IB Chemistry HLTopic 5 — How Much? Quantifying Chemical ChangePaper 1 & 2Practical skill~9 min read
Percentage Yield
Your calculation says 9.95 g. The balance says 8.50 g. Nothing has gone wrong — that is just what happens in a real lab. Percentage yield is the honest comparison between what you should have got and what you actually collected.
📚 What you need to know
Theoretical yield is the maximum possible mass, worked out from the limiting reactant.
Actual (experimental) yield is the mass you really collect and weigh.
% yield = actual ÷ theoretical × 100.
Both masses must be in the same unit and for the same substance.
The theoretical yield must come from the limiting reactant, never the excess one.
Yields are below 100% because of losses in transfer, side reactions, incomplete or reversible reactions, and purification.
A yield above 100% means the product is still wet or contains impurities.
The dashed section is the mark you are really being asked about: where did that 1.45 g go? Smoke escaping, powder left in the crucible, unburnt metal in the middle.
Where the theoretical yield comes from
The theoretical yield is not a measured thing. It is the answer to a reacting mass calculation, and it always starts from the limiting reactant. Start from the excess reactant and your theoretical yield will be too big, which makes your percentage yield too small — and both marks go.
Only the last box involves anything you measured in the lab. Everything before it is pure calculation from the equation.
A neat check on your own work: percentage yield should land somewhere between about 40% and 95% for a normal school experiment. If you get 12% or 340%, go back and look at which reactant you used and whether both masses are for the same substance.
Why yields are never perfect
Reason
What is actually happening
Typical example
Losses in transfer
Product stays on the glassware or filter paper
Crystals left in the beaker
Side reactions
Reactants make something else as well
Incomplete combustion giving CO
Incomplete reaction
The reaction stops before everything reacts
Unburnt metal in the middle of the lump
Reversible reaction
An equilibrium is reached, not completion
Making ammonia or an ester
Purification
Some product is lost in washing or recrystallising
Rinsing a solid with cold water
Reversible reactions are the honest ones. Making an ester is a classic low-yield reaction, not because you are clumsy, but because the reaction reaches equilibrium and stops. No amount of careful technique gets that one to 100%.
Worked examples
WORKED EXAMPLE
6.00 g of magnesium is burned in excess oxygen and 8.50 g of magnesium oxide is collected. Calculate the percentage yield. 2Mg(s) + O2(g) → 2MgO(s)
Step 1: oxygen is in excess, so magnesium is limitingn(Mg) = 6.00 ÷ 24.31 = 0.2468 molStep 2: mole ratio is 2 : 2, so 1 : 1n(MgO) = 0.2468 molStep 3: theoretical massm = 0.2468 × 40.31 = 9.949 gStep 4: compare with what was collected8.50 ÷ 9.949 × 100 = 85.4Percentage yield = 85.4%Some magnesium oxide always escapes as white smoke, and the lump often has unreacted metal at its core. 85% is a good result for this experiment.
WORKED EXAMPLE
4.00 g of zinc is added to 50.0 cm3 of 0.500 mol dm−3 copper(II) sulfate. 1.20 g of copper is filtered off and dried. Find the percentage yield. Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Step 1: check which reactant limitsn(Zn) = 4.00 ÷ 65.38 = 0.0612 mol n(CuSO₄) = 0.0500 × 0.500 = 0.0250 molStep 2: ratio is 1 : 1, so the smaller one limitsCuSO₄ is limiting at 0.0250 molStep 3: theoretical mass of copperm = 0.0250 × 63.55 = 1.589 gStep 4: percentage yield1.20 ÷ 1.589 × 100 = 75.5Percentage yield = 75.5%Start from the zinc instead and you would get a theoretical 3.89 g and a yield of 31% — a completely wrong answer from a correct-looking calculation.
WORKED EXAMPLE
0.100 mol of ethanol is heated with excess ethanoic acid to make ethyl ethanoate (M = 88.12 g mol−1). The reaction has a yield of 68%. What mass of ester is collected?
Step 1: work out the theoretical yield first
Ratio is 1 : 1 and the acid is in excess, so n(ester) = 0.100 molm = 0.100 × 88.12 = 8.812 gStep 2: now apply the yieldactual = 8.812 × 68 ÷ 100 = 5.992 g5.99 g of ethyl ethanoateThis is the equation used backwards: actual = theoretical × % yield ÷ 100. Esterification is reversible, so a yield below 70% is completely normal.
💡 Exam tip
Check for a limiting reactant first, every time. If the question gives two reactant amounts, it wants you to test them.
Label your two masses clearly as theoretical and actual before you divide. It stops the fraction going upside down.
Actual on top. If your answer is over 100% you have almost certainly flipped it.
Give a reason if asked why the yield is below 100%, and make it specific to that experiment rather than a general “losses”.
Percentage yield has no unit. Do not write grams after it.
Working backwards is fair game: theoretical × % ÷ 100 gives the actual mass.
⚠ Common mix-up
Using the excess reactant to find the theoretical yield.
Dividing theoretical by actual. Always actual ÷ theoretical.
Comparing a mass with a number of moles. Both values must be masses of the same substance.
Confusing percentage yield with atom economy. One is about how the experiment went, the other about the equation itself.
Blaming “human error” in an explanation. Name the real loss: product left on the filter paper, or an equilibrium.
Saying a low yield means the equation is wrong. A perfectly correct equation can still give a poor yield.
Percentage yield judges your technique. There is a second efficiency measure that judges the reaction itself, before you even step into the lab — and a reaction can score 95% on one and 40% on the other. Up next: Atom Economy.
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