Chemistry • Year 11 • Module 2 • Lesson 14
Percentage Yield & Percentage Purity
Apply purity and yield calculations to real industrial data, interpret graphs, compare processes, and practise the full purity–stoichiometry–yield sequence.
1. Interpret industrial process data
A chemical plant runs three reactions. The table below records the key data for each. 9 marks
| Reaction | Reactant sample (g) | % purity of sample | Theoretical yield (g) | Actual yield (g) | % yield |
|---|---|---|---|---|---|
| A: CaCO3 → CaO + CO2 | 50.0 | 90.0% | calculate | 22.7 | |
| B: Fe2O3 + 3CO → 2Fe + 3CO2 | 100.0 | 80.0% | calculate | 38.0 | |
| C: Mg + 2HCl → MgCl2 + H2 | 25.0 | 100% (pure) | calculate | 91.5 |
Molar masses: Ca=40.078, C=12.011, O=15.999, Fe=55.845, Mg=24.305, Cl=35.453.
1.1 For Reaction A: apply the purity correction, then complete the four-step stoichiometry to find the theoretical yield of CaO. Then calculate % yield. Show full working. 3 marks
1.2 For Reaction B: apply purity, complete stoichiometry, find theoretical yield of Fe, then calculate % yield. 3 marks
1.3 For Reaction C (pure Mg): calculate theoretical yield of MgCl2, then calculate % yield. 2 marks
1.4 Which reaction has the highest percentage yield? Suggest one reason why. 1 mark
2. Interpret a yield vs temperature graph — contact process
The graph below shows the equilibrium percentage yield of SO3 in the contact process (2SO2 + O2 ⇄ 2SO3) plotted against temperature at a fixed pressure. The industrial operating temperature is marked. 7 marks
Figure 2. Equilibrium % yield of SO3 from 2SO2 + O2 ⇄ 2SO3 as a function of temperature at constant pressure. Illustrative data based on industrial contact-process parameters.
2.1 Describe the trend shown in the graph. 2 marks
2.2 Read the approximate equilibrium yield at 450 °C. Then calculate the actual mass of SO3 expected from 100 g of pure SO2 reacting at 450 °C. (S=32.06, O=15.999) 3 marks
2.3 The Contact Process reaction is reversible: 2SO2 + O2 ⇄ 2SO3. A student claims: “The graph shows we can never achieve 100% yield even at 400 °C.” Using your knowledge of reversible reactions from the lesson, explain why the equilibrium % yield is less than 100% at all temperatures, and state what the graph tells you about how temperature affects the equilibrium position for this reaction. 2 marks
3. Predict and justify — the Haber process for ammonia
The Haber process: N2(g) + 3H2(g) ⇄ 2NH3(g). A chemical plant uses a 90.0% pure N2 feed (contaminated with Ar gas) and achieves a 25.0% yield per pass through the reactor. The plant starts with 56.0 kg of N2 feed. 6 marks
3.1 Calculate the mass of pure N2 available. Then calculate the theoretical yield of NH3. Then calculate the actual yield of NH3 per pass. Show the full sequence. (N=14.007, H=1.008) 4 marks
3.2 Predict what would happen to the actual yield per pass if the purity of the N2 feed were improved to 99.0% while the % yield remained 25.0%. Justify your prediction with a calculation. 2 marks
4. Compare percentage purity and percentage yield across five features
Complete the two-column table. For each feature, write a concise description that contrasts the two corrections. 10 marks (1 per cell)
| Feature | Percentage purity | Percentage yield |
|---|---|---|
| What it measures | ||
| Applied to reactant or product? | ||
| When to apply it in the calculation sequence | ||
| Formula | ||
| Australian context example |
Q1.1 — Reaction A: CaCO3 → CaO (3 marks)
m(pure CaCO3) = 50.0 × 0.900 = 45.0 g. n(CaCO3) = 45.0 ÷ 100.09 = 0.4496 mol. Ratio 1:1; n(CaO) = 0.4496 mol. MM(CaO) = 56.077; theoretical m(CaO) = 0.4496 × 56.077 = 25.2 g. % yield = (22.7 ÷ 25.2) × 100 = 90.1%. [1 purity, 1 stoichiometry, 1 % yield]
Q1.2 — Reaction B: Fe2O3 → Fe (3 marks)
m(pure Fe2O3) = 100.0 × 0.800 = 80.0 g. MM(Fe2O3) = 159.69; n = 80.0 ÷ 159.69 = 0.5010 mol. Ratio 1:2; n(Fe) = 1.002 mol. MM(Fe) = 55.845; theoretical = 1.002 × 55.845 = 55.9 g. % yield = (38.0 ÷ 55.9) × 100 = 68.0%. [1 purity, 1 stoichiometry, 1 % yield]
Q1.3 — Reaction C: Mg → MgCl2 (2 marks)
n(Mg) = 25.0 ÷ 24.305 = 1.028 mol. Ratio 1:1; n(MgCl2) = 1.028 mol. MM(MgCl2) = 95.211; theoretical = 1.028 × 95.211 = 97.9 g. % yield = (91.5 ÷ 97.9) × 100 = 93.5%. [1 stoichiometry, 1 % yield]
Q1.4 — Highest yield (1 mark)
Reaction C (93.5%). Possible reason: pure reactant means no moles are diverted to impurity reactions; the reaction (metal + acid) may also proceed with few side reactions under controlled conditions.
Q2.1 — Graph trend (2 marks)
As temperature increases from 400 °C to 700 °C, the equilibrium % yield of SO3 decreases from approximately 99% to 35% [trend]. The decrease is approximately linear over this range, with yield falling by about 2% per 10 °C rise [shape/rate detail].
Q2.2 — Actual yield at 450 °C (3 marks)
Reading: % yield at 450 °C ≈ 97%. n(SO2) = 100 ÷ 64.058 = 1.561 mol. Ratio SO2:SO3 = 1:1; n(SO3) = 1.561 mol. MM(SO3) = 80.057; theoretical m(SO3) = 1.561 × 80.057 = 124.9 g. Actual = 124.9 × (97 ÷ 100) = 121.2 g. [1 read, 1 theoretical, 1 actual]
Q2.3 — Reversible reactions and equilibrium yield (2 marks)
A reversible reaction reaches equilibrium before all the reactants are consumed; at equilibrium, both forward and reverse reactions are occurring simultaneously, so the conversion is never complete and yield is always below 100% [1]. The graph shows that increasing temperature decreases the equilibrium % yield — meaning higher temperatures shift the equilibrium position towards the reactants (away from SO3), producing less product at equilibrium [1].
Q3.1 — Haber process full sequence (4 marks)
m(pure N2) = 56 000 × 0.900 = 50 400 g [P]. n(N2) = 50 400 ÷ 28.014 = 1799 mol [Step 2]. Ratio N2:NH3 = 1:2; n(NH3) = 3598 mol [Step 3]. MM(NH3) = 17.031; theoretical = 3598 × 17.031 = 61 293 g ≈ 61.3 kg [Step 4]. Actual = 61 293 × 0.250 = 15 323 g ≈ 15.3 kg per pass. [1 purity, 1 stoich, 1 theoretical, 1 actual]
Q3.2 — Effect of improved purity (2 marks)
Prediction: actual yield per pass increases because more pure N2 is available to react [1]. At 99.0% purity: m(pure) = 56 000 × 0.990 = 55 440 g; n(N2) = 1979 mol; n(NH3) = 3958 mol; theoretical = 67 433 g; actual = 67 433 × 0.250 = 16 858 g ≈ 16.9 kg — an increase of ~1.6 kg compared to 90% purity, confirming purity improvement raises output [1].
Q4 — Compare and contrast table
What it measures: Purity — the fraction of a reactant sample that is the desired substance. Yield — the efficiency of a reaction (how much product was obtained versus the maximum possible). Applied to reactant or product? Purity — reactant (starting material). Yield — product (what is collected at the end). When to apply: Purity — BEFORE stoichiometry (Step P, before Step 2). Yield — AFTER stoichiometry (once theoretical yield is known). Formula: Purity — % purity = (pure mass ÷ sample mass) × 100; rearranged: m(pure) = m(sample) × (% ÷ 100). Yield — % yield = (actual ÷ theoretical) × 100. Australian context example: Purity — iron ore at Port Kembla: ore may contain silicates and other minerals alongside Fe2O3; the percentage of Fe2O3 is the purity. Yield — the fraction of theoretical iron output actually tapped from the blast furnace; losses occur due to incomplete reduction and slag carryover.