What is percentage yield?

$percentage yield = actual yieldtheoretical yield × 100$

What is atom economy?

$atom economy = molar mass of desired producttotal molar mass of all products × 100$

SCSA WACE 2021-style practice: Bromoethane is made by C_2H_5OH + HBr → C_2H_5Br + H_2O. In one preparation, 9.2\ g of ethanol reacts with excess HBr to give 15.3\ g of bromoethane. (Molar masses: C_2H_5OH=46.0, HBr=81.0, C_2H_5Br=109.0, H_2O=18.0\ g mol^-1.) (a) Calculate the theoretical yield of bromoethane. (b) Calculate the percentage yield. (c) Calculate the atom economy and comment on it.

A 7 mark question rewards the theoretical yield, the percentage yield, and the atom economy. **(a)** Moles of ethanol (limiting) = 9.246.0 = 0.20\ mol. The ratio is 1:1, so theoretical bromoethane = 0.20\ mol, mass = 0.20 × 109.0 = 21.8\ g. **(b)** $\%\,yield = (15.3)/(21.8) × 100 = 70.2\%.$ **(c)** Atom economy uses the molar masses from the equation: $atom economy = (109.0)/(109.0 + 18.0) × 100 = (109.0)/(127.0) × 100 = 85.8\%.$ 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\ g theoretical, 70.2\% yield, and 85.8\% atom economy with the by-product comment.

§-Syllabus dot point

WAChemistrySyllabus dot point

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.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

\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

\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.

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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20217 marksBromoethane 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 20235 marksA 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.