Percentage yield & atom economy (Chem only) – GCSE Chemistry exam questions with mark scheme

1Grade 6–7HigherTriple only5 marks

Calcium oxide is made by heating calcium carbonate.

$$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$$

$M_r$: $\text{CaCO}_3$ = 100, CaO = 56, $\text{CO}_2$ = 44

(a) Calculate the percentage atom economy for making calcium oxide. [2]

(b) 50.0 g of calcium carbonate was heated. 22.4 g of calcium oxide was produced. Calculate the percentage yield. [3]

(a) $\dfrac{56}{100} \times 100$ allow $\dfrac{56}{56+44} \times 100$(1)
(a) 56%(1)
(b) theoretical mass CaO $= \dfrac{50.0}{100} \times 56 = 28.0$ g(1)
(b) $\dfrac{22.4}{28.0} \times 100$(1)
(b) 80% allow ecf from incorrect theoretical mass(1)
Answer: (a) 56% (b) 80%
Examiner tip: Percentage yield is actual ÷ theoretical — the theoretical mass must come from the equation, not the 50.0 g you started with.

2Grade 1–3FoundationTriple only4 marks

A student made sodium chloride by neutralising sodium hydroxide solution with dilute hydrochloric acid and then evaporating the water.

$$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$

(a) Which statement describes atom economy? Tick (✓) one box. [1]

The mass of product actually made compared with the maximum mass that could be made	☐
A measure of the amount of starting materials that end up as useful products	☐
The total mass of the reactants compared with the total mass of the products	☐
How quickly the reactants are changed into useful products	☐

(b) The maximum theoretical mass of sodium chloride the student could make was 5.85 g.

The student actually made 4.33 g of sodium chloride.

Calculate the percentage yield. Give your answer to 3 significant figures. [2]

(a) A measure of the amount of starting materials that end up as useful products (second box only; more than one tick = 0)(1)
(b) $\dfrac{4.33}{5.85} \times 100$(1)
(b) 74.0 (%) an answer of 74 (%) without the required 3 s.f. scores 1 mark only (for the method); correct answer with no working scores 2(1)
(c) any one from: some sodium chloride was lost when it was transferred / left in the evaporating basin or some solution spat out / splashed when heated or some product lost during separation ignore 'human error' / 'not measured accurately' / 'reaction too slow'(1)
Answer: (a) second box (b) 74.0% (c) e.g. some sodium chloride was lost when transferring it or spat out during evaporation
Examiner tip: Percentage yield is always actual ÷ theoretical × 100 — dividing 5.85 by 4.33 gives over 100%, which should tell you it is upside-down.

3Grade 4–5Triple only5 marks

Ethanol can be made in two different ways.

Fermentation of glucose: $$\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$$

Hydration of ethene: $$\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$$

Relative formula masses ($M_r$): $\text{C}_6\text{H}_{12}\text{O}_6$ = 180, $\text{C}_2\text{H}_5\text{OH}$ = 46, $\text{CO}_2$ = 44

(a) Calculate the percentage atom economy for making ethanol by fermentation. [3]

(b) The atom economy for making ethanol by hydration of ethene is 100%. Explain why. [1]

(a) $2 \times 46 = 92$(1)
(a) $\dfrac{92}{180} \times 100$ allow $\dfrac{92}{92 + 88} \times 100$(1)
(a) 51.1 (%) allow 51 (%) or 51.11… ; 25.6 (%) from $\frac{46}{180} \times 100$ scores 2 marks (second and third marks by ecf)(1)
(b) ethanol is the only product or all the atoms in the reactants end up in the desired product or there are no waste products / by-products(1)
(c) any one from: less waste is produced or fewer raw materials / resources are needed (so it is more sustainable) or it is more economical / cheaper ignore 'faster' / 'more product' unqualified(1)
Answer: (a) 51.1% (b) ethanol is the only product, so all the reactant atoms end up in it (c) e.g. less waste / more sustainable use of resources
Examiner tip: In (a) the equation makes 2 molecules of ethanol — use 2 × 46 = 92 on top, or you will get half the right answer.

4Grade 6–7HigherTriple only6 marks

Ammonia is made from nitrogen and hydrogen.

$$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$$

Relative formula masses ($M_r$): $\text{N}_2$ = 28, $\text{NH}_3$ = 17

In one pass through a reactor, 56 kg of nitrogen reacted with excess hydrogen. Only 10.2 kg of ammonia was produced.

(a) Calculate the percentage yield of ammonia. [4]

(b) A student says:

“The atom economy for making ammonia is 100%, so the percentage yield should also be 100%.”

Explain why the student is wrong. [2]

(a) moles of $\text{N}_2 = \dfrac{56}{28} = 2$ (kmol) or 28 kg of $\text{N}_2$ → 34 kg of $\text{NH}_3$(1)
(a) maximum theoretical mass of $\text{NH}_3 = 2 \times 2 \times 17 = 68$ (kg) allow $\dfrac{56}{28} \times 34$(1)
(a) $\dfrac{10.2}{68} \times 100$ allow ecf from an incorrect theoretical mass(1)
(a) 15 (%) correct answer with no working scores 4; 30 (%) from 10.2/34 (1 : 2 ratio not used) scores 3 marks (M2 not awarded)(1)
(b) atom economy is 100% because ammonia is the only product / no atoms are wasted (it is a theoretical value from the equation)(1)
(b) but the reaction is reversible so it does not go to completion or some ammonia breaks back down into nitrogen and hydrogen or it reaches equilibrium allow some ammonia may be lost when it is separated(1)
Answer: (a) 15% (b) atom economy only depends on the equation (ammonia is the only product); the yield is low because the reaction is reversible and does not go to completion
Examiner tip: Atom economy and percentage yield measure different things: atom economy comes from the equation, yield comes from what was actually collected.

5Grade 6–7HigherTriple only7 marks

A student heated magnesium ribbon in a crucible with a lid. The student lifted the lid from time to time to let air in.

$$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$

The student repeated the experiment with different masses of magnesium. The table shows the results.

Mass of magnesium in g	0.24	0.36	0.48	0.60	0.72
Mass of magnesium oxide made in g	0.34	0.51	0.58	0.85	1.02

The graph shows these results. The line shows the maximum theoretical mass of magnesium oxide for each mass of magnesium.

(a) Use the graph to find the maximum theoretical mass of magnesium oxide that can be made from 0.60 g of magnesium. [1]

(b) Calculate the percentage yield when 0.60 g of magnesium was used. [2]

Give the mass of magnesium used for the anomalous result. Suggest one reason for this result. [2]

(d) Relative atomic masses ($A_r$): O = 16, Mg = 24

Show, using the equation, that the maximum theoretical mass of magnesium oxide made from 0.60 g of magnesium is 1.00 g. [2]

(a) 1.00 g allow 0.99–1.01 g(1)
(b) $\dfrac{0.85}{1.00} \times 100$ allow ecf from (a)(1)
(b) 85 (%)(1)
(c) 0.48 (g)(1)
(c) any one from: some magnesium oxide (smoke) escaped when the lid was lifted or not all the magnesium reacted / not heated for long enough / not enough air let in or some product was spilled / lost when transferring ignore 'human error' / 'measured wrongly' unqualified(1)
(d) moles of Mg $= \dfrac{0.60}{24} = 0.025$(1)
(d) moles of MgO = 0.025 (1 : 1 ratio) so mass $= 0.025 \times 40 = 1.00$ g allow $\dfrac{0.60}{48} \times 80 = 1.00$ g or $\dfrac{0.60}{24} \times 40$ for 2 marks(1)
Answer: (a) 1.00 g (b) 85% (c) 0.48 g; e.g. some magnesium oxide escaped when the lid was lifted (d) 0.60 ÷ 24 = 0.025 mol Mg → 0.025 mol MgO × 40 = 1.00 g
Examiner tip: Read the theoretical mass from the line, not from the crosses — the crosses are what was actually made.

6Grade 8–9HigherTriple only6 marks

Aspirin is made by reacting salicylic acid with ethanoic anhydride.

$$\text{C}_7\text{H}_6\text{O}_3 + \text{C}_4\text{H}_6\text{O}_3 \rightarrow \text{C}_9\text{H}_8\text{O}_4 + \text{CH}_3\text{COOH}$$

Substance	Formula	$M_r$
salicylic acid	$\text{C}_7\text{H}_6\text{O}_3$	138
ethanoic anhydride	$\text{C}_4\text{H}_6\text{O}_3$	102
aspirin	$\text{C}_9\text{H}_8\text{O}_4$	180
ethanoic acid	$\text{CH}_3\text{COOH}$	60

(a) Calculate the percentage atom economy for making aspirin. [2]

(b) A company needs to make 900 g of aspirin. The percentage yield of the process is 80%.

Calculate the mass of salicylic acid the company must use. Ethanoic anhydride is in excess. [4]

(a) $\dfrac{180}{138 + 102} \times 100$ or $\dfrac{180}{240} \times 100$(1)
(a) 75 (%)(1)
(b) theoretical mass of aspirin needed $= 900 \times \dfrac{100}{80} = 1125$ (g)(1)
(b) moles of aspirin $= \dfrac{1125}{180} = 6.25$ allow ecf(1)
(b) moles of salicylic acid = 6.25 (1 : 1 ratio) so mass $= 6.25 \times 138$ allow $1125 \times \dfrac{138}{180}$ for the second and third marks in (b)(1)
(b) 862.5 (g) allow 863 / 860 (g); 552 (g) from $900 \times 0.80$ (yield applied the wrong way) or 690 (g) (yield ignored) scores 3 marks (first mark in (b) not awarded; rest by ecf); correct answer with no working scores 4; allow yield applied at the end ($900 \div 180 = 5$ mol → $5 \times 138 = 690$ g → $690 \div 0.80 = 862.5$ g) for 4 marks(1)
Answer: (a) 75% (b) 862.5 g (≈ 863 g)
Examiner tip: If the yield is only 80%, you need MORE starting material than for 100% — divide by 0.80, don't multiply.

7Grade 6–7HigherTriple only6 marks

Hydrogen can be produced by two different methods.

Method A: reacting methane with steam: $$\text{CH}_4 + 2\text{H}_2\text{O} \rightleftharpoons \text{CO}_2 + 4\text{H}_2$$

Method B: electrolysis of water: $$2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$$

	Method A	Method B
Raw materials	methane (from natural gas) and water	water
Atom economy for hydrogen	15.4%	11.1%
Percentage yield	75%	98%
Conditions	850 °C, nickel catalyst	room temperature, electric current
Rate of production	fast	slow
By-product	carbon dioxide	oxygen
Energy source	burning natural gas	electricity (can be from wind or solar power)

Evaluate the two methods for producing hydrogen.

Use the information in the table and your own knowledge. Give a justified conclusion. [6]

Method A has a (slightly) higher atom economy, so a greater proportion of the reactant atoms ends up as hydrogen / less waste(1)
Method B has a higher percentage yield, so more of the maximum possible hydrogen is collected(1)
Method A has a lower yield because the reaction is reversible / reaches equilibrium (does not go to completion)(1)
Method A produces hydrogen faster, which suits large-scale production / gives more product per hour(1)
Method A needs a high temperature, so it uses a lot of energy (from burning natural gas, a fossil fuel)(1)
Method A uses methane / natural gas, which is a finite (non-renewable) resource; Method B uses water, which is plentiful / renewable(1)
the by-product of A, carbon dioxide, is a greenhouse gas that contributes to global warming / climate change(1)
the by-product of B, oxygen, is useful / can be sold (e.g. for hospitals or steel-making), so the low atom economy matters less(1)
Method B can use electricity from renewable sources so it can produce hydrogen with little or no carbon dioxide, but electricity is expensive(1)
justified conclusion, e.g. Method B is more sustainable because it uses a renewable raw material, has a useful by-product and need not release CO₂, even though it is slower and has a lower atom economy or Method A is better for cost / speed at present — either conclusion credited if justified by comparison of the data(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed evaluation that compares the two methods across several factors, including atom economy or yield and at least one sustainability factor (raw materials, energy or by-products), explaining why each factor matters. A conclusion is reached and is justified by the comparison.
Level 2	3–4	Some comparison of the two methods with at least one advantage and one disadvantage linked to a reason, for example a higher yield or a useful by-product. A conclusion may be given but is only partly justified.
Level 1	1–2	Simple statements taken from the table about one or both methods, with little or no explanation or comparison. Any conclusion is not supported.
No relevant content	0	No relevant content.

Answer: Evaluation weighing atom economy (A higher), yield (B higher; A reversible), rate (A faster), energy and raw materials (A finite fossil fuel, high temperature; B water, renewable electricity possible), by-products (CO₂ greenhouse gas vs useful oxygen), with a justified conclusion.
Examiner tip: Do not just copy the table — for each row say WHICH method is better and WHY that matters, then finish with a conclusion that weighs them up.

8Grade 4–5Triple only5 marks

A student made copper sulfate crystals by adding excess copper oxide to warm dilute sulfuric acid, filtering the mixture, then heating the filtrate and leaving it to crystallise.

The maximum theoretical mass of copper sulfate crystals was 12.5 g.

The student weighed the crystals in an evaporating basin.

Mass of empty evaporating basin in g	45.27
Mass of evaporating basin + dry crystals in g	54.02

(a) Calculate the percentage yield of copper sulfate crystals. [3]

(b) Give two stages in the method where some copper sulfate could have been lost. [2]

(a) mass of crystals $= 54.02 - 45.27 = 8.75$ (g)(1)
(a) $\dfrac{8.75}{12.5} \times 100$ allow ecf from an incorrect mass of crystals(1)
(a) 70 (%) allow 70.0 (%); correct answer with no working scores 3(1)
(b) any two from: some solution left in the filter paper / beaker / funnel (when filtering) · solution spat out of the basin while heating · some copper sulfate stayed dissolved in the solution left after crystallising · crystals left behind / stuck to the basin or filter paper when they were removed / dried(1)
(b) second correct stage ignore 'not all the acid reacted' (copper oxide was in excess) ignore 'human error' / 'spillage' unqualified by a stage(1)
Answer: (a) 70% (b) e.g. solution left on the filter paper; some copper sulfate remained dissolved after crystallisation
Examiner tip: Subtract the mass of the empty basin first — using 54.02 g as the actual mass gives a yield over 100%.

9Grade 4–5Triple only5 marks

A company makes titanium by reacting titanium(IV) chloride with sodium.

$$\text{TiCl}_4 + 4\text{Na} \rightarrow \text{Ti} + 4\text{NaCl}$$

Relative atomic masses ($A_r$): Na = 23, Cl = 35.5, Ti = 48

(a) Calculate the percentage atom economy for making titanium in this reaction. Give your answer to 3 significant figures. [4]

(b) The atom economy for this reaction cannot be changed. Suggest how the company could make the overall process less wasteful. [1]

(a) $M_r$ of $\text{TiCl}_4 = 48 + (4 \times 35.5) = 190$(1)
(a) total $M_r$ of reactants $= 190 + (4 \times 23) = 282$ allow total $M_r$ of products $= 48 + (4 \times 58.5) = 282$(1)
(a) $\dfrac{48}{282} \times 100$ allow ecf from an incorrect total(1)
(a) 17.0 (%) final answer must be given to 3 s.f.; an unrounded value in working (17.02…) followed by 17.0 is fine; a final answer of 17 (%) or 17.02 (%) scores 3 marks; 22.5 (%) from using only one Na ($48 \div 213$) scores 3 marks (M2 not awarded)(1)
(b) use / sell the sodium chloride (by-product) or electrolyse the sodium chloride to recover sodium (and chlorine) to reuse ignore 'recycle the titanium' / 'increase the yield'(1)
Answer: (a) 17.0% (b) e.g. sell the sodium chloride by-product or electrolyse it to recover sodium for reuse
Examiner tip: Multiply by the balancing numbers: 4 Na means 4 × 23 = 92 in the bottom of the fraction, not 23.

10Grade 8–9HigherTriple only7 marks

Iron can be made by reacting aluminium with iron(III) oxide.

$$2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$$

Relative atomic masses ($A_r$): O = 16, Al = 27, Fe = 56

(a) Calculate the percentage atom economy for making iron in this reaction. Give your answer to 3 significant figures. [2]

(b) A mixture of 5.4 g of aluminium and 20.0 g of iron(III) oxide was reacted. 9.8 g of iron was produced.

Calculate the percentage yield of iron. [5]

(a) $\dfrac{2 \times 56}{(2 \times 27) + 160} \times 100$ or $\dfrac{112}{214} \times 100$(1)
(a) 52.3 (%) final answer must be given to 3 s.f.; a final answer of 52 (%) or 52.34 (%) scores 1 mark(1)
(b) moles of Al $= \dfrac{5.4}{27} = 0.20$ and moles of $\text{Fe}_2\text{O}_3 = \dfrac{20.0}{160} = 0.125$(1)
(b) aluminium is the limiting reactant, because 0.20 mol Al reacts with only 0.10 mol $\text{Fe}_2\text{O}_3$ (so $\text{Fe}_2\text{O}_3$ is in excess)(1)
(b) moles of Fe = 0.20 (from Al, 1 : 1 ratio) allow ecf from wrong limiting reactant: 0.25 mol(1)
(b) maximum theoretical mass of Fe $= 0.20 \times 56 = 11.2$ (g) allow ecf: 0.25 × 56 = 14.0 g(1)
(b) $\dfrac{9.8}{11.2} \times 100 = 87.5$ (%) allow 88 (%); 70 (%) from using $\text{Fe}_2\text{O}_3$ as the limiting reactant scores 4 marks (limiting-reactant mark in (b) not awarded)(1)
Answer: (a) 52.3% (b) Al is limiting; theoretical Fe = 11.2 g; percentage yield = 87.5%
Examiner tip: Work out which reactant runs out first before finding the theoretical mass — using the iron(III) oxide here gives 14.0 g and the wrong yield.

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