Yield & atom economy

3 practice questions

Percentage yield, including gas measurements

Learning goalCompare the actual product with the theoretical amount from the limiting reactant.

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The theoretical yield is the maximum amount of desired product predicted by the balanced equation. The actual yield is what you recover from the experiment.

Calculate the theoretical yield from the limiting reactant.
Divide actual yield by theoretical yield and multiply by 100.
Compare matching quantities: mass with mass, amount with amount, or suitable gas volumes measured under matching conditions.

A reaction may be incomplete, competing reactions may form other products, or product may be lost during transfer and purification.

A result above 100% suggests an apparent product mass that is too high, for example because the product is wet or impure. It does not mean more pure product formed than the theoretical maximum.

Percentage yield = actual amount of product / theoretical amount of product × 100

Worked example

Calcium oxide yield

CaCO₃ → CaO + CO₂. 10.01 g CaCO₃ gives 4.49 g CaO. M_r(CaCO₃) = 100.1; M_r(CaO) = 56.1.

Calculate the percentage yield of calcium oxide.

Worked answer

n(CaCO₃) = m/Mᵣ
n(CaCO₃) = 10.01/100.1 = 0.100 mol

Theoretical CaO mass = 0.100 × 56.1 = 5.61 g

Percentage yield = (actual yield ÷ theoretical yield) × 100
Yield = 4.49/5.61 × 100

80.0% yield.

Worked example

Predict an actual mass

A process has theoretical product mass 12.0 g and expected yield 75.0%.

Calculate the expected actual mass of product.

Worked answer

Actual mass = 12.0 × 75.0/100

9.00 g product.

Worked example

Yield from measured oxygen

2H₂O₂ → 2H₂O + O₂. 20.0 cm³ of 0.500 mol dm⁻³ H₂O₂ releases 100 cm³ dry O₂ at 100 kPa and 300 K. R = 8.31 J mol⁻¹ K⁻¹.

Calculate the percentage yield of oxygen.

Worked answer

Route	Calculation	O₂ amount / mol
Theoretical from reactant	n(H₂O₂) = 0.500 × 0.0200 = 0.0100; then × 1/2	0.00500
Actual from gas	n = (100 000 × 1.00 × 10⁻⁴)/(8.31 × 300)	0.0040112…

Percentage yield = (actual yield ÷ theoretical yield) × 100
Yield = 0.0040112…/0.00500 × 100

80.2% yield.

EXAM TRAPWet or impure product can give an apparent yield above 100%. A gas leak makes the measured yield too low. If gas is collected over water, use the dry-gas pressure with any required correction supplied or justified.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 49

0.600 g Mg reacts with 40.0 cm³ of 0.500 mol dm⁻³ HCl. 220 cm³ H₂ is collected at 98.0 kPa and 298 K. Gas-volume uncertainty is ±2 cm³. Mg + 2HCl → MgCl₂ + H₂. Use Aᵣ(Mg) = 24.3 and R = 8.31 J mol⁻¹ K⁻¹.

Determine the limiting reactant and theoretical number of moles of H₂. (4 marks)
Calculate the actual number of moles of H₂ and percentage yield. (4 marks)
Calculate the percentage uncertainty in the gas volume. (1 mark)
State the effect of a gas leak on the measured yield. (1 mark)

Total: 10 marks

Show answer & marking points

(a) Compare the starting moles divided by the coefficient.

Quantity	Mg	HCl
Mass / g	0.600	Not needed
Aᵣ or Mᵣ (no unit)	24.3	Not needed
Concentration, c / mol dm⁻³	Solid	0.500
Volume, V / dm³	Solid	40.0/1000 = 0.0400
Amount, n / mol	n = m/Aᵣ 0.600/24.3 = 0.024691… [1]	n = cV 0.500 × 0.0400 = 0.0200 [1]
Coefficient	1	2
n ÷ coefficient / mol	0.024691…/1	0.0200/2 = 0.0100 (smaller)

HCl has the smaller n ÷ coefficient value, so HCl is limiting. [1] HCl : H₂ = 2 : 1; theoretical n(H₂) = 0.0200/2 = 0.0100 mol. [1]

Percentage yield = (actual yield ÷ theoretical yield) × 100
(b) p = 98000 Pa and V = 2.20 × 10⁻⁴ m³. [1] n(actual) = pV/RT [1] = 0.008706… mol. [1] Yield = 0.008706…/0.0100 × 100 = 87.1%. [1]

✓ DONE WHENCompare the actual product with the theoretical amount from the limiting reactant.

Atom economy and choosing a reaction route

Learning goalCalculate equation-based atom economy and distinguish it from experimental yield.

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Atom economy asks how much of the reactant mass is directed into the chosen desired product by the balanced equation. Percentage yield asks how much of that product is actually recovered.

Use the formula-mass contribution of the desired product, including its coefficient, divided by the total reactant formula-mass contributions, also including coefficients. Multiply by 100.

Conservation of mass means the denominator can also be the total formula-mass contribution of all products. Do not include catalysts, solvents or arbitrary excess quantities.

A high atom economy can reduce waste, but choosing a sustainable process also needs energy use, hazards, raw-material cost and supply, renewability, rate, yield, separation and recycling to be considered. A useful by-product may also affect the choice.

Atom economy = coefficient × M_r of desired product / Σ(coefficient × M_r of each reactant) × 100

M_r relative molecular or formula mass; no unit; Σ sum of all the indicated contributions.

Worked example

Sodium carbonate from hydrogen carbonate

2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O. Desired product: Na₂CO₃. Use M_r(NaHCO₃) = 84.0 and M_r(Na₂CO₃) = 106.0.

Calculate the percentage atom economy for producing sodium carbonate.

Worked answer

Atom economy = (total relative mass of desired product ÷ total relative mass of reactants) × 100
Atom economy = 106.0/(2 × 84.0) × 100

63.1%.

Worked example

Changing the desired product

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. A_r or M_r values: CaCO₃ 100.1, HCl 36.5, CaCl₂ 111.1, CO₂ 44.0.

Calculate the percentage atom economy first when CaCl₂ is the desired product, and then when CO₂ is the desired product.

Worked answer

Total reactant formula mass = 100.1 + 2(36.5) = 173.1

For CaCl₂: 111.1/173.1 × 100 = 64.2%

For CO₂: 44.0/173.1 × 100 = 25.4%

The same equation has a different atom economy for a different desired product.

Measure	What it compares	Source of information
Yield	Actual product with theoretical product	Experiment and stoichiometry
Atom economy	Desired-product mass with total stoichiometric reactant mass	Balanced equation

Worked example

Choose between routes

Route A: 100% atom economy, 70% yield, high energy demand and hazardous solvent. Route B: 65% atom economy, 95% yield, lower energy demand and a recoverable by-product.

Compare the two routes using atom economy, yield, energy demand and waste. Explain why atom economy alone cannot establish which route is preferable.

Worked answer

A incorporates more reactant mass into the desired product in principle. B recovers a greater fraction of its theoretical product and may have lower energy and waste-handling costs. A useful recommendation must state which priorities matter and use evidence from the data.

Neither percentage alone establishes the better route.

EXAM TRAPA reaction can have 100% atom economy and a low yield: the equation permits all atoms to enter the product, while the experiment may not produce or recover all of it.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 50

Two routes make ethanol. Each trial uses 1.00 mol of the limiting organic reactant; the other reactant is in excess. Route A: C₂H₄ + H₂O → C₂H₅OH. Route B: C₂H₅Cl + NaOH → C₂H₅OH + NaCl. Actual ethanol masses are 36.8 g for A and 41.4 g for B. Use Mᵣ: C₂H₄ 28.0, H₂O 18.0, C₂H₅Cl 64.5, NaOH 40.0, C₂H₅OH 46.0.

Calculate the atom economy for each route. (3 marks)
Calculate the percentage yield for each route. (3 marks)
Explain why catalysts, solvents and arbitrary excess quantities are not included in the atom-economy denominator. (2 marks)
Explain the effect on the calculated ethanol yield if the isolated product contains water. (1 mark)

Total: 9 marks

Show answer & marking points

Atom economy = (total relative mass of desired product ÷ total relative mass of reactants) × 100
(a) A: 46.0/(28.0 + 18.0) × 100 = 100%. [1] B: total reactant formula mass = 64.5 + 40.0 = 104.5 [1]; atom economy = 46.0/104.5 × 100 = 44.0%. [1]

(b) Theoretical ethanol mass = 46.0 g in each case. [1] A: 36.8/46.0 × 100 = 80.0%. [1] B: 41.4/46.0 × 100 = 90.0%. [1]

(c) Catalysts are regenerated and solvents are not consumed as stoichiometric reactants in these equations. [1] Atom economy uses the balanced-equation quantities that react, rather than any excess left over. [1]

(d) Water increases the apparent product mass, giving an overestimated yield, possibly above 100%. [1]

Exam-style question 51

Two routes make the same desired product. Each starts with 100.0 kg of a stoichiometric mixture of its reactants, excluding solvents, catalysts and inert material. The stated yield applies to the theoretical product amount.

Route	Atom economy	Yield	Energy / GJ t⁻¹	By-product information
X	92.0%	74.0%	8.0	Small amount; low toxicity
Y	68.0%	96.0%	4.5	Corrosive; largely recovered

Calculate the actual desired-product mass obtained by each route. (4 marks)
Recommend a route using the full data set, recognising how a different priority could alter the decision. (3 marks)
Explain why a high yield alone does not show that little waste forms. (2 marks)

Total: 9 marks

Show answer & marking points

(a) X: theoretical desired mass = 100.0 × 0.920 = 92.0 kg [1]; actual = 92.0 × 0.740 = 68.1 kg. [1] Y: theoretical = 100.0 × 0.680 = 68.0 kg [1]; actual = 68.0 × 0.960 = 65.3 kg. [1]

(b) A supported choice [1], relevant comparison using output/energy/by-product data [1], and recognition that a different priority could change the choice [1]. For example X gives more product per feed mass and less hazardous waste; Y uses less energy and recovers much of its corrosive by-product.

(c) Yield compares actual product with its theoretical maximum. [1] A route can still have poor atom economy and large stoichiometric by-products even if it gives a high yield. [1]

✓ DONE WHENCalculate equation-based atom economy and distinguish it from experimental yield.

Yield & atom economy: AQA A-level Chemistry