% yield = (actual yield ÷ theoretical yield) × 100%Theoretical yield = maximum possible mass calculated from stoichiometry
% purity = (mass of pure substance ÷ total mass of sample) × 100%
Worked example
A reaction theoretically produces 8.0 g of product. Only 6.2 g is obtained. Find the percentage yield.
% yield = (6.2 ÷ 8.0) × 100% = 77.5%
Reasons yield is less than 100%: reversible reactions don't go to completion; some product is lost during filtering/evaporation; side reactions consume reactants.
The Mole Triangle
moles = mass / Mr | moles = vol (dm³) / 24 | moles = c × V (dm³)
Mr in g/mol. Volume of gas at RTP = 24 dm3/mol. Concentration in mol/dm3.
Must-Know for Exam
moles = mass / Mr. Mr = sum of all Ar values in formula.
Gas at RTP: moles = volume (dm3) / 24. Remember: 1 dm3 = 1000 cm3.
Concentration: c = n/V. Volume MUST be in dm3. Convert cm3: divide by 1000.
Stoichiometry: use mole ratio from balanced equation to find moles of product from moles of reactant.
% yield = (actual / theoretical) x 100. % purity = (mass of pure substance / total mass) x 100.
Limiting reagent: the reactant that runs out first. Determines theoretical yield.
6. Common Exam Traps
Trap 1 — Always use moles, not mass, in ratios
You cannot use masses directly from a balanced equation. First convert mass to moles, apply the mole ratio from the equation, then convert back to mass. Skipping the mole step gives wrong answers.
Trap 2 — cm³ must become dm³
The concentration formula uses dm³. 25 cm³ = 0.025 dm³. Forgetting to divide by 1000 is the most common calculation error in titration questions.
Trap 3 — Mr not Ar for compounds
Use Ar for single elements; use Mr for compounds. Mr of NaCl = 23 + 35.5 = 58.5, not 23 (which is just Ar of Na).
Key Terms — Flashcard Review
Tap each card to reveal the definition.
Mole
1 mole = 6.02 x 10^23 particles (Avogadro's number). Amount of substance. Unit: mol.
Molar mass
Mass of 1 mole of substance in g/mol. Numerically equal to Mr. e.g. Mr of H2O = 18, so 18 g/mol.
Moles from mass
moles = mass (g) / molar mass (g/mol). Rearrange: mass = moles x Mr.
Molar volume of gas
At RTP: 1 mole of any gas occupies 24 dm3 (24 000 cm3). moles = volume (dm3) / 24.
Concentration
moles = concentration (mol/dm3) x volume (dm3). c = n/V. Volume in dm3 (divide cm3 by 1000).
Percentage yield
% yield = (actual yield / theoretical yield) x 100. Always less than 100% in practice.
🎯 Practice Quiz — Test Yourself
8 O Level-style questions on this topic. Select an answer to see instant feedback.
Question 1 of 8
Moles in 44 g of CO₂? (Mr = 44)
Explanation: moles = mass/Mr = 44/44 = 1 mol.
Question 2 of 8
Molar volume of gas at RTP (25°C, 1 atm):
Explanation: At RTP: 1 mol of any gas = 24 dm³. (At STP: 22.4 dm³/mol).
Explanation: Na:H₂ ratio = 2:1. So 4 mol Na → 2 mol H₂.
Question 5 of 8
% yield = 75%, theoretical yield = 20 g. Actual yield:
Explanation: Actual = (75/100) × 20 = 15 g.
Question 6 of 8
How many moles are in 11 g of CO2? (C=12, O=16)
Explanation: Mr of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol. moles = mass/Mr = 11/44 = 0.25 mol. Always calculate Mr first by adding up all Ar values in the formula.
Question 7 of 8
What volume of gas (at RTP) is produced by 0.5 mol of CO2?
Explanation: Volume = moles x 24 = 0.5 x 24 = 12 dm3 at RTP. All gases occupy 24 dm3 per mole at RTP (room temperature and pressure). This applies regardless of the identity of the gas.
Question 8 of 8
A solution has concentration 2 mol/dm3 and volume 250 cm3. The moles of solute present is:
Explanation: moles = c x V. V must be in dm3: 250 cm3 / 1000 = 0.25 dm3. moles = 2 x 0.25 = 0.5 mol. The most common error is forgetting to convert cm3 to dm3.