Percentage Yield and Atom Economy (Part 2/2) Revision Notes | A Level AQA Chemistry

1. Why Atom Economy Matters Beyond the Calculation

You already know how to calculate percentage yield — how much of the product you actually collect. Atom economy asks a different question: of all the atoms you put in, how many end up in the product you actually want? These two measures together give a complete picture of how efficient a reaction really is.

A reaction can have a 100% percentage yield and still be deeply wasteful. If half the atoms in your reactants end up as unwanted by-products, you have thrown away half your raw materials — and you still have to deal with the waste. That is the problem atom economy was designed to expose.

This distinction matters enormously in industry, where raw material costs, waste disposal, and environmental regulation all affect whether a process is commercially viable. Understanding atom economy is therefore not an abstract exercise — it is central to how chemists design sustainable processes.

2. The Atom Economy Formula

Notice that the formula uses the balanced equation — not experimental masses. This means atom economy is a fixed property of a given reaction pathway. You cannot improve it by being more careful in the lab; you can only improve it by choosing a different reaction route.

2a. Worked Example: Addition vs Substitution

The clearest way to see atom economy in action is to compare two different reaction types: addition reactions and substitution reactions. Addition reactions are the gold standard for atom economy; substitution reactions are typically much less efficient. Here is a concrete comparison.

Worked example: Calculating Atom Economy: Addition Reaction (Ethene + Bromine)

Calculate the atom economy for the addition of bromine to ethene to form 1,2-dibromoethane. The balanced equation is: $CH_2=CH_2(g) + Br_2(l) \rightarrow CH_2BrCH_2Br(l)$

1. $CH_2=CH_2(g) + Br_2(l) \rightarrow CH_2BrCH_2Br(l)$\nDesired product: $CH_2BrCH_2Br$ (1,2-dibromoethane) — In an addition reaction, both reactants combine to form a single product. There are no by-products. Check atom balance: C: 2 = 2 ✓, H: 4 = 4 ✓, Br: 2 = 2 ✓.
2. $M_r(CH_2BrCH_2Br) = (2 \times 12) + (4 \times 1) + (2 \times 80)$\n$= 24 + 4 + 160 = 188$ — C = 12, H = 1, Br = 80. Count each atom carefully from the formula.
3. $M_r(CH_2=CH_2) = (2 \times 12) + (4 \times 1) = 24 + 4 = 28$\n$M_r(Br_2) = 2 \times 80 = 160$\n$\sum M_r(\text{reactants}) = 28 + 160 = 188$ — Add the molecular masses of every reactant. Since stoichiometric coefficients are both 1 here, no further multiplication is needed.
4. $\text{Atom Economy} = \frac{188}{188} \times 100 = 100\%$ — Because this is an addition reaction with a single product, every atom from the reactants ends up in the product. Atom economy is always 100% for addition reactions that produce a single product.

Answer: Atom economy = 100%. This is expected for any addition reaction producing a single product, since all reactant atoms are incorporated into that product.

Now compare that with a substitution reaction, where a leaving group is displaced and exits as a by-product. The atom economy drops significantly — and the calculation shows exactly why.

Worked example: Calculating Atom Economy: Substitution Reaction (Production of Ethanol from Bromoethane)

Calculate the atom economy for the production of ethanol by nucleophilic substitution of bromoethane with aqueous sodium hydroxide. The balanced equation is: $CH_3CH_2Br + NaOH \rightarrow CH_3CH_2OH + NaBr$

1. Desired product: $CH_3CH_2OH$ (ethanol) By-product: $NaBr$ (sodium bromide) — In a substitution reaction, the leaving group (Br) exits with the sodium ion, forming a by-product. Not all atoms end up in the desired product.
2. $M_r(CH_3CH_2OH) = (2 \times 12) + (6 \times 1) + 16 = 24 + 6 + 16 = 46$ — Only the desired product goes in the numerator. The by-product is excluded from the top of the fraction.
3. $M_r(CH_3CH_2Br) = (2 \times 12) + (5 \times 1) + 80 = 24 + 5 + 80 = 109$ $M_r(NaOH) = 23 + 16 + 1 = 40$ $\sum M_r(\text{reactants}) = 109 + 40 = 149$ — Both reactants are included in the denominator, regardless of whether their atoms end up in the desired product.
4. $\text{Atom Economy} = \frac{46}{149} \times 100 = 30.9\%$ — Only about 31% of the reactant atoms end up in the desired product. The remaining ~69% become sodium bromide waste.
5. $M_r(NaBr) = 23 + 80 = 103$ $46 + 103 = 149$ ✓ — The masses of desired product and by-product must sum to the total reactant mass. This confirms the calculation is internally consistent.

Answer: Atom economy ≈ 30.9%. Roughly 69% of the reactant atoms are wasted as sodium bromide by-product.

3. Atom Economy and Reaction Type

There is a reliable pattern worth knowing: the type of reaction strongly predicts the atom economy you can expect. Addition reactions always achieve 100% atom economy because all reactant atoms combine into one product. Substitution, elimination, and other reactions that produce by-products will always have atom economies below 100%.

This pattern has direct implications for route selection in synthesis. When two routes to the same product exist — one via addition and one via substitution — the addition route will always have a higher atom economy, all else being equal. Chemists designing industrial processes must weigh this alongside other factors such as rate, yield, and cost of reagents.

4. Economic, Ethical, and Environmental Advantages of High Atom Economy

The AQA specification requires you to evaluate the advantages of high atom economy across three dimensions: economic, ethical, and environmental. These are not separate concerns — they are deeply interconnected. A process that wastes fewer atoms is simultaneously cheaper to run, less harmful to the environment, and more ethically defensible.

From an economic perspective, raw materials are the largest cost in most industrial chemical processes. A reaction with a high atom economy converts more of those raw materials into saleable product, reducing the cost per unit of product. Fewer by-products also means lower costs for waste treatment, disposal, or storage — all of which carry regulatory and financial burdens.

The environmental case is equally compelling. By-products that cannot be sold must be disposed of, and many chemical by-products are hazardous. High atom economy processes generate less waste per unit of product, reducing the risk of pollution, the energy required for waste treatment, and the carbon footprint of the overall process. This aligns with the principles of green chemistry, which seeks to design reactions that are inherently less wasteful rather than relying on end-of-pipe waste treatment.

Ethically, there is a responsibility to future generations to use finite raw material resources as efficiently as possible. Many chemical feedstocks are derived from non-renewable sources such as crude oil. A process with a higher atom economy extends the useful life of those resources and reduces the environmental legacy left by chemical manufacturing. Industries and governments increasingly recognise this through legislation, carbon pricing, and sustainability reporting requirements.

• Economic: Higher atom economy means more product per unit of raw material, reducing cost per tonne of product and lowering waste disposal expenditure.
• Economic: By-products that cannot be sold represent a direct financial loss; minimising them improves profitability.
• Environmental: Less waste generated per unit of product means lower risk of pollution and reduced energy demand for waste treatment.
• Environmental: Fewer hazardous by-products reduces the risk of environmental contamination from storage, transport, or accidental release.
• Ethical: More efficient use of finite, often non-renewable, raw materials is a responsibility to future generations.
• Ethical: Reducing chemical waste aligns with the principle that industry should not externalise environmental costs onto society.

5. Atom Economy vs Percentage Yield — A Critical Distinction

These two measures are frequently confused, and examiners test this distinction directly. Atom economy is a property of the reaction equation — it tells you what fraction of reactant atoms can theoretically end up in the desired product. Percentage yield is a property of the experiment — it tells you what fraction of the theoretical maximum product was actually obtained.

A reaction can have a high atom economy but a low percentage yield if, for example, the reaction does not go to completion or product is lost during purification. Conversely, a reaction can have a near-100% percentage yield but a very low atom economy if most of the reactant atoms end up as by-products. Both measures must be high for a process to be truly efficient.

6. Industrial Context and Route Selection

In industrial chemistry, atom economy is one of several criteria used to evaluate and select synthetic routes. A process with a high atom economy is not automatically the best choice — rate, selectivity, cost of reagents, energy requirements, and safety must all be considered. However, atom economy provides a rapid, equation-level filter that can rule out inherently wasteful routes before more detailed analysis begins.

Consider the industrial production of ethanol. The hydration of ethene — an addition reaction — has an atom economy of 100%, since water and ethene combine to give ethanol as the sole product. Fermentation of glucose, by contrast, produces ethanol and carbon dioxide, giving an atom economy well below 100%. From an atom economy perspective alone, hydration is the superior route. However, fermentation uses renewable feedstocks (sugars from plants) rather than ethene derived from crude oil, which introduces an ethical and environmental counterargument. This illustrates why atom economy must be evaluated alongside other factors.

Worked example: Atom Economy Comparison: Two Routes to Ethanol

Compare the atom economy of (a) hydration of ethene and (b) fermentation of glucose. Balanced equations: (a) $CH_2=CH_2(g) + H_2O(g) \rightarrow CH_3CH_2OH(g)$; (b) $C_6H_{12}O_6(aq) \rightarrow 2CH_3CH_2OH(aq) + 2CO_2(g)$

1. $M_r(CH_3CH_2OH) = (2 \times 12) + (6 \times 1) + 16 = 24 + 6 + 16 = 46$\n$M_r(CH_2=CH_2) = (2 \times 12) + (4 \times 1) = 28$\n$M_r(H_2O) = (2 \times 1) + 16 = 18$\n$\sum M_r(\text{reactants}) = 28 + 18 = 46$\n$\text{Atom Economy} = \frac{46}{46} \times 100 = 100\%$\nAtom balance check: C: 2 = 2 ✓, H: 4+2 = 6 ✓, O: 1 = 1 ✓ — Addition reaction: single product, so all reactant atoms end up in ethanol. Atom economy is 100%.
2. $M_r(C_6H_{12}O_6) = (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180$\nDesired product: $2 \times CH_3CH_2OH$, so $2 \times 46 = 92$\n$\text{Atom Economy} = \frac{92}{180} \times 100 = 51.1\%$ — The denominator is the molecular mass of glucose (the only reactant). The numerator uses 2 × 46 because two moles of ethanol are produced per mole of glucose. The remaining mass (180 − 92 = 88) corresponds to 2 × CO₂ (2 × 44 = 88), confirming the calculation.
3. $M_r(2 \times CO_2) = 2 \times (12 + 2 \times 16) = 2 \times 44 = 88$\n$92 + 88 = 180$ ✓ — The masses of desired product and by-product sum to the reactant mass, confirming internal consistency.
4. Hydration: 100% atom economy\nFermentation: 51.1% atom economy — Hydration is superior on atom economy grounds. However, fermentation uses renewable glucose from plant material, whereas hydration requires ethene from crude oil (non-renewable). A full evaluation must weigh atom economy against feedstock sustainability.

Answer: Hydration of ethene: atom economy = 100%. Fermentation of glucose: atom economy = 51.1%. Hydration is more atom-economical, but fermentation uses a renewable feedstock — both factors are relevant in a full industrial evaluation.

7. Exam Technique and Common Mistakes

Atom economy questions appear in calculation form, in 'evaluate' extended responses, and in 'suggest' questions where you must apply the concept to an unfamiliar context. Each command word demands a different approach, and the marks are distributed accordingly.

For 'calculate' questions, the mark scheme typically awards one mark for the correct formula or method, one mark for correct substitution (including stoichiometric coefficients), and one mark for the correct numerical answer with no unit required (atom economy is a percentage). Always show your working — a correct answer with no working risks losing method marks if the number is wrong.

For 'evaluate' or 'explain' questions about the advantages of high atom economy, the examiner is looking for specific, developed points — not vague statements. Writing 'it is better for the environment' earns no marks. Writing 'a higher atom economy produces less by-product waste per unit of desired product, reducing the risk of environmental pollution and lowering the cost of waste disposal' earns marks because it is specific, causal, and linked to a consequence.

8. Bringing It Together: A Final Retrieval Check

You have now covered both the calculation and the wider significance of atom economy. The two are inseparable in exam questions: a 'calculate' question tests the method, while an 'evaluate' question tests whether you understand what the number actually means for industry, society, and the environment.

Atom economy sits at the heart of sustainable chemistry. Every time a chemist selects a synthetic route, they are making a decision that has economic consequences for their employer, environmental consequences for the planet, and ethical consequences for society. The formula is straightforward; the reasoning behind it is what separates a good exam answer from an excellent one.