Percentage Yield & Purity
Section: 3. Stoichiometry | Syllabus: Cambridge IGCSE Chemistry (0620)
Percentage Yield
The theoretical yield is the maximum mass of product that stoichiometry predicts, assuming every reactant converts perfectly to product. The actual yield is the mass actually obtained in an experiment - and is always less than or equal to the theoretical yield.
Percentage yield = (actual yield ÷ theoretical yield) × 100
Real yields are usually below 100% because of:
- Incomplete or reversible reactions
- Unwanted side reactions
- Product lost during transfer, filtering, or purification
- Impure starting reactants
Small amounts of product are unavoidably lost during transfer and purification steps such as filtration
Worked Example: Recognising the Percentage Yield Formula
- Question: a student calculates that 10 g of magnesium oxide should be produced when a sample of magnesium is burned. After the reaction, the student determines the percentage yield. Which formula is used? [1]
- Answer: percentage yield = (mass of MgO formed ÷ mass of MgO expected) × 100 - i.e. actual yield ÷ theoretical yield × 100 [1]
Worked Example: Calculating Percentage Yield
- Question: 10 g of calcium carbonate decomposes on heating to produce 4.8 g of calcium oxide. Calculate the percentage yield. CaCO₃ → CaO + CO₂ (Ar: Ca = 40, C = 12, O = 16) [4]
- Answer: moles CaCO₃ = 10 ÷ 100 = 0.1 mol [1]; ratio 1:1, so moles CaO = 0.1 mol; theoretical yield = 0.1 × 56 = 5.6 g [1]; percentage yield = (4.8 ÷ 5.6) × 100 = 85.7% [2]
Percentage Purity
Purity is the proportion, by mass, of the desired substance within a sample - impure samples contain other, unwanted substances mixed in.
Percentage purity = (mass of pure substance ÷ total mass of sample) × 100
Impure reactants contain less of the actual reacting substance than their total mass suggests, so they produce a lower yield than an equal mass of the pure substance would.
Worked Example: Using Purity in a Calculation
- Question: what mass of 80% pure magnesium is needed to produce 4.0 g of magnesium oxide? 2Mg + O₂ → 2MgO (Ar: Mg = 24, O = 16) [4]
- Answer: moles MgO = 4.0 ÷ 40 = 0.1 mol; ratio 1:1, so moles Mg needed = 0.1 mol [1]; mass of pure Mg needed = 0.1 × 24 = 2.4 g [1]; since the sample is only 80% pure, mass of sample = (2.4 ÷ 80) × 100 = 3.0 g [2]
Common Mistakes
- ❌ Getting a percentage yield above 100%
- ✓ this is impossible - actual yield can never exceed theoretical yield, so a result over 100% signals an arithmetic error, most often the theoretical and actual yields swapped in the formula
- ❌ Not calculating the theoretical yield first
- ✓ percentage yield always needs a stoichiometric calculation (mass → moles → ratio → moles → mass) to find the theoretical yield before the percentage can be worked out
- ❌ Using the mass of an impure sample as if it were the mass of pure reactant
- ✓ when a reactant's purity is less than 100%, first find the mass of pure substance actually present before using it in a reacting-mass calculation
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