Atom Economy and Percentage Yield

In industrial and laboratory chemistry, it is not enough to simply make a product — you need to consider how efficiently you are making it. Two key measures of efficiency are atom economy and percentage yield. They answer different questions: atom economy asks "how much of the reactant atoms end up in the desired product?" while percentage yield asks "how much product did I actually obtain compared to the maximum possible?"

Atom Economy

Atom economy measures the proportion of reactant atoms that are converted into the desired product. It is a theoretical measure based on the balanced equation — it does not depend on how the reaction is actually carried out.

Atom economy (%) = (Mᵣ of desired product / sum of Mᵣ of all products) × 100

Alternatively (and equivalently), taking coefficients into account:

Atom economy (%) = (Mᵣ of desired product × its coefficient / sum of Mᵣ × coefficient for all products) × 100

Worked Example 1

Calculate the atom economy for the production of ethanol by fermentation.

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

Mᵣ of desired product (2 × C₂H₅OH) = 2 × 46.0 = 92.0 Mᵣ of all products = (2 × 46.0) + (2 × 44.0) = 92.0 + 88.0 = 180.0

Atom economy = (92.0 / 180.0) × 100 = 51.1%

This means that only about half of the reactant atoms end up in the ethanol. The rest become CO₂, which is a waste product (from the atom economy perspective).

Worked Example 2

Calculate the atom economy for the following addition reaction.

CH₂=CH₂ + HBr → CH₃CH₂Br

Mᵣ of desired product = 109.0 Sum of Mᵣ of all products = 109.0 (only one product)

Atom economy = (109.0 / 109.0) × 100 = 100%

Addition reactions always have 100% atom economy because there is only one product — all reactant atoms end up in the desired product.

Worked Example 3

Calculate the atom economy for producing iron from iron oxide using carbon monoxide.

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Desired product: 2Fe, Mᵣ = 2 × 56.0 = 112.0 All products: (2 × 56.0) + (3 × 44.0) = 112.0 + 132.0 = 244.0

Atom economy = (112.0 / 244.0) × 100 = 45.9%

Atom Economy by Reaction Type

Reaction type	Typical atom economy	Reason
Addition	100%	Only one product
Rearrangement	100%	Only one product
Substitution	< 100%	By-product formed
Elimination	< 100%	Small molecule lost (e.g. H₂O, HCl)
Fermentation	~51%	CO₂ is a by-product

Atom Economy and Green Chemistry

Atom economy is an important concept in green chemistry (sustainable chemistry). A high atom economy means:

Less waste is produced.
Fewer raw materials are needed.
Less energy is spent on separating products and disposing of by-products.
The process is more sustainable and cost-effective.

Industrial chemists prefer reactions with high atom economy. For example, the pharmaceutical industry has moved towards catalytic processes and addition reactions wherever possible, replacing older substitution-based synthesis routes that generate large amounts of waste.

Real-world example: The Haber process (N₂ + 3H₂ → 2NH₃) has 100% atom economy because ammonia is the only product. However, the percentage yield per pass is only about 15% — the unreacted gases are recycled to improve overall efficiency.

Percentage Yield

In practice, you rarely obtain the maximum theoretical amount of product. Percentage yield compares what you actually obtained with what you theoretically could have obtained:

Percentage yield (%) = (actual yield / theoretical yield) × 100

The theoretical yield is the maximum amount of product calculated from the balanced equation and the amount of limiting reagent used.

Worked Example 4

A student reacts 10.0 g of CaCO₃ with excess hydrochloric acid. They collect 4.00 g of CaCl₂. Calculate the percentage yield.

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Step 1: Moles of CaCO₃ = 10.0 / 100.0 = 0.100 mol

Step 2: From the equation, 1 mol CaCO₃ produces 1 mol CaCl₂. Theoretical moles of CaCl₂ = 0.100 mol

Step 3: Theoretical yield = 0.100 × 111.0 = 11.1 g

Step 4: Percentage yield = (4.00 / 11.1) × 100 = 36.0%

Worked Example 5

A chemist reacts 5.40 g of aluminium with excess copper(II) sulfate solution. They collect 14.2 g of copper. Calculate the percentage yield.

2Al(s) + 3CuSO₄(aq) → Al₂(SO₄)₃(aq) + 3Cu(s)

Step 1: Moles of Al = 5.40 / 27.0 = 0.200 mol

Step 2: Mole ratio Al : Cu = 2 : 3, so moles of Cu = 0.200 × 3/2 = 0.300 mol

Step 3: Theoretical yield of Cu = 0.300 × 63.5 = 19.05 g

Step 4: Percentage yield = (14.2 / 19.05) × 100 = 74.5%

Why Is the Actual Yield Less Than the Theoretical Yield?

There are several reasons why the actual yield is almost always less than 100%:

Reason	Explanation
Incomplete reaction	The reaction may not go to completion — an equilibrium may be established
Side reactions	Reactants may undergo unwanted alternative reactions
Product lost during transfer	Product is lost when transferring between containers, filtering, etc.
Product lost during purification	Washing, recrystallisation, and drying all result in some product loss
Impure reactants	If reactants contain impurities, less product forms than expected
Mechanical losses	Some product remains stuck to glassware

Important: Percentage yield should never exceed 100%. If a student calculates a yield > 100%, it usually means the product is impure (contains solvent or by-products) or there is a calculation error.

The Limiting Reagent

The limiting reagent is the reactant that is completely used up first, determining the maximum amount of product. The other reagent(s) are in excess.

Worked Example 6

2.00 g of sodium reacts with 3.00 g of chlorine. Which is the limiting reagent, and what mass of NaCl is formed?

2Na + Cl₂ → 2NaCl

Step 1: Moles of Na = 2.00 / 23.0 = 0.0870 mol Step 2: Moles of Cl₂ = 3.00 / 71.0 = 0.0423 mol

Step 3: From the equation, 2 mol Na reacts with 1 mol Cl₂. Moles of Na needed for 0.0423 mol Cl₂ = 2 × 0.0423 = 0.0845 mol. We have 0.0870 mol Na, which is more than 0.0845, so Cl₂ is the limiting reagent.

Step 4: Moles of NaCl = 2 × 0.0423 = 0.0845 mol (2:2 ratio with Na consumed) Mass of NaCl = 0.0845 × 58.5 = 4.95 g

Combining Atom Economy and Percentage Yield

To get the overall efficiency of a process, you can combine both measures:

Overall efficiency (%) = (atom economy / 100) × (percentage yield / 100) × 100

Worked Example 7

A reaction has an atom economy of 51.1% and a percentage yield of 80.0%. What is the overall efficiency?

Overall efficiency = (51.1/100) × (80.0/100) × 100 = 40.9%

This means only about 41% of the starting material atoms end up in the desired product in practice.

Common Mistakes

Mistake	Consequence	Fix
Using reactant Mᵣ instead of product Mᵣ in atom economy	Wrong atom economy	Use Mᵣ of products only (sum in denominator)
Confusing atom economy and percentage yield	They measure different things	Atom economy = theoretical (from equation); yield = experimental
Forgetting coefficients in atom economy	Incorrect Mᵣ sums	Multiply Mᵣ by coefficient for each product
Using excess reagent to calculate theoretical yield	Theoretical yield too high	Always use the limiting reagent
Getting percentage yield > 100%	Impure product or error	Check calculation; product may need purification

Comparison: Atom Economy vs Percentage Yield

Feature	Atom economy	Percentage yield
Based on	Balanced equation (theoretical)	Experimental results
Can it be improved by better technique?	No — it is fixed by the equation	Yes — careful work increases yield
Can it exceed 100%?	No	No (if it does, product is impure)
What it measures	Efficiency of atom usage	Efficiency of product collection
Affected by side reactions?	No	Yes
Affected by equilibrium?	No	Yes (incomplete reactions lower yield)

Understanding both atom economy and percentage yield — and the difference between them — is essential for evaluating chemical processes in industry and in the laboratory.

A-Level Deep Dive: Atom Economy and Percentage Yield

Spec mapping

Edexcel 9CH0 specification Topic 5 — Formulae, Equations and Amounts of Substance, sub-strand on atom economy and percentage yield, requires students to define and calculate atom economy ( $M_r$ of useful product / total $M_r$ of products × 100) and percentage yield (actual yield / theoretical yield × 100), and to discuss their importance for green chemistry and industrial sustainability (refer to the official specification document for exact wording). Examined across Papers 2 (organic synthesis context) and 3 (synoptic practical with CP14 esterification, CP6 distillation). The data sheet provides $A_r$ values; the formulae for atom economy and yield must be memorised.

Worked example with full mark scheme

Question (8 marks): The Williamson ether synthesis prepares ethoxyethane from sodium ethoxide and bromoethane:

$\text{C}_2\text{H}_5\text{ONa} + \text{C}_2\text{H}_5\text{Br} \rightarrow \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 + \text{NaBr}$ .

A student starts with 13.6 g of sodium ethoxide ( $M_r = 68.0$ ) and excess bromoethane, and isolates 9.40 g of ethoxyethane ( $M_r = 74.0$ ).

(a) Calculate the atom economy of the reaction. (2)

(b) Calculate the theoretical yield of ethoxyethane. (2)

(d) Identify two reasons why the percentage yield is below 100%. (2)

Solution with mark scheme:

(a) Atom economy:

Useful product: ethoxyethane, $M_r = 74.0$ .

Total $M_r$ of products: $74.0 + (23.0 + 79.9) = 74.0 + 102.9 = 176.9$ .

Atom economy = $74.0/176.9 \times 100 = 41.8\%$ .

M1 — applying the formula correctly. A1 — 41.8% (allow 41–42%).

(b) Theoretical yield:

$n(\text{NaOEt}) = 13.6/68.0 = 0.200\,\text{mol}$ .

1 : 1 stoichiometry: $n(\text{ether}) = 0.200\,\text{mol}$ .

$m_\text{theory} = 0.200 \times 74.0 = 14.8\,\text{g}$ .

M1 — moles and stoichiometry. A1 — 14.8 g.

M1 — applying yield formula. A1 — 63.5%.

(d) Two reasons:

Side reactions (e.g. elimination giving ethene + sodium ethoxide instead of substitution).
Loss during purification (e.g. distillation, recrystallisation).

B1 B1.

Total: 8 marks (M3 A3 B2).

Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (6 marks): Aspirin ( $M_r = 180.2$ ) is prepared from salicylic acid ( $M_r = 138.1$ ) and ethanoic anhydride ( $M_r = 102.1$ ):

$\text{C}_7\text{H}_6\text{O}_3 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_9\text{H}_8\text{O}_4 + \text{CH}_3\text{COOH}$ .