Yield and Atom Economy

This lesson covers percentage yield and atom economy — two important measures used to evaluate chemical reactions. Both are required by the Edexcel GCSE Chemistry specification (1CH0). Understanding these concepts is essential for questions about industrial chemistry and sustainability.

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Theoretical Yield and Actual Yield

The theoretical yield is the maximum mass of product that could be formed in a reaction, based on the balanced equation and the amount of limiting reagent used. It is the amount you calculate using reacting mass calculations.

The actual yield is the mass of product that you actually obtain when you carry out the reaction in practice.

In reality, the actual yield is always less than the theoretical yield.

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Percentage Yield

Percentage yield compares the actual yield to the theoretical yield:

$\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$percentage yield=theoretical yieldactual yield​×100

Key Points

• Percentage yield is always less than 100% in practice.
• A percentage yield of 100% would mean you obtained the maximum possible amount of product — this almost never happens.
• A higher percentage yield means less waste and a more efficient reaction.

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Why Is Yield Less Than 100%?

There are several reasons why the actual yield is lower than expected:

Reason	Explanation
Incomplete reaction	Not all of the reactants may react. The reaction may reach equilibrium before all reactants are consumed.
Side reactions	Reactants may undergo unwanted reactions producing different (by-)products.
Loss during transfer	Product may be lost when transferring between containers (e.g. some solution left in beakers).
Loss during purification	Filtering, washing and drying can cause some product to be lost.
Reversible reactions	If the reaction is reversible, products can react to re-form reactants.

Exam Tip: When explaining why yield is less than 100%, give specific reasons relevant to the reaction. Saying "product was lost" is too vague — say "product was lost during filtration" or "side reactions occurred".

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Worked Examples: Percentage Yield

Example 1

Question: A student calculated that 10.0 g of magnesium oxide should be produced. They actually obtained 8.5 g. Calculate the percentage yield.

Solution:

• percentage yield = (actual yield ÷ theoretical yield) × 100
• percentage yield = (8.5 ÷ 10.0) × 100
• percentage yield = 85%

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Example 2

Question: The theoretical yield of copper from reducing 15.9 g of copper oxide is 12.7 g. The student obtained 10.8 g. Calculate the percentage yield.

Solution:

• percentage yield = (10.8 ÷ 12.7) × 100
• percentage yield = 85.0% (to 3 significant figures)

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Example 3: Full Calculation

Question: 4.0 g of calcium reacts with excess water. The student collects 5.2 g of calcium hydroxide. Calculate the percentage yield.

Ca + 2H₂O → Ca(OH)₂ + H₂

Solution:

1. Mr of Ca = 40, Mr of Ca(OH)₂ = 40 + 2(16 + 1) = 40 + 34 = 74
2. Moles of Ca = 4.0 ÷ 40 = 0.1 mol
3. Ratio: 1 Ca → 1 Ca(OH)₂, so moles of Ca(OH)₂ = 0.1 mol
4. Theoretical yield = 0.1 × 74 = 7.4 g
5. Percentage yield = (5.2 ÷ 7.4) × 100 = 70.3%

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Atom Economy

Atom economy measures how much of the reactant atoms end up in the desired product (rather than in waste by-products).

$\text{atom economy} = \frac{M_r \text{ of desired product}}{\text{total } M_r \text{ of all products}} \times 100$atom economy=total Mr​ of all productsMr​ of desired product​×100

Key Points

• Atom economy is a theoretical calculation based on the equation — it does not depend on how much product you actually make.
• A reaction with high atom economy is more sustainable because less waste is produced.
• A reaction with low atom economy produces more waste by-products.

Exam Tip: Do not confuse atom economy with percentage yield. Atom economy is about the equation (theoretical). Percentage yield is about the experiment (practical). A reaction can have high atom economy but low percentage yield, or vice versa.

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Worked Examples: Atom Economy

Example 4

Question: Calculate the atom economy for producing hydrogen by the reaction of sodium with water.

2Na + 2H₂O → 2NaOH + H₂

• Mr of desired product (H₂) = 2
• Mr of all products = Mr of 2NaOH + Mr of H₂ = (2 × 40) + 2 = 80 + 2 = 82

$\text{atom economy} = \frac{2}{82} \times 100 = \textbf{2.4\%}$atom economy=822​×100=2.4%

This is a very low atom economy — most of the atoms end up in NaOH, not in the desired hydrogen.

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Example 5

Question: Calculate the atom economy for producing calcium oxide by heating calcium carbonate.

CaCO₃ → CaO + CO₂

• Mr of desired product (CaO) = 56
• Mr of all products = 56 + 44 = 100

$\text{atom economy} = \frac{56}{100} \times 100 = \textbf{56\%}$atom economy=10056​×100=56%

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Example 6: 100% Atom Economy

Question: Calculate the atom economy for producing magnesium oxide.

2Mg + O₂ → 2MgO

• Mr of desired product (2MgO) = 2 × 40 = 80
• Mr of all products = 80 (MgO is the only product)

$\text{atom economy} = \frac{80}{80} \times 100 = \textbf{100\%}$atom economy=8080​×100=100%

Addition reactions (where all atoms from the reactants end up in a single product) always have 100% atom economy.

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Why Does Atom Economy Matter?

Factor	High atom economy	Low atom economy
Waste produced	Less waste	More waste by-products
Cost	More cost-effective (less raw material wasted)	Less cost-effective
Environmental impact	Lower — less material to dispose of	Higher — more waste to treat or dispose of
Sustainability	More sustainable	Less sustainable

Industrial processes aim for high atom economy to:

• Reduce costs
• Minimise waste
• Reduce environmental impact
• Conserve resources

Exam Tip: Questions about atom economy often link to sustainability. Be prepared to explain why industries prefer reactions with high atom economy — think about waste, cost and the environment.

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Comparing Yield and Atom Economy