Percentage Yield, Purity and Gas Volume Calculations
Three short calculation types that build directly on the mole. Each has a single formula and a single common error.
Part 1: Percentage yield
% yield = (ACTUAL yield ÷ THEORETICAL yield) × 100
- The theoretical yield is the mass your calculation predicts, assuming the reaction goes perfectly
- The actual yield is the mass you really obtained
Worked example. A reaction should produce 8.0 g of product; 6.4 g is obtained.
% yield = (6.4 ÷ 8.0) × 100 = 80%
To find the theoretical yield, do a normal reacting-mass calculation:
balanced equation → moles of the given substance → mole ratio → moles of product → mass
Why yield is never 100%
- The reaction may be REVERSIBLE and not go to completion
- Some product is LOST during transfer, filtering or purification
- SIDE REACTIONS produce other products
- Some reactant may be impure
Percentage yield can NEVER exceed 100%. If yours does, you have divided the wrong way round or your theoretical yield is wrong.
A yield above 100% in practice usually means the product is still WET — the extra mass is water, not product.
Part 2: Percentage purity
% purity = (mass of PURE substance ÷ TOTAL mass of sample) × 100
Worked example. A 5.0 g sample contains 4.2 g of pure compound.
% purity = (4.2 ÷ 5.0) × 100 = 84%
Working backwards: to find the mass of pure substance in an impure sample:
mass of pure = (% purity ÷ 100) × total mass
Use the PURE mass in any mole calculation, not the total sample mass. If a question gives an impure sample, convert to the pure mass before finding moles — this is the step most often missed.
Assessing purity practically: a pure substance has a sharp, fixed melting point; impurities lower it and spread it over a range.
Part 3: Gas volume calculations
At room temperature and pressure (r.t.p.), ONE MOLE of ANY gas occupies 24 dm³ (24 000 cm³).
moles of gas = volume in dm³ ÷ 24 volume in dm³ = moles × 24
The molar gas volume is the SAME for every gas — it does not depend on which gas it is, or on its Mr. This follows from gas particles being far apart, so the particle size is irrelevant.
Worked example. What volume does 0.25 mol of CO₂ occupy at r.t.p.?
V = 0.25 × 24 = 6 dm³ (6000 cm³)
Worked example — a full reacting-volume question.
What volume of hydrogen forms when 4.8 g of magnesium reacts with excess acid?
Mg + 2HCl → MgCl₂ + H₂
- Moles of Mg = 4.8 ÷ 24 = 0.2 mol
- Ratio Mg : H₂ = 1 : 1, so moles of H₂ = 0.2 mol
- Volume = 0.2 × 24 = 4.8 dm³ (4800 cm³)
Always go via MOLES — you cannot convert mass directly to volume.
Check the units the question wants — dm³ or cm³. To convert dm³ → cm³, multiply by 1000.
Reacting volumes of gases
When all the substances are gases at the same temperature and pressure, the volumes are in the SAME RATIO as the moles in the balanced equation.
Example: N₂ + 3H₂ → 2NH₃
1 volume of nitrogen reacts with 3 volumes of hydrogen to give 2 volumes of ammonia.
So 10 cm³ of nitrogen needs 30 cm³ of hydrogen and gives 20 cm³ of ammonia.
This shortcut works ONLY for gases — you can use the coefficients directly as volume ratios, with no need to convert to moles at all.
Mistakes that cost marks
Dividing theoretical by actual in a yield calculation.
Getting a yield above 100% without questioning it.
Using the total sample mass instead of the pure mass.
Forgetting to multiply by 100 for a percentage.
Converting mass directly to gas volume without going via moles.
Using the wrong molar volume — it is 24 dm³, not 22.4 (that is at s.t.p., a different condition).
Mixing dm³ and cm³.
Applying the volume-ratio shortcut to solids or liquids.
Omitting units or the % sign.
Frequently asked questions
What is percentage yield? (actual ÷ theoretical) × 100.
How do I find the theoretical yield? By a reacting-mass calculation from the balanced equation.
Why is yield below 100%? Reversible reactions, losses in transfer, and side reactions.
Can yield exceed 100%? No — if it appears to, the product is probably still wet.
What is percentage purity? (mass of pure substance ÷ total mass) × 100.
Which mass do I use in a mole calculation? The pure mass, not the total sample mass.
What volume does one mole of gas occupy? 24 dm³ at r.t.p. — the same for any gas.
How do I find the volume of a gas from a mass? Convert to moles first, apply the mole ratio, then × 24.
How do I convert dm³ to cm³? Multiply by 1000.
Can I use volume ratios directly? Yes — but only when all the substances are gases at the same conditions.
Quick revision checklist
- I know % yield = actual ÷ theoretical × 100
- I can calculate a theoretical yield from an equation
- I can give three reasons yield is below 100%
- I know yield can never exceed 100%
- I know % purity = pure ÷ total × 100
- I use the pure mass in mole calculations
- I know 1 mole of gas = 24 dm³ at r.t.p.
- I know it is the same for every gas
- I always go via moles from mass to volume
- I can convert dm³ ↔ cm³
- I can use volume ratios for gas-only reactions
- I give units and the % sign
These notes cover percentage yield, purity and gas volume calculations in the Cambridge IGCSE Chemistry (0620) syllabus and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Chemistry lessons. These were among the less-covered subtopics in that set, so they are combined here and follow the syllabus closely rather than being padded. Always check the current syllabus and data booklet for your own exam series.
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