4.5 Atom Economy

A free GCSE Chemistry lesson for AQA (8462) from OBA School: calculating atom economy and using it, with other factors, to choose a reaction pathway. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

• Define atom economy and explain why it matters for sustainable development
• Calculate the atom economy of a reaction to make a given product from its balanced equation
• Explain why addition reactions have an atom economy of 100%
• Evaluate reaction pathways using atom economy, yield, rate, equilibrium position and the usefulness of by-products

Key words

• Atom economy: A measure of the amount of starting materials that end up as useful products. It is calculated from the balanced equation.
• Desired product: The substance you want to make in a reaction.
• By-product (waste product): Any other product formed at the same time as the desired product.
• Sustainable development: Meeting the needs of people today without stopping future generations from meeting their own needs.
• Addition reaction: A reaction in which two or more reactants join to make a single product.

What is atom economy?

Percentage yield tells you how much product you collected. Atom economy tells you something different: even with a perfect yield, what percentage of the mass of the reactants could ever end up in the product you want? The rest becomes by-products, which may be waste.

atom economy = (relative formula mass of desired product from equation ÷ sum of relative formula masses of all reactants from equation) × 100

Remember to multiply each Mr by its balancing number. Because mass is conserved, the total Mr of the reactants equals the total Mr of the products, so you can use either total on the bottom of the fraction.

Calculating atom economy

Worked example: making iron in a blast furnace

Fe₂O₃ + 3CO → 2Fe + 3CO₂
reactants: Mr Fe₂O₃ = 160; 3 × Mr CO = 3 × 28 = 84; total = 244
desired product: 2 × Ar Fe = 2 × 56 = 112
atom economy = 112 ÷ 244 × 100 = 45.9%

Answer: 45.9% (3 s.f.)

Worked example: making ethanol by fermentation

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
Mr glucose = 180; desired product = 2 × 46 = 92
atom economy = 92 ÷ 180 × 100 = 51.1%

Answer: 51.1%

Worked example: making ethanol from ethene

C₂H₄ + H₂O → C₂H₅OH
There is only one product, so every atom ends up in it: 46 ÷ (28 + 18) × 100 = 100%

Answer: 100%

Why atom economy matters

A reaction with a high atom economy turns more of the raw materials into useful products. This supports sustainable development because:

• fewer raw materials are used up, which matters when they come from finite resources such as crude oil;
• less waste is made, so less needs to be stored, treated or disposed of;
• the process is usually cheaper, so it is more profitable.

A low atom economy is less of a problem if the by-products can be sold or used. For example, if the carbon dioxide from fermentation is captured and sold, less is wasted.

If a by-product is hazardous, it must be made safe before it can be released into the air, rivers or landfill, which costs money. For example, sulfur dioxide made when metal sulfide ores are roasted causes acid rain, so it is removed from waste gases by reacting it with a base such as calcium carbonate.

Key fact: Addition reactions (such as ethene + steam, or nitrogen + hydrogen in the Haber process) have an atom economy of 100% because there is only one product. Many of them are reversible, so not all of the reactants react in one pass – the unreacted gases are recycled back into the reactor so they are not wasted.

Choosing a reaction pathway

Chemists look at several factors together when deciding how to make a product:

• Atom economy; What a chemist wants high, so little of the reactant mass is wasted
• Percentage yield; What a chemist wants high, so most of the possible product is collected
• Rate of reaction; What a chemist wants fast enough to make product quickly
• Equilibrium position; What a chemist wants for reversible reactions, lying towards the products
• By-products; What a chemist wants useful or saleable, not harmful
• Raw materials and energy; What a chemist wants renewable, cheap, and low temperatures and pressures where possible

Exam tip: In “evaluate” questions, compare the routes factor by factor, then give a clear final judgement that is backed up by the data.

Common mistakes

• Forgetting the balancing numbers: in Fe₂O₃ + 3CO the carbon monoxide counts as 3 × 28 = 84.
• Confusing atom economy with percentage yield. Atom economy comes from the equation; percentage yield comes from the experiment.
• Dividing by the Mr of the products you do not want, instead of the total of all reactants (or all products).
• Thinking a high yield means a high atom economy. A reaction can have a 95% yield and still make mostly waste.

Revision questions

Try these, then check your answers below each one.

Question 1. Lime is made by heating limestone: CaCO₃ → CaO + CO₂. (Mr: CaCO₃ = 100, CaO = 56)<br/>Calculate the atom economy for making calcium oxide.

Answer: 56 %

Working: 56 ÷ 100 × 100 = 56%.

Question 2. Which type of reaction always has an atom economy of 100%? (Choose from: Addition reactions with a single product, Thermal decomposition, Neutralisation, Displacement reactions.)

Answer: Addition reactions with a single product

Working: When reactants join to make only one product, every atom ends up in the desired product.

Question 3. True or false? A reaction with a high percentage yield must also have a high atom economy.

Answer: False

Working: False. A reaction can collect almost all its product (high yield) but still make lots of by-products (low atom economy).

Related lessons

• Previous lesson: 4.4 The Yield of a Chemical Reaction
• Next lesson: 4.6 Expressing Concentration
• In this chapter: 4.1 Relative Masses and Moles
• In this chapter: 4.2–4.3 Equations and Calculations
• In this chapter: 4.7–4.8 Titration and Titration Calculations

Continue in the full course

This lesson is section 4.5 (AGCCH0405) of our free GCSE Chemistry course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Chemistry course.