Percentage yield, composition and purity
Choose the correct reference amount for each percentage and connect it with a theoretical reaction calculation.
Section 3.3 · The mole and the Avogadro constant
Extended only (Supplement) · Chemistry 0620, 2026–2028. Core concentration units are introduced in a separate lesson.
Before you begin
- Calculate theoretical reacting masses with the limiting reactant.
- Find molar masses and use percentages of a whole.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which two quantities are compared in percentage yield?
What you will learn
- Calculate percentage yield, element composition and sample purity.
- Use stoichiometry to infer purity or interpret an unexpectedly high measured yield.
Yield compares actual product with a theoretical prediction
Percentage yield is actual product amount divided by theoretical product amount, multiplied by 100. Use masses or moles consistently for the same product. The theoretical prediction comes from the balanced reaction and the limiting reactant, assuming complete conversion. Loss during collection, incomplete reaction or unwanted side reactions can make the collected yield lower than this prediction.
A measured yield above 100% is not evidence that the reaction created extra matter. Check wet product, impurities, weighing errors or an incorrect theoretical calculation. The actual mass used for a valid yield comparison should represent the specified product. If yield is given, rearrange to actual amount=theoretical amount×yield/100; if actual and percentage are given, rearrange for theoretical amount.
Composition and purity use different denominators
Percentage by mass of an element in a compound is the total relative-mass contribution of that element in one formula unit divided by the compound's relative formula or molecular mass, multiplied by 100. Include all atoms and bracket multipliers. For CaCO₃ with relative formula mass 100, the oxygen contribution is 3×16=48, so oxygen is 48% by mass. This is not a percentage of oxygen atoms.
Percentage purity is pure substance mass divided by total sample mass, multiplied by 100. If an impurity is stated to be inert, the reacted amount can identify the mass of the pure component. For impure CaCO₃, use the measured CO₂ amount and the 1:1 equation to find the CaCO₃ amount, then convert to mass and divide by original sample mass. That inference assumes complete reaction and correct collection or measurement of the gas; incomplete reaction or gas loss would distort the apparent purity.
Pause and explain
A measured product yield is 110%. Which is a sensible first interpretation?
Worked example
A 10.0 g sample of impure CaCO₃, with inert impurity, reacts completely and produces 1.92 dm³ CO₂ at r.t.p. Find percentage purity. Use CaCO₃→CaO+CO₂, Ca=40, C=12, O=16 and 24 dm³/mol.
Show the worked solution
- CO₂ amount is 1.92/24=0.0800 mol.
- The 1:1 ratio gives 0.0800 mol CaCO₃, with mass 0.0800×100=8.00 g.
- Purity is 8.00/10.0×100=80.0%; the other 2.00 g is inert impurity under the stated assumptions.
Answer 80.0% pure CaCO₃.
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10.0 g pure CaCO₃ decomposes completely in theory by CaCO₃→CaO+CO₂. A student collects 4.20 g dry pure CaO. Calculate theoretical CaO mass and percentage yield using Ca=40, C=12 and O=16.
Give me a hint
Find carbonate moles and apply the 1:1 ratio before comparing product masses.
Compare my reasoning
- CaCO₃ amount is 10.0/100=0.100 mol, predicting 0.100 mol CaO.
- Theoretical CaO mass is 0.100×56=5.60 g.
- Yield is 4.20/5.60×100=75.0%.
5.60 g theoretical CaO; 75.0% yield.
Look for these in your work
- The theoretical denominator is calculated for the product.
- Actual and theoretical masses refer to dry pure CaO.
Find percentage by mass of nitrogen in NH₄NO₃ using N=14, H=1 and O=16. Then find the nitrogen mass in 20.0 g pure NH₄NO₃.
Give me a hint
Count both nitrogen atoms before dividing their contribution by the full relative formula mass.
Compare my reasoning
- Mᵣ=2×14+4×1+3×16=80; nitrogen contribution is 28.
- Nitrogen percentage is 28/80×100=35%.
- Nitrogen mass in 20.0 g is 20.0×0.35=7.00 g.
35% nitrogen by mass; 7.00 g nitrogen in the sample.
Look for these in your work
- Both nitrogen atoms are included.
- Mass percentage is used rather than atom fraction.
A 5.00 g sample containing CaCO₃ and inert impurity gives 0.960 dm³ CO₂ at r.t.p. after complete reaction. Infer its purity. A second sample's apparent purity is 105%; explain why that result needs investigation.
Give me a hint
Use n=V/24 and the 1:1 carbonate-to-CO₂ ratio; purity cannot exceed the mass of the whole sample.
Compare my reasoning
- CO₂ amount is 0.960/24=0.0400 mol, so reacted CaCO₃ mass is 0.0400×100=4.00 g.
- Purity is 4.00/5.00×100=80.0%.
- An apparent 105% exceeds the whole sample; check gas measurement, conditions, sample mass and whether an impurity also produces CO₂ before interpreting it as purity.
80.0% purity; 105% signals an invalid assumption or measurement/calculation problem.
Look for these in your work
- The pure mass is inferred through a balanced reaction.
- The interpretation checks both experimental data and the inert-impurity assumption.
Common mistakes
- Using starting reactant mass as the denominator for product yield.
- Confusing percentage of atoms with percentage by mass.
- Using pure mass instead of total sample mass as the purity denominator.
- Interpreting wet or impure product mass above the theoretical yield as a successful extra yield.
What can you explain now?
The theoretical product mass is 12.0 g and the dry pure product obtained is 9.0 g. Find percentage yield.
Compare with the explanation
75%
9.0/12.0×100=75%; numerator and denominator are masses of the same product.
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Make it stick
Write the three percentage formulas from memory and name each denominator. Calculate a yield and a purity in separate examples, stating the assumptions behind the theoretical prediction.
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.