Percentage Yield and Atom Economy

This lesson covers percentage yield and atom economy as required by the Edexcel GCSE Combined Science specification (1SC0). You will learn how to calculate both values and understand why they are important in industrial chemistry.

---

Theoretical Yield vs Actual Yield

Term	Definition
Theoretical yield	The maximum mass of product calculated from the balanced equation and the amounts of reactants used.
Actual yield	The mass of product actually obtained from the experiment.

The actual yield is almost always less than the theoretical yield.

---

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%

• Percentage yield is always between 0% and 100%.
• A yield of 100% means every atom of reactant was converted to the desired product — this is almost never achieved in practice.

---

Worked Examples — Percentage Yield

Example 1

Question: The theoretical yield of copper sulfate from a reaction is 16.0 g. A student obtains 12.0 g. What is the percentage yield?

$\text{percentage yield} = \frac{12.0}{16.0} \times 100 = 75.0\%$percentage yield=16.012.0​×100=75.0%

Example 2

Question: 10.0 g of calcium carbonate is heated. The theoretical yield of calcium oxide is 5.6 g. The student obtains 4.2 g. Calculate the percentage yield.

$\text{percentage yield} = \frac{4.2}{5.6} \times 100 = 75.0\%$percentage yield=5.64.2​×100=75.0%

Example 3

Question: A reaction has a theoretical yield of 24 g of magnesium oxide but only 19.2 g is obtained. Calculate the percentage yield.

$\text{percentage yield} = \frac{19.2}{24} \times 100 = 80.0\%$percentage yield=2419.2​×100=80.0%

Example 4 — Working Backwards

Question: A reaction has a percentage yield of 60%. If the theoretical yield is 50 g, what mass of product was actually obtained?

$\text{actual yield} = \frac{60}{100} \times 50 = 30 \text{ g}$actual yield=10060​×50=30 g

Exam Tip: If a question asks you to calculate percentage yield, you often need to first calculate the theoretical yield using a reacting masses calculation, then use the percentage yield formula.

---

Why Is Percentage Yield Less Than 100%?

There are several reasons why the actual yield is less than the theoretical yield:

Reason	Explanation
Incomplete reaction	Not all reactants are converted to products — the reaction may reach equilibrium or not go to completion.
Side reactions	Unwanted reactions produce by-products instead of the desired product.
Loss during transfer	Product is lost when transferring between containers, filtering, or pouring.
Loss during purification	Some product is lost during washing, crystallisation, or drying.
Impure reactants	If reactants contain impurities, less product is formed than expected.

Exam Tip: In a 3-mark question about why yield is less than 100%, give three different reasons from the list above. Do not just repeat the same idea in different words.

---

Atom Economy

Atom economy measures how much of the total mass of reactants ends up as the desired product (rather than waste by-products).

$\text{atom economy} = \frac{M_r \text{ of desired product(s)}}{\text{sum of } M_r \text{ of all products}} \times 100\%$atom economy=sum of Mr​ of all productsMr​ of desired product(s)​×100%

Note: Some exam boards define atom economy using the total $M_r$Mr​ of reactants in the denominator. For Edexcel, the formula uses the sum of $M_r$Mr​ of all products (which equals the total $M_r$Mr​ of reactants, by conservation of mass). Both give the same answer.

---

Worked Examples — Atom Economy

Example 1 — Thermal Decomposition of Calcium Carbonate

$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$CaCO3​→CaO+CO2​

Desired product: CaO ($M_r$Mr​ = 56)

Sum of $M_r$Mr​ of all products: 56 + 44 = 100

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

This means 56% of the reactant mass becomes the useful product, and 44% becomes waste ($\text{CO}_2$CO2​).

Example 2 — Making Hydrogen from Zinc and Acid

$\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$Zn+2HCl→ZnCl2​+H2​

Desired product: $\text{H}_2$H2​ ($M_r$Mr​ = 2)

Sum of $M_r$Mr​ of all products: $M_r$Mr​ of $\text{ZnCl}_2$ZnCl2​ + $M_r$Mr​ of $\text{H}_2$H2​ = 136 + 2 = 138

$\text{atom economy} = \frac{2}{138} \times 100 = 1.4\%$atom economy=1382​×100=1.4%

This is a very low atom economy — most of the mass ends up as zinc chloride, not hydrogen.

Example 3 — Neutralisation (High Atom Economy)

$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$NaOH+HCl→NaCl+H2​O

If both products are useful (e.g. sodium chloride is the desired product and water is harmless):

Desired product: NaCl ($M_r$Mr​ = 58.5)

Sum of $M_r$Mr​ of all products: 58.5 + 18 = 76.5

$\text{atom economy} = \frac{58.5}{76.5} \times 100 = 76.5\%$atom economy=76.558.5​×100=76.5%

But if both products are considered useful:

$\text{atom economy} = \frac{76.5}{76.5} \times 100 = 100\%$atom economy=76.576.5​×100=100%

Exam Tip: A reaction with only one product always has 100% atom economy, because no atoms are wasted on by-products. Addition reactions in organic chemistry often achieve this.

---

Why Atom Economy Matters in Industry

Factor	High Atom Economy	Low Atom Economy
Waste	Less waste produced	More waste — costly to dispose of
Sustainability	More sustainable — better use of resources	Less sustainable
Cost	Lower raw material costs per unit of product	Higher raw material costs
Environment	Less environmental impact	More by-products to deal with

Chemical companies aim for reactions with high atom economy to:

• Reduce waste disposal costs.
• Maximise profit by converting more reactant into saleable product.
• Meet environmental regulations.

---

Percentage Yield vs Atom Economy

flowchart LR
    subgraph PY ["Percentage Yield"]
        direction TB
        A1["Actual yield"] -->|"÷"| A2["Theoretical yield"]
        A2 -->|"× 100"| A3["% Yield"]
    end
    subgraph AE ["Atom Economy"]
        direction TB
        B1["M_r of desired product"] -->|"÷"| B2["Sum of M_r of all products"]
        B2 -->|"× 100"| B3["% Atom Economy"]
    end
    PY ~~~ AE
    style A3 fill:#3b82f6,color:#fff,stroke:#2563eb
    style B3 fill:#10b981,color:#fff,stroke:#059669

	Percentage Yield	Atom Economy
Measures	How much product you actually get vs the maximum possible	How much of the reactant mass becomes useful product
Can be improved by	Better technique, purer reactants, more careful transfer	Choosing a different reaction pathway with fewer by-products
Depends on	Experimental conditions and technique	The balanced equation only — it is a theoretical value

---

Common Mistakes to Avoid

Mistake	Correction
Confusing yield with atom economy	Yield = actual vs theoretical amount. Atom economy = useful product vs all products.
Giving percentage yield > 100%	If your answer is > 100%, check your calculation — it should be impossible (unless the product is impure/wet)
Not calculating theoretical yield first	You often need a reacting masses calculation before you can find percentage yield
Forgetting to identify the desired product	The question must tell you which product is desired for atom economy

---

Summary

• Percentage yield = (actual yield ÷ theoretical yield) × 100%.
• Yield is always ≤ 100%. It is less due to: incomplete reaction, side reactions, losses during transfer/purification.
• Atom economy = ($M_r$Mr​ of desired product ÷ sum of $M_r$Mr​ of all products) × 100%.
• High atom economy means less waste — this is important for sustainability and industrial costs.
• Reactions with only one product have 100% atom economy.
• Yield depends on experimental technique; atom economy depends on the reaction itself.

---

Extended Worked Examples

Percentage yield and atom economy questions very often come in pairs: an exam might first ask you to calculate the theoretical yield by reacting masses, then use it to get percentage yield, then finally compute the atom economy. Work step by step, one formula at a time.