Percentage Yield and Atom Economy — Formulas and Worked Examples
Percentage yield and atom economy are two different questions about the same reaction. Percentage yield asks how much of what was possible did you actually get? Atom economy asks how much of the mass you put in ends up in the product you wanted? Students routinely mix them up, and a reaction can score 95% on one and 30% on the other. This guide gives both formulas, four fully worked examples, and the errors that cost marks in Class 12 board papers and in entrance exams.
The two formulas
What each term means
- Actual yield — the mass of pure product you really isolated, weighed in the lab. Always given in the question.
- Theoretical yield — the mass you would get if every molecule of the limiting reagent converted perfectly. You must calculate this from the balanced equation.
- Coefficient — the balancing number in front of a formula. It appears in both calculations and is the most commonly forgotten piece.
- Desired product — atom economy needs you to choose which product you actually want. Everything else counts as waste, even if it is useful elsewhere.
Both answers are percentages, so both are unitless. Atomic masses used below: H = 1.008, C = 12.011, N = 14.007, O = 15.999, Na = 22.990, Ca = 40.078, Br = 79.904 g/mol.
Worked example 1 — percentage yield of quicklime
25.0 g of calcium carbonate is heated: CaCO₃ → CaO + CO₂. The student isolates 12.0 g of CaO. Find the percentage yield.
Step 1 — molar masses.
M(CaCO₃) = 40.078 + 12.011 + (3 × 15.999) = 40.078 + 12.011 + 47.997 = 100.086 g/mol
M(CaO) = 40.078 + 15.999 = 56.077 g/mol
Step 2 — moles of the limiting reagent.
n(CaCO₃) = 25.0 ÷ 100.086 = 0.24979 mol
Step 3 — theoretical yield. The coefficients are 1 : 1, so
n(CaO) = 0.24979 mol.
Theoretical mass = 0.24979 × 56.077 = 14.007 g
Step 4 — percentage yield.
(12.0 ÷ 14.007) × 100 = 85.7%
Worked example 2 — atom economy of the same reaction
Notice that this calculation does not use the 25.0 g or the 12.0 g at all. Atom economy is a property of the equation, not of your lab technique.
Reactant total = M(CaCO₃) = 100.086 g/mol
Desired product = CaO = 56.077 g/mol
Atom economy = (56.077 ÷ 100.086) × 100 = 56.03%
Cross-check. The other product, CO₂, has M = 12.011 + 31.998 = 44.009 g/mol, giving (44.009 ÷ 100.086) × 100 = 43.97%. 56.03 + 43.97 = 100.00% — every atom is accounted for. If your two figures do not add to 100, the equation is unbalanced or a molar mass is wrong.
So this reaction can be run with an excellent 85.7% yield and still throw away 44% of the starting mass as carbon dioxide. That is exactly why both numbers are reported.
Worked example 3 — the Haber process, 100% atom economy
N₂ + 3H₂ → 2NH₃. Every atom on the left appears in the only product, so the atom economy is 100% by inspection. Now the yield: 28.0 g of nitrogen is reacted with excess hydrogen and 25.0 g of ammonia is collected.
M(N₂) = 2 × 14.007 = 28.014 g/mol
M(NH₃) = 14.007 + (3 × 1.008) = 14.007 + 3.024 = 17.031 g/mol
n(N₂) = 28.0 ÷ 28.014 = 0.99950 mol
Coefficients are 1 N₂ : 2 NH₃, so n(NH₃) = 2 × 0.99950 = 1.99900 mol
Theoretical mass = 1.99900 × 17.031 = 34.045 g
Percentage yield = (25.0 ÷ 34.045) × 100 = 73.4%
This is the opposite pattern to example 2: perfect atom economy, moderate yield. Addition and combination reactions always have 100% atom economy because there is only one product.
Worked example 4 — a substitution with poor atom economy
C₂H₅Br + NaOH → C₂H₅OH + NaBr. The wanted product is ethanol.
M(C₂H₅Br) = (2 × 12.011) + (5 × 1.008) + 79.904 = 24.022 + 5.040 + 79.904 = 108.966
M(NaOH) = 22.990 + 15.999 + 1.008 = 39.997
Reactant total = 108.966 + 39.997 = 148.963 g/mol
M(C₂H₅OH) = (2 × 12.011) + (6 × 1.008) + 15.999 = 24.022 + 6.048 + 15.999 = 46.069
Atom economy = (46.069 ÷ 148.963) × 100 = 30.93%
Cross-check. M(NaBr) = 22.990 + 79.904 = 102.894, and 46.069 + 102.894 = 148.963 — equal to the reactant total, as mass conservation requires.
Even a flawless 100% yield here would still send about 69% of the input mass to waste as sodium bromide. This single comparison is the whole argument for green chemistry, and it is a favourite short-answer question.
Percentage yield vs atom economy — side by side
| Percentage yield | Atom economy | |
|---|---|---|
| Measures | How well the reaction was run | How efficient the reaction route is in principle |
| Needs lab data? | Yes — the actual mass isolated | No — only the balanced equation |
| Improved by | Better technique, purer reagents, less loss on transfer | Choosing a different reaction pathway entirely |
| Typical exam wording | "Calculate the percentage yield" | "Calculate the percentage atom economy for the formation of …" |
Common mistakes that cost marks
- Using an unbalanced equation. Both formulas depend on coefficients. Balance first, every single time.
- Ignoring the limiting reagent. Theoretical yield is set by the reagent that runs out first, not by whichever mass the question mentions first.
- Reporting a yield above 100%. That is not a good result; it means the product is still wet, or contains unreacted starting material. Say so in the answer — examiners award the mark for the explanation.
- Putting actual masses into the atom-economy formula. Atom economy uses molar masses only.
- Forgetting that atom economy needs a stated desired product. The same equation gives a different atom economy for each product.
- Rounding early. Keep the full mole value through the working and round only the final percentage.
Where these appear in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11–12 | Stoichiometry numericals; green-chemistry short answers |
| JEE/NEET | Limiting reagent combined with percentage yield in one question |
| IIT-JAM / CUET-PG | Multi-step synthesis yields; overall yield across two steps |
| GATE / CSIR-NET | Process efficiency, E-factor and green-chemistry metrics |
For a two-step synthesis, the overall yield is the product of the step yields, not their average: 80% followed by 70% gives 0.80 × 0.70 = 0.56, i.e. 56%.
Do the stoichiometry without slips. The calculator suite has no dedicated yield tool, but the two steps that go wrong most often — working out molar masses and converting mass to moles — each have their own calculator inside it.
Open the ABC Chemistry Calculator Suite →Preparing for Class 11–12 boards? ABC Chemistry runs Class 11–12 chemistry coaching at its Gurugram centre and online classes across India — details at abcchemistry.in.