Quantitative chemistry · GCSE Chemistry
Atom economy and yield
GCSE Chemistry revision on atom economy and percentage yield: two different formulae, why they are not the same, and why industry wants both high yield and high atom economy.
Percentage yield compares what you got with what the equation said you could get. Atom economy asks what fraction of the product atoms are the product you wanted. Do not mix the formulae.
The important bits
What you need to know
- 1
Percentage yield = (actual mass of product / theoretical mass of product) × 100. Theoretical mass comes from a mole calculation assuming complete reaction.
- 2
Yield is less than 100% because of incomplete reaction, side reactions, loss on filtering or transferring, and reversible reactions that reach equilibrium.
- 3
Atom economy = (Mᵣ of desired product / sum of Mᵣ of all products) × 100. Use the balanced equation. You do not need an experimental mass.
- 4
Addition reactions can have 100% atom economy because there is only one product. For example, adding H₂ to an alkene to make an alkane (Triple organic link).
- 5
Reactions that also make a waste salt or small molecule (H₂O, CO₂, HCl) have atom economy below 100%. The waste still has mass even if you never weigh it.
- 6
Industry wants high atom economy (less waste, cheaper disposal, greener) and high percentage yield (more of the desired product from the feedstock), plus low energy cost.
- 7
A reaction can have 100% atom economy and a poor yield if it is slow or reversible. A reaction can have a high yield of a desired product and still waste most of the atoms as something else.
- 8
Higher tier on Combined and Chemistry. Foundation may meet yield in words. Always show which formula you used; they are not interchangeable.
Quotations worth analysing
Short evidence. Real method.
“% yield = (actual mass / theoretical mass) × 100”
Actual is what you weighed. Theoretical is what the mole calculation predicted. If actual > theoretical, you have wet crystals or a calculation error — yield cannot exceed 100% for a pure dry product.
“atom economy = (Mᵣ desired product / Σ Mᵣ all products) × 100”
Only products in the denominator, not reactants. Use the big numbers: 2NaCl means 2 × 58.5. Desired product is the one named in the question.
“CaCO₃ → CaO + CO₂”
If you want CaO, atom economy = 56/100 × 100 = 56%. If you wanted CO₂, it would be 44%. The equation did not change; the desired product did.
Go deeper
Yield is experimental; atom economy is theoretical
If the question gives an actual mass and enough data to find a theoretical mass, it is yield. Work the mole calculation first, then divide. If the question only gives an equation and asks how efficiently atoms are used, it is atom economy — no weighing required. Students put actual mass into the atom-economy formula, or put Mᵣ values into yield without converting to the theoretical mass of that specific product. Keep a heading in your working: “theoretical mass” versus “atom economy”. Loss of product on a filter paper cuts yield; it does not change atom economy, because atom economy assumes every reactant atom ends up in some product as written.
Go deeper
Why both numbers matter in a chemical factory
A process with 90% yield but 40% atom economy still throws away more than half of the atoms as waste that must be sold, burned or dumped. A process with 100% atom economy but 20% yield wastes feedstock and energy because most of the reactants never become product in the time allowed — you might recycle unreacted starting materials if the reaction is reversible (Haber process). Evaluation questions want: high atom economy reduces waste and environmental impact; high yield reduces cost of reactants; catalysts and conditions improve rate and sometimes yield; recycling leftover reactants improves effective yield. “Greener” is not a full answer without atoms or energy.
Go deeper
Worked pair so the formulae cannot swap in your head
CH₄ + 2O₂ → CO₂ + 2H₂O. If the desired product is CO₂, atom economy = 44 / (44 + 36) × 100 = 55%. If 8.0 g of methane (0.50 mol) produces 11.0 g of CO₂, theoretical CO₂ is 0.50 × 44 = 22.0 g, so percentage yield = (11.0/22.0) × 100 = 50%. Same reaction, two different numbers answering two different questions. Wet product would inflate actual mass and fake a higher yield; drying crystals matters. Side reactions that make carbon (soot) cut the yield of CO₂ without changing the atom-economy formula of the equation you wrote for complete combustion.
See the idea in action
The reaction NaOH + HCl → NaCl + H₂O is used to make sodium chloride. Mᵣ(NaCl) = 58.5, Mᵣ(H₂O) = 18. Atom economy for NaCl = 58.5 / (58.5 + 18) × 100 = 76.5%. A student should theoretically obtain 5.85 g of NaCl but collects 5.12 g of dry crystals. Percentage yield = (5.12 / 5.85) × 100 = 87.5%. The missing mass is not a change in atom economy; it is loss on transfer or incomplete drying/crystallisation. If the crystals were damp, the apparent yield could wrongly exceed the dry theoretical mass.
Exam technique
Turn knowledge into marks
Read the question: actual and theoretical masses mean yield; formulae only mean atom economy. Show Mᵣ totals. Yield cannot be over 100% for a pure dry product. Name a reason yield is low: loss, incomplete reaction, or equilibrium.
Common mistakes
Do not give these marks away
- 01
Using the yield formula for atom economy, or putting reactants in the atom-economy denominator.
- 02
Reporting a yield over 100% without questioning wet product or a calculation error.
- 03
Saying a catalyst increases atom economy, or that atom economy depends on how carefully you filter.
Which statement is correct?
APercentage yield and atom economy always have the same value
BAtom economy is calculated from the balanced equation; percentage yield needs an actual mass of product
CA catalyst increases atom economy
DAtom economy uses the mass of reactants in the denominator
Show the answer
Atom economy is calculated from the balanced equation; percentage yield needs an actual mass of product. Atom economy is theoretical and uses Mᵣ of products. Yield compares experimental mass with the theoretical mass from a mole calculation.
Quick questions
If this is the bit you searched
What is the difference between percentage yield and atom economy GCSE?
Percentage yield compares actual product mass with theoretical mass from the equation. Atom economy is the percentage of the total product Mᵣ that is the desired product. One is experimental; one is theoretical.
How do you calculate percentage yield?
Work out the theoretical mass using moles, then (actual mass / theoretical mass) × 100. Use the same units of mass on top and bottom.
How do you calculate atom economy?
(Mᵣ of the desired product ÷ sum of Mᵣ of all products) × 100, using the balanced equation. Include the big numbers in front of formulae.
Why is a high atom economy important in industry?
More of the atoms end up in the useful product, so there is less waste to dispose of. That cuts cost and environmental impact. High yield is still needed as well.