Percentage yield and atom economyEdexcel International A Level Chemistry: Revision notes
Section 1
Percentage yield
The theoretical yield is the maximum mass (or amount) of product that a balanced equation predicts. The actual yield is the mass actually obtained. Percentage yield = actual yield / theoretical yield x 100.
Example: 2.43 g of Mg gives, in theory, 0.100 mol MgO = 4.03 g. An actual yield of 3.60 g is 3.60 / 4.03 x 100 = 89.3%. Calculate the theoretical yield from the limiting reagent: moles of limiting reagent, ratio from the equation, then mass.
Percentage yield cannot normally exceed 100%. If you get more than 100%, you have inverted the ratio.
Section 2
Why yields are below 100%
Practical yields are lower than theoretical because of: incomplete reactions (reversible reactions, not enough time); side reactions giving other products; loss of product during transfer, filtering, washing or recrystallisation; product lost as smoke or vapour; and impure reactants.
When you are asked to suggest a reason, relate it to the experiment given. Avoid vague answers such as 'human error'. Name the process: 'some solid was left on the filter paper' or 'some product escaped as smoke when the lid was lifted'.
Section 3
Atom economy
Percentage atom economy = molar mass of the desired product / sum of the molar masses of all products x 100%. It is calculated from the balanced equation (using the number of moles of each substance), not from experimental masses, and it is the same as the mass of desired product divided by the total mass of reactants.
Example: CH4 + H2O → CO + 3H2. The desired product is 3H2 = 6.0; the total is 28.0 + 6.0 = 34.0, so the atom economy is 17.6%. An addition reaction with only one product has an atom economy of 100%. A high atom economy means less waste and better use of resources.
Remember to multiply each molar mass by its coefficient in the equation (for example 3H2 is 3 x 2.0).
Section 4
Yield versus atom economy
The two measures answer different questions. Atom economy is a property of the equation (the efficiency of the reaction design in theory). Percentage yield is a property of the experiment (how much was obtained in practice).
A reaction can have a 100% yield but a low atom economy (it still makes by-products), and a reaction with 100% atom economy can have a low yield. Industry considers both, along with by-product value, energy use and hazards. Overall product per mass of reactant is roughly atom economy x yield. By-products that are useful, or unreacted reactants that can be recycled, improve the efficiency of a process.
Do not say that a higher yield means a higher atom economy. They are independent.
Section 5
Determining formulae and confirming equations by experiment
Experiments can determine a formula from masses. Example: burn 1.215 g Mg to give 2.015 g oxide. Oxygen gained = 0.800 g = 0.0500 mol; Mg = 1.215 / 24.3 = 0.0500 mol, so the ratio is 1 : 1 and the formula is MgO. Equations can be confirmed by comparing the measured and predicted amounts of a product.
Evaluating data: if the oxygen is lower than expected, some product may have been lost as smoke or not all the metal reacted. Improve by heating to constant mass (reheat and reweigh until the mass does not change), keeping the lid on as far as possible, and weighing the crucible accurately.
Must Know
- percentage yield = actual / theoretical x 100
- atom economy = molar mass of desired product / sum of molar masses of all products x 100
- yield is experimental, atom economy comes from the equation
- reasons for low yield: incomplete reaction, side reactions, losses
- heat to constant mass and use mass changes to find or check a formula
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Percentage yield and atom economy
- A student heats 2.43 g of magnesium ribbon in a crucible with a loose lid, lifting the lid from time to time, until the reaction seems to have stopped: 2Mg + O₂ → 2MgO. The white solid left in the crucible has a mass of 3.60 g. Relative atomic masses: O = 16.0, Mg = 24.3.Use the student's masses to calculate the ratio of the amount of oxygen to the amount of magnesium in the product, and suggest why it differs from the ratio in MgO.2 marks
- Hydrogen can be manufactured by steam reforming of methane: CH₄(g) + H₂O(g) → CO(g) + 3H₂(g). Relative atomic masses: H = 1.0, C = 12.0, O = 16.0. Molar masses: CH₄ = 16.0, H₂O = 18.0, CO = 28.0, H₂ = 2.0 g mol⁻¹. Another route to hydrogen is the electrolysis of water: 2H₂O(l) → 2H₂(g) + O₂(g).Calculate the percentage atom economy of making hydrogen by the electrolysis of water, and state how it compares with steam reforming.2 marks
- A company compares two ways of making ethanol, C₂H₅OH (molar mass 46.0 g mol⁻¹). Method 1 is the hydration of ethene. Method 2 is the fermentation of glucose, C₆H₁₂O₆ (molar mass 180.0 g mol⁻¹): C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g). In one batch of method 2, fermentation of 450 kg of glucose gives 168 kg of ethanol.Calculate the percentage atom economy of method 2 for making ethanol.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).