Percentage Yield, Atom Economy and Titrations — AQA A-level Chemistry

SPEC 3.1.2.5 - AMOUNT OF SUBSTANCE

Percentage Yield, Atom Economy & Titrations

Measuring reaction efficiency, sustainability, and volumetric analysis techniques

Why Efficiency Matters

In chemistry, we rarely get as much product as we expect. Understanding why this happens and how to measure efficiency is essential for both exams and real-world chemistry.

Percentage Yield

Measures how much product you actually obtained compared to what you theoretically could have made.

Based on experimental data

Atom Economy

Measures how much of the reactant atoms end up in the desired product rather than waste.

Based on the balanced equation

Key Distinction

Percentage yield = experimental (what actually happened)

Atom economy = theoretical (what the equation predicts)

Percentage Yield

Definition

Percentage yield compares the actual yield (mass of product obtained) to the theoretical yield (maximum possible mass from stoichiometry).

Percentage Yield = \frac{Actual Yield}{Theoretical Yield} \times 100

Calculating Theoretical Yield

Calculate moles of the limiting reactant:

n = \frac{m}{M _{r}}

Use the mole ratio from the balanced equation to find moles of product

Calculate mass of product:

m = n \times M_{r}

TermMeaning

Actual YieldThe mass of product you actually obtained from the experiment

Theoretical YieldThe maximum mass of product calculated from the balanced equation (assuming 100% conversion)

Important

Percentage yield is always ≤ 100%. If you calculate a value greater than 100%, you have made an error (or the product is impure/wet).

Why Percentage Yield is Less Than 100%

In practice, we never achieve 100% yield. Here are the main reasons:

ReasonExplanation

Incomplete ReactionsReversible reactions reach equilibrium before full conversion; some reactants remain unreacted

Side ReactionsReactants may form unwanted by-products instead of the desired product

Loss During TransferProduct left behind on glassware, filter paper, or during pouring

Loss During PurificationSome product lost during filtration, recrystallisation, or evaporation

Measurement ErrorsInaccurate weighing or volume measurements affect calculations

Exam Tip

When asked to explain a low percentage yield, give specific reasons relevant to the reaction type, not just "experimental error".

Worked Example01

Calcium carbonate decomposes on heating:

CaCO₃ → CaO + CO₂

A student heated 5.00 g of calcium carbonate and obtained 2.52 g of calcium oxide. Calculate the percentage yield.

(Aᵣ: Ca = 40, C = 12, O = 16)

1

Calculate moles of CaCO₃:

n = \frac{5.00}{40 + 12 + 48} = \frac{5.00}{100} = 0.0500 mol

2

From equation: 1 mol CaCO₃ → 1 mol CaO, so theoretical moles of CaO = 0.0500 mol

3

Theoretical yield of CaO:

m = 0.0500 \times (40 + 16) = 0.0500 \times 56 = 2.80 g

4

Percentage yield:

\frac{2.52}{2.80} \times 100 = 90.0%

Answer: Percentage yield = 90.0%

Worked Example02

The Haber process synthesises ammonia:

N₂ + 3H₂ → 2NH₃

Calculate the actual mass of ammonia produced if the theoretical yield is 34.0 g and the percentage yield is 77.7%.

1

Rearrange the percentage yield formula:

Actual yield = \frac{% yield}{100} \times Theoretical yield

2

Substitute values:

Actual yield = \frac{77.7}{100} \times 34.0

3

Calculate:

Actual yield = 0.777 \times 34.0 = 26.4 g

Answer: Actual yield = 26.4 g

Multi-Step Reaction Yields

Key Principle

For reactions with multiple steps, the overall yield is found by multiplying the yields of each step together.

Overall Yield = Yield_{1} \times Yield_{2} \times Yield_{3} \times ...

Example Calculation

A 4-step synthesis where each step has 80% yield:

0.80 \times 0.80 \times 0.80 \times 0.80 = 0.41 = 41%

Even with good individual yields, the overall yield drops dramatically!

Number of StepsEach Step 90%Each Step 80%Each Step 70%

2 steps81%64%49%

3 steps73%51%34%

4 steps66%41%24%

5 steps59%33%17%

Industry Implication

This is why chemists prefer shorter synthesis routes with fewer steps, even if individual reactions are more complex.

Worked Example03

A pharmaceutical is synthesised in a 3-step process. The percentage yields for each step are:

Step 1: 80% Step 2: 60% Step 3: 75%

Calculate the overall percentage yield for the synthesis.

1

Convert percentages to decimals: 80% = 0.80, 60% = 0.60, 75% = 0.75

2

Multiply all yields together:

0.80 \times 0.60 \times 0.75

3

Calculate:

0.80 \times 0.60 = 0.48

, then

0.48 \times 0.75 = 0.36

4

Convert back to percentage:

0.36 \times 100 = 36%

Answer: Overall percentage yield = 36%

Atom Economy

Definition

Atom economy measures what proportion of the reactant atoms end up in the desired product. It is calculated from the balanced equation only.

Atom Economy = \frac{M _{r} of desired product}{\sum M _{r} of all reactants} \times 100

Include Coefficients!

You must include stoichiometric coefficients when calculating Mᵣ values:

For Fe₂O₃ + 3CO → 2Fe + 3CO₂:

Mᵣ of desired product = 2 × 56 = 112 (not just 56)

Mᵣ of 3CO = 3 × 28 = 84 (not just 28)

Why It Matters

High atom economy means:

Less waste produced → more sustainable

Lower costs (fewer raw materials wasted)

Fewer disposal and environmental issues

This is a key principle of Green Chemistry.

When is Atom Economy 100%?

Atom economy is 100% when there is only one product (the desired product). This happens in:

Addition Reactions

All atoms from reactants end up in a single product.C₂H₄ + H₂O → C₂H₅OHEthene + Water → Ethanol (100% atom economy)

Substitution Reactions

These produce by-products, so atom economy is less than 100%.C₂H₅OH + HBr → C₂H₅Br + H₂OWater is a by-product → atom economy < 100%

Exam Tip

If asked to compare reaction routes, the one with fewer by-products (or more useful by-products) has better atom economy.

Worked Example04

Iron is extracted from iron(III) oxide using carbon monoxide:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Calculate the atom economy for the production of iron.

(Aᵣ: Fe = 56, O = 16, C = 12)

1

Mᵣ of desired product (2Fe):

2 \times 56 = 112

2

Mᵣ of Fe₂O₃:

(2 \times 56) + (3 \times 16) = 112 + 48 = 160

3

Mᵣ of 3CO:

3 \times (12 + 16) = 3 \times 28 = 84

4

Sum of Mᵣ of all reactants:

160 + 84 = 244

5

Atom economy:

\frac{112}{244} \times 100 = 45.9%

Answer: Atom economy = 45.9%

Worked Example05

Ethanol is produced by fermentation of glucose:

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

Calculate the atom economy for the production of ethanol.

(Aᵣ: C = 12, H = 1, O = 16)

1

Mᵣ of C₂H₅OH:

(2 \times 12) + (6 \times 1) + 16 = 24 + 6 + 16 = 46

2

Mᵣ of desired product (2C₂H₅OH):

2 \times 46 = 92

3

Mᵣ of glucose (C₆H₁₂O₆):

(6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180

4

Atom economy:

\frac{92}{180} \times 100 = 51.1%

Answer: Atom economy = 51.1%

Note: Although fermentation has moderate atom economy, CO₂ is the only by-product and the process uses renewable glucose.

Yield vs Atom Economy

Both are important measures of efficiency, but they tell us different things:

AspectPercentage YieldAtom Economy

What it measuresEfficiency of the experimentEfficiency of the reaction design

Based onExperimental dataBalanced equation only

Can be improved byBetter technique, optimising conditionsChoosing a different reaction pathway

Maximum value100% (in practice, always less)100% (addition reactions)

Key Insight

A reaction can have high yield but low atom economy (wasteful reaction done well) or high atom economy but low yield (good reaction done poorly). Industry aims for both to be high.

Water of Crystallisation

Definition

Water of crystallisation is water that is chemically bonded within the crystal structure of a compound. Compounds containing water of crystallisation are called hydrated compounds.

TermMeaningExample

HydratedContains water of crystallisationCuSO₄·5H₂O (blue crystals)

AnhydrousContains no water of crystallisationCuSO₄ (white powder)

Key Points

The dot (·) separates the salt formula from the water molecules

Compounds can have different degrees of hydration (e.g., CoCl₂·6H₂O or CoCl₂·2H₂O)

Heating a hydrated salt drives off water, leaving the anhydrous form

This process is reversible: adding water reforms the hydrated compound

Method to Find x in M·xH₂O

Find mass of water lost: mass(hydrated) − mass(anhydrous)

Calculate moles of anhydrous salt: n = mass ÷ Mᵣ

Calculate moles of water: n = mass ÷ 18

Find simplest ratio: divide both by the smallest value

Worked Example06

11.25 g of hydrated copper sulfate, CuSO₄·xH₂O, is heated until a constant mass of 7.19 g is reached.

Calculate the formula of the hydrated copper(II) sulfate.

(Aᵣ: Cu = 63.5, S = 32, O = 16, H = 1)

Answer: The formula is CuSO₄·5H₂O

Introduction to Titrations

Definition

A titration is a quantitative technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration (a standard solution).

In an acid-base titration:

A measured volume of one solution is placed in a conical flask (using a pipette )

The other solution is added gradually from a burette until the reaction is complete

An indicator shows when the endpoint is reached (colour change)

IndicatorIn AcidIn AlkaliBest For

PhenolphthaleinColourlessPinkStrong acid + strong base

Methyl orangeRedYellowStrong acid + weak base

Standard Solutions

Definition

A standard solution is a solution of accurately known concentration, prepared by dissolving a precise mass of solute in a volumetric flask.

Preparing a standard solution:

1

Calculate the mass of solute needed using: mass = moles × Mᵣ

2

Weigh the solid accurately using a 3 d.p. balance

3

Dissolve in a small amount of distilled water in a beaker

4

Transfer quantitatively to a volumetric flask (rinse beaker)

5

Add distilled water up to the graduation mark

6

Stopper and invert to mix thoroughly

Common Error

Never heat a volumetric flask - the calibration is only accurate at room temperature (usually 20°C).

Obtaining Concordant Results

For reliable titration data, results must be concordant (within 0.10 cm³ of each other).

Procedure

Perform a rough titration first (add titrant quickly to find approximate endpoint)

Repeat titrations carefully, adding dropwise near the endpoint

Record all burette readings to 2 decimal places (±0.05 cm³)

Continue until at least 2-3 concordant results are obtained

Calculate the mean titre using concordant values only

TitrationInitial (cm³)Final (cm³)Titre (cm³)

Rough0.0024.8024.80 (not used)

10.0024.3524.35

20.5024.9024.40

31.2025.5524.35

Mean titre = (24.35 + 24.40 + 24.35) ÷ 3 = 24.37 cm³

Titration Calculations

The key formulas you need for titration calculations:

Key Formulas

n = c \times V

where n = moles, c = concentration (mol dm⁻³), V = volume (dm³)

Unit Conversion

Volumes are often given in cm³. To convert to dm³:

V (dm^{3}) = \frac{V (cm ^{3} )}{1000}

Using Mole Ratios

EquationRatioTo Find Moles of AcidHCl + NaOH → ...1:1n(acid) = n(alkali)H₂SO₄ + 2NaOH → ...1:2n(acid) = n(alkali) ÷ 2H₃PO₄ + 3NaOH → ...1:3n(acid) = n(alkali) ÷ 3

Method

Write the balanced equation

Calculate moles of the known substance: n = c × V

Use mole ratio from equation to find moles of unknown

Calculate concentration or mass of unknown

Worked Example07

25.0 cm³ of sodium hydroxide solution was neutralised by 23.50 cm³ of 0.100 mol dm⁻³ hydrochloric acid.

NaOH + HCl → NaCl + H₂O

Calculate the concentration of the sodium hydroxide solution.

1

Moles of HCl:

n = c \times V = 0.100 \times \frac{23.50}{1000} = 0.00235 mol

2

From equation: 1 mol NaOH reacts with 1 mol HCl (1:1 ratio)

3

Moles of NaOH = 0.00235 mol

4

Concentration of NaOH:

c = \frac{n}{V} = \frac{0.00235}{25.0/1000} = \frac{0.00235}{0.0250} = 0.0940 mol dm^{- 3}

Answer: [NaOH] = 0.0940 mol dm⁻³

Worked Example08

25.0 cm³ of sulfuric acid was neutralised by 28.40 cm³ of 0.150 mol dm⁻³ sodium hydroxide.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Calculate the concentration of the sulfuric acid.

1

Moles of NaOH:

n = 0.150 \times \frac{28.40}{1000} = 0.00426 mol

2

From equation: 2 mol NaOH reacts with 1 mol H₂SO₄ (2:1 ratio)

3

Moles of H₂SO₄ =

\frac{0.00426}{2} = 0.00213 mol

4

Concentration of H₂SO₄:

c = \frac{0.00213}{0.0250} = 0.0852 mol dm^{- 3}

Answer: [H₂SO₄] = 0.0852 mol dm⁻³

Practice

Question 1: Aspirin (C₉H₈O₄) is synthesised from salicylic acid and ethanoic anhydride:

C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + CH₃COOH

A student obtained 1.62 g of aspirin from 1.50 g of salicylic acid. Calculate:

(a) The theoretical yield of aspirin

(b) The percentage yield

(Mᵣ: salicylic acid = 138, aspirin = 180)

a)

Moles of salicylic acid = 1.50 ÷ 138 = 0.01087 mol

Mole ratio 1:1, so theoretical moles of aspirin = 0.01087 mol

Theoretical yield = 0.01087 × 180 = 1.96 g

b)

Percentage yield = (1.62 ÷ 1.96) × 100 = 82.7%

Practice

Question 2: Titanium is extracted using the Kroll process:

TiCl₄ + 2Mg → Ti + 2MgCl₂

Calculate the atom economy for the production of titanium.

(Aᵣ: Ti = 48, Cl = 35.5, Mg = 24)

1

Mᵣ of Ti (desired product) = 48

2

Mᵣ of TiCl₄ = 48 + (4 × 35.5) = 48 + 142 = 190

3

Mᵣ of 2Mg = 2 × 24 = 48

4

Sum of Mᵣ of reactants = 190 + 48 = 238

5

Atom economy = (48 ÷ 238) × 100 = 20.2%

Practice

Question 3: 20.0 cm³ of potassium hydroxide solution required 18.75 cm³ of 0.120 mol dm⁻³ nitric acid for neutralisation.

KOH + HNO₃ → KNO₃ + H₂O

Calculate the concentration of the potassium hydroxide solution in mol dm⁻³.

1

Moles of HNO₃ = 0.120 × (18.75 ÷ 1000) = 0.00225 mol

2

Mole ratio KOH:HNO₃ = 1:1

3

Moles of KOH = 0.00225 mol

4

Concentration of KOH = 0.00225 ÷ (20.0 ÷ 1000) = 0.00225 ÷ 0.0200 = 0.113 mol dm⁻³

Summary

Key Formulas

ConceptFormula

Percentage Yield

\frac{Actual Yield}{Theoretical Yield} \times 100

Multi-Step YieldOverall = Yield₁ × Yield₂ × Yield₃ × ...

Atom Economy

\frac{M _{r} of desired product}{\sum M _{r} of all reactants} \times 100

Water of CrystallisationFind moles of salt and water, then simplest ratio

Moles (from concentration)

n = c \times V

(V in dm³)

Volume conversiondm³ = cm³ ÷ 1000

Key Exam Skills

Percentage yield is experimental; atom economy is theoretical

Multi-step yields: multiply individual yields (e.g., 80% × 80% = 64%)

Addition reactions have 100% atom economy (one product only)

Hydrated salts use dot notation (e.g., CuSO₄·5H₂O)

Always check mole ratios in titration calculations

Concordant titres must be within 0.10 cm³