The big idea: A balanced equation tells you the most product you could ever make. In reality you almost always get less, and some of the reactant atoms end up in by-products you don't want.
Two separate measurements capture this:
- Percentage yield — how much product you actually got compared with the theoretical maximum. - Percentage atom economy — what fraction of the reactant atoms end up in the desired product.
They answer different questions, so a reaction can score well on one and badly on the other.
Define every term: - Theoretical yield — the amount of product predicted from the balanced equation if the limiting reactant reacted completely. - Actual yield — the amount of product you really obtain (always less, due to side reactions, reversible reactions, or losses on purification). - Desired product — the one you actually want; the rest are by-products. - Green chemistry — designing reactions that waste fewer atoms and resources.
Percentage yield compares what you got with what you could have got. Both yields must be in the same units — both masses, or both moles.
- amount (or mass) of product actually obtained
- amount (or mass) predicted by the mole ratio if the reaction went perfectly
Worked example — percentage yield from masses
A student predicts a theoretical yield of 12.5 g of aspirin but isolates only 9.6 g after purification. Calculate the percentage yield.
Solution
- Formula first — both masses are already the desired product, so use them directly:
- Substitute the two masses:
- Work it out:
Final answer
Percentage yield = 76.8% (≈ 77%).
Worked example — mass of product at a known yield
Burning a fuel could theoretically produce 0.40 mol of water, but the reaction runs at 85% yield. Calculate the amount, in moles, of water actually formed.
Solution
- Formula first — rearrange for the actual yield:
- Substitute the values:
- Work it out — keep the unit:
Final answer
0.34 mol of water is actually formed.
Know your predicted grade
Take timed mock exams and get detailed feedback on every answer. See exactly where you're losing marks.
Atom economy asks a different question: of all the atoms you put in as reactants, what fraction ends up in the product you actually want? Use molar masses, and add up every reactant — even ones whose atoms are wasted.
- molar mass of the wanted product (g mol⁻¹)
- sum of the molar masses of all reactants used (g mol⁻¹)
Two shortcuts: - Because mass is conserved, M(all reactants) also equals M(desired product) + M(all by-products) — so you can sum the products instead if that's easier. - An addition reaction (everything combines into one product) has 100% atom economy — there are no by-products.
Worked example — atom economy of a synthesis
Hydrogen is made by reacting methane with steam: CH4 + H2O → CO + 3H2. Calculate the percentage atom economy for producing hydrogen, H2. (M: CH4 = 16.05, H2O = 18.02, H2 = 2.02 g mol⁻¹.)
Solution
- Formula first — the desired product is the 3 mol of H2:
- Top line = molar mass of the H2 produced (×3 for the coefficient):
- Bottom line = sum of all reactant molar masses:
- Substitute and work it out:
Final answer
Percentage atom economy = 17.8% — most of the atoms end up in the by-product CO, not in the hydrogen.
How this is tested: Yield and atom economy appear on both papers, and the link to green chemistry is a favourite extended-response idea.
- Paper 1A (MCQ): a one-step '% atom economy of this synthesis' or 'which route has 100% atom economy'. - Paper 2: a part-marks '% yield from these masses', often followed by a discuss part on why a high atom economy makes an industrial process more sustainable.
The classic trap: putting only one reactant on the bottom of the atom-economy fraction, or mixing up yield and atom economy.
Score every mark: (1) Write the formula before the numbers. (2) For atom economy, multiply each molar mass by its coefficient and sum all reactants. (3) For the 'why it matters' part, link low atom economy to wasted atoms, more by-products and more waste to treat.
IB-style question — ethanol synthesis (a)
Ethanol can be made directly by hydration of ethene: C2H4 + H2O → C2H5OH. (a) Calculate the percentage atom economy for this preparation of ethanol. (M: C2H4 = 28.06, H2O = 18.02, C2H5OH = 46.08 g mol⁻¹.)
Solution
- Formula first:
- Only one product forms, so its molar mass is the top line:
- Sum both reactant molar masses for the bottom line:
- Substitute:
Final answer
Percentage atom economy = 100% — it is an addition reaction, so every reactant atom ends up in the ethanol.
IB-style question — ethanol synthesis (b)
(b) A second industrial route to ethanol gives a much lower atom economy. Discuss why a high percentage atom economy is an advantage for an industrial process. [2]
How to score the marks
- [1] Higher atom economy means a greater proportion of the reactant atoms end up in the desired product, so fewer atoms are wasted as by-products.
- [2] This makes the process more sustainable / economical: less raw material is wasted and there is less waste by-product to separate, treat or dispose of (lower cost and environmental impact).
Final answer
A high atom economy means most reactant atoms become the desired product, so fewer atoms are wasted as by-products. The process is therefore more sustainable and economical — less raw material is wasted and there is less waste to treat or dispose of.