A reaction has a theoretical yield of 40 g but only 32 g is obtained. Calculate the percentage yield. [1 mark]

SQA SQA Higher 2018-style practice: Ammonia is made by N_2 + 3H_2 → 2NH_3. In a process, 14.0 kg of nitrogen reacted completely and 13.6 kg of ammonia was collected. Calculate (a) the theoretical mass of ammonia and (b) the percentage yield. (GFM: N_2 = 28.0, NH_3 = 17.0.)

Markers reward moles of nitrogen, the mole ratio, the theoretical mass, and the percentage yield. Moles of nitrogen (working in mol using kg and kg mol^-1, or convert to g): $n(N_2) = (14000)/(28.0) = 500 mol$ From the equation, 1 mol N_2 gives 2 mol NH_3: $n(NH_3) = 2 × 500 = 1000 mol$ Theoretical mass of ammonia: $m = nM = 1000 × 17.0 = 17000 g = 17.0 kg$ Percentage yield: $(13.6)/(17.0) × 100 = 80.0\%$ A common loss is forgetting the 1 : 2 mole ratio between nitrogen and ammonia.

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How do chemists make industrial processes as efficient as possible?

Calculations of percentage yield and atom economy, the use of an excess reagent, identifying the limiting reactant, and the factors that make an industrial process efficient and sustainable.

An SQA Higher Chemistry answer on getting the most from reactants, covering calculations of percentage yield and atom economy, the use of an excess reagent, identifying the limiting reactant, and the factors that make an industrial process efficient and sustainable.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this key area is asking
Percentage yield
Atom economy
Excess and limiting reactants
Worked example: percentage yield with a limiting reactant
Efficient and sustainable processes
Examples in context
Try this

What this key area is asking

The SQA wants you to calculate percentage yield and atom economy, identify the limiting reactant and the reactant in excess, and explain the factors that make an industrial process efficient and sustainable. The two calculations here, percentage yield and atom economy, both appear regularly, and candidates lose marks by confusing them, so keep their definitions sharply separate.

Percentage yield

The theoretical mass is calculated from the balanced equation using mole ratios. The actual yield is always less because of side reactions, incomplete reactions and losses in handling.

Atom economy

Excess and limiting reactants

Adding one reactant in excess ensures the other, often more expensive, reactant is completely used up. The reactant that is used up first is the limiting reactant, and it determines the maximum amount of product. To find the limiting reactant, compare the moles of each reactant against the balanced equation.

Worked example: percentage yield with a limiting reactant

Worked example

$24.0 \text{ g}$ of magnesium reacts with $40.0 \text{ g}$ of oxygen to form magnesium oxide: $2Mg + O_2 \rightarrow 2MgO$ . If $36.0 \text{ g}$ of $MgO$ is actually collected, find the percentage yield. ( $GFM$ : $Mg = 24.5$ , $O_2 = 32.0$ , $MgO = 40.5$ .)

Step 1: Moles of each reactant

n(Mg) = \frac{24.0}{24.5} = 0.980 \text{ mol} \qquad n(O_2) = \frac{40.0}{32.0} = 1.25 \text{ mol}

Step 2: Identify the limiting reactant

The ratio needed is $2 \text{ Mg} : 1 \text{ O}_2$ . The $0.980 \text{ mol}$ of $Mg$ needs only $0.490 \text{ mol}$ of $O_2$ , but $1.25 \text{ mol}$ is present, so oxygen is in excess and magnesium is limiting.

Step 3: Theoretical mass of product

$2 \text{ mol } Mg$ gives $2 \text{ mol } MgO$ , so $0.980 \text{ mol } Mg$ gives $0.980 \text{ mol } MgO$ :

m = nM = 0.980 \times 40.5 = 39.7 \text{ g}

Step 4: Percentage yield

\frac{36.0}{39.7} \times 100 = 90.7\%

The limiting reactant, not the excess, always sets the theoretical yield.

Efficient and sustainable processes

An efficient industrial process aims to:

maximise percentage yield and atom economy,
recycle unreacted reactants and any catalysts,
use cheap, abundant and renewable feedstocks where possible,
minimise energy use and waste, reducing cost and environmental impact.

Examples in context

The drive for high atom economy is reshaping the pharmaceutical industry, where traditional drug syntheses often had atom economies below $20\%$ , meaning most of the reactant mass became waste solvent and by-product. Green-chemistry redesigns, using catalytic addition reactions instead of substitution, cut that waste dramatically. The hydration of ethene to ethanol, with its $100\%$ atom economy, is favoured over fermentation when a pure, fast supply is needed. In ammonia manufacture, the unreacted nitrogen and hydrogen leaving the Haber reactor are recycled back through the converter, raising the overall conversion well above the single-pass equilibrium yield, a classic example of recycling improving efficiency without changing the equilibrium itself.

Try this

Q1. A reaction has a theoretical yield of $40 \text{ g}$ but only $32 \text{ g}$ is obtained. Calculate the percentage yield. [1 mark]

Cue. $\frac{32}{40} \times 100 = 80\%$ .

Q2. State one reason a reactant might be added in excess in an industrial process. [1 mark]

Cue. To ensure the more expensive or valuable reactant is completely used up.

Q3. For $CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$ , calculate the atom economy for $CH_3Cl$ . ( $GFM$ : $CH_4 = 16.0$ , $Cl_2 = 71.0$ , $CH_3Cl = 50.5$ .) [2 marks]

Cue. $\frac{50.5}{16.0 + 71.0} \times 100 = \frac{50.5}{87.0} \times 100 = 58.0\%$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20184 marksAmmonia is made by

N_2 + 3H_2 \rightarrow 2NH_3

. In a process,

14.0 \text{ kg}

of nitrogen reacted completely and

13.6 \text{ kg}

of ammonia was collected. Calculate (a) the theoretical mass of ammonia and (b) the percentage yield. (

GFM

N_2 = 28.0

NH_3 = 17.0

Show worked answer →

Markers reward moles of nitrogen, the mole ratio, the theoretical mass, and the percentage yield.

Moles of nitrogen (working in mol using kg and $\text{kg mol}^{-1}$ , or convert to g):

n(N_2) = \frac{14000}{28.0} = 500 \text{ mol}

From the equation, $1 \text{ mol } N_2$ gives $2 \text{ mol } NH_3$ :

n(NH_3) = 2 \times 500 = 1000 \text{ mol}

Theoretical mass of ammonia:

m = nM = 1000 \times 17.0 = 17000 \text{ g} = 17.0 \text{ kg}

Percentage yield:

\frac{13.6}{17.0} \times 100 = 80.0\%

A common loss is forgetting the $1 : 2$ mole ratio between nitrogen and ammonia.

SQA Higher 20203 marksEthene reacts with steam to make ethanol:

C_2H_4 + H_2O \rightarrow C_2H_5OH

. (a) Calculate the atom economy for this reaction. (b) State what the high atom economy tells you about the reaction. (

GFM

C_2H_4 = 28.0

H_2O = 18.0

C_2H_5OH = 46.0

Show worked answer →

The reaction is an addition, so all reactant atoms end up in the product.

(a) Atom economy is the mass of desired product over the total mass of reactants:

\text{atom economy} = \frac{46.0}{28.0 + 18.0} \times 100 = \frac{46.0}{46.0} \times 100 = 100\%

(b) A $100\%$ atom economy means every atom of the reactants ends up in the desired product, so there is no waste by-product. This makes the reaction efficient and sustainable.

Markers reward recognising that addition reactions have $100\%$ atom economy and linking this to "no waste".

Related dot points

Sources & how we know this

SQA Higher Chemistry Course Specification — SQA (2018)