This free empirical formula worksheet takes Grade 11 students from percent composition to empirical and molecular formulas, finishing with a hydrate challenge.
Students calculate percent composition, simplify formulas, find empirical formulas from percentage and mass data, scale up to molecular formulas, and fix two classic rounding errors. The worked answer key shows every mole ratio, and every answer has been solved and checked.
- Grade 11 Chemistry
- 30 marks
- 40–50 minutes
- Worked answer key
What's in This Empirical Formula Worksheet
The sheet covers the whole topic in the order it's usually taught.
| Part | Skill practised | Marks |
|---|
| A | Percent composition | 4 |
| B | Simplify to the empirical formula | 4 |
| C | Empirical formula from percent or mass data | 10 |
| D | Molecular formula from molar mass | 6 |
| E | Full problem: data → empirical → molecular | 2 |
| F | Spot the mistake | 2 |
| G | Challenge: hydrates | 2 |
It supports the Ontario Grade 11 chemistry curriculum expectations in the Quantities in Chemical Reactions unit.
Percent Composition
Percent composition tells you what fraction of a compound's mass comes from each element:
% element = (mass of that element in 1 mol ÷ molar mass of the compound) × 100
For water, H2O = 18.02 g/mol and the hydrogen contributes 2.02 g, so %H = 2.02 ÷ 18.02 × 100 = 11.2% and %O = 88.8%. The percentages should always add to 100%. If students need a refresher on molar mass, the mole conversion worksheet covers it.
How to Find an Empirical Formula
Every problem in Part C of this empirical formula worksheet uses the same four steps:
- Assume 100 g. Each percentage becomes a mass in grams. If you're given masses, use them directly.
- Convert to moles. Divide each mass by the element's molar mass.
- Divide by the smallest. This gives the mole ratio.
- Make whole numbers. If a ratio ends in .5, multiply everything by 2; if it ends in .33 or .67, multiply by 3; if it ends in .25 or .75, multiply by 4.
Example: a compound is 52.9% Al and 47.1% O. Al: 52.9 ÷ 26.98 = 1.961 mol. O: 47.1 ÷ 16.00 = 2.944 mol. Dividing by 1.961 gives 1 : 1.50, and multiplying by 2 gives 2 : 3, so the formula is Al2O3.
From Empirical to Molecular Formula
The molecular formula is a whole-number multiple of the empirical formula. Divide the molar mass by the empirical formula mass to find the multiple, then multiply every subscript. For CH2O (30.03 g/mol) with a molar mass of 180.18 g/mol, 180.18 ÷ 30.03 = 6, so the molecular formula is C6H12O6, which is glucose. For more on the theory, see empirical formula on Wikipedia.
Empirical Formula Practice Problems
These match the empirical formula worksheet PDF. Use the molar masses printed on the sheet: H 1.01, C 12.01, N 14.01, O 16.00, Na 22.99, Mg 24.31, Al 26.98, S 32.07, Cl 35.45, K 39.10, Ca 40.08, Fe 55.85, Cu 63.55 g/mol.
Part A – Percent Composition
- Find the percent composition of H2O.
- Find the percent composition of CO2.
- Find the percent composition of NaCl.
- Find the percent composition of CaCO3.
Part B – Write the Empirical Formula
- C6H12O6
- H2O2
- C2H6
- N2O4
Part C – Empirical Formula From Data
- A compound contains 40.0% C, 6.7% H and 53.3% O. Find its empirical formula.
- A compound contains 75.0% C and 25.0% H. Find its empirical formula.
- A compound contains 2.43 g of Mg combined with 1.60 g of O. Find its empirical formula.
- A compound contains 52.9% Al and 47.1% O. Find its empirical formula.
- A compound contains 63.6% N and 36.4% O. Find its empirical formula.
Part D – Molecular Formula
- Empirical formula CH2O, molar mass 180.18 g/mol. Find the molecular formula.
- Empirical formula CH, molar mass 78.12 g/mol. Find the molecular formula.
- Empirical formula NO2, molar mass 92.02 g/mol. Find the molecular formula.
Part E – Put It All Together
- A hydrocarbon is 85.6% carbon and 14.4% hydrogen by mass. Its molar mass is 56.12 g/mol. Find its empirical and molecular formulas.
Part F – Spot the Mistake
- A student finds a mole ratio of C 1 : H 1.33 and writes the empirical formula as CH. Explain the error and correct it.
- A student writes the empirical formula of C6H12O6 as C3H6O3. Explain the error and correct it.
Part G – Challenge: Hydrates
- A 2.50 g sample of hydrated copper(II) sulfate, CuSO4·xH2O, is heated until all the water is driven off. 1.60 g of anhydrous CuSO4 remains. Find x.
- What is the percent by mass of water in CuSO4·5H2O?
Empirical Formula Worksheet Answer Key
Every answer shows the mole ratio, so students can see exactly where a calculation went wrong. The printable key is on pages 3–4 of the PDF.
Part A answers (Q1–4)
- H2O: %H = 11.2%, %O = 88.8%
- CO2: %C = 27.3%, %O = 72.7%
- NaCl: %Na = 39.3%, %Cl = 60.7%
- CaCO3: %Ca = 40.0%, %C = 12.0%, %O = 48.0%
Part B answers (Q5–8)
- C6H12O6 → CH2O
- H2O2 → HO
- C2H6 → CH3
- N2O4 → NO2
Part C answers (Q9–13)
- C: 40.0 ÷ 12.01 = 3.331; H: 6.7 ÷ 1.01 = 6.634; O: 53.3 ÷ 16.00 = 3.331 → ÷ 3.331 → 1.00 : 1.99 : 1.00 → CH2O
- C: 75.0 ÷ 12.01 = 6.245; H: 25.0 ÷ 1.01 = 24.752 → ÷ 6.245 → 1.00 : 3.96 → CH4
- Mg: 2.43 ÷ 24.31 = 0.100; O: 1.60 ÷ 16.00 = 0.100 → ÷ 0.100 → 1.00 : 1.00 → MgO
- Al: 52.9 ÷ 26.98 = 1.961; O: 47.1 ÷ 16.00 = 2.944 → ÷ 1.961 → 1.00 : 1.50 × 2 → Al2O3
- N: 63.6 ÷ 14.01 = 4.540; O: 36.4 ÷ 16.00 = 2.275 → ÷ 2.275 → 2.00 : 1.00 → N2O
Part D answers (Q14–16)
- CH2O = 30.03 g/mol; 180.18 ÷ 30.03 = 6 → C6H12O6
- CH = 13.02 g/mol; 78.12 ÷ 13.02 = 6 → C6H6
- NO2 = 46.01 g/mol; 92.02 ÷ 46.01 = 2 → N2O4
Part E answer (Q17)
- C: 85.6 ÷ 12.01 = 7.127; H: 14.4 ÷ 1.01 = 14.257 → ÷ 7.127 → 1.00 : 2.00 → empirical CH2 (14.03 g/mol); 56.12 ÷ 14.03 = 4 → molecular C4H8
Part F answers (Q18–19)
- 1.33 is not close to a whole number; multiply both by 3 to get 3 : 4. Correct: C3H4
- That ratio can still be divided by 3. Correct: CH2O
Part G answers (Q20–21)
- water = 0.90 g ÷ 18.02 = 0.0499 mol; CuSO4 = 1.60 ÷ 159.62 = 0.0100 mol; ratio = 4.98 → x = 5 (CuSO4·5H2O)
- 5 × 18.02 ÷ 249.72 × 100 = 36.1%
5 Common Empirical Formula Mistakes
Part F of the empirical formula worksheet targets the first two.
- Rounding too early. A ratio of 1 : 1.33 is not 1 : 1. Multiply by 3 to get 3 : 4.
- Not simplifying fully. C3H6O3 still divides by 3. The empirical formula of glucose is CH2O.
- Dividing percentages by atomic number instead of molar mass.
- Dividing by the largest mole value instead of the smallest, which gives fractions below 1.
- Forgetting the water in hydrates. The mass lost on heating is the water. Convert it to moles and compare with moles of the anhydrous salt.
How to Use This Empirical Formula Worksheet in Class
- Day 1: Parts A and B after introducing percent composition.
- Day 2: Part C, working the first problem together on the board.
- Day 3: Parts D–G, ideally before a hydrate lab so students predict the formula first.
Follow up with the stoichiometry quiz, or use our free chemistry calculators to check molar masses.
Frequently Asked Questions
Is this empirical formula worksheet free?
Yes. The worksheet and answer key are free to download and print as a PDF, with no account needed.
What is the difference between an empirical and a molecular formula?
The empirical formula is the simplest whole-number ratio of atoms. The molecular formula is the actual number of atoms in one molecule. Glucose has the molecular formula C6H12O6 and the empirical formula CH2O.
How do you find an empirical formula from percent composition?
Assume a 100 g sample so each percent becomes grams, convert grams to moles, divide every mole value by the smallest, and multiply to whole numbers if needed.
What do you do if the mole ratio is 1.5 or 1.33?
Multiply every ratio by the same number to reach whole numbers: by 2 for .5, by 3 for .33 or .67, and by 4 for .25 or .75.
What grade is this empirical formula worksheet for?
It is written for Grade 11 chemistry and also works as review for Grade 12 and first-year college chemistry.
More Free Grade 11 Chemistry Resources
Ready to practise? Download the free empirical formula worksheet PDF (use the Download button above), print it double-sided, and keep the answer key pages for marking.