SACE Board SACE 2022-style practice: Ethene oxide is made by direct oxidation: 2C_2H_4 + O_2 → 2C_2H_4O. An older route was C_2H_4 + Cl_2 + H_2O → C_2H_4O + 2HCl. (a) Calculate the atom economy of each route for ethene oxide. (b) State which route is greener and justify your answer. (M: C_2H_4O = 44.0, HCl = 36.5\ g mol^-1.)

Atom economy = M(desired product)M(all products) × 100. Route 1: only product is C_2H_4O, so atom economy = 44.044.0 × 100 = 100\%. (2 marks) Route 2: products are C_2H_4O (44.0) and 2HCl (2 × 36.5 = 73.0): atom economy = 44.044.0 + 73.0 × 100 = 44.0117.0 × 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 HCl. (1 mark)

§-Syllabus dot point

SAChemistrySyllabus dot point

How can chemical processes be designed to reduce waste and environmental impact?

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

The principles of green chemistry, the difference between atom economy and percentage yield, and how to evaluate a process for waste and sustainability, with fully worked SACE-style atom-economy and yield calculations.

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
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

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

Atom economy

Atom economy versus percentage yield

These two measures are often confused but answer different questions.

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.

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 practice questions

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

SACE 20225 marksEthene 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 20204 marksIn 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)