Green chemistry principles (SACE Stage 2 Chemistry)

Syllabus dot point

“Apply green chemistry principles, including atom economy, to evaluate the efficiency and sustainability of chemical processes.”

8 min readUpdated 2 Jun 2026

Quick answer

Green chemistry designs processes that minimise waste, hazard and energy use. A central measure is atom economy, the percentage of reactant atoms that end up in the desired product: $\dfrac{M(\text{desired product})}{M(\text{all products})} \times 100$ . A reaction with one product has $100\%$ atom economy; one that makes by-products wastes those atoms. Atom economy is a property of the balanced equation and is distinct from percentage yield, which measures how much product is actually obtained.

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What this dot point is asking
Lead worked calculation
The principles of green chemistry
Atom economy
Atom economy versus percentage yield
Evaluating a process
Why catalysts are green
Why it matters for managing processes
Exam-style questions

What this dot point is asking

SACE expects you to calculate atom economy and percentage yield, distinguish them clearly, and use green chemistry principles to evaluate which of two routes is more sustainable.

Lead worked calculation

Comparing two routes by atom economy

Compare the atom economy for making hydrogen by two routes. Route A (steam reforming): $\text{CH}_4 + 2\text{H}_2\text{O} \rightarrow \text{CO}_2 + 4\text{H}_2$ . Route B (electrolysis of water): $2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$ . ( $M$ : $\text{H}_2 = 2.0$ , $\text{CO}_2 = 44.0$ , $\text{O}_2 = 32.0\ \text{g mol}^{-1}$ .)

Route A atom economy

Desired product is $4\text{H}_2 = 8.0$ ; all products $= 8.0 + 44.0 = 52.0$ .

\text{atom economy} = \frac{8.0}{52.0} \times 100 = 15.4\%

Route B atom economy

Desired product is $2\text{H}_2 = 4.0$ ; all products $= 4.0 + 32.0 = 36.0$ .

\text{atom economy} = \frac{4.0}{36.0} \times 100 = 11.1\%

Interpret carefully

By atom economy alone, neither is high, and route A looks slightly better. But green evaluation is broader: route B can use renewable electricity and emits no $\text{CO}_2$ , while route A produces greenhouse $\text{CO}_2$ . Atom economy is one criterion among several, not the whole picture.

The principles of green chemistry

Green chemistry

Green chemistry is the design of chemical products and processes that reduce or eliminate the use and generation of hazardous and wasteful substances. Its principles include: prevent waste rather than treat it; maximise atom economy; use safer solvents and reagents; design for energy efficiency (mild conditions); use renewable feedstocks; use catalysts rather than stoichiometric reagents; and design products that degrade safely.

Atom economy

\text{atom economy} = \frac{M(\text{desired product})}{M(\text{total products})} \times 100

A high atom economy means most reactant atoms become useful product and little is wasted as by-product. Addition reactions (one product) reach

100\%

; substitution and elimination reactions lose atoms to by-products and score lower.

Atom economy versus percentage yield

These two measures are often confused but answer different questions.

The crucial distinction

Percentage yield asks: of the maximum product the equation allows, how much did I actually get? It is reduced by losses, side reactions and reaching equilibrium. Atom economy asks: of all the atoms in the reactants, what fraction the equation sends into my desired product? It is fixed by the balanced equation regardless of how the reaction is run. A reaction can have $100\%$ atom economy but $50\%$ yield, or vice versa. A truly green process needs both high.

Evaluating a process

To judge how green a process is, consider more than one number:

Atom economy and yield: high values mean efficient use of feedstock and less waste.
Hazard: are the reagents, solvents or products toxic or corrosive?
Energy: does it need extreme temperature or pressure (high energy cost and emissions)?
Feedstock: renewable or from finite fossil resources?
By-products: are they useful, recyclable, or harmful waste?
Catalysis: catalysts cut energy use and avoid stoichiometric reagent waste.

Why catalysts are green

Catalysts let reactions run at lower temperatures and pressures, cutting energy use and emissions, and they replace stoichiometric reagents that would otherwise become waste. Because they are regenerated, a small amount serves repeatedly. This is why the catalyst principle is central to green process design and links directly to the Haber process and catalytic converters.

Treating atom economy and yield as the same

The most common SACE error is conflating atom economy with percentage yield. Atom economy is calculated from the balanced equation (molar masses of products) and is unaffected by how well the reaction runs; percentage yield is calculated from actual versus theoretical mass of product. Another error is judging "greenness" by atom economy alone, while ignoring energy use, hazard and renewable feedstocks, which can outweigh a high atom economy.

Why it matters for managing processes

Green chemistry reframes process design around waste prevention and sustainability rather than yield alone. Using atom economy alongside yield, energy and hazard assessments lets chemists choose genuinely cleaner routes, the modern standard for industrial chemistry and a direct application of managing chemical processes responsibly.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2022-style5 marks

Ethene oxide is made by direct oxidation:

2\text{C}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{C}_2\text{H}_4\text{O}

. An older route was

\text{C}_2\text{H}_4 + \text{Cl}_2 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_4\text{O} + 2\text{HCl}

. (a) Calculate the atom economy of each route for ethene oxide. (b) State which route is greener and justify your answer. (

M

\text{C}_2\text{H}_4\text{O} = 44.0

\text{HCl} = 36.5\ \text{g mol}^{-1}

Show worked answer →

Atom economy $= \dfrac{M(\text{desired product})}{M(\text{all products})} \times 100$ .

Route 1: only product is $\text{C}_2\text{H}_4\text{O}$ , so atom economy $= \dfrac{44.0}{44.0} \times 100 = 100\%$ . (2 marks)

Route 2: products are $\text{C}_2\text{H}_4\text{O}$ ( $44.0$ ) and $2\text{HCl}$ ( $2 \times 36.5 = 73.0$ ): atom economy $= \dfrac{44.0}{44.0 + 73.0} \times 100 = \dfrac{44.0}{117.0} \times 100 = 37.6\%$ . (2 marks)

(b) Route 1 is greener: its $100\%$ atom economy means all reactant atoms end up in the product with no waste by-product, whereas route 2 wastes most atoms as $\text{HCl}$ . (1 mark)

SACE 2020-style4 marks

In a synthesis,

25.0\ \text{g}

of reactant (

M = 100.0\ \text{g mol}^{-1}

) gave

18.3\ \text{g}

of product (

M = 122.0\ \text{g mol}^{-1}

) via a

1:1

reaction. (a) Calculate the percentage yield. (b) Explain the difference between percentage yield and atom economy.

Show worked answer →

(a) $n(\text{reactant}) = \dfrac{25.0}{100.0} = 0.250\ \text{mol}$ , so theoretical $n(\text{product}) = 0.250\ \text{mol}$ and theoretical mass $= 0.250 \times 122.0 = 30.5\ \text{g}$ . Percentage yield $= \dfrac{18.3}{30.5} \times 100 = 60.0\%$ . (2 marks)

(b) Percentage yield measures how much product is actually obtained compared with the theoretical maximum (affected by losses and incomplete reaction). Atom economy is a property of the equation: the proportion of reactant atoms that end up in the desired product. A reaction can have $100\%$ atom economy but a low yield, or high yield but poor atom economy. (2 marks)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board