S4 Chemistry Practice Paper: The Mole, Formulae and Reaction Stoichiometry (100 marks)

2026-11-10

S4 Chemistry – The mole and stoichiometry

S4 Chemistry paper: 6 questions, 30 parts, 100 marks, 75 min. Covers the mole, formulae, stoichiometry, limiting reagents, yield, titration and gas volume.

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Section A – The mole and the Avogadro constant

Show the formula, substitution and unit in each calculation.

#	Question	Answer and marking notes
1	Calculate the molar mass of hydrated copper(II) sulphate, CuSO4·5H2O. (Ar: Cu 63.5, S 32, O 16, H 1.)	63.5 + 32 + 4 x 16 + 5 x 18 = 249.5 g/mol. (3 marks: anhydrous part 2; water of crystallisation 1)
2	A sample of the compound has a mass of 24.95 g. Find the number of moles in the sample.	n = m / M = 24.95 / 249.5 = 0.100 mol. (3 marks: formula 1; substitution 1; answer with unit 1)
3	Find the number of water molecules in the sample. (Avogadro constant = 6.02 x 10^23 /mol.)	Moles of water = 0.100 x 5 = 0.500 mol; number = 0.500 x 6.02 x 10^23 = 3.01 x 10^23. (4 marks: moles of water 2; multiplication 1; answer 1)
4	Find the mass of water of crystallisation in the sample.	m = n x M = 0.500 x 18 = 9.0 g. (4 marks: moles 2; molar mass 1; answer with unit 1)
5	Find the percentage by mass of water of crystallisation in the sample.	9.0 / 24.95 x 100 = 36.1%. (4 marks: mass ratio 2; multiply by 100 1; significant figures 1)

Section B – Empirical and molecular formulae

All percentage compositions are by mass.

#	Question	Answer and marking notes
6	An organic compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Using a 100 g sample, calculate the mole ratio of the elements.	C: 40.0 / 12 = 3.33 mol; H: 6.7 / 1 = 6.7 mol; O: 53.3 / 16 = 3.33 mol; ratio 1.00 : 2.01 : 1.00. (4 marks: each mole value 1; ratio 1)
7	Hence write the empirical formula of the compound.	The mole ratio is about 1 : 2 : 1, so the empirical formula is CH2O. (3 marks: simplified ratio 2; formula 1)
8	Calculate the empirical formula mass.	12 + 2 x 1 + 16 = 30.0. (3 marks: carbon 1; hydrogen 1; oxygen 1)
9	The relative molecular mass of the compound is 180. Find the factor relating the molecular formula to the empirical formula.	n = 180 / 30.0 = 6. (3 marks: formula 1; substitution 1; answer 1)
10	Write the molecular formula of the compound.	Multiply the empirical formula by 6: C6H12O6. (4 marks: carbon 1; hydrogen 2; oxygen 1)

Section C – Reaction stoichiometry and gas volume

Molar volume of gas at room temperature and pressure = 24.0 dm3/mol.

#	Question	Answer and marking notes
11	13.0 g of zinc reacts with excess dilute hydrochloric acid: Zn + 2HCl -> ZnCl2 + H2. Find the number of moles of zinc. (Ar: Zn 65.0)	n = 13.0 / 65.0 = 0.200 mol. (3 marks: formula 1; substitution 1; answer with unit 1)
12	Find the number of moles of hydrogen produced.	From the equation, 1 mol of Zn gives 1 mol of H2, so n(H2) = 0.200 mol. (3 marks: ratio 2; answer 1)
13	Find the volume of hydrogen at room temperature and pressure, in dm3 and in cm3.	V = 0.200 x 24.0 = 4.80 dm3 = 4800 cm3. (4 marks: formula 1; dm3 answer 2; conversion 1)
14	Find the number of moles of hydrochloric acid required.	1 mol of Zn needs 2 mol of HCl, so n(HCl) = 0.400 mol. (3 marks: ratio 2; answer 1)
15	The acid used has a concentration of 2.0 mol/dm3. Find the volume of acid required, in cm3.	V = n / c = 0.400 / 2.0 = 0.200 dm3 = 200 cm3. (3 marks: formula 1; dm3 1; conversion 1)

Section D – Limiting reagent and percentage yield

Synthesis of ammonia: N2 + 3H2 -> 2NH3.

#	Question	Answer and marking notes
16	28.0 g of nitrogen takes part in the reaction. Find the number of moles of nitrogen.	n = 28.0 / 28.0 = 1.00 mol. (3 marks: molar mass 1; calculation 1; answer with unit 1)
17	4.0 g of hydrogen is also present. Identify the limiting reagent and explain your answer.	n(H2) = 4.0 / 2.0 = 2.00 mol; 1.00 mol of N2 needs 3.00 mol of H2 but only 2.00 mol is present, so hydrogen is the limiting reagent. (4 marks: moles of H2 1; required amount 2; conclusion 1)
18	Find the theoretical number of moles and the mass of ammonia formed. (Mr(NH3) = 17.0)	n(NH3) = 2 x 2.00 / 3 = 1.33 mol; m = 1.33 x 17.0 = 22.7 g. (4 marks: moles 2; mass 1; unit 1)
19	The experiment gives 18.1 g of ammonia. Calculate the percentage yield.	18.1 / 22.7 x 100 = 79.9%. (4 marks: ratio of masses 2; multiply by 100 1; answer 1)
20	Give two reasons why the yield is less than 100%.	The reaction is reversible and reaches equilibrium before completion; some ammonia dissolves in water or is lost during transfer and collection. (2 marks: one per reason)

Section E – Concentration, dilution and titration

Titration reaction: 2NaOH + H2SO4 -> Na2SO4 + 2H2O.

#	Question	Answer and marking notes
21	25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution is used. Find the number of moles of sodium hydroxide.	n = c x V = 0.100 x 0.0250 = 0.00250 mol. (3 marks: volume in dm3 1; formula 1; answer 1)
22	12.5 cm3 of sulphuric acid reaches the end point. Find the number of moles of sulphuric acid.	From the equation 2 mol of NaOH react with 1 mol of H2SO4: n = 0.00250 / 2 = 0.00125 mol. (3 marks: ratio 2; answer 1)
23	Find the concentration of the sulphuric acid.	c = n / V = 0.00125 / 0.0125 = 0.100 mol/dm3. (4 marks: volume 1; formula 1; answer 1; unit 1)
24	10.0 cm3 of the acid is diluted to 250 cm3. Find the new concentration.	Moles before dilution = 0.100 x 0.0100 = 0.00100 mol; c = 0.00100 / 0.250 = 0.00400 mol/dm3. (3 marks: moles unchanged 2; answer 1)
25	Find the mass of sulphuric acid in the 250 cm3 of diluted solution. (Mr(H2SO4) = 98.0)	n = 0.00400 x 0.250 = 0.00100 mol; m = 0.00100 x 98.0 = 0.0980 g. (3 marks: moles 1; mass 1; unit 1)

Section F – Molar volume of gas and percentage purity

The impurities do not react with the acid.

#	Question	Answer and marking notes
26	5.00 g of an impure calcium carbonate sample reacts with excess dilute hydrochloric acid, giving 1.00 dm3 of carbon dioxide at room temperature and pressure. Find the number of moles of carbon dioxide.	n = V / 24.0 = 1.00 / 24.0 = 0.0417 mol. (3 marks: formula 1; substitution 1; answer 1)
27	From CaCO3 + 2HCl -> CaCl2 + H2O + CO2, find the number of moles of calcium carbonate in the sample.	1 mol of CO2 comes from 1 mol of CaCO3, so n(CaCO3) = 0.0417 mol. (3 marks: ratio 2; answer 1)
28	Find the mass of calcium carbonate in the sample. (Mr(CaCO3) = 100)	m = 0.0417 x 100 = 4.17 g. (4 marks: formula 1; substitution 1; answer 1; unit 1)
29	Calculate the percentage purity of the sample.	4.17 / 5.00 x 100 = 83.3%. (3 marks: mass ratio 2; answer 1)
30	If the gas is collected in a syringe, explain why the measured volume may be slightly low and suggest one improvement.	Some carbon dioxide dissolves in water and some gas remains in the syringe and tubing; use a gas syringe flushed with the gas before collection, or allow the apparatus to return to room temperature before reading the volume. (3 marks: reason 2; improvement 1)

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