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IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 3: Stoichiometry -- Mock Exam 1
1 hour 15 minutes
80
7
75:00
0620

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Question 1 -- Chemical Formulae and Balancing Equations
Total: 12 marks
A chemistry teacher in Mumbai demonstrates several reactions to her class and asks them to write balanced equations.
(a) [4]
Write the chemical formula for each of the following compounds.

(i) Magnesium oxide
(ii) Calcium hydroxide
(iii) Aluminium sulfate
(iv) Ammonium nitrate
Model Answer -- 1(a)
Magnesium oxide: MgO [1]
Calcium hydroxide: Ca(OH)2 [1]
Aluminium sulfate: Al2(SO4)3 [1]
Ammonium nitrate: NH4NO3 [1]
(b) [4]
Balance the following equations.

(i) ___ Fe + ___ O2 → ___ Fe2O3 [1]

(ii) ___ Na + ___ H2O → ___ NaOH + ___ H2 [1]

(iii) ___ CaCO3 + ___ HCl → ___ CaCl2 + ___ H2O + ___ CO2 [1]

(iv) ___ C2H6 + ___ O2 → ___ CO2 + ___ H2O [1]
Model Answer -- 1(b)
4Fe + 3O2 → 2Fe2O3 [1]
2Na + 2H2O → 2NaOH + H2 [1]
CaCO3 + 2HCl → CaCl2 + H2O + CO2 [1]
2C2H6 + 7O2 → 4CO2 + 6H2O [1]
(c) [2]
(i) State what each of the following state symbols represents: (s), (l), (g), (aq). [1]

(ii) Write a balanced equation with state symbols for the reaction of hydrochloric acid with sodium hydroxide solution. [1]
Model Answer -- 1(c)
(s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous / dissolved in water [1]
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) [1]
(d) [2]
Consider the equation: 2Mg(s) + O2(g) → 2MgO(s)

(i) State the ratio of moles of magnesium to moles of oxygen in this reaction. [1]

(ii) Explain, in terms of atoms and molecules, what this equation tells us. [1]
Model Answer -- 1(d)
The mole ratio of magnesium to oxygen is 2 : 1 [1]
Two atoms of magnesium react with one molecule of oxygen to form two formula units of magnesium oxide [1]
Question 2 -- Relative Atomic Mass and Relative Molecular Mass
Total: 12 marks
A student at a school in London is studying relative masses and how they are used in chemical calculations.
(a) [3]
(i) Define relative atomic mass (Ar). [2]

(ii) State what relative molecular mass (Mr) means. [1]
Model Answer -- 2(a)
Relative atomic mass is the weighted average mass of naturally occurring atoms of an element [1]
on a scale where one atom of carbon-12 has a mass of exactly 12 [1]
Relative molecular mass is the sum of all the relative atomic masses of the atoms in a molecule [1]
(b) [4]
Calculate the relative molecular mass (Mr) of each of the following compounds.

(i) Sulfuric acid, H2SO4 [1]

(ii) Calcium carbonate, CaCO3 [1]

(iii) Copper(II) sulfate pentahydrate, CuSO4·5H2O [1]

(iv) Aluminium hydroxide, Al(OH)3 [1]
Model Answer -- 2(b)
H2SO4: (2×1) + 32 + (4×16) = 98 [1]
CaCO3: 40 + 12 + (3×16) = 100 [1]
CuSO4·5H2O: 64 + 32 + (4×16) + 5×(2×1 + 16) = 250 [1]
Al(OH)3: 27 + 3×(16 + 1) = 78 [1]
(c) [3]
Calculate the percentage by mass of nitrogen in ammonium nitrate, NH4NO3.
Model Answer -- 2(c)
Mr of NH4NO3 = 14 + (4×1) + 14 + (3×16) = 80 [1]
Mass of N in formula = 2 × 14 = 28 [1]
% N = (28 / 80) × 100 = 35.0% [1]
(d) [2]
Chlorine has two isotopes: 35Cl (75.0%) and 37Cl (25.0%).

Calculate the relative atomic mass of chlorine. Give your answer to one decimal place.
Model Answer -- 2(d)
Ar = (75.0 × 35 + 25.0 × 37) / 100 = (2625 + 925) / 100 [1]
Ar = 3550 / 100 = 35.5 [1]
Question 3 -- The Mole and Avogadro Constant
Total: 12 marks
A student in Delhi is learning about the mole concept and how to convert between mass, moles and number of particles.
(a) [2]
(i) State what is meant by the term mole. [1]

(ii) State the value of the Avogadro constant and its significance. [1]
Model Answer -- 3(a)
A mole is the amount of substance that contains 6.02 × 1023 particles (atoms, molecules, ions or electrons) [1]
The Avogadro constant is 6.02 × 1023 per mol; it is the number of particles in one mole of any substance [1]
(b) [4]
Calculate the number of moles in each of the following.

(i) 11.2 g of iron, Fe [1]

(ii) 4.4 g of carbon dioxide, CO2 [1]

(iii) 0.78 g of aluminium oxide, Al2O3 [2]
Model Answer -- 3(b)
Moles of Fe = 11.2 / 56 = 0.2 mol [1]
Mr of CO2 = 12 + (2×16) = 44; Moles = 4.4 / 44 = 0.1 mol [1]
Mr of Al2O3 = (2×27) + (3×16) = 102 [1]
Moles = 0.78 / 102 = 0.00765 mol (or 7.65 × 10−3 mol) [1]
(c) [2]
Calculate the mass of 0.25 mol of calcium carbonate, CaCO3.
Model Answer -- 3(c)
Mr of CaCO3 = 40 + 12 + (3×16) = 100 [1]
Mass = moles × Mr = 0.25 × 100 = 25.0 g [1]
(d) [4]
(i) Calculate the number of molecules in 0.5 mol of water. [1]

(ii) Calculate the number of atoms in 0.5 mol of water. [1]

(iii) A sample contains 1.806 × 1024 atoms of neon. Calculate the number of moles of neon in the sample. [2]
Model Answer -- 3(d)
Molecules = 0.5 × 6.02 × 1023 = 3.01 × 1023 molecules [1]
Each H2O has 3 atoms, so total atoms = 3 × 3.01 × 1023 = 9.03 × 1023 atoms [1]
Moles = number of atoms / Avogadro constant [1]
Moles = 1.806 × 1024 / 6.02 × 1023 = 3.0 mol [1]
Question 4 -- Reacting Masses
Total: 12 marks
In a laboratory in Pune, students carry out experiments to investigate reacting masses in chemical reactions.
(a) [3]
Zinc reacts with hydrochloric acid as shown:

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Calculate the mass of zinc chloride produced when 13.0 g of zinc reacts with excess hydrochloric acid.
Model Answer -- 4(a)
Moles of Zn = 13.0 / 65 = 0.2 mol [1]
From equation: 1 mol Zn produces 1 mol ZnCl2, so moles of ZnCl2 = 0.2 mol [1]
Mr of ZnCl2 = 65 + (2×35.5) = 136; Mass = 0.2 × 136 = 27.2 g [1]
(b) [4]
Magnesium reacts with oxygen:

2Mg(s) + O2(g) → 2MgO(s)

A student heats 4.8 g of magnesium in a crucible with 4.8 g of oxygen.

(i) Calculate the moles of magnesium and the moles of oxygen. [2]

(ii) Determine which reagent is in excess and calculate the mass of magnesium oxide produced. [2]
Model Answer -- 4(b)
Moles of Mg = 4.8 / 24 = 0.2 mol [1]
Moles of O2 = 4.8 / 32 = 0.15 mol [1]
Ratio needed is 2 mol Mg : 1 mol O2. For 0.2 mol Mg, we need 0.1 mol O2. We have 0.15 mol O2, so oxygen is in excess and magnesium is the limiting reagent. [1]
0.2 mol Mg → 0.2 mol MgO; Mass = 0.2 × 40 = 8.0 g [1]
(c) [3]
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass.

Determine the empirical formula of this compound.
Model Answer -- 4(c)
C: 40.0 / 12 = 3.33   H: 6.7 / 1 = 6.7   O: 53.3 / 16 = 3.33 [1] -- divide by Ar
Divide by smallest (3.33): C = 1, H = 2.01, O = 1 [1] -- ratio
Empirical formula = CH2O [1]
(d) [2]
The compound in part (c) has a relative molecular mass of 180.

Determine its molecular formula.
Model Answer -- 4(d)
Empirical formula mass of CH2O = 12 + 2 + 16 = 30; n = 180 / 30 = 6 [1]
Molecular formula = C6H12O6 (glucose) [1]
Question 5 -- Percentage Yield and Percentage Purity
Total: 10 marks
A pharmaceutical company in Hyderabad manufactures calcium oxide by heating limestone (calcium carbonate) in a kiln.
(a) [4]
CaCO3(s) → CaO(s) + CO2(g)

A kiln is loaded with 500 g of calcium carbonate.

(i) Calculate the theoretical yield of calcium oxide. [2]

(ii) The actual yield of calcium oxide is 252 g. Calculate the percentage yield. [2]
Model Answer -- 5(a)
Moles of CaCO3 = 500 / 100 = 5.0 mol [1]
1 mol CaCO3 → 1 mol CaO; mass CaO = 5.0 × 56 = 280 g (theoretical) [1]
% yield = (actual / theoretical) × 100 [1]
% yield = (252 / 280) × 100 = 90.0% [1]
(b) [2]
Suggest two reasons why the percentage yield of calcium oxide is less than 100%.
Model Answer -- 5(b)
The reaction may be incomplete / not all the CaCO3 may have decomposed [1]
Some product may have been lost during transfer / mechanical losses / side reactions may have occurred [1]
(c) [4]
A sample of impure iron ore has a mass of 20.0 g. It contains iron(III) oxide, Fe2O3, and sand (silicon dioxide, SiO2) as an impurity. When the sample is reacted with excess hydrochloric acid, only the Fe2O3 reacts. The mass of pure Fe2O3 in the sample is 16.0 g.

(i) Calculate the percentage purity of the iron ore sample. [1]

(ii) Calculate the mass of iron that could be obtained from the 16.0 g of Fe2O3. [3]

Fe2O3 + 3CO → 2Fe + 3CO2
Model Answer -- 5(c)
% purity = (16.0 / 20.0) × 100 = 80.0% [1]
Mr of Fe2O3 = (2×56) + (3×16) = 160; Moles = 16.0 / 160 = 0.1 mol [1]
From equation: 1 mol Fe2O3 → 2 mol Fe; so moles of Fe = 0.2 mol [1]
Mass of Fe = 0.2 × 56 = 11.2 g [1]
Question 6 -- Concentration and Volume Calculations
Total: 12 marks
A technician at a laboratory in Birmingham prepares solutions of known concentration for use in titration experiments.
(a) [2]
(i) State the formula linking concentration (in mol/dm³), number of moles and volume (in dm³). [1]

(ii) Convert 250 cm³ to dm³. [1]
Model Answer -- 6(a)
concentration = moles / volume (in dm³) or c = n / V [1]
250 cm³ = 250 / 1000 = 0.250 dm³ [1]
(b) [3]
A technician dissolves 4.0 g of sodium hydroxide (NaOH) in water and makes the solution up to 500 cm³.

Calculate the concentration of the solution in mol/dm³.
Model Answer -- 6(b)
Mr of NaOH = 23 + 16 + 1 = 40; Moles = 4.0 / 40 = 0.1 mol [1]
Volume = 500 / 1000 = 0.5 dm³ [1]
Concentration = 0.1 / 0.5 = 0.2 mol/dm³ [1]
(c) [4]
The sodium hydroxide solution from part (b) is used in a titration with sulfuric acid:

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

It is found that 25.0 cm³ of the NaOH solution is exactly neutralised by 20.0 cm³ of the sulfuric acid.

Calculate the concentration of the sulfuric acid in mol/dm³.
Model Answer -- 6(c)
Moles NaOH = 0.2 × (25.0/1000) = 0.2 × 0.025 = 0.005 mol [1]
From equation: 2 mol NaOH reacts with 1 mol H2SO4, so moles H2SO4 = 0.005 / 2 = 0.0025 mol [1]
Volume H2SO4 = 20.0 / 1000 = 0.020 dm³ [1]
Concentration = 0.0025 / 0.020 = 0.125 mol/dm³ [1]
(d) [3]
(i) State the formula to convert concentration from mol/dm³ to g/dm³. [1]

(ii) Calculate the concentration of the sulfuric acid solution from part (c) in g/dm³. [2]
Model Answer -- 6(d)
concentration (g/dm³) = concentration (mol/dm³) × Mr [1]
Mr of H2SO4 = (2×1) + 32 + (4×16) = 98 [1]
Concentration = 0.125 × 98 = 12.25 g/dm³ [1]
Question 7 -- Gas Volumes at RTP
Total: 10 marks
Students at a school in Cambridge collect gases during experiments at room temperature and pressure (RTP).
(a) [2]
(i) State the molar volume of any gas at RTP. [1]

(ii) State the formula linking volume of gas, number of moles and molar volume. [1]
Model Answer -- 7(a)
The molar volume of any gas at RTP is 24 dm³/mol (or 24000 cm³/mol) [1]
volume = moles × molar volume or moles = volume / 24 [1]
(b) [3]
Calcium carbonate reacts with dilute hydrochloric acid:

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

Calculate the volume of carbon dioxide gas produced at RTP when 10.0 g of calcium carbonate reacts with excess hydrochloric acid.
Model Answer -- 7(b)
Moles of CaCO3 = 10.0 / 100 = 0.1 mol [1]
From equation: 1 mol CaCO3 → 1 mol CO2, so moles CO2 = 0.1 mol [1]
Volume = 0.1 × 24 = 2.4 dm³ (or 2400 cm³) [1]
(c) [3]
Magnesium reacts with dilute sulfuric acid:

Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

A student collects 600 cm³ of hydrogen gas at RTP. Calculate the mass of magnesium that reacted.
Model Answer -- 7(c)
Moles of H2 = 600 / 24000 = 0.025 mol (or 0.6 / 24 = 0.025) [1]
From equation: 1 mol Mg → 1 mol H2, so moles Mg = 0.025 mol [1]
Mass of Mg = 0.025 × 24 = 0.60 g [1]
(d) [2]
In the reaction: N2(g) + 3H2(g) → 2NH3(g)

(i) State the ratio of volumes of nitrogen : hydrogen : ammonia in this reaction (at the same temperature and pressure). [1]

(ii) Calculate the volume of ammonia produced when 6.0 dm³ of nitrogen reacts with excess hydrogen. [1]
Model Answer -- 7(d)
Volume ratio N2 : H2 : NH3 = 1 : 3 : 2 (same as mole ratio for gases at same T and P) [1]
Volume NH3 = 6.0 × 2 = 12.0 dm³ [1]

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IGCSE Chemistry 0620 -- Paper 4 Theory (Extended) -- Topic 3: Stoichiometry -- Mock Exam 1
For examination preparation only. Not an official Cambridge Assessment paper.