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Percent Yield Worksheet with Answers (Grade 11 Chemistry)

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📄 Percent Yield Worksheet with Answers (Grade 11 Chemistry)

A 17-question Grade 11 chemistry worksheet on percent composition and percent yield.

This free percent yield worksheet gives Grade 11 students focused practice with the two "percent" calculations in chemistry: percent composition by mass and percent yield.

Students find the percent of each element in a compound, use lab data to identify an oxide, calculate theoretical yield from a balanced equation, and then compare it with what was really collected. Every answer on this percent yield worksheet is checked to 3 significant figures.

  • Grade 11 Chemistry
  • 30 marks
  • 45–60 minutes
  • Worked answer key

What's in This Percent Yield Worksheet

This percent yield worksheet starts with percent composition, which uses the same molar mass skills as theoretical yield. Each question has ruled space for working.

PartSkill practisedMarks
APercent composition from a chemical formula8
BUsing percent composition: lab data and mass of an element6
CTheoretical yield and percent yield8
DWorking backwards from a percent yield4
EExplaining yields below and above 100%2
FSpot the mistake2

The percent yield worksheet supports the Ontario Grade 11 Quantities in Chemical Reactions unit, alongside the mole conversion worksheet and the empirical formula worksheet.

Percent Composition on the Percent Yield Worksheet

% element = (mass of that element in 1 mol ÷ molar mass of the compound) × 100

Percent composition tells you what fraction of a compound's mass comes from each element, and Part A of the percent yield worksheet practises it with four compounds. For water, H2O, the molar mass is 18.02 g/mol and oxygen contributes 16.00 g, so water is 88.8% oxygen by mass, even though only one of its three atoms is oxygen.

In the lab, the same idea runs the other way: % element = mass of element ÷ mass of sample × 100. Question 5 uses this to identify a copper oxide.

Theoretical Yield and Percent Yield

percent yield = actual yield ÷ theoretical yield × 100
  • Theoretical yield is the maximum mass of product predicted by the balanced equation, starting from the limiting reactant.
  • Actual yield is the mass really collected in the lab.
  • Percent yield compares the two. Real reactions fall short because of incomplete reactions, side reactions and losses during filtering.

Part C of the percent yield worksheet always names one reactant in excess. When both masses are given, use the limiting reactant worksheet method. For more background, see yield in chemistry on Wikipedia.

Worked Example: Question 8

80.0 g of Fe2O3 reacts with excess carbon monoxide, and 49.7 g of iron is collected.

  1. Grams to moles: 80.0 g ÷ 159.70 g/mol = 0.501 mol Fe2O3.
  2. Mole ratio: Fe2O3 + 3CO → 2Fe + 3CO2, so 1 mol Fe2O3 gives 2 mol Fe: 1.00 mol Fe.
  3. Moles to grams: 1.00 mol × 55.85 g/mol = 56.0 g Fe (theoretical yield).
  4. Percent yield: 49.7 ÷ 56.0 × 100 = 88.8%.

Percent Composition and Percent Yield Practice Problems

These match the percent yield worksheet PDF. Use the molar masses printed on the sheet and give answers to 3 significant figures.

Part A – Percent Composition from a Formula

  1. Find the percent composition of NH3.
  2. Find the percent composition of Fe2O3.
  3. Find the percent composition of Ca(OH)2.
  4. Find the percent composition of KNO3.

Part B – Using Percent Composition

  1. A 4.00 g sample of a copper oxide contains 3.20 g of copper. Find the percent copper. Is the oxide CuO or Cu2O? (2 marks)
  2. What mass of iron is in 50.0 g of iron(III) oxide, Fe2O3? (2 marks)
  3. Ammonium nitrate, NH4NO3, is used as a fertilizer. What mass of nitrogen does a 25.0 g scoop contain? (2 marks)

Part C – Theoretical Yield and Percent Yield

  1. In a blast furnace, 80.0 g of Fe2O3 reacts with excess carbon monoxide and 49.7 g of iron is collected. Find the theoretical yield and the percent yield. (2 marks)
    Fe2O3 + 3CO → 2Fe + 3CO2
  2. 4.60 g of sodium reacts with excess chlorine gas and 10.4 g of NaCl is collected. Find the theoretical yield and the percent yield. (2 marks)
    2Na + Cl2 → 2NaCl
  3. Yeast ferments 90.0 g of glucose and 37.5 g of ethanol, C2H5OH, is collected. Find the theoretical yield and the percent yield. (2 marks)
    C6H12O6 → 2C2H5OH + 2CO2
  4. 6.80 g of hydrogen peroxide decomposes and 2.70 g of oxygen gas is collected. Find the theoretical yield and the percent yield. (2 marks)
    2H2O2 → 2H2O + O2

Part D – Working Backwards

  1. A reaction has a theoretical yield of 40.0 g and usually gives a 75.0% yield. What mass of product should a student expect to collect? (2 marks)
  2. A chemist needs 10.0 g of NaCl. The reaction 2Na + Cl2 → 2NaCl gives an 80.0% yield. What mass of sodium should she start with? (2 marks)

Part E – Explain the Yield

  1. Give two reasons the actual yield of a reaction is usually less than the theoretical yield.
  2. A student's dried precipitate gives a percent yield of 112%. Suggest two possible causes.

Part F – Spot the Mistake

  1. A student finds the percent oxygen in water as 1 atom ÷ 3 atoms × 100 = 33.3%. Explain the error and correct it.
  2. A student collects 15.3 g of product with a theoretical yield of 18.0 g and writes 18.0 ÷ 15.3 × 100 = 118%. Explain the error and correct it.

Percent Yield Worksheet Answer Key

Each percent yield worksheet answer shows the working. The printable key is on pages 3–4 of the PDF.

Part A answers (Q1–4)
  1. M = 17.04 g/mol. N: 14.01 ÷ 17.04 × 100 = 82.2%; H: 3 × 1.01 ÷ 17.04 × 100 = 17.8%
  2. M = 159.70 g/mol. Fe: 2 × 55.85 ÷ 159.70 × 100 = 69.9%; O: 3 × 16.00 ÷ 159.70 × 100 = 30.1%
  3. M = 74.10 g/mol. Ca: 40.08 ÷ 74.10 × 100 = 54.1%; O: 2 × 16.00 ÷ 74.10 × 100 = 43.2%; H: 2 × 1.01 ÷ 74.10 × 100 = 2.7%
  4. M = 101.11 g/mol. K: 39.10 ÷ 101.11 × 100 = 38.7%; N: 14.01 ÷ 101.11 × 100 = 13.9%; O: 3 × 16.00 ÷ 101.11 × 100 = 47.5%
Part B answers (Q5–7)
  1. 3.20 ÷ 4.00 × 100 = 80.0% Cu. CuO is 79.9% Cu and Cu2O is 88.8% Cu, so the oxide is CuO.
  2. % Fe = 2 × 55.85 ÷ 159.70 × 100 = 69.9%; 50.0 g × 0.6994 = 35.0 g Fe
  3. % N = 2 × 14.01 ÷ 80.06 × 100 = 35.0%; 25.0 g × 0.3500 = 8.75 g N (there are 2 N atoms)
Part C answers (Q8–11)
  1. 80.0 g ÷ 159.70 = 0.501 mol Fe2O3; × 2/1 = 1.00 mol Fe; × 55.85 = 56.0 g Fe = theoretical yield 56.0 g. 49.7 ÷ 56.0 × 100 = 88.8%
  2. 4.60 g ÷ 22.99 = 0.200 mol Na; × 2/2 = 0.200 mol NaCl; × 58.44 = 11.7 g NaCl = theoretical yield 11.7 g. 10.4 ÷ 11.7 × 100 = 88.9%
  3. 90.0 g ÷ 180.18 = 0.500 mol C6H12O6; × 2/1 = 0.999 mol C2H5OH; × 46.08 = 46.0 g C2H5OH = theoretical yield 46.0 g. 37.5 ÷ 46.0 × 100 = 81.5%
  4. 6.80 g ÷ 34.02 = 0.200 mol H2O2; × 1/2 = 0.0999 mol O2; × 32.00 = 3.20 g O2 = theoretical yield 3.20 g. 2.70 ÷ 3.20 × 100 = 84.4%
Part D answers (Q12–13)
  1. actual = 75.0% × 40.0 g = 0.750 × 40.0 = 30.0 g
  2. theoretical needed = 10.0 ÷ 0.800 = 12.5 g NaCl; 12.5 ÷ 58.44 = 0.214 mol NaCl = 0.214 mol Na (1 : 1); × 22.99 = 4.92 g Na
Part E answers (Q14–15)
  1. Any two: the reaction is incomplete or reversible; side reactions make other products; product is lost when filtering, transferring or rinsing.
  2. The product was still wet, or it contains impurities such as unreacted reactant; a weighing error is also possible.
Part F answers (Q16–17)
  1. Percent composition is by mass, not by number of atoms: % O = 16.00 ÷ 18.02 × 100 = 88.8%.
  2. The fraction is upside down. Percent yield = actual ÷ theoretical × 100 = 15.3 ÷ 18.0 × 100 = 85.0%.

5 Common Mistakes on a Percent Yield Worksheet

  1. Flipping the fraction. Percent yield is actual ÷ theoretical. Theoretical ÷ actual gives an impossible answer over 100%.
  2. Counting atoms instead of mass. Percent composition is by mass, so oxygen is 88.8% of water, not 33.3%.
  3. Forgetting subscripts. Fe2O3 has two iron atoms, and NH4NO3 has two nitrogen atoms. Multiply before dividing.
  4. Skipping the mole ratio. Grams of reactant never equal grams of product. Go through moles.
  5. Rounding too early. Keep extra digits in the middle of a calculation and round only the final answer.

How to Use This Percent Yield Worksheet in Class

  • Day 1: Parts A and B after a lesson on molar mass.
  • Day 2: Model question 8, then students try 9–11 in pairs.
  • Day 3: Parts D–F before a yield lab.

After the percent yield worksheet, try the stoichiometry quiz and the stoichiometry calculation quiz, both auto-graded.

Frequently Asked Questions

Is this percent yield worksheet free?

Yes. The worksheet and its worked answer key are free to download and print as a PDF, with no account needed.

How do you calculate percent yield?

Divide the actual yield (what you collected) by the theoretical yield (what the balanced equation predicts) and multiply by 100.

What is the difference between percent yield and percent composition?

Percent composition describes a compound: the percent by mass of each element in it. Percent yield describes a reaction: how much product you actually made compared with the maximum possible.

Can percent yield be over 100%?

Not for a pure, dry product. A result over 100% usually means the product was still wet or contained impurities.

More Free Grade 11 Chemistry Resources

Ready to practise? Download the free percent yield worksheet PDF (use the Download button above), print it double-sided, and keep the answer key pages for marking.

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