IGCSE Chemistry 0620 Paper 4 Practice Paper: Moles, Stoichiometry and Solutions (100 marks, 60 min)

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 60 minutes | Total: 100 marks | Parts: 29

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Not allowed

Questions

Question 1 (18 marks) – Moles and the Avogadro constant

(a) A magnesium ribbon has a mass of 4.8 g. The relative atomic mass of magnesium is 24. Calculate the number of moles of magnesium in the ribbon. (3 marks)

(b) Using the Avogadro constant 6.02 x 10^23 per mol, calculate the number of magnesium atoms in the ribbon. (3 marks)

(c) Calculate the mass of one magnesium atom in grams. (2 marks)

(d) A bottle contains 11.0 g of carbon dioxide (Mr = 44). Calculate (i) the number of moles of carbon dioxide, (ii) the number of carbon dioxide molecules, (iii) the total number of oxygen atoms. (4 marks)

(e) Calculate the mass of 0.10 mol of hydrated copper(II) sulfate, CuSO4.5H2O (Mr = 250). (3 marks)

(f) Compare the number of atoms in 4.8 g of magnesium and 4.8 g of carbon (Ar = 12), and explain your answer. (3 marks)

Question 2 (17 marks) – Empirical and molecular formulae

(a) A sample of an oxide of copper has a mass of 3.20 g and contains 2.56 g of copper, the rest being oxygen. Calculate the percentage by mass of copper and of oxygen in the sample. (4 marks)

(b) Using the data above, calculate the mole ratio of copper to oxygen in the sample (Cu = 64, O = 16). (4 marks)

(c) Hence write the empirical formula of this copper oxide and state which copper oxide it is. (4 marks)

(d) Another compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, and has a relative molecular mass of 180. Determine its empirical formula and molecular formula. (5 marks)

Question 3 (17 marks) – Thermal decomposition of calcium carbonate and yield

(a) Calculate the relative formula mass of calcium carbonate, CaCO3 (Ca = 40, C = 12, O = 16). (3 marks)

(b) 25.0 g of calcium carbonate is completely decomposed. Calculate the number of moles of calcium carbonate used. (4 marks)

(c) The equation is CaCO3 -> CaO + CO2. Calculate the theoretical mass of calcium oxide formed (Mr(CaO) = 56). (4 marks)

(d) Calculate the volume of carbon dioxide produced at room temperature and pressure (molar volume 24 dm3/mol). (3 marks)

(e) In the experiment only 11.2 g of calcium oxide is obtained. Calculate the percentage yield of the reaction. (3 marks)

Question 4 (16 marks) – Titration and solution concentration

(a) 25.0 cm3 of sodium hydroxide solution has a concentration of 0.100 mol/dm3. Calculate the number of moles of sodium hydroxide in it. (3 marks)

(b) The equation is H2SO4 + 2NaOH -> Na2SO4 + 2H2O. Calculate the number of moles of sulfuric acid needed for complete neutralisation. (4 marks)

(c) The sulfuric acid used has a concentration of 0.0500 mol/dm3. Calculate the volume of acid needed, in cm3. (4 marks)

(d) Calculate the concentration of the sodium hydroxide solution in g/dm3 (Mr(NaOH) = 40). (2 marks)

(e) If the same solution is made up to 250 cm3, calculate the mass of sodium hydroxide it contains. (3 marks)

Question 5 (16 marks) – Limiting reactant and gas volumes

(a) The equation for making ammonia is N2 + 3H2 -> 2NH3. State the mole ratio of nitrogen to hydrogen to ammonia. (3 marks)

(b) 6.0 dm3 of nitrogen is mixed with 15.0 dm3 of hydrogen at the same temperature and pressure. Deduce which gas is the limiting reactant, supporting your answer with a calculation. (5 marks)

(c) Calculate the volume of ammonia that can be formed under these conditions. (4 marks)

(d) Calculate the volume of nitrogen left after the reaction, assuming it goes to completion. (4 marks)

Question 6 (16 marks) – Water of crystallisation and hydrated salts

(a) 5.00 g of blue hydrated copper(II) sulfate is heated to constant mass, leaving 3.20 g of white anhydrous copper(II) sulfate. Calculate the mass of water lost. (3 marks)

(b) Calculate the number of moles of water lost (H2O = 18). (4 marks)

(c) Calculate the number of moles of anhydrous copper(II) sulfate (CuSO4 = 160). (4 marks)

(d) Hence determine the value of x in CuSO4.xH2O. (3 marks)

(e) State the relative formula mass of this hydrated salt. (2 marks)

Answers and marking notes

Question 1 – Moles and the Avogadro constant

(a) n = m / Ar = 4.8 / 24 = 0.20 mol (3; subtract 1 for 4.8 x 24 or a missing unit). [3 marks]

(b) Number of atoms = 0.20 x 6.02 x 10^23 = 1.20 × 10²³ (3; subtract 1 for a wrong power or significant figures). [3 marks]

(c) Mass = 24 / (6.02 x 10^23) = 3.99 × 10⁻²³ g (2). [2 marks]

(d) (i) n = 11.0 / 44 = 0.250 mol (1); (ii) molecules = 0.250 x 6.02 x 10^23 = 1.51 × 10²³ (2); (iii) oxygen atoms = 2 x 1.51 × 10²³ = 3.01 × 10²³ (1). [4 marks]

(e) m = n x Mr = 0.10 x 250 = 25 g (3; using Mr = 160 scores 1 only). [3 marks]

(f) Moles of carbon = 4.8 / 12 = 0.40 mol, which is greater than 0.20 mol of magnesium (2); for equal masses the number of moles is inversely proportional to Ar, so carbon has more atoms (1). No mark for “carbon has more” alone. [3 marks]

Question 2 – Empirical and molecular formulae

(a) Mass of oxygen = 3.20 – 2.56 = 0.64 g (1); copper = 2.56 / 3.20 x 100 = 80.0% (2); oxygen = 0.64 / 3.20 x 100 = 20.0% (1). [4 marks]

(b) n(Cu) = 2.56 / 64 = 0.040 mol; n(O) = 0.64 / 16 = 0.040 mol (1 each); mole ratio = 0.040 : 0.040 = 1 : 1 (2). [4 marks]

(c) Empirical formula CuO (2); it is copper(II) oxide, since copper is +2 in CuO (2). Award 3 if CuO is written with no explanation. [4 marks]

(d) Mole ratio C : H : O = 40.0/12 : 6.7/1 : 53.3/16 = 3.33 : 6.7 : 3.33 (2), giving the empirical formula CH2O (1); empirical mass = 30, 180 / 30 = 6 (1), so the molecular formula is C6H12O6 (1). [5 marks]

Question 3 – Thermal decomposition of calcium carbonate and yield

(a) Mr = 40 + 12 + 3 x 16 = 100 (3; answer alone with no addition shown scores 2). [3 marks]

(b) n = 25.0 / 100 = 0.250 mol (4; a correct method with a wrong Mr scores 2). [4 marks]

(c) The ratio is 1 : 1, so n(CaO) = 0.250 mol (2); m = 0.250 x 56 = 14.0 g (2). [4 marks]

(d) V = 0.250 x 24 = 6.0 dm3 (3; subtract 1 for omitting x 24 or a wrong unit). [3 marks]

(e) Percentage yield = 11.2 / 14.0 x 100 = 80.0% (3; an inverted fraction scores 0). [3 marks]

Question 4 – Titration and solution concentration

(a) n = c x V = 0.100 x 25.0 / 1000 = 0.0025 mol (3; omitting the division by 1000 scores 1). [3 marks]

(b) Mole ratio H2SO4 : NaOH = 1 : 2 (2); n(H2SO4) = 0.00250 / 2 = 0.00125 mol (2). [4 marks]

(c) V = n / c = 0.00125 / 0.0500 = 0.025 dm3 (2) = 25.0 cm3 (2). [4 marks]

(d) 0.100 x 40 = 4.0 g/dm3 (2). [2 marks]

(e) m = 4.00 x 250 / 1000 = 1.00 g (3; using 250 without dividing by 1000 scores 1). [3 marks]

Question 5 – Limiting reactant and gas volumes

(a) N2 : H2 : NH3 = 1 : 3 : 2 (3; correct numbers in the wrong order score 1). [3 marks]

(b) 6.0 dm3 of nitrogen needs 6.0 x 3 = 18.0 dm3 of hydrogen (2); only 15.0 dm3 is available, so hydrogen is in short supply (2); hence hydrogen is the limiting reactant (1). No marks for the answer alone. [5 marks]

(c) From hydrogen: V(NH3) = 15.0 x 2 / 3 = 10.0 dm3 (4; using nitrogen and obtaining 12 dm3 scores 2). [4 marks]

(d) Nitrogen used = 15.0 / 3 = 5.0 dm3 (2); remaining = 6.0 – 5.0 = 1.0 dm3 (2). [4 marks]

Question 6 – Water of crystallisation and hydrated salts

(a) m(water) = 5.00 – 3.20 = 1.80 g (3; an answer of 3.20 g scores 0). [3 marks]

(b) n = 1.80 / 18 = 0.100 mol (4; using 5.00 as the numerator scores 1). [4 marks]

(c) n = 3.20 / 160 = 0.0200 mol (4). [4 marks]

(d) x = 0.100 / 0.0200 = 5 (3; an inverted ratio giving 0.2 scores 0). [3 marks]

(e) Mr = 160 + 5 x 18 = 250 (2; writing 160 alone scores 0). [2 marks]

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