IB Chemistry SLTopic 5 — Quantifying Chemical ChangePaper 1 & 2Core skill~11 min read
Percentage Yield
The equation tells you the most you could possibly get. The balance tells you what you actually got. Percentage yield is the gap between those two numbers, and every gram of it is a story about something that went wrong.
📚 What you need to know
Theoretical yield is the maximum product obtainable, calculated from the limiting reactant assuming perfect conversion.
Actual (experimental) yield is what you collect and weigh.
% yield = (actual ÷ theoretical) × 100, with both in the same units.
Yields fall short because of incomplete or reversible reactions, side reactions, and losses during transfer and purification.
A yield cannot exceed 100%. If it appears to, the product is wet or impure.
Percentage yield says nothing about how much waste a reaction makes — that is atom economy.
The theoretical yield is not something you measure — it is calculated, and it always starts from the limiting reactant. That is worth stating plainly, because a question that gives you two reactant masses is asking you to do the limiting-reactant test first, whether it says so or not.
Three of these four boxes are calculated. Only “actual yield” comes off a balance.
Units must match before you divide. Grams with grams, moles with moles. Mixing a mass with a number of moles gives a percentage that means nothing, and it is a surprisingly easy slip when a question quotes them differently.
Where the rest of it went
Losses are not one thing. In an exam, naming a specific mechanism scores; saying “some was lost” usually does not.
The four common explanations, and when each applies:
The reaction is reversible and reaches equilibrium before completion — the classic reason an industrial yield is capped, as in the Haber process.
Side reactions consume reactant to make something you did not want.
Losses on transfer — solid left in the flask, on the filter paper, or in the funnel.
Losses on purification — some product stays dissolved in the solvent during recrystallisation, or is discarded with impurities.
A yield above 100% is not a triumph; it is a warning. It means the “product” weighed more than it should, so it contains something else — usually water that was not dried off, or unreacted starting material. The correct response is to dry it again to constant mass.
WORKED EXAMPLE
5.00 g of calcium carbonate was heated strongly and 2.50 g of calcium oxide was collected. Calculate the percentage yield. CaCO3(s) → CaO(s) + CO2(g)
Step 1 — moles of the limiting reactantOnly one reactant, so it is limiting by default.n = 5.00 ÷ 100.09 = 0.04996 molStep 2 — theoretical yield1 : 1, so n(CaO) = 0.04996 mol0.04996 × 56.08 = 2.802 gStep 3 — compare(2.50 ÷ 2.802) × 100 = 89.2% yield = 89.2%Likely causes here: the decomposition did not go to completion, or some solid was left in the crucible.
WORKED EXAMPLE
10.0 g of zinc was added to excess hydrochloric acid and 3.20 dm3 of hydrogen was collected at STP. Calculate the percentage yield. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Step 1 — the acid is in excess, so zinc is limitingn(Zn) = 10.0 ÷ 65.38 = 0.1530 molStep 2 — theoretical yield, as a gas volumeZn : H₂ is 1 : 1, so n(H₂) = 0.1530 molV = 0.1530 × 22.7 = 3.472 dm³Step 3 — compare, same units(3.20 ÷ 3.472) × 100 = 92.2% yield = 92.2%Yields work with any measure of amount, not just mass — volumes and moles are equally valid, as long as both figures are the same kind.
WORKED EXAMPLE
A process makes calcium oxide from calcium carbonate with a yield of 85.0%. What mass of calcium carbonate is needed to produce 50.0 g of calcium oxide?
Step 1 — work back to the theoretical yield50.0 g is only 85.0% of what the equation promised.theoretical = 50.0 ÷ 0.850 = 58.82 gStep 2 — moles of CaO needed58.82 ÷ 56.08 = 1.049 molStep 3 — back through the ratio to the reactant1 : 1, so n(CaCO₃) = 1.049 mol1.049 × 100.09 = 105.0105 g of CaCO₃Divide by the yield, never multiply. Multiplying by 0.850 gives 42.5 g and would leave you badly short.
💡 Exam tip
Find the limiting reactant first whenever two reactant quantities are given.
Put the actual yield on top. Inverting the fraction gives an answer above 100%, which should ring alarm bells.
Working backwards from a required product mass? Divide by the fractional yield.
Explaining a low yield? Give a specific reason — reversible reaction, side product, loss on filtration — not just “experimental error”.
Both figures must be the same quantity in the same units before you divide.
⚠️ Common mix-up
Calculating the theoretical yield from the excess reactant instead of the limiting one.
Dividing theoretical by actual and reporting a yield over 100%.
Multiplying by the yield when the question asks how much reactant is needed.
Confusing yield with atom economy. A reaction can go to completion and still waste most of its mass.
Blaming “human error”. It earns nothing; name the physical loss instead.
Up next: Atom Economy — a completely different question about the same reaction: not how much of the possible product you got, but how much of what you bought was ever going to be product at all.
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