Percentage Yield and Atom Economy [chemistry only] — AQA GCSE Chemistry

AQA 8462 — 4.3.3.1 & 4.3.3.2

⬢ CHEMISTRY ONLY — NOT IN COMBINED SCIENCE ⬢

Percentage Yield & Atom Economy

Two ways of measuring how good a reaction is. Yield asks "how much did we actually make?". Atom economy asks "how much of our starting material ended up where we wanted it to?". Both are calculation topics — and both have one specific examiner trap each.

What You'll Learn

Part 1 — Percentage Yield (4.3.3.1)

Calculate % yield from actual and theoretical masses, rearrange the formula to find any unknown, and explain why yield is always less than 100% — naming specific reasons.

Part 2 — Atom Economy (4.3.3.2)

Calculate atom economy from M_r values, and explain why high atom economy matters for sustainable industry. HT Compare two reaction pathways and justify which is preferable using the spec's five factors.

Why It Matters

These are both calculation questions on Paper 1. Together they are worth around 5–8 marks per series. The single most-marked error: confusing the two formulas with each other. Get the difference clear early.

What Is Percentage Yield?

When you carry out a reaction in the lab or in industry, the actual mass of product is almost always less than the maximum mass calculated from the balanced equation. Percentage yield measures how close you came to that maximum.

Percentage Yield

The actual mass of product obtained divided by the theoretical (maximum) mass, multiplied by 100.

% yield = (actual mass of product ÷ maximum theoretical mass) × 100

Key Idea

Yield is always < 100% in practice. A "100% yield" means every reactant atom converted perfectly to product with no losses — never happens in real reactions.

Why Yield Is Always Less Than 100%

This is a common 2-mark question. The AQA spec lists three reasons — students lose marks by giving vague answers or only one reason. Memorise these three:

Reason 1 — Reaction Is Reversible

Some reactions do not go to completion because they are reversible. The reaction reaches an equilibrium where some product re-forms reactants — so not all reactant has been converted to the product.

Reason 2 — Product Lost in Separation

Some product is lost when it is separated from the reaction mixture. For example, during filtration, evaporation, or transferring between containers, some product is left behind in the filter paper, beaker, or evaporating dish.

Reason 3 — Side Reactions

Some reactants react in ways different to the expected reaction, producing unintended by-products instead of the desired product. This reduces the amount of desired product formed.

Examiner Alert — Cross-cutting

Mark scheme rejects: "the reaction wasn't complete" on its own — too vague. Pair it with a reason: "the reaction is reversible so not all reactants converted", or "some product was lost during filtration / transfer". The three reasons above are exactly the ones listed in spec 4.3.3.1 — quote them, don't paraphrase.

Worked Example 1

1

Calculate Percentage Yield

A student produces 9.0 g of copper sulfate. The theoretical maximum mass is 12.5 g. Calculate the percentage yield. [2 marks]

Step 1 — Identify values: actual = 9.0 g; theoretical = 12.5 g

Step 2 — Apply formula: % yield = (actual ÷ theoretical) × 100

Step 3 — Substitute: = (9.0 ÷ 12.5) × 100

% yield = 72.0%

Mark Scheme

2 marks: 1 for the method (actual ÷ theoretical, then × 100); 1 for the correct value with %. Always include the % symbol — that's the unit.

Worked Example 2 — Rearranged

Real exam context (P1 Jun24 Q01.5): you're told the % yield and the theoretical mass, and asked to find the actual mass produced. You need to rearrange the formula.

2

Find Actual Mass from % Yield

The maximum theoretical mass of salt is 12.5 g. The percentage yield is 92.8%. Calculate the mass of salt actually produced. [3 marks]

Step 1 — Start with formula: 92.8 = (mass produced ÷ 12.5) × 100

Step 2 — Rearrange: mass produced = (92.8 ÷ 100) × 12.5

Step 3 — Calculate: = 0.928 × 12.5

mass produced = 11.6 g

Examiner Alert — P1 Jun24 Q01.5 Calculator Trap

Examiner report (Jun24): "Some students used their calculator incorrectly, using both the % button and also dividing by 100, giving an answer that was a factor of 100 too low. They seemed unfamiliar with the idea that the symbol '%' means divided by 100." If you use the calculator's % button, do not also divide by 100. Either type "92.8 ÷ 100 × 12.5" OR use the % button on 92.8 and multiply by 12.5 — never both.

Mark Scheme (P1 Jun24 Q01.5)

M1 = correct setup of formula; M2 = correct rearrangement (mass produced = % yield ÷ 100 × theoretical); M3 = correct value (11.6 g). Showing each line of working insures the method marks even if the arithmetic slips.

What Is Atom Economy?

A reaction can have a high yield but still be wasteful — if half the atoms in your reactants end up in a useless by-product, you're wasting feedstock. Atom economy (also called atom utilisation) measures the proportion of starting atoms that end up in the desired product.

Atom Economy

The relative formula mass (M_r) of the desired product divided by the sum of M_r values of all the reactants, multiplied by 100.

atom economy = (M_r of desired product ÷ sum of M_r of all reactants) × 100

Reactants or Products?

The AQA spec defines the formula using reactants in the denominator. Mark schemes (e.g. P1 Jun23 Q09.6) use "total M_r of reactants". Because mass is conserved (sum of reactants = sum of products), using all products gives the same numerical answer — but stick to the reactants version to match the spec wording.

Yield vs Atom Economy

Yield = how much you actually made vs how much you could have made. Atom economy = how efficiently the atoms of your reactants are used. A reaction can have 100% yield but a terrible atom economy if it produces lots of waste by-products.

Worked Example 3 — Atom Economy

3

Calculate Atom Economy

Calculate the atom economy for producing hydrogen from the reaction:

Zn + 2HCl → ZnCl₂ + H₂

A_r: Zn = 65, H = 1, Cl = 35.5. Desired product: H₂. [3 marks]

Step 1 — M_r of desired product: M_r(H₂) = 2 × 1 = 2

Step 2 — Sum of M_r of all reactants: Zn + 2HCl = 65 + 2(1 + 35.5) = 65 + 73 = 138

Step 3 — Apply formula: atom economy = (2 ÷ 138) × 100

atom economy = 1.45% (very low — most atoms end up in ZnCl₂ by-product)

Mark Scheme

M1 = correct M_r of desired product (2); M2 = correct sum of all reactants' M_r (138); M3 = correct % (1.45). The denominator is all reactants (per AQA spec). Using all products gives the same answer because mass is conserved.

Atom Economy Visualised

An atom economy of 1.45% doesn't sound dramatically bad until you see what it looks like. Each bar below shows the same reactant mass (representing 100% of the starting atoms) split into useful product (purple) and waste atoms (grey).

What This Shows

Reaction 1's 1.45% is a barely-visible sliver — almost all the reactant atoms end up in the unwanted ZnCl₂. This is why industry doesn't make hydrogen this way — they'd waste 98.55% of the metal feedstock. Reaction 3 is only 3.6 percentage points better than Reaction 2, but at industrial scale this matters.

Examiner Alert — Backward Atom Economy

Real exam context (P1 Jun23 Q09.6, 4 marks). The question gives you the atom economy as a number and asks you to work backwards to find an unknown A_r.

"Metal X is extracted from XO₃ by: XO₃ + 3H₂ → X + 3H₂O. The atom economy is 77.3%. Calculate the relative atomic mass of X." 4 marks

Examiner Alert — P1 Jun23 Q09.6

Examiner report flags two recurring errors:

(1) Many students misread M_r(XO₃) as A_r(X) × 48 instead of A_r(X) + 3(16) = A_r(X) + 48. The "3" in XO₃ means three oxygen atoms, not "multiply".

(2) Students set up the algebra correctly but couldn't rearrange. The "shortcut" method below avoids algebra.

The Shortcut Method (Mark Scheme Approved)

Identify the "waste" M_r in the reactants: the reactant atoms that don't end up in the desired product. The 3 H₂ reactants → become 3 H₂O products (waste). The 3 O atoms in XO₃ also end up in waste H₂O. Total waste M_r = 3 × M_r(H₂O) = 3 × 18 = 54.

If atom economy = 77.3%, then the waste = 100 − 77.3 = 22.7% of the total reactant mass.

So 22.7% of total reactants = 54, meaning total M_r of reactants = 54 ÷ 0.227 = 238.

Total reactants = M_r(XO₃) + M_r(3H₂) = (A_r(X) + 48) + 6 = A_r(X) + 54.

So A_r(X) = 238 − 54 = 184.

Why Atom Economy Matters

Industry doesn't pick a reaction just because it works. They pick reactions with high atom economy — and here's why:

1. Less Waste

High atom economy means fewer by-products are made. Less material to dispose of, less environmental impact.

2. More Sustainable

Raw materials are used efficiently. A 90% atom economy reaction uses 9× less feedstock per kg of product than a 10% reaction.

Spec quote (4.3.3.2)

"It is important for sustainable development and for economic reasons to use reactions with high atom economy." Use this exact framing — sustainability + economics — in any "why high AE matters" answer.

Choosing a Reaction Pathway HT

The spec restricts the "explain why a particular pathway is chosen" skill to HT only. The spec lists exactly five factors to weigh up (and these are the ones the mark scheme credits):

Atom economy

Yield

Rate

Equilibrium position

Usefulness of by-products

Each is a possible mark in a comparison question. Don't add factors outside this list (e.g. "energy cost" / "raw material cost") — they're not in the spec for this section.

Common Errors at a Glance

TopicLoses the markEarns the mark

Yield vs atom economyUsing one formula for the other (mixing up actual mass with M_r).Yield = mass-based (actual ÷ theoretical). AE = M_r-based (desired ÷ all reactants).

M_r of XO₃A_r(X) × 48 (wrong — "3" is the number of O atoms).A_r(X) + (3 × 16) = A_r(X) + 48.

"Why yield < 100%""The reaction wasn't complete." (too vague, no reason why)."The reaction is reversible" / "Product was lost in filtration".

Atom economy denominatorUsing only the by-product M_r, or only the desired-product M_r.Sum of all reactants' M_r (per spec wording — same answer as all products' M_r by mass conservation).

Final answerNo % symbol; or % > 100% (formula inverted).Quote with %; check sense (yield & AE both ≤ 100%).

PEE Framework — Choosing a Method HT

Question (HT only): A chemical can be made by two different reactions. Method A has an atom economy of 100%. Method B has an atom economy of 78%. Both have similar yields, rates, and equilibrium positions. Suggest which method a manufacturer should use, and explain why.3 marks

P — PointThe manufacturer should use Method A, which has the higher atom economy (100% vs 78%).

E — EvidenceA higher atom economy means more of the reactant atoms end up in the desired product, with fewer wasted in by-products.

E — ExplanationThis makes Method A more sustainable (less waste / by-product) and means raw materials are used more efficiently — which is what the spec lists as the two reasons for preferring high atom economy.

Mark Scheme Sketch

M1 = identify the higher-AE method; M2 = link to less waste / more efficient use of atoms; M3 = consequence (sustainability / less by-product). Each idea = one mark. The question stem usually tells you to ignore other factors (yield, rate, etc.) by stating they are "similar" or "the same".

Practice Question 1

A student carries out a reaction. The maximum theoretical mass of product is 80.0 g. The actual mass obtained is 64.0 g.

(a) Calculate the percentage yield. 2 marks

(b) Give two reasons why the percentage yield is less than 100%. 2 marks

(a) M1

Use the formula: % yield = (actual ÷ theoretical) × 100

(a) M2

= (64.0 ÷ 80.0) × 100 = 80.0%

•

Part (a) answer: 80.0%

(b) M1

The reaction may be reversible — at equilibrium, not all reactants have been converted to products.

(b) M2

Some product is lost during separation / filtration / transfer between containers.

•

(Other valid reason from the spec: some reactants react in ways different to the expected reaction — i.e. side reactions producing unwanted products.)

Practice Question 2

Copper can be extracted from copper(II) oxide by two methods:

Method 1: 2CuO + C → 2Cu + CO₂

Method 2: CuO + H₂ → Cu + H₂O

A_r: Cu = 63.5, O = 16, C = 12, H = 1. Desired product = Cu.

(a) Calculate the atom economy of each method. 4 marks

(b) Suggest which method has the better atom economy and give a reason. 2 marksHT

(a) M1

Method 1: useful M_r = 2 × 63.5 = 127. Sum of all reactants M_r = 2(CuO) + C = 2(63.5+16) + 12 = 159 + 12 = 171.

(a) M2

Atom economy (Method 1) = (127 ÷ 171) × 100 = 74.3%

(a) M3

Method 2: useful M_r = 63.5. Sum of all reactants M_r = CuO + H₂ = 79.5 + 2 = 81.5.

(a) M4

Atom economy (Method 2) = (63.5 ÷ 81.5) × 100 = 77.9%

(b) M1

Method 2 has the higher atom economy (77.9% vs 74.3%).

(b) M2

A higher proportion of the reactant atoms end up in the desired product (Cu); less mass is wasted as the by-product (water vs carbon dioxide), so it is more sustainable.

Summary

Percentage Yield (4.3.3.1)

% yield = (actual mass ÷ theoretical mass) × 100. Always < 100% in practice. Be ready to rearrange to find actual mass given % yield (P1 Jun24 Q01.5). The spec lists three reasons for < 100%: reversible reaction, product lost in separation, and reactants reacting in unintended ways (side reactions).

Atom Economy (4.3.3.2)

atom economy = (M_r of desired product ÷ sum of M_r of all reactants) × 100. A measure of how efficiently atoms of reactants end up in the desired product. The spec says high atom economy is important for sustainable development and for economic reasons. HT Comparing pathways requires weighing five factors: AE, yield, rate, equilibrium position, usefulness of by-products.

Yield vs Atom Economy

Yield = mass-based. Atom economy = M_r-based. A reaction can have 100% yield but bad atom economy if it makes lots of by-product.

Most Costly Error

M_r(XO₃) ≠ A_r(X) × 48. It's A_r(X) + (3 × 16). The "3" is the count of oxygen atoms, not a multiplier.

Exam Tip

Label every step. "M_r(desired) = …", "Sum of reactants' M_r = …", "Atom economy = … × 100 = …%". For % yield, always quote the % symbol on the answer. Calculator trap (Jun24 Q01.5): if you press the % button, do not also divide by 100. For "why < 100%" questions, give two of the three spec reasons.