Empirical Formula from Experimental Data

Empirical formula calculations work backwards from experimental data (masses or percentages of each element) to find the simplest whole-number ratio of atoms in a compound. This is a core Extended tier skill that frequently appears on Paper 4.

The method

1. If given percentages, treat them as masses in grams (assume a 100 g sample).
2. Divide each mass by the Ar of that element to find moles.
3. Divide every mole value by the smallest mole value to get a ratio.
4. If the ratio is not in whole numbers, multiply all values by the smallest factor that makes them whole numbers.
5. Write the empirical formula using the whole-number ratio.

The formula

moles of each element = mass of element / Ar

Then divide all by the smallest to get the ratio.

Worked examples

Example 1: From mass data (two elements)

Question: A compound contains 2.4 g of carbon and 0.8 g of hydrogen. Find its empirical formula. (Ar: C = 12, H = 1)

Working:

• Moles of C = 2.4 / 12 = 0.2
• Moles of H = 0.8 / 1 = 0.8
• Divide by smallest (0.2): C = 1, H = 4
• Ratio = 1 : 4

Answer: Empirical formula = CH4

Examiner note: The smallest mole value is 0.2. Dividing both by 0.2 gives a clean whole-number ratio.

Example 2: From percentage data

Question: A compound is 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula. (Ar: C = 12, H = 1, O = 16)

Working:

• Treat percentages as masses: C = 40 g, H = 6.7 g, O = 53.3 g
• Moles of C = 40 / 12 = 3.33
• Moles of H = 6.7 / 1 = 6.7
• Moles of O = 53.3 / 16 = 3.33
• Divide by smallest (3.33): C = 1, H = 2.01, O = 1
• Round to whole numbers: C = 1, H = 2, O = 1

Answer: Empirical formula = CH2O

Examiner note: The 2.01 for hydrogen rounds to 2. Small rounding discrepancies from experimental data are expected.

Example 3: Ratio requiring multiplication

Question: A compound contains 1.12 g of iron and 0.48 g of oxygen. Find its empirical formula. (Ar: Fe = 56, O = 16)

Working:

• Moles of Fe = 1.12 / 56 = 0.02
• Moles of O = 0.48 / 16 = 0.03
• Divide by smallest (0.02): Fe = 1, O = 1.5
• Multiply by 2 to get whole numbers: Fe = 2, O = 3

Answer: Empirical formula = Fe2O3

Examiner note: A ratio of 1:1.5 is not whole numbers. Multiply both by 2 to get 2:3.

Example 4: From combustion data

Question: 4.6 g of an organic compound containing only carbon, hydrogen and oxygen was burned. It produced 8.8 g of CO2 and 5.4 g of H2O. Find the empirical formula. (Ar: C = 12, H = 1, O = 16)

Working:

• Mass of C in CO2: Mr of CO2 = 44. Mass of C = (12/44) x 8.8 = 2.4 g
• Mass of H in H2O: Mr of H2O = 18. Mass of H = (2/18) x 5.4 = 0.6 g
• Mass of O = 4.6 - 2.4 - 0.6 = 1.6 g
• Moles of C = 2.4 / 12 = 0.2
• Moles of H = 0.6 / 1 = 0.6
• Moles of O = 1.6 / 16 = 0.1
• Divide by smallest (0.1): C = 2, H = 6, O = 1

Answer: Empirical formula = C2H6O

Examiner note: Combustion analysis requires you to extract the mass of each element from the products before applying the standard method.

Example 5: Metal oxide from reaction data

Question: 0.54 g of aluminium was heated in excess oxygen. 1.02 g of aluminium oxide was produced. Find the empirical formula. (Ar: Al = 27, O = 16)

Working:

• Mass of Al = 0.54 g
• Mass of O = 1.02 - 0.54 = 0.48 g
• Moles of Al = 0.54 / 27 = 0.02
• Moles of O = 0.48 / 16 = 0.03
• Divide by smallest (0.02): Al = 1, O = 1.5
• Multiply by 2: Al = 2, O = 3

Answer: Empirical formula = Al2O3

Examiner note: The mass of oxygen is found by subtracting the mass of aluminium from the mass of the oxide.

Example 6: Three-element compound from percentages

Question: A compound is 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen. Find its empirical formula. (Ar: C = 12, H = 1, O = 16)

Working:

• Moles of C = 52.2 / 12 = 4.35
• Moles of H = 13.0 / 1 = 13.0
• Moles of O = 34.8 / 16 = 2.175
• Divide by smallest (2.175): C = 2, H = 5.98, O = 1
• Round: C = 2, H = 6, O = 1

Answer: Empirical formula = C2H6O

Examiner note: The same compound can have the same empirical formula but different molecular formulae. You need the Mr to distinguish between ethanol (C2H6O) and dimethyl ether (C2H6O).

Common mistakes

1. Dividing by the largest value instead of the smallest. Always divide all mole values by the smallest mole value.
2. Not multiplying to get whole numbers. If you get a ratio like 1:1.5 or 1:1.33, multiply through by 2 or 3 respectively.
3. Rounding 1.5 to 2. A ratio of 1.5 is not close enough to 2. It should be multiplied by 2 to give 3. Only round values very close to a whole number (like 1.98 or 3.02).
4. Forgetting to find the mass of oxygen by difference. In combustion or reaction problems, oxygen mass is usually found by subtraction.
5. Confusing empirical and molecular formulae. The empirical formula gives only the simplest ratio.

When this appears

Empirical formula questions are Extended tier content and appear on Paper 4. They build on percentage composition and relative formula mass. The next step is often finding the molecular formula. See the stoichiometry topic for context.