32 + (4 × 16)) = 54 + 3 × 96 = 54 + 288 = 342.

Q9. How many moles of water molecules are there in 36 g of water, H2O? (Mr of H2O = 18)

  • (A) 2 mol
  • (B) 0.5 mol
  • (C) 18 mol
  • (D) 36 mol
Show Answer

Answer: A — moles = mass / Mr = 36 / 18 = 2 mol.

Q10. How many atoms in total are there in 1 mol of methane molecules, CH4? (Avogadro constant = 6.02 × 1023 per mol)

  • (A) 6.02 × 1023
  • (B) 3.01 × 1024
  • (C) 1.20 × 1024
  • (D) 3.01 × 1023
Show Answer

Answer: B — each CH4 molecule contains 5 atoms, so total atoms = 5 × 6.02 × 1023 = 3.01 × 1024 atoms.

Section 2: Reacting Masses & Limiting Reactants

Q11. Magnesium reacts with hydrochloric acid: Mg + 2HCl → MgCl2 + H2. What mass of magnesium chloride is formed from 2.4 g of magnesium? (Ar: Mg = 24, Cl = 35.5)

  • (A) 4.75 g
  • (B) 9.5 g
  • (C) 19 g
  • (D) 95 g
Show Answer

Answer: B — moles of Mg = 2.4 / 24 = 0.1 mol, and the ratio Mg : MgCl2 is 1 : 1, so 0.1 mol of MgCl2 forms. Mr of MgCl2 = 24 + (2 × 35.5) = 95, so mass = 0.1 × 95 = 9.5 g.

Q12. Calcium carbonate reacts with dilute hydrochloric acid to give carbon dioxide. What mass of CO2 is produced from 5 g of CaCO3? (Mr: CaCO3 = 100, CO2 = 44)

  • (A) 1.1 g
  • (B) 4.4 g
  • (C) 2.2 g
  • (D) 0.22 g
Show Answer

Answer: C — moles of CaCO3 = 5 / 100 = 0.05 mol, and the ratio CaCO3 : CO2 is 1 : 1, so mass of CO2 = 0.05 × 44 = 2.2 g.

Q13. A reaction has a theoretical yield of 40 g, but only 30 g of product is actually collected. What is the percentage yield?

  • (A) 75%
  • (B) 25%
  • (C) 133%
  • (D) 60%
Show Answer

Answer: A — percentage yield = (actual yield / theoretical yield) × 100 = (30 / 40) × 100 = 75%.

Q14. Hydrogen burns in oxygen: 2H2 + O2 → 2H2O. A mixture contains 0.6 mol of H2 and 0.4 mol of O2. How many moles of water are formed?

  • (A) 0.3 mol
  • (B) 0.4 mol
  • (C) 0.8 mol
  • (D) 0.6 mol
Show Answer

Answer: D — 0.6 mol of H2 needs only 0.3 mol of O2 (ratio 2 : 1), so H2 is the limiting reactant and the oxygen is in excess. Water formed = 0.6 mol, because the ratio H2 : H2O is 1 : 1.

Q15. Zinc reacts with hydrochloric acid: Zn + 2HCl → ZnCl2 + H2. What mass of hydrogen gas is produced from 6.5 g of zinc? (Ar: Zn = 65, H = 1)

  • (A) 2.0 g
  • (B) 1.0 g
  • (C) 0.1 g
  • (D) 0.2 g
Show Answer

Answer: D — moles of Zn = 6.5 / 65 = 0.1 mol, and the ratio Zn : H2 is 1 : 1, so 0.1 mol of H2 forms. Mass = 0.1 × 2 = 0.2 g.

Q16. Magnesium burns in oxygen: 2Mg + O2 → 2MgO. What mass of magnesium oxide is formed when 4.8 g of magnesium burns completely? (Ar: Mg = 24, O = 16)

  • (A) 4.8 g
  • (B) 6.4 g
  • (C) 8.0 g
  • (D) 16 g
Show Answer

Answer: C — moles of Mg = 4.8 / 24 = 0.2 mol, and the ratio Mg : MgO is 1 : 1, so 0.2 mol of MgO forms. Mr of MgO = 24 + 16 = 40, so mass = 0.2 × 40 = 8.0 g.

Q17. In an experiment, the theoretical yield of a product is 8.0 g and the actual yield is 6.4 g. What is the percentage yield?

  • (A) 125%
  • (B) 80%
  • (C) 20%
  • (D) 64%
Show Answer

Answer: B — percentage yield = (6.4 / 8.0) × 100 = 80%. (Dividing the wrong way round gives 125%, which is impossible for a yield.)

Q18. Ammonia is made from nitrogen and hydrogen: N2 + 3H2 → 2NH3. A reactor contains 0.5 mol of N2 and 1.2 mol of H2. Which reactant is limiting, and how many moles of NH3 are formed?

  • (A) H2 is limiting; 0.8 mol of NH3
  • (B) N2 is limiting; 1.0 mol of NH3
  • (C) H2 is limiting; 1.2 mol of NH3
  • (D) N2 is limiting; 0.8 mol of NH3
Show Answer

Answer: A — 0.5 mol of N2 would need 1.5 mol of H2 (ratio 1 : 3), but only 1.2 mol is present, so H2 is limiting. NH3 formed = 1.2 × (2 / 3) = 0.8 mol.

Q19. On heating, calcium carbonate decomposes: CaCO3 → CaO + CO2. What mass of calcium oxide is formed from 20 g of CaCO3? (Mr: CaCO3 = 100, CaO = 56)

  • (A) 5.6 g
  • (B) 11.2 g
  • (C) 22.4 g
  • (D) 8.8 g
Show Answer

Answer: B — moles of CaCO3 = 20 / 100 = 0.2 mol, and the ratio CaCO3 : CaO is 1 : 1, so mass of CaO = 0.2 × 56 = 11.2 g. (8.8 g is the mass of the CO2 given off.)

Q20. Methane burns completely: CH4 + 2O2 → CO2 + 2H2O. What mass of oxygen is needed to burn 8 g of methane completely? (Mr: CH4 = 16, O2 = 32)

  • (A) 16 g
  • (B) 64 g
  • (C) 32 g
  • (D) 8 g
Show Answer

Answer: C — moles of CH4 = 8 / 16 = 0.5 mol, and the ratio CH4 : O2 is 1 : 2, so 1.0 mol of O2 is needed. Mass = 1.0 × 32 = 32 g.


Section 3: Solutions, Concentrations & Titrations

Q21. A solution contains 0.5 mol of solute dissolved in 250 cm3 of solution. What is its concentration in mol/dm3?

  • (A) 0.002 mol/dm3
  • (B) 0.5 mol/dm3
  • (C) 2.0 mol/dm3
  • (D) 125 mol/dm3
Show Answer

Answer: C — first convert the volume: 250 cm3 = 0.25 dm3. Concentration = moles / volume = 0.5 / 0.25 = 2.0 mol/dm3. (Using cm3 directly wrongly gives 0.002.)

Q22. 8 g of sodium hydroxide is dissolved to make 200 cm3 of solution. What is the concentration in g/dm3?

  • (A) 1.6 g/dm3
  • (B) 40 g/dm3
  • (C) 4 g/dm3
  • (D) 0.04 g/dm3
Show Answer

Answer: B — 200 cm3 = 0.2 dm3, so concentration = mass / volume = 8 / 0.2 = 40 g/dm3.

Q23. How many moles of hydrochloric acid are there in 25 cm3 of 0.2 mol/dm3 HCl?

  • (A) 5 mol
  • (B) 0.05 mol
  • (C) 0.5 mol
  • (D) 0.005 mol
Show Answer

Answer: D — moles = concentration × volume in dm3 = 0.2 × (25 / 1000) = 0.2 × 0.025 = 0.005 mol.

Q24. In a titration, 20.0 cm3 of sodium hydroxide solution neutralises 25.0 cm3 of 0.100 mol/dm3 hydrochloric acid. NaOH + HCl → NaCl + H2O. What is the concentration of the NaOH?

  • (A) 0.080 mol/dm3
  • (B) 0.125 mol/dm3
  • (C) 0.250 mol/dm3
  • (D) 0.100 mol/dm3
Show Answer

Answer: B — moles of HCl = 0.100 × 0.025 = 0.0025 mol, and the ratio is 1 : 1, so moles of NaOH = 0.0025 mol. Concentration = 0.0025 / 0.020 = 0.125 mol/dm3.

Q25. Sulfuric acid reacts with sodium hydroxide: H2SO4 + 2NaOH → Na2SO4 + 2H2O. What volume of 0.200 mol/dm3 NaOH is needed to neutralise 25.0 cm3 of 0.100 mol/dm3 H2SO4?

  • (A) 12.5 cm3
  • (B) 50.0 cm3
  • (C) 25.0 cm3
  • (D) 10.0 cm3
Show Answer

Answer: C — moles of H2SO4 = 0.100 × 0.025 = 0.0025 mol, and the ratio acid : NaOH is 1 : 2, so moles of NaOH = 0.005 mol. Volume = 0.005 / 0.200 = 0.025 dm3 = 25.0 cm3.

Q26. 50 cm3 of a 2.0 mol/dm3 solution is diluted with water to a total volume of 250 cm3. What is the new concentration?

  • (A) 0.4 mol/dm3
  • (B) 0.2 mol/dm3
  • (C) 1.0 mol/dm3
  • (D) 10 mol/dm3
Show Answer

Answer: A — the amount of solute does not change: moles = 2.0 × 0.050 = 0.1 mol. New concentration = 0.1 / 0.250 = 0.4 mol/dm3.

Q27. What mass of sodium chloride must be dissolved to make 500 cm3 of a 0.200 mol/dm3 solution? (Mr of NaCl = 58.5)

  • (A) 11.7 g
  • (B) 2.93 g
  • (C) 5.85 g
  • (D) 58.5 g
Show Answer

Answer: C — moles needed = 0.200 × 0.500 = 0.100 mol, so mass = 0.100 × 58.5 = 5.85 g.

Q28. A solution of potassium nitrate has a concentration of 0.500 mol/dm3. What is its concentration in g/dm3? (Mr of KNO3 = 101)

  • (A) 50.5 g/dm3
  • (B) 101 g/dm3
  • (C) 202 g/dm3
  • (D) 5.05 g/dm3
Show Answer

Answer: A — concentration in g/dm3 = concentration in mol/dm3 × Mr = 0.500 × 101 = 50.5 g/dm3.

Q29. In a titration, 25.0 cm3 of calcium hydroxide solution neutralises 20.0 cm3 of 0.100 mol/dm3 hydrochloric acid. Ca(OH)2 + 2HCl → CaCl2 + 2H2O. What is the concentration of the Ca(OH)2?

  • (A) 0.080 mol/dm3
  • (B) 0.020 mol/dm3
  • (C) 0.400 mol/dm3
  • (D) 0.040 mol/dm3
Show Answer

Answer: D — moles of HCl = 0.100 × 0.020 = 0.002 mol, and the ratio Ca(OH)2 : HCl is 1 : 2, so moles of Ca(OH)2 = 0.001 mol. Concentration = 0.001 / 0.025 = 0.040 mol/dm3.

Q30. 5.3 g of sodium carbonate, Na2CO3, is dissolved to make 250 cm3 of solution. What is the concentration in mol/dm3? (Mr of Na2CO3 = 106)

  • (A) 0.050 mol/dm3
  • (B) 0.200 mol/dm3
  • (C) 0.800 mol/dm3
  • (D) 21.2 mol/dm3
Show Answer

Answer: B — moles = 5.3 / 106 = 0.05 mol, and 250 cm3 = 0.25 dm3, so concentration = 0.05 / 0.25 = 0.200 mol/dm3.

Section 4: Gas Volumes, Empirical Formula & Percentage Composition

Q31. What is the volume of 0.5 mol of oxygen gas at room temperature and pressure (RTP)? (1 mol of gas occupies 24 dm3 at RTP)

  • (A) 12 dm3
  • (B) 24 dm3
  • (C) 48 dm3
  • (D) 6 dm3
Show Answer

Answer: A — gas volume = moles × 24 dm3 = 0.5 × 24 = 12 dm3.

Q32. How many moles of carbon dioxide gas are there in 6 dm3 of the gas at RTP? (1 mol of gas occupies 24 dm3 at RTP)

  • (A) 4 mol
  • (B) 0.25 mol
  • (C) 0.5 mol
  • (D) 144 mol
Show Answer

Answer: B — moles = volume / 24 = 6 / 24 = 0.25 mol.

Q33. What is the mass of 48 dm3 of hydrogen gas at RTP? (1 mol of gas occupies 24 dm3 at RTP; Mr of H2 = 2)

  • (A) 2 g
  • (B) 24 g
  • (C) 96 g
  • (D) 4 g
Show Answer

Answer: D — moles = 48 / 24 = 2 mol, so mass = 2 × 2 = 4 g.

Q34. A hydrocarbon contains 80% carbon and 20% hydrogen by mass. What is its empirical formula? (Ar: C = 12, H = 1)

  • (A) CH4
  • (B) C2H6
  • (C) CH3
  • (D) CH
Show Answer

Answer: C — moles: C = 80 / 12 = 6.67, H = 20 / 1 = 20. Dividing both by 6.67 gives the ratio C : H = 1 : 3, so the empirical formula is CH3.

Q35. A compound contains 2.4 g of carbon, 0.4 g of hydrogen and 3.2 g of oxygen. What is its empirical formula? (Ar: C = 12, H = 1, O = 16)

  • (A) CH4O
  • (B) CHO
  • (C) C2H4O2
  • (D) CH2O
Show Answer

Answer: D — moles: C = 2.4 / 12 = 0.2, H = 0.4 / 1 = 0.4, O = 3.2 / 16 = 0.2. The ratio C : H : O = 1 : 2 : 1, so the empirical formula is CH2O. (C2H4O2 has the same ratio but is not the simplest whole-number formula.)

Q36. What is the percentage by mass of oxygen in sulfuric acid, H2SO4? (Mr of H2SO4 = 98)

  • (A) 32.7%
  • (B) 65.3%
  • (C) 16.3%
  • (D) 2.0%
Show Answer

Answer: B — mass of oxygen in one formula unit = 4 × 16 = 64, so percentage = (64 / 98) × 100 = 65.3%.

Q37. What is the percentage by mass of calcium in calcium carbonate, CaCO3? (Ar: Ca = 40; Mr of CaCO3 = 100)

  • (A) 12%
  • (B) 48%
  • (C) 40%
  • (D) 60%
Show Answer

Answer: C — percentage of Ca = (40 / 100) × 100 = 40%.

Q38. Calcium oxide is made by heating calcium carbonate: CaCO3 → CaO + CO2. What is the atom economy for the desired product, CaO? (Mr: CaCO3 = 100, CaO = 56, CO2 = 44)

  • (A) 44%
  • (B) 100%
  • (C) 80%
  • (D) 56%
Show Answer

Answer: D — atom economy = (Mr of desired product / total Mr of all reactants) × 100 = (56 / 100) × 100 = 56%. The remaining 44% is the CO2 by-product.

Q39. Magnesium reacts with excess dilute hydrochloric acid: Mg + 2HCl → MgCl2 + H2. What volume of hydrogen gas is produced from 2.4 g of magnesium at RTP? (Ar: Mg = 24; 1 mol of gas occupies 24 dm3 at RTP)

  • (A) 2.4 dm3
  • (B) 24 dm3
  • (C) 1.2 dm3
  • (D) 0.24 dm3
Show Answer

Answer: A — moles of Mg = 2.4 / 24 = 0.1 mol, and the ratio Mg : H2 is 1 : 1, so 0.1 mol of H2 forms. Volume = 0.1 × 24 = 2.4 dm3.

Q40. A compound has the empirical formula CH2O and a relative molecular mass of 180. What is its molecular formula? (Ar: C = 12, H = 1, O = 16)

  • (A) C3H6O3
  • (B) C6H12O6
  • (C) C2H4O2
  • (D) C12H24O12
Show Answer

Answer: B — the empirical formula mass is 12 + 2 + 16 = 30, and 180 / 30 = 6, so multiply CH2O by 6 to get C6H12O6.

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