Percentage yield, purity and composition
Introduced| English | Meaning |
|---|---|
| yield/jiːld/ | The amount of product obtained in a reaction, often expressed as a percentage of the theoretical amount. |
| purity/ˈpjʊərɪti/ | purity |
Why can collected product weigh more than the prediction?
- A student weighs crystals before they are dry and calculates a yield 产率 above 100%. The extra water is not extra product from the equation.
- Supplement: percentage yield, percentage purity 纯度 and percentage composition answer different questions. Choose the correct whole for the denominator.
Percentage yield compares actual with theoretical
- Percentage yield = actual mass of product ÷ theoretical mass of the same product × 100. The theoretical amount comes from the balanced equation and the limiting reactant.
- If 6.0 g is predicted and 4.5 g is collected, yield $=4.5/6.0\times100=75\%$. Loss during transfer, incomplete reaction or competing reactions can reduce yield.
The theoretical product mass is 8.0 g and 6.4 g is collected. Find the percentage yield.
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Yield=6.4/8.0×100=80%.
Percentage purity compares pure with total sample
- Percentage purity = mass of the pure substance ÷ mass of the impure sample × 100. An 8.0 g sample containing 6.0 g of the wanted substance is $6.0/8.0\times100=75\%$ pure.
- To recover the pure mass, multiply sample mass by purity ÷ 100. If 20 g of impure CaCO₃ is 80% pure, only 16 g of CaCO₃ is available to react, assuming the impurity is unreactive.
An impure 12.0 g sample contains 9.0 g of the pure substance. Find its percentage purity.
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Purity=9.0/12.0×100=75%.
Percentage composition counts an element's mass
- Percentage by mass of an element = its total $A_r$ contribution in the formula ÷ the compound's $M_r$ × 100. Count repeated atoms before forming the fraction.
- In H₂O, oxygen contributes 16 out of $M_r=18$, so oxygen percentage $=16/18\times100=88.9\%$ to three significant figures. The two hydrogens together contribute the remaining 11.1%.
For CaCO₃ → CaO + CO₂, use $M_r$: CaCO₃ = 100, CO₂ = 44. A 20 g sample at 80% purity contains 16 g carbonate. Theoretical CO₂ mass $=16/100\times44=7.04\ \mathrm{g}$. If 5.28 g CO₂ is collected, yield $=5.28/7.04\times100=75\%$. The 80% purity is used before predicting product; 75% yield compares collected product with that prediction.
Find the percentage by mass of carbon in CO₂. Use Ar: C=12, O=16. Give your answer to one decimal place.
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Mr=44. Carbon percentage=12/44×100=27.2727…%, or 27.3%.
Match each percentage to its denominator.
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Choose the whole named by the question before dividing.
Investigate an impossible-looking result
- A true yield for a correctly identified pure, dry product cannot exceed 100% for the stated reaction. An apparent value above 100% suggests water, contamination, weighing error or an incorrect theoretical calculation.
- Do not simply replace the result with 100%. Check the measurement and assumptions. A high purity also does not prove a high yield: a small amount of very pure product is possible.
Yield uses theoretical product as the whole. Purity uses total sample as the whole. Composition uses the compound's total relative mass. Reversing numerator and denominator can give a plausible-looking but wrong percentage.
A calculated yield is 112% after crystals were weighed before drying. What is the best next step?
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Surface or retained water adds mass that is not the dry product. An apparent yield above 100% needs investigation.
Explain what the percentage means
- Name the requested percentage and write its numerator and denominator in words before substituting numbers.
- State any assumption about impurities and product dryness. Keep unrounded values until the final step when a purity calculation feeds a yield calculation.
Same percentage method, different wholes: collected ÷ predicted; pure ÷ sample; element contribution ÷ formula mass. The context determines which fraction is valid.
A 25.0 g sample is 80% CaCO₃ and 20% unreactive impurity. CaCO₃ → CaO + CO₂. Use Mr(CaCO₃)=100 and Mr(CO₂)=44. Find theoretical CO₂ mass in g.
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Pure carbonate=25.0×0.80=20.0 g. n=20.0/100=0.200 mol, so CO₂ mass=0.200×44=8.80 g.
A reaction has theoretical yield 8.80 g CO₂ after allowing for purity. Only 6.60 g is collected. Find its percentage yield.
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6.60/8.80×100=75%. Do not apply the purity percentage a second time.
A student says a 99% pure product must have a 99% yield. Explain why that conclusion does not follow.
Compare the reasoning and each criterion; equivalent wording is valid.