Paper 4 Stoichiometry Structured Practice

Five structured questions in the style of IGCSE Chemistry Paper 4, totalling 27 marks. Write your answers in full before checking the mark scheme.

Question 1 (6 marks)

Calcium carbonate reacts with dilute hydrochloric acid:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

(Ar: Ca = 40, C = 12, O = 16, H = 1, Cl = 35.5)

(a) Calculate the relative formula mass of CaCO₃. [1]

(b) Calculate the mass of calcium chloride produced when 25.0 g of calcium carbonate reacts completely with excess hydrochloric acid. [3]

(c) In the experiment, only 22.2 g of calcium chloride was obtained. Calculate the percentage yield. [2]

Mark scheme

(a) Mr of CaCO₃ = 40 + 12 + (3 x 16) = 100 [1]

(b) Moles of CaCO₃ = 25.0 / 100 = 0.25 mol [1] From the equation, 1 mol CaCO₃ produces 1 mol CaCl₂, so moles of CaCl₂ = 0.25 mol [1] Mr of CaCl₂ = 40 + (2 x 35.5) = 111 Mass of CaCl₂ = 0.25 x 111 = 27.75 g [1]

(c) Percentage yield = (actual yield / theoretical yield) x 100 [1] = (22.2 / 27.75) x 100 = 80.0% [1]

Total: 6 marks

Question 2 (5 marks)

A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass.

(Ar: C = 12, H = 1, O = 16)

(a) Determine the empirical formula of the compound. [3]

(b) The relative molecular mass of the compound is 46. Determine the molecular formula. [2]

Mark scheme

(a) C: 52.2/12 = 4.35; H: 13.0/1 = 13.0; O: 34.8/16 = 2.175 [1] Divide by smallest (2.175): C = 2, H = 6 (accept 5.98 rounded), O = 1 [1] Empirical formula = C₂H₆O [1]

(b) Empirical formula mass of C₂H₆O = (2 x 12) + (6 x 1) + 16 = 46 [1] Molecular formula mass / empirical formula mass = 46 / 46 = 1 Molecular formula = C₂H₆O (same as empirical formula) [1]

Total: 5 marks

Question 3 (6 marks)

A student dissolves 4.0 g of sodium hydroxide (NaOH) in water to make 500 cm³ of solution.

(Ar: Na = 23, O = 16, H = 1)

(a) Calculate the number of moles of NaOH dissolved. [2]

(b) Calculate the concentration of the solution in mol/dm³. [2]

(c) Calculate the concentration of the solution in g/dm³. [2]

Mark scheme

(a) Mr of NaOH = 23 + 16 + 1 = 40 [1] Moles = mass / Mr = 4.0 / 40 = 0.10 mol [1]

(b) Volume in dm³ = 500 / 1000 = 0.50 dm³ [1] Concentration = moles / volume = 0.10 / 0.50 = 0.20 mol/dm³ [1]

(c) Concentration in g/dm³ = mass / volume = 4.0 / 0.50 [1] = 8.0 g/dm³ [1]

Total: 6 marks

Question 4 (5 marks)

Magnesium reacts with oxygen: 2Mg + O₂ → 2MgO

(Ar: Mg = 24, O = 16)

A student heats 4.8 g of magnesium ribbon in a crucible until it stops burning.

(a) Calculate the number of moles of magnesium used. [1]

(b) Using the molar ratio, calculate the number of moles of magnesium oxide formed. [1]

(c) Calculate the expected mass of magnesium oxide. [1]

(d) The student obtains 7.6 g of magnesium oxide. Suggest why the actual mass is less than the theoretical mass. [2]

Mark scheme

(a) Moles of Mg = 4.8 / 24 = 0.20 mol [1]

(b) From the equation, 2 mol Mg produces 2 mol MgO (ratio 1:1), so moles of MgO = 0.20 mol [1]

(c) Mr of MgO = 24 + 16 = 40. Mass = 0.20 x 40 = 8.0 g [1]

(d) Some magnesium oxide may have been lost as white smoke/fumes escaped from the crucible [1]. Some magnesium may not have reacted completely / the lid was lifted allowing product to escape [1].

Total: 5 marks

Question 5 (5 marks)

Iron ore contains iron(III) oxide (Fe₂O₃) mixed with impurities. A 200 g sample of ore contains 80.0 g of Fe₂O₃.

(Ar: Fe = 56, O = 16)

(a) Calculate the percentage purity of the ore. [1]

(b) Calculate the mass of iron that can be obtained from 80.0 g of Fe₂O₃. [3]

(c) State one reason why the actual mass of iron obtained may be less than your calculated answer. [1]

Mark scheme

(a) Percentage purity = (80.0 / 200) x 100 = 40.0% [1]

(b) Mr of Fe₂O₃ = (2 x 56) + (3 x 16) = 160 [1] Moles of Fe₂O₃ = 80.0 / 160 = 0.50 mol From Fe₂O₃ → 2Fe, moles of Fe = 2 x 0.50 = 1.0 mol [1] Mass of Fe = 1.0 x 56 = 56.0 g [1]

(c) Some iron may be lost during the process / incomplete reduction / impurities may interfere with the reaction [1].

Total: 5 marks