Percentage yield and atom economy: WACE Year 12 Chemistry

Syllabus dot point

“Calculate percentage yield and atom economy and use them to evaluate the efficiency of a chemical synthesis”

8 min readUpdated 2 Jun 2026

Quick answer

Percentage yield compares how much product you actually obtained with the maximum (theoretical) amount the equation predicts. Atom economy measures what fraction of the mass of all the reactants ends up in the desired product. They are different efficiency measures: a reaction can have a high yield but poor atom economy if it produces large by-products.

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 actual mass of product obtained expressed as a percentage of the theoretical (maximum) mass predicted by the balanced equation.

\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

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

The proportion of the total mass of reactants that ends up in the desired product, calculated from the balanced equation.

\text{atom economy} = \frac{\text{molar mass of desired product}}{\text{total molar mass of all products}} \times 100

(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.

Yield and atom economy are different

Percentage yield depends on how the reaction is carried out (it can be improved by technique); atom economy depends only on the balanced equation (it is fixed by the chemistry). A reaction can have a high yield and still waste atoms in by-products. A good green synthesis has both high yield and high atom economy.

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.

Calculating both measures

Question

A reaction has the equation $\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$ . A chemist obtains $36.8$ g of ethanol from a reaction with a theoretical yield of $46.0$ g. Calculate the percentage yield and the atom economy. ( $M(\text{C}_2\text{H}_5\text{OH}) = 46.0$ g mol $^{-1}$ .)

Solution

Percentage yield:

\frac{36.8}{46.0} \times 100 = 80.0\%

Atom economy: this is an addition reaction with only one product, ethanol, so every atom of the reactants ends up in the product:

\text{atom economy} = \frac{46.0}{46.0} \times 100 = 100\%

The high atom economy reflects that addition reactions produce no by-product, while the 80 percent yield reflects practical losses.

Working with the limiting reagent

Almost every percentage-yield calculation hinges on correctly identifying the limiting reagent, because the theoretical yield is set by whichever reactant runs out first. The reliable method is to convert the given masses of each reactant to moles, then divide each by its coefficient in the balanced equation; the smallest value identifies the limiting reagent. The theoretical yield of product is then found from that reagent using the mole ratio, converted back to mass with the molar mass. A common trap is to assume the reactant present in the smaller mass is limiting, but a low-molar-mass reactant can supply more moles, so always compare in moles, not grams. Once the limiting reagent and theoretical mass are fixed, the actual mass collected divided by that theoretical mass gives the percentage yield.

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.

Confusing yield with atom economy

The headline error is treating the two measures as the same; yield is about how much product you got versus the maximum, while atom economy is about how many atoms are wasted as by-product. Another is calculating theoretical yield without identifying the limiting reagent. Always state which measure you are calculating and use the correct formula.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style7 marks

Bromoethane is made by

\text{C}_2\text{H}_5\text{OH} + \text{HBr} \rightarrow \text{C}_2\text{H}_5\text{Br} + \text{H}_2\text{O}

. In one preparation,

9.2\ \text{g}

of ethanol reacts with excess

\text{HBr}

to give

15.3\ \text{g}

of bromoethane. (Molar masses:

\text{C}_2\text{H}_5\text{OH}=46.0

\text{HBr}=81.0

\text{C}_2\text{H}_5\text{Br}=109.0

\text{H}_2\text{O}=18.0\ \text{g mol}^{-1}

.) (a) Calculate the theoretical yield of bromoethane. (b) Calculate the percentage yield. (c) Calculate the atom economy and comment on it.

Show worked answer →

A 7 mark question rewards the theoretical yield, the percentage yield, and the atom economy.

(a) Moles of ethanol (limiting) $= \dfrac{9.2}{46.0} = 0.20\ \text{mol}$ . The ratio is 1:1, so theoretical bromoethane $= 0.20\ \text{mol}$ , mass $= 0.20 \times 109.0 = 21.8\ \text{g}$ .

(b)

\%\,\text{yield} = \frac{15.3}{21.8} \times 100 = 70.2\%.

(c) Atom economy uses the molar masses from the equation:

\text{atom economy} = \frac{109.0}{109.0 + 18.0} \times 100 = \frac{109.0}{127.0} \times 100 = 85.8\%.

The atom economy is below

100\%

because the substitution produces a water by-product; the

14.2\%

of mass lost to water is inherent to this route, separate from the practical losses that lowered the yield to

70\%

Markers reward $21.8\ \text{g}$ theoretical, $70.2\%$ yield, and $85.8\%$ atom economy with the by-product comment.

WACE 2023-style5 marks

A target molecule can be made by an addition reaction (one product) or a condensation reaction (product plus water). (a) State which route has the higher atom economy and why. (b) The condensation route has a percentage yield of

95\%

and the addition route

70\%

. Discuss which route is more efficient overall.

Show worked answer →

A 5 mark answer needs the atom-economy comparison and a reasoned overall judgement.

(a) The addition route has the higher atom economy, because it has a single product so all reactant atoms are incorporated ( $100\%$ in principle). The condensation route loses some mass as a water by-product, so its atom economy is below $100\%$ .

(b) "Efficiency" depends on which measure matters. The addition route wastes no atoms but converts only $70\%$ of the theoretical product, so material is lost in practice. The condensation route converts $95\%$ but inherently discards atoms as water. Overall, the addition route is usually preferred on green-chemistry grounds (no by-product to treat, higher atom economy), but if the addition reagents are far more expensive or the lost $30\%$ is costly, the higher-yielding condensation route could be more efficient economically. A full judgement weighs atom economy, yield, reagent cost, and waste treatment together.

Markers reward addition having higher atom economy with the single-product reason, and a balanced discussion weighing atom economy against the differing yields.

Practise this

Sources & how we know this

Chemistry ATAR course: syllabus and support materials — SCSA
Chemistry: past ATAR course examinations — SCSA