IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 3: Stoichiometry -- Mock Exam 2

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
A_r values: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, Al = 27, S = 32, Cl = 35.5, K = 39, Ca = 40, Fe = 56, Cu = 64, Zn = 65, Br = 80, Ag = 108
Molar volume of gas at RTP = 24 dm³/mol
Avogadro constant = 6.02 × 10²³ /mol
Your answers will be automatically graded when you submit.

Question Navigation

Question 1 -- Formulae from Ionic Charges and Valencies

Total: 12 marks

A student in Kolkata is learning to write chemical formulae using ionic charges and valencies.

(a) [4]

Use ionic charges to work out the formula of each compound.

(i) Potassium bromide (K⁺ and Br⁻)
(ii) Iron(III) chloride (Fe³⁺ and Cl⁻)
(iii) Zinc nitrate (Zn²⁺ and NO₃⁻)
(iv) Calcium phosphate (Ca²⁺ and PO₄³⁻)

Model Answer -- 1(a)

Potassium bromide: KBr [1]

Iron(III) chloride: FeCl₃ [1]

Zinc nitrate: Zn(NO₃)₂ [1]

Calcium phosphate: Ca₃(PO₄)₂ [1]

⚠ If you missed marks here: Balance the charges to zero: Ca²⁺ and PO₄³⁻ need 3 calciums (+6) for 2 phosphates (−6), giving Ca₃(PO₄)₂; Ca₂(PO₄)₃ has the numbers swapped. Keep the brackets round a group ion that is multiplied, and remember Fe³⁺ needs THREE chloride ions: Zn(NO₃)₂ and FeCl₃, not ZnNO₃ or FeCl.

1 mark -- KBr correctly written 1 mark -- FeCl3 correctly written 1 mark -- Zn(NO3)2 correctly written 1 mark -- Ca3(PO4)2 correctly written

(b) [3]

Copper(II) carbonate decomposes on heating.

(i) Write the word equation for this decomposition. [1]

(ii) Write a balanced symbol equation for this reaction. [1]

(iii) Add state symbols to your equation. [1]

Model Answer -- 1(b)

Copper(II) carbonate → copper(II) oxide + carbon dioxide [1]

CuCO₃ → CuO + CO₂ [1]

CuCO₃(s) → CuO(s) + CO₂(g) [1]

⚠ If you missed marks here: Heating a carbonate gives the metal OXIDE and carbon dioxide, so "copper + carbon dioxide" or CuCO₃ → Cu + CO₃ loses the mark (CO₃ is not a substance on its own). For the state symbols, both copper compounds are solids (s) and carbon dioxide is a gas (g); there is no water here, so (aq) is wrong.

1 mark -- copper carbonate produces copper oxide carbon dioxide 1 mark -- CuCO3 CuO CO2 balanced 1 mark -- correct state symbols solid solid gas

Balance the following equations.

(i) ___ Al + ___ HCl → ___ AlCl₃ + ___ H₂ [1]

(ii) ___ Fe₂O₃ + ___ C → ___ Fe + ___ CO₂ [1]

(iii) ___ C₃H₈ + ___ O₂ → ___ CO₂ + ___ H₂O [1]

Model Answer -- 1(c)

2Al + 6HCl → 2AlCl₃ + 3H₂ [1]

2Fe₂O₃ + 3C → 4Fe + 3CO₂ [1]

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O [1]

⚠ If you missed marks here: For aluminium, 3HCl gives only 1½ H₂, so double everything to remove the half: 2Al + 6HCl → 2AlCl₃ + 3H₂. For propane, balance C, then H, then O last: 3CO₂ + 4H₂O hold 6 + 4 = 10 oxygen atoms, so 5O₂ – and never change a subscript to make the numbers fit.

1 mark -- 2Al 6HCl 2AlCl3 3H2 balanced 1 mark -- 2Fe2O3 3C 4Fe 3CO2 balanced 1 mark -- C3H8 5O2 3CO2 4H2O balanced

(d) [2]

(i) State the law of conservation of mass. [1]

(ii) Explain why a balanced equation must have equal numbers of each type of atom on both sides. [1]

Model Answer -- 1(d)

In a chemical reaction, matter is neither created nor destroyed / the total mass of reactants equals the total mass of products [1]

Because atoms are rearranged in a reaction but not created or destroyed, so each atom present in the reactants must appear in the products [1]

⚠ If you missed marks here: "Energy cannot be created or destroyed" is the conservation of ENERGY and does not score; this law says the total mass of the products equals the total mass of the reactants. For (ii) the reason is atoms: they are only rearranged in a reaction, never made or destroyed, so each type must appear in equal numbers on both sides.

1 mark -- matter neither created nor destroyed mass reactants equals products 1 mark -- atoms rearranged not created destroyed present reactants appear products

Question 2 -- Moles, Molar Mass and Avogadro Calculations

Total: 12 marks

A chemistry laboratory in Singapore has a set of samples for students to analyse.

(a) [4]

Calculate the number of moles in each of the following.

(i) 5.6 g of potassium hydroxide, KOH [1]

(ii) 9.8 g of sulfuric acid, H₂SO₄ [1]

(iii) 21.6 g of silver bromide, AgBr [2]

Model Answer -- 2(a)

M_r of KOH = 39 + 16 + 1 = 56; Moles = 5.6 / 56 = 0.1 mol [1]

M_r of H₂SO₄ = 2 + 32 + 64 = 98; Moles = 9.8 / 98 = 0.1 mol [1]

M_r of AgBr = 108 + 80 = 188 [1]

Moles of AgBr = 21.6 / 188 = 0.115 mol (3 s.f.) [1]

⚠ If you missed marks here: Use relative atomic masses, not proton numbers: K is 39 (not 19), Ag is 108 and Br is 80, so AgBr = 188. Then 21.6 ÷ 188 = 0.1149…, which is 0.115 mol to 3 significant figures – the same number of significant figures as the 21.6 g you started from.

1 mark -- KOH Mr 56 moles 0.1 1 mark -- H2SO4 Mr 98 moles 0.1 1 mark -- AgBr Mr 188 1 mark -- AgBr moles 0.115

(b) [3]

(i) Calculate the number of molecules in 8.8 g of carbon dioxide, CO₂. [2]

(ii) Calculate the mass of one molecule of water, H₂O. Give your answer in standard form. [1]

Model Answer -- 2(b)

Moles of CO₂ = 8.8 / 44 = 0.2 mol [1]

Molecules = 0.2 × 6.02 × 10²³ = 1.204 × 10²³ [1]

Mass of 1 molecule of H₂O = 18 / (6.02 × 10²³) = 2.99 × 10⁻²³ g [1]

⚠ If you missed marks here: Find the moles first (8.8 ÷ 44 = 0.2 mol), then multiply by 6.02 × 10²³; multiplying the 8.8 g directly gives 5.30 × 10²⁴, far too many. For one water molecule, DIVIDE the molar mass by the Avogadro constant, 18 ÷ (6.02 × 10²³) = 2.99 × 10⁻²³ g – a single molecule has a tiny mass, so a positive power of ten means the division is upside down.

1 mark -- moles CO2 0.2 1 mark -- molecules 1.204 times 10 power 23 1 mark -- mass one water molecule 2.99 times 10 power negative 23 grams

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass.

(i) Determine the empirical formula of this hydrocarbon. [2]

(ii) The M_r of the hydrocarbon is 28. Determine its molecular formula. [1]

Model Answer -- 2(c)

C: 85.7/12 = 7.14 H: 14.3/1 = 14.3 Ratio: 7.14/7.14 = 1, 14.3/7.14 = 2 [1]

Empirical formula = CH₂ [1]

Empirical formula mass = 14; n = 28/14 = 2; Molecular formula = C₂H₄ (ethene) [1]

⚠ If you missed marks here: Divide each percentage by its A_r, then by the smallest: 85.7 ÷ 12 = 7.14 and 14.3 ÷ 1 = 14.3 give 1 : 2, so CH₂. For (ii), CH₂ cannot be the molecular formula because its mass is only 14; 28 ÷ 14 = 2 doubles BOTH subscripts, giving C₂H₄.

1 mark -- divide percentages atomic masses correct ratio 1 mark -- empirical formula CH2 1 mark -- molecular formula C2H4 ethene

(d) [2]

Calculate the percentage by mass of oxygen in copper(II) sulfate, CuSO₄.

Model Answer -- 2(d)

M_r of CuSO₄ = 64 + 32 + (4×16) = 160 [1]

% O = (64/160) × 100 = 40.0% [1]

⚠ If you missed marks here: CuSO₄ contains FOUR oxygen atoms, so the mass of oxygen is 4 × 16 = 64; using a single oxygen gives 10%. Divide by the M_r of the whole compound, 64 + 32 + 64 = 160: 64 ÷ 160 × 100 = 40.0%.

1 mark -- Mr CuSO4 equals 160 1 mark -- percentage oxygen 40 percent

Question 3 -- Reacting Masses and Stoichiometric Calculations

Total: 12 marks

A steel works in Sheffield uses the blast furnace process to extract iron from iron ore.

(a) [4]

Iron is extracted from haematite, Fe₂O₃, by reduction with carbon monoxide:

Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g)

Calculate the mass of iron that can be obtained from 800 g of iron(III) oxide.

Model Answer -- 3(a)

M_r of Fe₂O₃ = (2×56) + (3×16) = 160 [1]

Moles of Fe₂O₃ = 800 / 160 = 5.0 mol [1]

From equation: 1 mol Fe₂O₃ → 2 mol Fe; so moles of Fe = 5.0 × 2 = 10.0 mol [1]

Mass of Fe = 10.0 × 56 = 560 g [1]

⚠ If you missed marks here: Each Fe₂O₃ contains TWO iron atoms, so 5.0 mol of the oxide gives 10.0 mol of iron – missing the × 2 gives 280 g. Check the M_r as well: (2 × 56) + (3 × 16) = 160; using one Fe (104) throws every later step off.

1 mark -- Mr Fe2O3 160 1 mark -- moles Fe2O3 5.0 1 mark -- mole ratio 1 to 2 moles iron 10 1 mark -- mass iron 560 grams

(b) [4]

A student adds 6.5 g of zinc to a solution containing 7.3 g of hydrochloric acid (HCl).

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

(i) Calculate the moles of zinc and the moles of HCl. [2]

(ii) Determine which reagent is the limiting reagent. Explain your answer. [1]

(iii) Calculate the mass of hydrogen gas produced. [1]

Model Answer -- 3(b)

Moles of Zn = 6.5 / 65 = 0.1 mol [1]

Moles of HCl = 7.3 / 36.5 = 0.2 mol [1]

Ratio needed: 1 mol Zn : 2 mol HCl. We have 0.1 mol Zn and 0.2 mol HCl. Exact stoichiometric ratio -- neither is in excess (or both just used up). [1]

0.1 mol Zn → 0.1 mol H₂; Mass = 0.1 × 2 = 0.2 g [1]

⚠ If you missed marks here: Compare moles against the 1 : 2 ratio, not with each other: 0.1 mol Zn needs exactly 0.2 mol HCl, so neither is in excess – calling HCl "in excess" because 0.2 > 0.1 loses the mark. Hydrogen gas is H₂ (M_r 2), so 0.1 mol is 0.2 g, not 0.1 g.

1 mark -- moles zinc 0.1 1 mark -- moles HCl 0.2 1 mark -- stoichiometric ratio neither excess both used 1 mark -- mass hydrogen 0.2 grams

Sodium carbonate reacts with hydrochloric acid:

Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

Calculate the mass of sodium carbonate needed to produce 5.85 g of sodium chloride.

Model Answer -- 3(c)

M_r of NaCl = 23 + 35.5 = 58.5; Moles NaCl = 5.85 / 58.5 = 0.1 mol [1]

From equation: 1 mol Na₂CO₃ → 2 mol NaCl; so moles Na₂CO₃ = 0.1 / 2 = 0.05 mol [1]

M_r of Na₂CO₃ = (2×23) + 12 + (3×16) = 106 [1]

Mass = 0.05 × 106 = 5.3 g [1]

⚠ If you missed marks here: One Na₂CO₃ makes TWO NaCl, so you need half as many moles of carbonate: 0.1 ÷ 2 = 0.05 mol – using 0.1 mol gives 10.6 g, exactly double. Na₂CO₃ also has two sodiums: (2 × 23) + 12 + 48 = 106, not 83.

1 mark -- moles NaCl 0.1 mol 1 mark -- mole ratio 1 to 2 moles Na2CO3 0.05 1 mark -- Mr Na2CO3 106 1 mark -- mass Na2CO3 5.3 grams

Question 4 -- Percentage Yield and Percentage Purity

Total: 12 marks

A cement factory near Chennai processes limestone to produce quicklime (calcium oxide) and uses the product in construction.

(a) [4]

CaCO₃(s) → CaO(s) + CO₂(g)

250 g of calcium carbonate is heated strongly. The actual yield of calcium oxide is 119 g.

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

(ii) Calculate the percentage yield. [2]

Model Answer -- 4(a)

Moles CaCO₃ = 250 / 100 = 2.5 mol [1]

Moles CaO = 2.5 mol; Theoretical mass = 2.5 × 56 = 140 g [1]

% yield = (actual / theoretical) × 100 [1]

% yield = (119 / 140) × 100 = 85.0% [1]

⚠ If you missed marks here: The theoretical yield is the mass of CaO that 250 g of CaCO₃ could give, 2.5 mol × 56 = 140 g – not 250 g, because the CO₂ escapes. Percentage yield = actual ÷ theoretical × 100 = 119 ÷ 140 × 100 = 85.0%; a result of 118% means the fraction is upside down.

1 mark -- moles CaCO3 2.5 1 mark -- theoretical yield 140 grams 1 mark -- percentage yield formula correct 1 mark -- percentage yield 85 percent

(b) [4]

A 10.0 g sample of impure calcium carbonate is reacted with excess hydrochloric acid. The volume of carbon dioxide collected at RTP is 1920 cm³.

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

Calculate the percentage purity of the calcium carbonate sample.

Model Answer -- 4(b)

Moles of CO₂ = 1920 / 24000 = 0.08 mol [1]

From equation: 1 mol CaCO₃ → 1 mol CO₂; so moles CaCO₃ = 0.08 mol [1]

Mass of pure CaCO₃ = 0.08 × 100 = 8.0 g [1]

% purity = (8.0 / 10.0) × 100 = 80.0% [1]

⚠ If you missed marks here: Convert the volume first: 1920 ÷ 24 000 = 0.08 mol – dividing by 24 gives 80 mol, which could never come from 10 g. The CO₂ tells you how much PURE CaCO₃ reacted (0.08 × 100 = 8.0 g), so purity = 8.0 ÷ 10.0 × 100 = 80.0%.

1 mark -- moles CO2 0.08 1 mark -- moles CaCO3 0.08 ratio 1 mark -- mass pure CaCO3 8.0 grams 1 mark -- percentage purity 80 percent

Give two reasons why the percentage yield in an industrial process is often less than 100%.

Model Answer -- 4(c)

Some product is lost during transfer between containers / purification / filtration [1]

The reaction may be reversible / incomplete, or side reactions may produce unwanted by-products [1]

⚠ If you missed marks here: "Human error" and "mistakes in measuring" are too vague to score. Name real causes: product lost during transfer, filtration or purification; the reaction not going to completion or being reversible; or side reactions making unwanted by-products.

1 mark -- product lost transfer purification filtration 1 mark -- reversible incomplete side reactions by-products

(d) [2]

A factory must make 28.0 tonnes of calcium oxide each day. Its process has a percentage yield of 85.0%.

CaCO₃(s) → CaO(s) + CO₂(g)

Calculate the mass of calcium carbonate, in tonnes, that the factory must heat each day. Give your answer to three significant figures.

Model Answer -- 4(d)

At 100% yield: 56 t of CaO comes from 100 t of CaCO₃, so 28.0 t of CaO needs 28.0 × 100 / 56 = 50.0 t of CaCO₃ (or: CaO to be made in theory = 28.0 / 0.850 = 32.9 t) [1]

Allowing for the 85.0% yield: 50.0 / 0.850 = 58.8 t of CaCO₃ (or 32.9 × 100 / 56 = 58.8 t) [1]

⚠ If you missed marks here: Work in tonnes exactly as you would in grams: 56 t of CaO comes from 100 t of CaCO₃, so 28.0 t needs 50.0 t at a 100% yield (multiplying by 56/100 instead of 100/56 gives 15.7 t, less than the mass of CaO you are trying to make). Because only 85.0% of the product is obtained, the factory must heat MORE than this, so DIVIDE by 0.850 to get 58.8 t; multiplying (42.5 t) gives less limestone than even a perfect process would need.

1 mark -- 50.0 t at 100% (or 32.9 t) 1 mark -- correct answer 58.8 t

Question 5 -- Gas Volume Calculations at RTP

Total: 10 marks

In a school laboratory in Bristol, students investigate reactions that produce gases and measure their volumes at room temperature and pressure (RTP).

(a) [3]

Sodium reacts with water:

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Calculate the volume of hydrogen gas produced at RTP when 2.3 g of sodium reacts with excess water.

Model Answer -- 5(a)

Moles of Na = 2.3 / 23 = 0.1 mol [1]

From equation: 2 mol Na → 1 mol H₂; so moles H₂ = 0.1 / 2 = 0.05 mol [1]

Volume = 0.05 × 24 = 1.2 dm³ (or 1200 cm³) [1]

⚠ If you missed marks here: Two sodium atoms give only ONE H₂ molecule, so halve the moles: 0.1 mol Na → 0.05 mol H₂ → 0.05 × 24 = 1.2 dm³. Carrying 0.1 mol straight through gives 2.4 dm³, exactly double, and "1.2 cm³" is the right number with the wrong unit.

1 mark -- moles sodium 0.1 1 mark -- mole ratio 2 to 1 moles H2 0.05 1 mark -- volume 1.2 dm3 or 1200 cm3

(b) [3]

A student collects 4800 cm³ of carbon dioxide gas at RTP from the reaction of calcium carbonate with hydrochloric acid.

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Calculate the mass of calcium carbonate that reacted.

Model Answer -- 5(b)

Moles of CO₂ = 4800 / 24000 = 0.2 mol [1]

From equation: 1 mol CaCO₃ → 1 mol CO₂; so moles CaCO₃ = 0.2 mol [1]

Mass = 0.2 × 100 = 20.0 g [1]

⚠ If you missed marks here: Divide cm³ by 24 000 (not 24) to get moles: 4800 ÷ 24 000 = 0.2 mol. Then convert that to CaCO₃ using M_r = 100 – multiplying by 44 gives the mass of the CO₂ (8.8 g), which is not what was asked.

1 mark -- moles CO2 0.2 1 mark -- moles CaCO3 0.2 ratio 1 mark -- mass CaCO3 20 grams

In the combustion of methane:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

(i) State the ratio of volumes of CH₄ : O₂ : CO₂ (at the same temperature and pressure). [1]

(ii) Calculate the volume of oxygen needed to completely burn 120 cm³ of methane. [1]

Model Answer -- 5(c)

Volume ratio CH₄ : O₂ : CO₂ = 1 : 2 : 1 [1]

Volume of O₂ = 120 × 2 = 240 cm³ [1]

⚠ If you missed marks here: Gas volumes react in the same ratio as the numbers in the equation, so 120 cm³ of methane needs 2 × 120 = 240 cm³ of oxygen; halving it (60 cm³) reads the ratio backwards. There is no need to convert to moles or use the 24 dm³ molar volume – doing so only adds chances to slip.

1 mark -- ratio 1 to 2 to 1 1 mark -- volume oxygen 240 cm3

(d) [2]

(i) Explain why one mole of any gas occupies the same volume at the same temperature and pressure. [1]

(ii) State the conditions for RTP (room temperature and pressure). [1]

Model Answer -- 5(d)

At the same temperature and pressure, equal numbers of moles of gas contain the same number of molecules, and the volume of a gas depends on the number of molecules (not their size) since gas molecules are far apart [1]

RTP conditions: approximately 20 °C (293 K) and 1 atmosphere (101 kPa) [1]

⚠ If you missed marks here: Gas molecules are NOT all the same size – the reason is that they are so far apart that their own size hardly matters, so the volume depends only on the NUMBER of molecules, and one mole of any gas contains the same number. For (ii) give both a temperature and a pressure (about 20°C and 1 atmosphere); "room conditions" alone scores nothing.

1 mark -- equal moles same number molecules volume depends number molecules not size far apart 1 mark -- approximately 20 degrees celsius 1 atmosphere

Question 6 -- Titration and Concentration Calculations

Total: 12 marks

A student in Bangalore carries out a titration to find the concentration of a hydrochloric acid solution.

(a) [3]

The student dissolves 10.6 g of anhydrous sodium carbonate (Na₂CO₃) in water and makes the volume up to 1.00 dm³.

(i) Calculate the concentration of the sodium carbonate solution in mol/dm³. [2]

(ii) Calculate the concentration in g/dm³. [1]

Model Answer -- 6(a)

M_r of Na₂CO₃ = (2×23) + 12 + (3×16) = 106; Moles = 10.6 / 106 = 0.1 mol [1]

Concentration = 0.1 / 1.00 = 0.1 mol/dm³ [1]

Concentration in g/dm³ = 10.6 / 1.00 = 10.6 g/dm³ [1]

⚠ If you missed marks here: Check the M_r: Na₂CO₃ has two sodiums, (2 × 23) + 12 + 48 = 106, so 10.6 g is 0.100 mol, and in 1.00 dm³ that is 0.100 mol/dm³. The g/dm³ value is simply the mass dissolved per dm³, 10.6 g/dm³ – there is no need to divide by M_r again.

1 mark -- Mr Na2CO3 106 moles 0.1 1 mark -- concentration 0.1 mol per dm3 1 mark -- concentration 10.6 grams per dm3

(b) [5]

25.0 cm³ of the sodium carbonate solution is titrated with hydrochloric acid:

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

The mean titre is 25.0 cm³ of HCl.

(i) Calculate the moles of sodium carbonate in 25.0 cm³ of solution. [1]

(ii) Use the mole ratio to find the moles of HCl. [1]

(iii) Calculate the concentration of HCl in mol/dm³. [1]

(iv) Calculate the concentration of HCl in g/dm³. [2]

Model Answer -- 6(b)

Moles Na₂CO₃ = 0.1 × (25.0/1000) = 0.0025 mol [1]

From equation: 1 mol Na₂CO₃ reacts with 2 mol HCl; moles HCl = 0.0025 × 2 = 0.005 mol [1]

Concentration HCl = 0.005 / 0.025 = 0.2 mol/dm³ [1]

M_r of HCl = 1 + 35.5 = 36.5 [1]

Concentration = 0.2 × 36.5 = 7.3 g/dm³ [1]

⚠ If you missed marks here: Each Na₂CO₃ reacts with TWO HCl, so the acid moles are double: 0.0025 × 2 = 0.005 mol – forgetting this gives 0.1 mol/dm³ instead of 0.2. For g/dm³, HCl is 1 + 35.5 = 36.5, so 0.2 × 36.5 = 7.3 g/dm³; using 35.5 gives 7.1.

1 mark -- moles Na2CO3 0.0025 1 mark -- mole ratio 1 to 2 moles HCl 0.005 1 mark -- concentration HCl 0.2 mol per dm3 1 mark -- Mr HCl 36.5 1 mark -- concentration HCl 7.3 grams per dm3

A student needs to prepare 500 cm³ of 0.05 mol/dm³ NaOH from a stock solution of 0.5 mol/dm³ NaOH.

(i) Calculate the number of moles of NaOH needed in the diluted solution. [1]

(ii) Calculate the volume of stock solution required. [2]

(iii) Describe briefly how the student would prepare the diluted solution. [1]

Model Answer -- 6(c)

Moles needed = 0.05 × 0.5 = 0.025 mol [1]

Volume of stock = moles / concentration = 0.025 / 0.5 = 0.05 dm³ [1]

= 50 cm³ [1]

Measure 50 cm³ of the stock solution using a measuring cylinder or pipette, transfer to a 500 cm³ volumetric flask, and add distilled water up to the 500 cm³ mark [1]

⚠ If you missed marks here: Moles needed = 0.05 × 0.500 dm³ = 0.025 mol, and the stock volume is 0.025 ÷ 0.5 = 0.05 dm³ = 50 cm³; dividing by 0.05 (the diluted concentration) gives 500 cm³, the whole flask. In the method, make the 50 cm³ UP TO the 500 cm³ mark of a volumetric flask with distilled water – adding 500 cm³ of water to it gives 550 cm³ and the wrong concentration.

1 mark -- moles needed 0.025 1 mark -- volume stock solution calculated correctly 1 mark -- 50 cm3 1 mark -- measure stock solution add distilled water volumetric flask

Question 7 -- Water of Crystallisation and Mixed Calculations

Total: 10 marks

A student in Manchester heats hydrated copper(II) sulfate crystals to determine the number of water molecules in the formula.

(a) [2]

(i) Define the term water of crystallisation. [1]

(ii) State the difference between a hydrated salt and an anhydrous salt. [1]

Model Answer -- 7(a)

Water of crystallisation is water molecules that are chemically bonded / incorporated into the crystal structure of a salt [1]

A hydrated salt contains water of crystallisation; an anhydrous salt has had the water removed / contains no water of crystallisation [1]

⚠ If you missed marks here: Water of crystallisation is part of the crystal itself – water molecules held in its structure – not dampness on the surface or water used to dissolve the salt. An anhydrous salt contains NO water of crystallisation; "a salt that has been dried" does not say that the water inside the crystal has gone.

1 mark -- water molecules chemically bonded crystal structure salt 1 mark -- hydrated contains water crystallisation anhydrous water removed

(b) [5]

A student heats 6.25 g of hydrated copper(II) sulfate (CuSO₄·nH₂O). After heating until constant mass, the residue has a mass of 4.00 g.

Determine the value of n.

Model Answer -- 7(b)

Mass of water lost = 6.25 − 4.00 = 2.25 g [1]

Moles of CuSO₄ = 4.00 / 160 = 0.025 mol [1]

Moles of H₂O = 2.25 / 18 = 0.125 mol [1]

Ratio: CuSO₄ : H₂O = 0.025 : 0.125 [1]

Divide by smallest: 1 : 5, so n = 5 -- the formula is CuSO₄·5H₂O [1]

⚠ If you missed marks here: The 4.00 g residue is ANHYDROUS CuSO₄ (M_r 160), so 4.00 ÷ 160 = 0.025 mol – using 6.25 g or M_r 250 mixes up the hydrated and anhydrous salts. Then n = moles of water ÷ moles of CuSO₄ = 0.125 ÷ 0.025 = 5; dividing the other way gives 0.2, which cannot be a number of water molecules.

1 mark -- mass water 2.25 grams 1 mark -- moles CuSO4 0.025 1 mark -- moles H2O 0.125 1 mark -- ratio CuSO4 H2O calculated 1 mark -- value 5 formula CuSO4 5H2O

(i) Calculate the percentage by mass of water in CuSO₄·5H₂O. [2]

(ii) State one observation the student would make during the heating. [1]

Model Answer -- 7(c)

M_r of CuSO₄·5H₂O = 160 + (5×18) = 250 [1]

% water = (90/250) × 100 = 36.0% [1]

The blue crystals turn to a white powder (as the hydrated salt loses its water of crystallisation) [1]

⚠ If you missed marks here: The salt holds FIVE waters, so the water mass is 5 × 18 = 90 and 90 ÷ 250 × 100 = 36.0% (using one water gives 7.2%). On heating, blue crystals turn to a WHITE powder; "turns blue" is the reverse change (adding water to anhydrous copper(II) sulfate), and "it changes colour" without naming the colours does not score.

1 mark -- Mr CuSO4 5H2O equals 250 1 mark -- percentage water 36 percent 1 mark -- blue crystals turn white powder

Self-Assessment

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