How do chemists measure how efficient a synthesis is in terms of product obtained and atoms used?
Calculate percentage yield and atom economy and use them to evaluate the efficiency of a chemical synthesis
A focused answer to the WACE Year 12 Chemistry dot point on percentage yield and atom economy, how each is calculated, what they measure, and why both matter when evaluating a synthesis, with a worked example and common exam mistakes.
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What this dot point is asking
Two distinct measures judge how efficient a synthesis is. Confusing them is a frequent exam error, so it is worth being precise about what each one means.
Percentage yield
The theoretical yield is calculated from the limiting reagent using the mole ratio. Yields are below 100 percent in practice because of incomplete reactions (equilibria), side reactions, and losses during purification and transfer.
Atom economy
(equivalently, divided by the total molar mass of reactants used). Atom economy is fixed by the equation: a reaction that produces large by-product molecules has a low atom economy no matter how carefully it is run.
Why both matter
A high-yield reaction that produces a large by-product wastes raw material and creates waste to dispose of, even if almost all the limiting reagent is converted. Addition reactions have high atom economy (no by-product), while substitution and condensation reactions inevitably have lower atom economy because they release a small molecule. Industry seeks reactions with both high yield and high atom economy to minimise cost and waste, which links directly to green chemistry.
Why this matters
Yield and atom economy are the quantitative language of green and industrial chemistry. They let chemists compare alternative synthesis routes, justify choosing an addition route over a substitution route, and connect organic synthesis to the sustainability themes in the final part of Unit 4.