How do we calculate percentage yield, atom economy and gas volumes?
Yield and gas calculations: percentage yield and why yields are below 100 percent, atom economy and sustainability, and calculating gas volumes using the molar volume of a gas.
A focused answer to Edexcel GCSE Chemistry topic 5 (separate chemistry), covering how to calculate percentage yield and why it is below 100 percent, atom economy and its link to sustainability, and calculating the volume of a gas using the molar gas volume of 24 dm3 per mole at room temperature and pressure.
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What this dot point is asking
Edexcel wants you (in separate chemistry) to calculate percentage yield and explain why yields are below 100 percent, calculate atom economy and link it to sustainability, and calculate the volume of a gas using the molar gas volume of dm/mol at room temperature and pressure. These are high-value calculation marks on the separate paper.
Percentage yield
The yield is almost always below 100 percent, for several reasons:
- Product is lost during transfer, filtering or purification.
- The reaction may be incomplete or reversible, so not all reactant becomes product.
- Reactants may be impure, so there is less than expected.
- Side reactions form unwanted products.
No atoms are ever destroyed, so a low yield is about losses and incomplete reaction, not loss of mass overall.
Atom economy
A reaction with a high atom economy wastes few atoms, so it produces less waste, uses raw materials more efficiently, and is more sustainable and often more profitable. A reaction can have a high yield but a low atom economy if it makes a lot of unwanted by-product, so industry aims for both. Reactions that make only one product have an atom economy of 100 percent, because every atom in the reactants ends up in the useful product; reactions that make several products always have a lower atom economy, and a company may try to sell the by-products to reduce waste and improve the overall efficiency.
Gas volumes and the molar volume
Equal numbers of moles of any gas occupy the same volume under the same conditions. At room temperature and pressure (rtp) the molar gas volume is dm/mol:
To find a gas volume from a mass of reactant: convert the mass to moles, use the mole ratio to find the moles of gas, then multiply by dm/mol.
Try this
Q1. A reaction has a theoretical yield of g and an actual yield of g. Calculate the percentage yield. [2 marks]
- Cue. .
Q2. Calculate the volume of mol of oxygen at rtp. [2 marks]
- Cue. dm.
Q3. Explain why a high atom economy is good for sustainability. [2 marks]
- Cue. More of the reactant mass becomes useful product, so there is less waste and raw materials are used more efficiently.
Exam-style practice questions
Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
Edexcel 20194 marksIn a reaction the theoretical yield of product is g, but only g is actually obtained. Calculate the percentage yield, and give two reasons why the yield is less than 100 percent.Show worked answer →
A 4-mark yield calculation with reasons.
Percentage yield (actual yield theoretical yield) (1 mark for the formula, 1 mark for the value). Two reasons (1 mark each, any two): some product is lost when it is transferred between containers or filtered; the reaction may be incomplete or reversible; reactants may be impure; or there may be unwanted side reactions producing other products.
Markers reward the correct formula (actual over theoretical) and two valid reasons; "atoms are destroyed" is wrong.
Edexcel 20213 marksCalculate the volume of carbon dioxide produced at room temperature and pressure when mol of calcium carbonate decomposes: . The molar volume of a gas at room temperature and pressure is dm/mol.Show worked answer →
A 3-mark gas-volume calculation.
The ratio of to is , so moles of mol (1 mark). Volume of gas moles molar volume dm (1 mark for the method, 1 mark for the answer with units).
Markers reward using the mole ratio to find the moles of gas, then multiplying by dm/mol.
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Sources & how we know this
- Pearson Edexcel GCSE (9-1) Chemistry (1CH0) specification — Pearson Edexcel (2016)