Chemical synthesis and analysis: WACE Year 12 Chemistry

Syllabus dot point

“Evaluate chemical synthesis using percentage yield and atom economy and green chemistry principles, and interpret instrumental analysis data to identify substances”

9 min readUpdated 2 Jun 2026

Quick answer

Chemical synthesis is judged on efficiency: percentage yield compares the mass of product actually obtained with the theoretical maximum, while atom economy measures how much of the reactant mass ends up in the desired product. Green chemistry aims to maximise both and reduce waste, and instrumental techniques such as mass spectrometry, infrared and NMR spectroscopy identify and measure substances.

What this dot point is asking

Designing a chemical synthesis means choosing reactions that convert available starting materials into a target product efficiently, safely and with minimal waste. Two quantitative measures judge that efficiency, and modern practice is guided by green chemistry.

Percentage yield

The percentage yield compares the amount of product actually obtained with the maximum predicted by stoichiometry (the theoretical yield):

\%\,\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

Real syntheses never reach 100 percent because of side reactions, incomplete reactions (equilibria), losses during purification, and impurities in the reactants.

Percentage yield

\%\,\text{yield} = \frac{\text{actual mass (or moles) of product}}{\text{theoretical mass (or moles) of product}} \times 100

The theoretical yield is found from the limiting reagent using mole ratios.

Atom economy

Atom economy measures how much of the total mass of all reactants ends up in the desired product, rather than in by-products:

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

A reaction can have a high yield but poor atom economy if it produces large amounts of unwanted by-products. Addition reactions tend to have high atom economy (everything ends up in one product); substitution and elimination reactions have lower atom economy because they release small molecules.

Yield and atom economy are different

Percentage yield measures how much product you actually got versus the maximum possible. Atom economy measures how much of the reactant mass is built into the wanted product by the reaction itself. A reaction can score well on one and poorly on the other.

Green chemistry

Green chemistry designs syntheses to reduce environmental impact: maximising atom economy and yield, using renewable feedstocks, avoiding toxic or hazardous reagents and solvents, using catalysts rather than stoichiometric reagents, minimising energy use, and designing products and by-products that are safe and degradable. The goal is to prevent waste rather than to clean it up afterwards.

Instrumental analysis

Chemists identify and measure substances using instrumental techniques. The WACE course expects you to interpret data from:

Mass spectrometry (MS): measures the mass of the molecule (molecular ion peak gives the molar mass) and the masses of fragments, which help deduce structure.
Infrared (IR) spectroscopy: absorption bands at characteristic wavenumbers reveal functional groups (for example a broad band near $3300\ \text{cm}^{-1}$ for O-H, a strong band near $1700\ \text{cm}^{-1}$ for C=O).
Nuclear magnetic resonance (NMR) spectroscopy: the number of signals, their chemical shifts and the relative peak areas indicate the different proton (or carbon) environments and how many of each.

Combining these techniques lets a chemist confirm the identity and purity of a synthesised product.

Calculating percentage yield

Question

A student reacts ethanol with excess ethanoic acid and obtains $6.6\ \text{g}$ of ethyl ethanoate ( $M = 88\ \text{g mol}^{-1}$ ) from $0.10\ \text{mol}$ of ethanol. Find the percentage yield.

Solution

Ethanol is the limiting reagent (acid is in excess), and the mole ratio of ethanol to ester is $1:1$ , so the theoretical yield is $0.10\ \text{mol}$ of ester.

Theoretical mass $= 0.10 \times 88 = 8.8\ \text{g}$ .

\%\,\text{yield} = \frac{6.6}{8.8} \times 100 = 75\%

Esterification is reversible, so a yield below 100 percent is expected.

When answering synthesis questions in the WACE examination, identify the limiting reagent before computing the theoretical yield, keep yield and atom economy separate, and quote IR and NMR evidence specifically (state the wavenumber or shift and the group it indicates).

Mixing up yield and atom economy

A common error is calculating atom economy when the question asks for percentage yield, or vice versa; they answer different questions. Another frequent slip is using the wrong reactant as the basis: the theoretical yield must come from the limiting reagent. A third is giving vague spectroscopy answers; markers want the specific band or shift and the functional group it shows.

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 2022-style7 marks

Ethene can be converted to ethanol by two routes: (i) hydration,

\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}

; (ii) reaction with concentrated sulfuric acid then water, which overall also gives ethanol but produces some by-products. (a) Calculate the atom economy of route (i). (b) A hydration plant uses

2.80\ \text{kg}

of ethene and obtains

3.45\ \text{kg}

of ethanol. Calculate the percentage yield. (

M(\text{C}_2\text{H}_4)=28.0

M(\text{C}_2\text{H}_5\text{OH})=46.0\ \text{g mol}^{-1}

.) (c) State why route (i) is preferred on green-chemistry grounds.

Show worked answer →

A 7 mark question rewards the atom economy, the yield, and the green-chemistry comparison.

(a) In route (i) the only product is ethanol, so all reactant mass ends up in the product:

\text{atom economy} = \frac{46.0}{28.0 + 18.0} \times 100 = \frac{46.0}{46.0} \times 100 = 100\%.

(b) Moles of ethene $= \dfrac{2800}{28.0} = 100\ \text{mol}$ , so theoretical ethanol $= 100\ \text{mol}$ , theoretical mass $= 100 \times 46.0 = 4600\ \text{g} = 4.60\ \text{kg}$ .

\%\,\text{yield} = \frac{3.45}{4.60} \times 100 = 75\%.

(c) Route (i) is a direct addition with $100\%$ atom economy (no by-products) and uses a reusable acid catalyst rather than stoichiometric concentrated sulfuric acid, so it generates less waste and uses fewer hazardous reagents, both key green-chemistry aims.

Markers reward $100\%$ atom economy, $75\%$ yield (from the limiting ethene), and a green-chemistry reason citing atom economy or waste.

WACE 2020-style6 marks

An organic compound has a molecular ion peak at

m/z = 60

in its mass spectrum, a broad infrared absorption near

3000\ \text{cm}^{-1}

together with a strong band near

1710\ \text{cm}^{-1}

, and two peaks in its proton NMR. Deduce the structure and justify each piece of evidence.

Show worked answer →

A 6 mark question rewards using each technique to build and confirm the structure.

The molecular ion at $m/z = 60$ gives a molar mass of $60\ \text{g mol}^{-1}$ , consistent with $\text{C}_2\text{H}_4\text{O}_2$ (ethanoic acid, $2(12) + 4(1) + 2(16) = 60$ ).

The broad infrared band near $3000\ \text{cm}^{-1}$ indicates an $\text{O-H}$ of a carboxylic acid (broadened by hydrogen bonding), and the strong band near $1710\ \text{cm}^{-1}$ indicates a C=O carbonyl, together pointing to a $\text{-COOH}$ group.

The proton NMR shows two signals, consistent with two hydrogen environments: the $\text{-CH}_3$ protons and the acidic $\text{-COOH}$ proton.

All three lines of evidence agree on ethanoic acid, $\text{CH}_3\text{COOH}$ .

Markers reward $M = 60$ giving the formula, the $\text{O-H}$ and C=O assignments, the two NMR environments, and the consistent conclusion of ethanoic acid.

Practise this

Sources & how we know this

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