5.5.2 · D5Green Chemistry & Sustainability
Question bank — Atom economy
True or false — justify
A reaction with 100% yield must have high atom economy.
False — yield only says you got all the product the equation allows; the equation itself may dump half the atoms as by-product (e.g. the 58% substitution). These are two independent axes.
A reaction with 100% atom economy must have high yield.
False — atom economy is fixed by the equation, but the reaction might barely proceed (equilibrium, slow kinetics, losses). You can have 100% AE and 5% yield.
Atom economy can exceed 100%.
False — the numerator (desired product mass) is a part of the denominator (all products mass), so the fraction can never exceed 1; 100% is the ceiling, reached only when there is no by-product.
Two reactions making the same product always have the same atom economy.
False — different routes produce different by-products, so the denominator changes. A metal made by addition vs by displacement (with a salt by-product) differ hugely.
Changing the temperature of a reaction changes its atom economy.
False — temperature affects rate and yield, but atom economy depends only on the balanced equation, which temperature does not alter.
If you double every coefficient in the equation, the atom economy doubles.
False — doubling scales numerator and denominator equally, so the ratio is unchanged. Atom economy is a proportion, immune to overall scaling.
A decomposition reaction (one reactant → several products) can have 100% atom economy.
True only if every product is wanted (or the extras are massless, like heat); false the moment there is even one unwanted by-product, since that mass sits in the denominator alone. See the single-product edge case below for the 100% limit.
Using a catalyst improves atom economy.
False directly — a catalyst isn't consumed and doesn't appear in the balanced equation, so it can't change AE; but it can enable a cleaner route with fewer by-products, which raises AE indirectly.
Water counted as a by-product always lowers atom economy.
True — water leaving as H₂O carries mass that isn't in your product, so the denominator grows; a condensation reaction is penalised exactly for this lost H and O.
Spot the error
"AE = mass of desired product ÷ mass of reactants used up in the lab."
The error is mixing lab reality with equation theory — AE uses molar masses from the balanced equation, not measured lab masses. That confuses it with yield.
"For , AE ."
Missing the stoichiometric coefficients: it must be . Forgetting to multiply by is the classic slip.
"To find AE, put the wanted product on top and the reactants underneath — but I'll use CO₂ from the products for the extra mass."
The error is mixing sides: use all-products in the denominator or all-reactants, never one species from each. Mixing double-counts or drops mass.
"This reaction gives 82% yield, so 18% of the atoms became waste by-product."
Wrong — the missing 18% is unreacted starting material and handling losses, not by-product. By-product mass is an atom-economy question, not a yield question.
"AE is low, so the reaction is uneconomical — we should scrap it."
Too hasty — if the "by-product" is itself sellable (or water, easily discarded), low nominal AE may be fine. Judge with E-factor and whether products are useful, not AE alone.
"An addition reaction always beats a substitution on atom economy."
Usually true, but stated too absolutely — addition has no by-product by design, so it's the gold standard; the trap is treating "usually" as "by definition always" without checking the equation.
"Since mass is conserved, atom economy is always 100%."
The error confuses total mass conservation (always true — nothing vanishes) with the fraction that is desired. Conserved atoms can still all leave as unwanted by-product.
Why questions
Why does an addition reaction give 100% atom economy?
Because every reactant atom combines into a single product molecule, so there is no by-product term in the denominator — desired mass equals total product mass.
Why does a substitution reaction cap below 100%?
Substitution kicks one atom/group out as a separate molecule (e.g. HCl), so that mass is built into the equation as unavoidable waste — no lab technique can recover it.
Why can atom economy use reactant masses OR product masses in the denominator?
By conservation of mass the two totals are equal, so both denominators give the identical number; pick whichever side is easier to add up.
Why is high atom economy considered "green"?
Fewer atoms wasted means less landfill/effluent, fewer raw materials bought, and cheaper separation — smaller footprint at the equation level. It's Principle #2.
Why isn't atom economy alone enough to judge a green process?
It ignores solvents, energy, unreacted reagents and real losses; metrics like E-factor capture the whole process mass, not just the balanced equation.
Why do we weight each species by its coefficient before summing?
Because "" means two iron atoms are produced per cycle; using one would count only half the actual mass and give a wrong fraction.
Edge cases
If a reaction has two products and BOTH are wanted, what is the atom economy?
100% — both go in the numerator as "desired," leaving nothing in the denominator-only waste term. AE is about unwanted mass, not the number of products.
A reaction produces only heat and a single solid product from a single solid reactant. What's its atom economy?
100% — heat has no mass, so the sole product carries all the atoms; there is no by-product mass in the denominator. This is exactly the "100% decomposition" limit hinted at in the true/false section.
A "waste" by-product later becomes sellable. What happens to the reported atom economy?
It rises — reclassifying that mass from waste to desired moves it from denominator-only into the numerator, increasing the useful fraction.
What is the atom economy of a reaction where the desired product mass is zero (all reactant leaves as by-product)?
0% — the numerator is zero, so none of the input atoms reach the wanted product; the reaction makes only waste with respect to that target.
Can atom economy be negative?
No — both masses are positive quantities and the numerator is a subset of the denominator, so the ratio lies strictly between 0% and 100%. Negative is physically meaningless here.
If a catalyst appears in the equation you wrote but not in the net balanced equation, do you include it?
No — a true catalyst is regenerated and cancels out of the net equation, so it contributes to neither numerator nor denominator of atom economy.
Recall A note on the molar masses used
Standard atomic masses (rounded to reasonable precision) give Fe ≈ 55.8, C ≈ 12.0, O ≈ 16.0, so CO₂ ≈ 44.0 g mol⁻¹. These are the IUPAC standard atomic-weight values rounded to one decimal — enough for AE to the nearest 0.1%. Rounding choice doesn't change any conceptual answer above.