Real reactions never give 100% yield, and real samples are rarely pure. Percentage yield tells you how efficient a reaction was after it happened. Percentage purity tells you how much of a reactant sample is actually usable before you start. These are different corrections applied at different stages — and confusing them is one of the most reliable ways to lose marks.
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A chemistry student reacts iron oxide ore in a blast furnace to extract iron, but collects less iron than their calculations predicted. What are two different reasons this could happen — one related to the quality of the starting ore, and one related to what happens during the reaction itself?
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📚 Core Content
Wrong: The mole is a measure of mass.
Right: The mole is a measure of amount of substance; one mole contains Avogadro's number of particles.
The theoretical yield is the maximum mass of product that stoichiometry predicts. In practice, the actual yield is always less than — or at best equal to — the theoretical yield. The percentage yield expresses how much of the possible product was actually obtained.
Common reasons yield falls short:
A sample of a substance often contains impurities — other compounds mixed in. The percentage purity tells you what fraction of the sample is actually the substance of interest. You must account for purity before performing stoichiometric calculations, because you can only react the pure portion.
Given an impure reactant sample → find pure mass → then proceed with Steps 2–4.
Given actual mass of product collected → compare to theoretical yield from stoichiometry.
🧮 Worked Examples
🧪 Activities
1 Theoretical yield of NaCl = 11.7 g. Actual yield collected = 9.36 g. Calculate % yield.
2 A 40.0 g sample of NaOH is 95.0% pure. What mass of pure NaOH is available for reaction?
3 A 90.0% pure sample of CaCO₃ (25.0 g) is heated: CaCO₃ → CaO + CO₂. Calculate the expected mass of CaO produced. (Ca=40.078, C=12.011, O=15.999)
4 In WE2, theoretical yield of CaO = 11.2 g and actual = 9.80 g. A student calculates % yield = (11.2 ÷ 9.80) × 100 = 114%. What error did they make?
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| Reaction | Sample data | Actual yield collected |
|---|---|---|
| 2SO₂ + O₂ → 2SO₃ | 32.0 g SO₂, 96.0% pure | 36.0 g SO₃ |
| N₂ + 3H₂ → 2NH₃ | 14.0 g N₂, 100% pure 8.00 g H₂, 100% pure |
14.5 g NH₃ |
| CuO + H₂ → Cu + H₂O | 20.0 g CuO, 92.0% pure | 13.0 g Cu |
S=32.06, O=15.999, N=14.007, H=1.008, Cu=63.546
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At the start of this lesson, you thought about why a student might collect less iron from an iron oxide ore than predicted — one reason related to the ore quality, and one related to the reaction itself.
Both reasons have names in chemistry: percentage purity accounts for impurities in the starting ore (only the pure Fe₂O₃ fraction contributes moles), and percentage yield accounts for losses during the reaction (incomplete conversion, side reactions, product lost in transfer). These two corrections are applied at opposite ends of the stoichiometric calculation — purity before, yield after.
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✍️ Short Answer
6. A 78.0% pure sample of Fe₂O₃ (40.0 g) is reduced in a blast furnace: Fe₂O₃ + 3CO → 2Fe + 3CO₂. The reaction has a 92.0% yield. (a) Calculate the theoretical yield of iron. (b) Calculate the actual mass of iron produced. (Fe=55.845, O=15.999) 5 MARKS
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7. A student reacts 20.0 g of a 95.0% pure sample of Mg with excess HCl: Mg + 2HCl → MgCl₂ + H₂. They collect 16.5 g of MgCl₂. (a) Calculate the theoretical yield of MgCl₂. (b) Calculate the percentage yield. (Mg=24.305, Cl=35.453) 5 MARKS
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8. Explain the difference between percentage purity and percentage yield. In your answer, state what each measures and at which stage of a stoichiometric calculation each is applied. 4 MARKS
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9. Two industrial processes both produce sulfuric acid (H₂SO₄) via SO₃. Process A starts with 96.0% pure SO₂ (50.0 g sample) and achieves 85.0% yield. Process B starts with 100% pure SO₂ (50.0 g) but achieves only 70.0% yield. The reaction step is SO₃ + H₂O → H₂SO₄ where SO₃ comes from SO₂. Using only the data given, calculate which process produces more H₂SO₄. (S=32.06, O=15.999, H=1.008) Note: assume 1 mol SO₂ → 1 mol SO₃ → 1 mol H₂SO₄. 6 MARKS
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10. A chemical engineer argues: "For industrial purposes, optimising percentage yield is always more important than optimising percentage purity of the starting material." Using chemical reasoning, evaluate this claim. In your response, consider: (a) how each factor affects the final product quantity, and (b) a scenario where improving purity of the reactant might be the better investment. 5 MARKS
Write a structured response addressing both parts:
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Reaction 1 (2SO₂ + O₂ → 2SO₃):
m(pure SO₂) = 32.0 × 0.960 = 30.72 g; n = 30.72 ÷ 64.058 = 0.4796 mol Ratio SO₂:SO₃ = 1:1; n(SO₃) = 0.4796 mol; MM(SO₃) = 80.057 Theoretical = 0.4796 × 80.057 = 38.4 g; % yield = (36.0 ÷ 38.4) × 100 = 93.8%Reaction 2 (N₂ + 3H₂ → 2NH₃, Haber — identify LR first):
n(N₂) = 14.0 ÷ 28.014 = 0.4998; ÷1 = 0.4998 n(H₂) = 8.00 ÷ 2.016 = 3.968; ÷3 = 1.323 N₂ is LR (0.4998 < 1.323). Ratio N₂:NH₃ = 1:2; n(NH₃) = 0.4998 × 2 = 0.9996 mol; MM(NH₃) = 17.031 Theoretical = 0.9996 × 17.031 = 17.0 g; % yield = (14.5 ÷ 17.0) × 100 = 85.3%Reaction 3 (CuO + H₂ → Cu + H₂O):
m(pure CuO) = 20.0 × 0.920 = 18.4 g; MM(CuO) = 79.545; n = 18.4 ÷ 79.545 = 0.2313 mol Ratio 1:1; n(Cu) = 0.2313 mol; MM(Cu) = 63.546; theoretical = 0.2313 × 63.546 = 14.7 g % yield = (13.0 ÷ 14.7) × 100 = 88.4%1. C — 80.0%. % yield = (12.0 ÷ 15.0) × 100 = 80.0%. Option A swaps actual and theoretical.
2. B — 46.0 g. m(pure) = 50.0 × 0.920 = 46.0 g.
3. D — Purity before stoichiometry (corrects what goes in); yield after (measures what came out).
4. A — 10.6 g. m(pure CaCO₃) = 30.0 × 0.800 = 24.0 g; n = 24.0 ÷ 100.09 = 0.2398 mol; ratio 1:1; n(CO₂) = 0.2398; m = 0.2398 × 44.009 = 10.6 g. Option B uses full 30.0 g without purity correction.
5. B — % yield > 100% is impossible in a real reaction. The most common cause is swapping actual and theoretical in the formula.
6. C — The student's suggestion of adding more reactant may increase the absolute amount of product, but it does not address why yield is below 100%. If the loss is due to a reversible reaction (equilibrium), side reactions, or product lost during workup, simply using more reactant will not improve the process efficiency or yield percentage. The correct approach is to diagnose the cause of yield loss and address it specifically.
7. B — 31.6 g. Working backwards: we need 25.0 g of actual product. Actual = theoretical × yield ÷ 100, so theoretical needed = 25.0 ÷ 0.900 = 27.78 g. This theoretical comes from the pure NaCl in the sample: pure mass needed = 27.78 g (since it's a 1:1 conservation here — the NaCl itself is the product from the pure portion). Sample mass = 27.78 ÷ 0.880 = 31.6 g.
Q6 (5 marks):
(a) m(pure Fe₂O₃) = 40.0 × 0.780 = 31.2 g n(Fe₂O₃) = 31.2 ÷ 159.69 = 0.1954 mol; ratio 1:2; n(Fe) = 0.3908 mol Theoretical m(Fe) = 0.3908 × 55.845 = 21.8 g (b) Actual m(Fe) = 21.8 × (92.0 ÷ 100) = 20.1 gQ7 (5 marks):
(a) m(pure Mg) = 20.0 × 0.950 = 19.0 g; n(Mg) = 19.0 ÷ 24.305 = 0.7817 mol Ratio Mg:MgCl₂ = 1:1; n(MgCl₂) = 0.7817 mol MM(MgCl₂) = 24.305 + 2(35.453) = 95.211; m(theoretical) = 0.7817 × 95.211 = 74.4 g (b) % yield = (16.5 ÷ 74.4) × 100 = 22.2%Note: this low yield would suggest product was lost during collection — discuss with students.
Q8 (4 marks): Percentage purity measures what fraction of a reactant sample consists of the desired substance, expressed as a percentage. It is a property of the starting material and is applied before stoichiometric calculations — the pure mass (m(sample) × purity ÷ 100) is used as the input for converting to moles. Percentage yield measures how efficient a reaction was — specifically, what fraction of the theoretically possible product was actually obtained. It is calculated after the stoichiometric calculation, by dividing the experimentally measured mass by the theoretical yield and multiplying by 100. The two corrections act at opposite ends of the calculation and cannot be interchanged.
Q9 (6 marks):
Process A: m(pure SO₂) = 50.0 × 0.960 = 48.0 g; MM(SO₂) = 64.058; n = 48.0÷64.058 = 0.7493 mol ratio 1:1; n(H₂SO₄) = 0.7493 mol; MM(H₂SO₄) = 98.072; theoretical = 0.7493×98.072 = 73.5 g Actual A = 73.5 × 0.850 = 62.5 g Process B: m(pure SO₂) = 50.0 g; n = 50.0÷64.058 = 0.7806 mol Theoretical = 0.7806×98.072 = 76.6 g; Actual B = 76.6 × 0.700 = 53.6 gProcess A (62.5 g) produces more H₂SO₄ than Process B (53.6 g), even though Process A uses impure reagent, because its higher yield (85% vs 70%) more than compensates for the lower purity (96% vs 100%). This illustrates that both factors must be considered together.
Q10 (5 marks): (a) Both percentage purity and percentage yield reduce the final product quantity. Low purity reduces the effective number of moles entering the reaction; low yield reduces how much product is recovered. Mathematically: actual product = theoretical yield × (% yield ÷ 100), and theoretical yield is calculated from pure moles = sample mass × (% purity ÷ 100) ÷ MM. Both corrections multiply together to determine actual output — neither is inherently more important. (b) A scenario where improving purity may be the better investment: if impurities in the reactant are chemically reactive and cause side reactions with the product or catalyst, they may be the primary cause of yield loss. In this case, purifying the starting material would simultaneously improve both input moles and the yield. For example, iron impurities in an acid solution may catalyse decomposition of the product. The engineer's claim is too absolute — the relative importance of each factor depends on the specific reaction, the nature of the impurities, and the economics of purification vs. reactant cost.
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