IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 3: Stoichiometry -- Challenge Paper

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions -- Challenge Paper

This paper contains harder Cambridge-style questions for students aiming at grades A* and A.
Answer all questions in the spaces provided.
Show all working for calculations. Marks will be deducted for missing working.
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, Ba = 137, Pb = 207
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 -- Empirical and Molecular Formula Determination

Total: 12 marks

An analytical chemist at the Indian Institute of Technology, Bombay, analyses unknown organic compounds using combustion analysis.

(a) [5]

When 4.50 g of an organic compound X (containing C, H and O only) is completely burned in excess oxygen, it produces 6.60 g of CO₂ and 2.70 g of H₂O.

(i) Calculate the mass of carbon in the sample. [1]

(ii) Calculate the mass of hydrogen in the sample. [1]

(iii) Calculate the mass of oxygen in the sample. [1]

(iv) Determine the empirical formula of compound X. [2]

Model Answer -- 1(a)

Moles of CO₂ = 6.60/44 = 0.15 mol; Mass of C = 0.15 × 12 = 1.80 g [1]

Moles of H₂O = 2.70/18 = 0.15 mol; Mass of H = 0.15 × 2 = 0.30 g [1]

Mass of O = 4.50 − 1.80 − 0.30 = 2.40 g [1]

C: 1.80/12 = 0.15 H: 0.30/1 = 0.30 O: 2.40/16 = 0.15 [1]

Ratio = 0.15 : 0.30 : 0.15 = 1 : 2 : 1 → Empirical formula = CH₂O [1]

⚠ If you missed marks here: The classic slips are finding mass of C as 6.60 × 12/44 but forgetting hydrogen is doubled – each H₂O contains 2 H, so mass H = 0.15 × 2 = 0.30 g, not 0.15 g. If your oxygen mass was wrong, remember you can only get it by subtraction (4.50 − 1.80 − 0.30) – you cannot use the oxygen in the CO₂ because that came from the air. For the final ratio, divide each element's MASS by its A_r first; dividing the raw masses by each other gives a wrong ratio.

1 mark -- mass carbon 1.80 grams from CO2 1 mark -- mass hydrogen 0.30 grams from H2O 1 mark -- mass oxygen 2.40 grams subtraction 1 mark -- moles calculated divide atomic masses 1 mark -- empirical formula CH2O

(b) [3]

Compound X has a relative molecular mass of 60.

(i) Determine the molecular formula of compound X. [1]

(ii) Suggest a possible name for compound X. [1]

(iii) Write the balanced equation for the complete combustion of compound X. [1]

Model Answer -- 1(b)

Empirical formula mass = 12 + 2 + 16 = 30; n = 60/30 = 2; Molecular formula = C₂H₄O₂ [1]

Ethanoic acid / acetic acid (CH₃COOH) [1]

C₂H₄O₂ + 2O₂ → 2CO₂ + 2H₂O [1]

⚠ If you missed marks here: If you wrote CH₂O as the molecular formula, you forgot to compare M_r = 60 with the empirical formula mass (30): n = 60/30 = 2, so every subscript doubles to C₂H₄O₂. On the equation, a common slip is forgetting that X already contains 2 oxygen atoms – count O on both sides before choosing the O₂ coefficient (only 2O₂ is needed, not 3).

1 mark -- molecular formula C2H4O2 1 mark -- ethanoic acid acetic acid 1 mark -- combustion equation balanced 2O2 2CO2 2H2O

A hydrated salt has the formula MgSO₄·nH₂O. A 12.30 g sample of this salt is heated until all the water is driven off. The residue has a mass of 6.00 g.

Determine the value of n.

Model Answer -- 1(c)

Mass of water = 12.30 − 6.00 = 6.30 g [1]

Moles of MgSO₄ = 6.00 / 120 = 0.05 mol (M_r = 24 + 32 + 64 = 120) [1]

Moles of H₂O = 6.30 / 18 = 0.35 mol [1]

n = 0.35 / 0.05 = 7; Formula is MgSO₄·7H₂O (Epsom salt) [1]

⚠ If you missed marks here: The residue mass 6.00 g is the anhydrous MgSO₄ – if you divided 12.30 g (the whole hydrated salt) by 120 you got the wrong moles. Check your M_r: SO₄ is 32 + 64 = 96, so MgSO₄ = 120 (a common error is using only one O). And n = moles water ÷ moles salt, in that order – dividing the other way gives 1/7 and should tell you something is upside down.

1 mark -- mass water 6.30 grams 1 mark -- moles MgSO4 0.05 1 mark -- moles H2O 0.35 1 mark -- value 7 MgSO4 7H2O

Question 2 -- Multi-step Stoichiometry with Limiting Reagents

Total: 12 marks

An industrial chemist in Manchester designs a two-step process to produce iron(II) sulfate crystals from scrap iron.

(a) [5]

Iron reacts with dilute sulfuric acid:

Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g)

A student adds 5.6 g of iron to 200 cm³ of 0.4 mol/dm³ sulfuric acid.

(i) Calculate the moles of iron. [1]

(ii) Calculate the moles of sulfuric acid. [1]

(iii) Identify the limiting reagent. Explain your reasoning. [1]

(iv) Calculate the maximum mass of iron(II) sulfate (FeSO₄) that can be produced. [2]

Model Answer -- 2(a)

Moles of Fe = 5.6 / 56 = 0.1 mol [1]

Moles of H₂SO₄ = 0.4 × (200/1000) = 0.4 × 0.2 = 0.08 mol [1]

Mole ratio is 1:1. We have 0.1 mol Fe but only 0.08 mol H₂SO₄. Sulfuric acid is the limiting reagent because there is less of it relative to the stoichiometry. [1]

Moles FeSO₄ = 0.08 mol (based on limiting reagent); M_r FeSO₄ = 56 + 32 + 64 = 152 [1]

Mass = 0.08 × 152 = 12.16 g [1]

⚠ If you missed marks here: If your acid moles came out as 80, you forgot to convert 200 cm³ to dm³ (divide by 1000 first: 0.4 × 0.2 = 0.08 mol). The limiting reagent is decided by comparing MOLES against the 1:1 ratio, not masses – 0.08 mol acid runs out before 0.1 mol Fe. If your mass was 15.2 g, you based the product on the iron (0.1 mol) instead of the limiting acid (0.08 mol).

1 mark -- moles iron 0.1 1 mark -- moles H2SO4 0.08 1 mark -- sulfuric acid limiting reagent less relative stoichiometry 1 mark -- moles FeSO4 0.08 Mr 152 1 mark -- mass FeSO4 12.16 grams

(b) [3]

(i) Calculate the moles of iron that actually reacted. [1]

(ii) Calculate the mass of iron remaining unreacted. [1]

(iii) The student filters the solution to remove excess iron. Calculate the volume of hydrogen gas produced at RTP. [1]

Model Answer -- 2(b)

Moles of Fe reacted = 0.08 mol (same as H₂SO₄ since 1:1 ratio) [1]

Excess Fe = 0.1 − 0.08 = 0.02 mol; Mass = 0.02 × 56 = 1.12 g [1]

Volume H₂ = 0.08 × 24 = 1.92 dm³ [1]

⚠ If you missed marks here: Everything in this part must be built on the LIMITING reagent from (a): only 0.08 mol Fe reacts because only 0.08 mol acid exists. If you got 2.4 dm³ of hydrogen, you used the full 0.1 mol of iron – the 0.02 mol excess Fe never reacts, which is exactly the 1.12 g left over.

1 mark -- moles iron reacted 0.08 1 mark -- excess iron 0.02 mol mass 1.12 grams 1 mark -- volume hydrogen 1.92 dm3

Aluminium reacts with iron(III) oxide in the thermite reaction:

2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(l)

A mixture contains 10.8 g of aluminium and 48.0 g of iron(III) oxide.

(i) Determine which reagent is limiting. [2]

(ii) Calculate the mass of iron produced. [1]

(iii) Calculate the mass of the excess reagent remaining. [1]

Model Answer -- 2(c)

Moles Al = 10.8/27 = 0.4 mol; Moles Fe₂O₃ = 48.0/160 = 0.3 mol [1]

Ratio needed: 2 mol Al : 1 mol Fe₂O₃. For 0.4 mol Al, need 0.2 mol Fe₂O₃. Have 0.3 mol, so Al is limiting. [1]

0.4 mol Al → 0.4 mol Fe; Mass Fe = 0.4 × 56 = 22.4 g [1]

Fe₂O₃ used = 0.4/2 = 0.2 mol; Excess = 0.3 − 0.2 = 0.1 mol; Mass = 0.1 × 160 = 16.0 g [1]

⚠ If you missed marks here: The trap here is comparing raw moles: 0.3 < 0.4 makes Fe₂O₃ look limiting, but the ratio is 2Al : 1Fe₂O₃ – 0.4 mol Al only needs 0.2 mol of the oxide, so Al runs out first. Always divide each amount by its coefficient before comparing. If your iron mass was 33.6 g, you used 0.3 mol Fe₂O₃ × 2 instead of following the limiting Al (0.4 mol Al → 0.4 mol Fe = 22.4 g).

1 mark -- moles aluminium 0.4 moles Fe2O3 0.3 1 mark -- aluminium limiting reagent explained 1 mark -- mass iron 22.4 grams 1 mark -- excess Fe2O3 16.0 grams remaining

Question 3 -- Back Titration and Gravimetric Analysis

Total: 12 marks

A pharmaceutical laboratory in Cambridge uses titration techniques to analyse the purity of an antacid tablet containing calcium carbonate.

(a) [7]

A 1.50 g antacid tablet is dissolved in 50.0 cm³ of 0.500 mol/dm³ hydrochloric acid (an excess). The excess acid is then titrated with 0.200 mol/dm³ sodium hydroxide solution. 15.0 cm³ of NaOH is needed to neutralise the excess acid.

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

(i) Calculate the total moles of HCl added. [1]

(ii) Calculate the moles of excess HCl (that reacted with NaOH). [1]

(iii) Calculate the moles of HCl that reacted with CaCO₃. [1]

(iv) Calculate the moles of CaCO₃ in the tablet. [1]

(v) Calculate the mass of CaCO₃ in the tablet. [1]

(vi) Calculate the percentage purity of the tablet. [1]

(vii) Explain why a back titration was used instead of a direct titration. [1]

Model Answer -- 3(a)

Total moles HCl = 0.500 × (50.0/1000) = 0.025 mol [1]

Moles excess HCl = moles NaOH = 0.200 × (15.0/1000) = 0.003 mol [1]

Moles HCl reacted with CaCO₃ = 0.025 − 0.003 = 0.022 mol [1]

From equation: 2 mol HCl : 1 mol CaCO₃; Moles CaCO₃ = 0.022 / 2 = 0.011 mol [1]

Mass CaCO₃ = 0.011 × 100 = 1.10 g [1]

% purity = (1.10 / 1.50) × 100 = 73.3% [1]

A back titration was used because CaCO₃ is insoluble in water / does not dissolve easily, so a direct titration with an indicator would not give a sharp end point. By adding excess acid first and back-titrating, all the CaCO₃ reacts. [1]

⚠ If you missed marks here: Back titrations go wrong in two places: mixing up WHICH acid the NaOH measures (it neutralises only the leftover 0.003 mol, so the acid that reacted with the tablet is 0.025 − 0.003 = 0.022 mol – if you titrated "against the CaCO₃" directly you skipped the subtraction), and forgetting the 2:1 ratio (0.022 mol HCl ÷ 2, not × 2 – doubling gives a purity over 100%, which should ring alarm bells). For the last mark, "it's more accurate" scores nothing – you must say CaCO₃ is insoluble so a direct titration gives no sharp end point.

1 mark -- total moles HCl 0.025 1 mark -- excess HCl moles NaOH 0.003 1 mark -- moles HCl reacted CaCO3 0.022 1 mark -- moles CaCO3 0.011 1 mark -- mass CaCO3 1.10 grams 1 mark -- percentage purity 73.3 1 mark -- CaCO3 insoluble direct titration no sharp end point excess acid back-titrating

(b) [5]

A student adds excess barium chloride solution to 25.0 cm³ of 0.100 mol/dm³ sodium sulfate solution:

BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

The precipitate of barium sulfate is filtered, dried and weighed.

(i) Calculate the expected mass of barium sulfate precipitate. [3]

(ii) The student obtains 0.52 g. Calculate the percentage yield. [2]

Model Answer -- 3(b)

Moles Na₂SO₄ = 0.100 × 0.025 = 0.0025 mol [1]

1 mol Na₂SO₄ → 1 mol BaSO₄; moles BaSO₄ = 0.0025 mol [1]

M_r BaSO₄ = 137 + 32 + 64 = 233; Mass = 0.0025 × 233 = 0.583 g [1]

% yield = (0.52 / 0.583) × 100 [1]

= 89.2% [1]

⚠ If you missed marks here: If your moles were 2.5, you forgot 25.0 cm³ = 0.025 dm³. Check the M_r of BaSO₄: 137 + 32 + (4 × 16) = 233 – a frequent slip is counting only one or two oxygens. For percentage yield it is always actual ÷ theoretical (0.52/0.583); if you got 112% you inverted the fraction – a yield above 100% from a normal experiment is a sign to flip it.

1 mark -- moles Na2SO4 0.0025 1 mark -- moles BaSO4 0.0025 ratio 1 mark -- Mr BaSO4 233 mass 0.583 grams 1 mark -- percentage yield formula applied 1 mark -- percentage yield 89.2

Question 4 -- Gas Volume Calculations and Mixed Reactions

Total: 12 marks

A research team at Imperial College London investigates gas volumes produced in various reactions.

(a) [4]

During the electrolysis of water:

2H₂O(l) → 2H₂(g) + O₂(g)

(i) If 3.6 g of water is completely decomposed, calculate the volume of hydrogen gas produced at RTP. [2]

(ii) Calculate the volume of oxygen gas produced at RTP. [1]

(iii) Calculate the total number of gas molecules produced. [1]

Model Answer -- 4(a)

Moles of H₂O = 3.6 / 18 = 0.2 mol [1]

2 mol H₂O → 2 mol H₂; moles H₂ = 0.2; Volume = 0.2 × 24 = 4.8 dm³ [1]

2 mol H₂O → 1 mol O₂; moles O₂ = 0.1; Volume = 0.1 × 24 = 2.4 dm³ [1]

Total gas moles = 0.2 + 0.1 = 0.3; Molecules = 0.3 × 6.02 × 10²³ = 1.806 × 10²³ [1]

⚠ If you missed marks here: The 2:2:1 ratio catches people: H₂ moles EQUAL the water moles (0.2), but O₂ is HALF (0.1) – if your oxygen volume was 4.8 dm³ you treated the ratio as 1:1. For the molecule count, use the TOTAL gas moles (0.2 + 0.1 = 0.3); multiplying only 0.2 by 6.02 × 10²³ misses the oxygen molecules.

1 mark -- moles H2O 0.2 1 mark -- volume hydrogen 4.8 dm3 1 mark -- volume oxygen 2.4 dm3 1 mark -- total molecules 1.806 times 10 power 23

(b) [4]

A student adds 100 cm³ of 1.0 mol/dm³ hydrochloric acid to excess magnesium carbonate:

MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g)

(i) Calculate the moles of HCl used. [1]

(ii) Calculate the moles of CO₂ produced. [1]

(iii) Calculate the volume of CO₂ at RTP. [1]

(iv) Calculate the mass of MgCO₃ that reacted. [1]

Model Answer -- 4(b)

Moles HCl = 1.0 × (100/1000) = 0.1 mol [1]

2 mol HCl → 1 mol CO₂; moles CO₂ = 0.1 / 2 = 0.05 mol [1]

Volume CO₂ = 0.05 × 24 = 1.2 dm³ (or 1200 cm³) [1]

Moles MgCO₃ = 0.05 mol; M_r = 24 + 12 + 48 = 84; Mass = 0.05 × 84 = 4.2 g [1]

⚠ If you missed marks here: If your CO₂ volume was 2.4 dm³, you missed the 2HCl : 1CO₂ ratio – the acid moles must be halved (0.1 ÷ 2 = 0.05), not carried straight through. Also check M_r of MgCO₃ = 24 + 12 + 48 = 84; using 3 × 16 = 48 for the oxygens is where slips like 68 come from.

1 mark -- moles HCl 0.1 1 mark -- moles CO2 0.05 1 mark -- volume CO2 1.2 dm3 1 mark -- mass MgCO3 4.2 grams

50 cm³ of propane (C₃H₈) is mixed with 300 cm³ of oxygen and sparked:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

(i) Calculate the volume of oxygen needed to react with 50 cm³ of propane. [1]

(ii) State which gas is in excess and calculate the volume of excess gas remaining after the reaction. [2]

(iii) Calculate the total volume of gases after the reaction (at RTP, water is a liquid). [1]

Model Answer -- 4(c)

O₂ needed = 50 × 5 = 250 cm³ [1]

O₂ supplied = 300 cm³, needed = 250 cm³. Oxygen is in excess. [1]

Excess O₂ = 300 − 250 = 50 cm³ [1]

CO₂ produced = 50 × 3 = 150 cm³; Total gas = 150 + 50 = 200 cm³ (H₂O is liquid at RTP) [1]

⚠ If you missed marks here: With gases you can use the equation coefficients as VOLUME ratios directly (Avogadro's law) – no need to convert to moles. The big trap is the last mark: if you got 350 or 400 cm³ you counted the water as a gas, but at RTP the 4H₂O is liquid, so the final mixture is only 150 cm³ CO₂ + 50 cm³ unreacted O₂.

1 mark -- oxygen needed 250 cm3 1 mark -- oxygen excess identified 1 mark -- excess oxygen 50 cm3 1 mark -- total gas volume 200 cm3 CO2 plus excess O2

Question 5 -- Percentage Yield in Multi-step Synthesis

Total: 10 marks

A chemical plant in Gujarat manufactures sodium sulfate (Na₂SO₄) through a two-step process.

(a) [4]

Step 1: 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

The plant uses 400 g of NaOH and excess H₂SO₄.

(i) Calculate the theoretical yield of Na₂SO₄. [3]

(ii) The actual yield from Step 1 is 639 g. Calculate the percentage yield. [1]

Model Answer -- 5(a)

Moles NaOH = 400 / 40 = 10.0 mol [1]

2 mol NaOH → 1 mol Na₂SO₄; moles Na₂SO₄ = 10.0 / 2 = 5.0 mol [1]

M_r Na₂SO₄ = 46 + 32 + 64 = 142; Theoretical mass = 5.0 × 142 = 710 g [1]

% yield = (639 / 710) × 100 = 90.0% [1]

⚠ If you missed marks here: If your theoretical yield was 1420 g, you missed the 2NaOH : 1Na₂SO₄ ratio – 10 mol NaOH gives only 5 mol of product, so the moles must be halved. Also check M_r of Na₂SO₄: there are TWO sodiums (2 × 23 = 46), so 46 + 32 + 64 = 142, not 119. Percentage yield is actual ÷ theoretical (639/710), never the other way round.

1 mark -- moles NaOH 10 1 mark -- moles Na2SO4 5.0 ratio 1 mark -- theoretical mass 710 grams 1 mark -- percentage yield 90 percent

(b) [3]

In an industrial process, three consecutive steps each have a percentage yield:
Step 1: 90%, Step 2: 85%, Step 3: 95%

(i) Calculate the overall percentage yield for all three steps combined. [2]

(ii) Explain why multi-step syntheses typically give lower overall yields. [1]

Model Answer -- 5(b)

Overall yield = (90/100) × (85/100) × (95/100) × 100 [1]

= 0.90 × 0.85 × 0.95 × 100 = 72.675% ≈ 72.7% [1]

Each step introduces losses (transfer, incomplete reaction, purification), and these losses multiply together, making the overall yield significantly less than any individual step yield [1]

⚠ If you missed marks here: The usual error is AVERAGING the yields ((90 + 85 + 95)/3 = 90%) instead of MULTIPLYING them as fractions: 0.90 × 0.85 × 0.95 × 100 = 72.7%. Each step only keeps a fraction of what the previous step passed on. For the explanation mark, "some product is lost" alone is thin – say the losses at each step multiply together, so the overall yield ends up below every individual step's yield.

1 mark -- multiply fractional yields together 1 mark -- overall yield 72.7 percent 1 mark -- losses multiply together overall yield less than individual steps

Compare the atom economy of two methods of producing copper:

Method A: CuO + H₂ → Cu + H₂O
Method B: CuCO₃ → CuO + CO₂ then CuO + C → Cu + CO

(i) Calculate the atom economy of Method A for producing copper. [1]

(ii) Calculate the atom economy of the second step of Method B for producing copper. [1]

(iii) State which method has a higher atom economy and why this matters. [1]

Model Answer -- 5(c)

Method A: atom economy = (64 / (64 + 18)) × 100 = (64/82) × 100 = 78.0% [1]

Method B step 2: atom economy = (64 / (64 + 28)) × 100 = (64/92) × 100 = 69.6% [1]

Method A has higher atom economy (78.0% vs 69.6%). Higher atom economy means less waste is produced per unit of desired product, which is better for the environment and more cost-effective. [1]

⚠ If you missed marks here: Atom economy = (M_r of DESIRED product ÷ total M_r of ALL products) × 100 – a common slip is putting the reactant masses in the denominator or forgetting the waste product (H₂O = 18 in Method A, CO = 28 in Method B). Note this is about the balanced equation, not experimental yield – mixing up atom economy with percentage yield loses the mark. For (iii), "less waste per unit of product" is the phrase the examiner wants, not just "it's better".

1 mark -- method atom economy 78 percent 1 mark -- method step atom economy 69.6 percent 1 mark -- higher atom economy less waste environment cost-effective

Question 6 -- Complex Concentration and Dilution Problems

Total: 12 marks

A water treatment plant near Lagos tests the quality of drinking water by analysing dissolved ions.

(a) [3]

A solution contains 7.1 g/dm³ of dissolved chlorine gas (Cl₂).

(i) Convert this concentration to mol/dm³. [1]

(ii) Calculate the number of molecules of Cl₂ in 500 cm³ of this solution. [2]

Model Answer -- 6(a)

M_r of Cl₂ = 71; Concentration = 7.1 / 71 = 0.1 mol/dm³ [1]

Moles in 500 cm³ = 0.1 × 0.5 = 0.05 mol [1]

Molecules = 0.05 × 6.02 × 10²³ = 3.01 × 10²² [1]

⚠ If you missed marks here: To convert g/dm³ to mol/dm³ you divide by M_r – and Cl₂ is a molecule, so M_r = 71, not 35.5 (using 35.5 gives 0.2 mol/dm³, exactly double). Then 500 cm³ must become 0.5 dm³ before multiplying by concentration; if your molecule count was 6.02 × 10²² you used 0.1 mol instead of 0.05.

1 mark -- concentration 0.1 mol per dm3 1 mark -- moles 0.05 in 500 cm3 1 mark -- molecules 3.01 times 10 power 22

(b) [5]

A student titrates 20.0 cm³ of potassium hydroxide solution with 0.150 mol/dm³ sulfuric acid. The mean titre is 18.0 cm³.

2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)

(i) Calculate the concentration of the KOH solution in mol/dm³. [3]

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

(iii) Calculate the mass of KOH dissolved in 250 cm³ of the solution. [1]

Model Answer -- 6(b)

Moles H₂SO₄ = 0.150 × (18.0/1000) = 0.0027 mol [1]

Moles KOH = 0.0027 × 2 = 0.0054 mol (2:1 ratio) [1]

Concentration KOH = 0.0054 / 0.020 = 0.27 mol/dm³ [1]

M_r KOH = 56; Concentration = 0.27 × 56 = 15.12 g/dm³ [1]

Mass in 250 cm³ = 15.12 × 0.25 = 3.78 g [1]

⚠ If you missed marks here: Two ratio traps: the KOH moles are DOUBLE the acid moles (2KOH : 1H₂SO₄, so 0.0027 × 2 = 0.0054) – dividing by 2 instead gives a concentration of 0.0675 mol/dm³. And when finding the concentration, divide by the KOH volume (20.0 cm³ = 0.020 dm³), not the titre volume 18.0. For g/dm³ just multiply mol/dm³ by M_r = 56 (39 + 16 + 1).

1 mark -- moles H2SO4 0.0027 1 mark -- moles KOH 0.0054 ratio 2 to 1 1 mark -- concentration KOH 0.27 mol per dm3 1 mark -- concentration 15.12 grams per dm3 1 mark -- mass KOH 3.78 grams in 250 cm3

A chemist prepares a serial dilution starting with 1.00 mol/dm³ HCl.

At each step, 10.0 cm³ of solution is taken and made up to 100.0 cm³ with distilled water.

(i) Calculate the concentration after the first dilution. [1]

(ii) Calculate the concentration after the second dilution. [1]

(iii) Calculate the concentration after the fourth dilution. [1]

(iv) How many dilution steps are needed to reduce the concentration below 0.001 mol/dm³? [1]

Model Answer -- 6(c)

Each dilution is ×10 (10/100). After 1st: 1.00 / 10 = 0.100 mol/dm³ [1]

After 2nd: 0.100 / 10 = 0.0100 mol/dm³ [1]

After 4th: 1.00 / 10⁴ = 1.00 × 10⁻⁴ mol/dm³ [1]

After 3rd: 0.001; after 4th: 0.0001. Need 4 dilutions (3 gives exactly 0.001, 4 gives below) [1]

⚠ If you missed marks here: Taking 10 cm³ up to 100 cm³ is a ×10 dilution each time (concentration ÷ 10), not subtracting or halving – after n steps the concentration is 1.00/10ⁿ. The final part is a wording trap: 3 dilutions give EXACTLY 0.001 mol/dm³, and "below 0.001" means you need one more, so the answer is 4.

1 mark -- first dilution 0.100 mol dm3 1 mark -- second dilution 0.0100 1 mark -- fourth dilution 0.0001 1 mark -- four dilutions needed below 0.001

Question 7 -- Thermal Decomposition and Integrated Stoichiometry

Total: 10 marks

A student at a school in Edinburgh investigates the thermal decomposition of metal carbonates by measuring gas volumes and mass changes.

(a) [5]

A student heats 2.74 g of an unknown Group II metal carbonate, MCO₃. The metal carbonate decomposes completely:

MCO₃(s) → MO(s) + CO₂(g)

The mass of the residue (MO) is 1.54 g.

(i) Calculate the mass of CO₂ lost. [1]

(ii) Calculate the moles of CO₂ produced. [1]

(iii) State the moles of MCO₃ that decomposed. [1]

(iv) Calculate the M_r of MCO₃. [1]

(v) Identify the metal M. Show your working. [1]

Model Answer -- 7(a)

Mass CO₂ = 2.74 − 1.54 = 1.20 g [1]

Moles CO₂ = 1.20 / 44 = 0.02727 mol [1]

From equation: 1 mol MCO₃ → 1 mol CO₂; so moles MCO₃ = 0.02727 mol [1]

M_r of MCO₃ = 2.74 / 0.02727 = 100.4 ≈ 100 [1]

A_r of M = 100 − 12 − 48 = 40; M = calcium (Ca) The carbonate is CaCO₃ [1]

⚠ If you missed marks here: The mass LOST (2.74 − 1.54 = 1.20 g) is the CO₂ – using the residue mass 1.54 g here is the classic slip. For M_r, divide the ORIGINAL carbonate mass by the moles (2.74/0.02727 = 100), not the residue mass – and it is mass ÷ moles, not moles ÷ mass. Finally subtract the whole CO₃ group (12 + 48 = 60) to get A_r = 40, calcium; check M is actually in Group II before writing it down.

1 mark -- mass CO2 1.20 grams 1 mark -- moles CO2 0.0273 1 mark -- moles MCO3 same 0.0273 1 mark -- Mr MCO3 approximately 100 1 mark -- metal calcium Ar 40 CaCO3

(b) [5]

The calcium oxide (CaO) residue from part (a) is dissolved in 100 cm³ of 0.500 mol/dm³ hydrochloric acid (excess):

CaO(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l)

(i) Calculate the moles of CaO. [1]

(ii) Calculate the moles of HCl that reacted with the CaO. [1]

(iii) Calculate the total moles of HCl originally present. [1]

(iv) Calculate the moles of HCl remaining unreacted. [1]

(v) Calculate the concentration of the resulting solution in mol/dm³ (assume volume remains 100 cm³). [1]

Model Answer -- 7(b)

Moles CaO = 1.54 / 56 = 0.0275 mol [1]

From equation: 1 mol CaO reacts with 2 mol HCl; moles HCl reacted = 0.0275 × 2 = 0.055 mol [1]

Total moles HCl = 0.500 × 0.100 = 0.050 mol [1]

Moles HCl unreacted = 0.050 − 0.055 = −0.005. This shows the HCl is actually the limiting reagent (not excess as stated). If all CaO reacted: moles HCl remaining = 0; some CaO remains undissolved (0.050/2 = 0.025 mol CaO reacts). [1]

The resulting solution contains CaCl₂: moles = 0.025; concentration CaCl₂ = 0.025 / 0.100 = 0.25 mol/dm³. No excess HCl in solution. [1]

⚠ If you missed marks here: This part rewards NOTICING the contradiction: the CaO needs 0.055 mol HCl but only 0.500 × 0.100 = 0.050 mol exists, so the subtraction goes negative – if you wrote "−0.005 mol remaining" without comment, that is where the mark went. A negative moles answer always means the "excess" reagent is really limiting. The final concentration is then based on the CaCl₂ actually formed (0.050/2 = 0.025 mol in 0.100 dm³ = 0.25 mol/dm³), with zero HCl left.

1 mark -- moles CaO 0.0275 1 mark -- moles HCl needed 0.055 1 mark -- total moles HCl 0.050 1 mark -- HCl limiting reagent identified insufficient excess 1 mark -- CaCl2 concentration 0.25 mol per dm3

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