Percentage Yield & Percentage Purity
Real reactions never give 100% yield, and real samples are rarely pure. Percentage yield tells you how efficient a reaction was. Percentage purity tells you how much of a reactant sample is actually usable. They're different corrections applied at different stages.
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Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Applied BEFORE stoichiometry — on the reactant
⚠️ Purity is a property of the starting material — apply it at the beginning. Yield is a measure of how the reaction performed — calculate it at the end. The order matters every time.
Key facts
- % yield = (actual ÷ theoretical) × 100
- % purity = (pure mass ÷ sample mass) × 100
- Purity applied before stoichiometry (on reactant)
- Yield calculated after stoichiometry (on product)
- Both expressed as a percentage
Concepts
- Why actual yield is always ≤ theoretical yield
- Why purity reduces available moles before calculation
- Reasons yield falls short in practice
- Industrial significance of optimising both
Skills
- Calculate % yield from actual and theoretical masses
- Apply purity correction to find m(pure) before stoichiometry
- Solve combined problems: purity correction then yield
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.
Why Is Yield Less Than 100%?
Side reactions — reactants are consumed forming unwanted by-products.
Incomplete reaction — not all reactant molecules collide successfully.
Product loss — during transfer, filtration, evaporation, or purification steps.
Impure reactants — impurities reduce the effective amount of reactant available.
% yield = (actual yield ÷ theoretical yield) × 100; always ≤ 100%. Theoretical yield comes from stoichiometry (using LR moles); actual yield is measured experimentally. Yield < 100% due to equilibrium, side reactions, product loss, or impure reactants.
Pause — copy the highlighted formula into your book before moving on.
Did you get this? True or false: a percentage yield greater than 100% is possible if the experiment is very efficient.
We just saw that % yield compares actual to theoretical product after a reaction. That raises a question: what if the starting reactant itself isn't pure — how do you account for impurities before calculating yield? This card answers it → by applying percentage purity before stoichiometry.
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.
Rearranged to find usable mass: m(pure) = m(sample) × (% purity ÷ 100)
Example: 50.0 g sample of NaOH, 96.0% pure.
m(pure NaOH) = 50.0 × (96.0 ÷ 100) = 48.0 g
Only 48.0 g is available to react — use this in Step 2 of stoichiometry.
When to Apply Each Correction
- Given an impure reactant sample → find pure mass
- Then proceed with Steps 2–4 of stoichiometry
- m(pure) = m(sample) × (% purity ÷ 100)
- Given actual mass of product collected
- Compare to theoretical yield from stoichiometry
- % yield = (actual ÷ theoretical) × 100
% purity = (mpure ÷ msample) × 100; m(pure) = m(sample) × (% purity ÷ 100). Apply purity before stoichiometry — use m(pure) in Step 2, not m(sample). Combined order: purity → moles → mole ratio → theoretical yield → % yield. Purity is a property of the reactant; yield is efficiency of the reaction.
Add the highlighted rule to your notes before the check below.
Quick check: When should you apply the % purity correction in a combined problem?
Fill the blanks: drag each token into the matching blank.
Purity is a property of the ___ and is applied ___ stoichiometry. Yield is a property of the ___ and is calculated ___ stoichiometry.
Worked examples · reveal as you go
In the reaction 2SO₂ + O₂ → 2SO₃, 8.00 g of SO₃ is theoretically possible but only 6.40 g is collected. Calculate the percentage yield.
CaCO₃ → CaO + CO₂. 20.0 g of CaCO₃ is heated and 9.80 g of CaO is collected. Calculate the percentage yield of CaO. (Ca=40.078, C=12.011, O=15.999)
MM(CaO) = 56.077; m(CaO) theoretical = 0.1998 × 56.077 = 11.2 g
An impure sample of CaCO₃ (85.0% pure) is heated: CaCO₃ → CaO + CO₂. What mass of CO₂ is produced from 20.0 g of the sample? (Ca=40.078, C=12.011, O=15.999)
An 88.0% pure sample of iron ore (Fe₂O₃) is reduced: Fe₂O₃ + 3CO → 2Fe + 3CO₂. 50.0 g of the ore is used, and 24.0 g of iron is collected. Calculate (a) the theoretical yield of iron, (b) the percentage yield. (Fe=55.845, O=15.999)
Lock-in task: Why is it impossible for a real chemistry experiment to give a percentage yield greater than 100%? Give one practical reason a reported value > 100% would mean the experiment was flawed.
A student reacts 5.00 g of CaCO₃ with excess HCl and collects 1.85 g of CO₂. The theoretical yield (from stoichiometry) is 2.20 g. The student gets 84.1% yield. Predict the single most likely reason their yield was below 100% (think: where could CO₂ go?).
How close was your prediction?
Good calibration — you'd score the explanation mark in an exam.
Worth remembering: for any gas product, ask "how does it escape?" before listing other causes.
Common errors · the 3 traps that cost marks
Confusing % purity with % yield — applying them at the wrong stage
Purity is a property of the starting material — it must be applied at the very beginning to find how much pure reactant is available. Yield is a measure of how efficient the reaction was — it's calculated at the end by comparing actual to theoretical product. Students who apply yield to the reactant, or purity to the product, get nonsensical answers.
Fix: Ask yourself — "Is this about the reactant before the reaction (purity) or the product after the reaction (yield)?" Purity = before, yield = after.
Using sample mass instead of pure mass in stoichiometry
If 50 g of an 80% pure sample is used, the correct input for Step 2 (n = m ÷ MM) is 40 g — not 50 g. Using 50 g ignores the 20% impurity and overestimates the moles available. The answer will be 25% too high.
Fix: The purity step produces m(pure). This — not the sample mass — goes into Step 2. Write "m(pure) = ..." explicitly before calculating moles.
Claiming % yield can exceed 100%
If a calculated % yield comes out above 100%, there is an error somewhere. In a real reaction, you cannot produce more product than stoichiometry allows. Likely cause: theoretical and actual were swapped, or sample mass used instead of pure mass.
Fix: If % yield > 100%, stop and check. Likely causes: sample mass used instead of pure mass for theoretical yield, or actual and theoretical swapped in the formula.
Quick-fire practice · 5 reps +2 XP per reveal
Theoretical yield of NaCl = 11.7 g. Actual yield collected = 9.36 g. Calculate % yield.
A 40.0 g sample of NaOH is 95.0% pure. What mass of pure NaOH is available for reaction?
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)
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?
A 92.0% pure CuO sample (20.0 g) is reduced with H₂: CuO + H₂ → Cu + H₂O. The reaction has 88.4% yield. Calculate the actual mass of Cu collected. (Cu = 63.546, O = 15.999)
Earlier you were asked: Why might a student collect less iron from iron oxide ore than predicted — one reason about the ore, one about the reaction?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q1. 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)
Q2. 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)
Q3. 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.
Q4. 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.
Q5. 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.
📖 Comprehensive answers (click to reveal)
Multiple choice — drill bank
MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.
Short answer model answers
Q1 (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 g
Q2 (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.
Q3 (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.
Q4 (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 g
Process 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%).
Q5 (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.
Five timed questions on percentage yield and percentage purity. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaClimb platforms, hit checkpoints, and answer yield/purity questions. Quick recall, lighter than the boss.
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