Calculate reacting masses and percentage yield
Use a balanced equation to convert between reactant and product amounts, and compare actual yield with the theoretical amount.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Use the data and constants supplied for the question, show mole-ratio working, and report a justified number of significant figures. Written practice is self-checked
Before you begin
- Use common chemical symbols and formulae, and rearrange a simple equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What ratio do balanced equation coefficients directly give?
What you will learn
- Calculate reacting masses using formulae and balanced equations
- Calculate percentage yield and distinguish theoretical from actual product
The equation coefficients give mole ratios
A balanced equation relates amounts of reacting species. Convert the known mass to moles, use the coefficient ratio to find the required amount, and then convert to the requested mass. Coefficients are ratios of entities and moles, not generally ratios of masses.
The theoretical yield is the amount predicted if the limiting reactant reacts completely according to the equation. Percentage yield = actual yield ÷ theoretical yield × 100%. The actual yield can be lower because reaction may be incomplete, side reactions may occur or product may be lost during separation; do not claim a specific cause without evidence.
Keep units and significant figures visible
Write the balanced equation and show the mole conversion before substituting. Use molar masses from the formula and supplied Ar data. Round only the final result, and match the precision requested or justified by the data.
If more than one reactant amount is supplied, first identify the limiting reactant by comparing available mole amounts with their equation ratios. The theoretical product amount is based on the limiting reactant, not whichever reactant has the smaller mass.
Pause and explain
How is percentage yield calculated?
Worked example
2Mg + O₂ → 2MgO. What mass of MgO forms from 4.80 g Mg if oxygen is in excess? Use Ar(Mg,O)=24.0,16.0.
Show the worked solution
- n(Mg) = 4.80/24.0 = 0.200 mol.
- The 2:2 ratio gives n(MgO)=0.200 mol.
- Mr(MgO)=40.0; mass = 0.200×40.0.
Answer 8.00 g MgO theoretically.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
N₂ + 3H₂ → 2NH₃. Calculate the maximum mass of NH₃ from 5.60 g N₂ with excess H₂. Use Mr(N₂)=28.0 and Mr(NH₃)=17.0.
Give me a hint
Convert N₂ to moles and apply the 1:2 equation ratio.
Compare my reasoning
- n(N₂)=5.60/28.0=0.200 mol.
- n(NH₃)=2×0.200=0.400 mol.
- Mass NH₃=0.400×17.0.
6.80 g NH₃ theoretically.
Look for these in your work
- Moles of limiting reactant calculated
- Coefficient ratio applied
- Product mass converted correctly
A reaction has a theoretical yield of 12.0 g and produces 9.30 g of dry product. Calculate percentage yield.
Give me a hint
Use actual over theoretical, then multiply by 100.
Compare my reasoning
- Identify actual = 9.30 g and theoretical = 12.0 g.
- Calculate 9.30/12.0.
- Convert the ratio to a percentage.
77.5%.
Look for these in your work
- Correct numerator and denominator
- Percentage conversion correct
A student calculates a product mass from the smaller reactant mass without comparing mole ratios. Explain the flaw and give the correct limiting-reactant method.
Give me a hint
Mass alone does not reveal which reactant is present in the smaller stoichiometric amount.
Compare my reasoning
- Convert each reactant mass to moles.
- Compare each amount with its balanced coefficient ratio.
- Use the reactant that runs out first to calculate theoretical product.
The smaller mass is not necessarily the limiting reactant because the substances have different molar masses and equation ratios. Convert both to moles, compare available amounts with their coefficients, and use the limiting reactant for the theoretical yield.
Look for these in your work
- Mass-versus-amount flaw identified
- Coefficient comparison described
- Theoretical-yield basis stated
Common mistakes
- Using a mass ratio instead of a mole ratio
- Using the excess reactant to set theoretical yield
- Calculating yield as theoretical divided by actual
- Rounding intermediate amounts too soon
What can you explain now?
What ratio do balanced equation coefficients directly give?
Compare with the explanation
Mole ratios
Coefficients compare numbers of particles and therefore amounts in moles.
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Make it stick
What is the sequence from a measured reactant mass to a theoretical product yield?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.