Empirical & Molecular Formula
Section: 3. Stoichiometry | Syllabus: Cambridge IGCSE Chemistry (0620)
Calculating Empirical Formula from Mass Data
Given the mass (or percentage by mass) of each element in a compound, the empirical formula can be calculated using moles.
- Write down the mass of each element
- Divide each mass by its Ar to get moles
- Divide all the mole values by the smallest one, to get the simplest ratio
- If the ratio contains decimals, multiply every value by the same number to get whole numbers
- Write the empirical formula
Worked Example: From Mass Data
- Question: a compound contains 2.4 g of carbon and 0.8 g of hydrogen. Find its empirical formula. (Ar: C = 12, H = 1) [3]
- Answer: moles C = 2.4 ÷ 12 = 0.2 mol; moles H = 0.8 ÷ 1 = 0.8 mol [1]; divide by smallest (0.2): C = 1, H = 4 [1]; empirical formula = CH₄ [1]
Worked Example: Needing to Multiply Up
- Question: a compound contains 6.4 g of sulfur and 4.8 g of oxygen. Find its empirical formula. (Ar: S = 32, O = 16) [3]
- Answer: moles S = 6.4 ÷ 32 = 0.2 mol; moles O = 4.8 ÷ 16 = 0.3 mol [1]; divide by smallest (0.2): S = 1, O = 1.5 [1]; multiply both by 2 to clear the decimal: empirical formula = S₂O₃ [1]
Common multipliers: if you get 0.5, multiply by 2; if you get 0.33 or 0.67, multiply by 3; if you get 0.25 or 0.75, multiply by 4.
Calculating from Percentage Composition
If given percentages instead of masses, treat each percentage as if it were a mass in grams (i.e. assume a 100 g sample) - the method is otherwise identical.
Worked Example: From Percentage Composition
- Question: a compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula. (Ar: C = 12, H = 1, O = 16) [3]
- Answer: treat % as g: moles C = 40÷12 = 3.33; moles H = 6.7÷1 = 6.7; moles O = 53.3÷16 = 3.33 [1]; divide by smallest (3.33): C = 1, H = 2, O = 1 [1]; empirical formula = CH₂O [1]
Calculating Empirical Formula from Combustion Data
In a combustion experiment, all the carbon in a hydrocarbon ends up in the CO₂ produced, and all the hydrogen ends up in the H₂O produced.
- moles of C in the original compound = moles of CO₂ produced
- moles of H in the original compound = 2 × moles of H₂O produced (since each H₂O molecule contains 2 H atoms)
Combustion analysis measures the mass of CO₂ and H₂O produced, giving the moles of C and H in the original compound
Worked Example: From Combustion Data
- Question: complete combustion of a hydrocarbon produces 4.4 g of CO₂ and 1.8 g of H₂O. Find its empirical formula. (Ar: C = 12, H = 1, O = 16) [4]
- Answer: moles CO₂ = 4.4 ÷ 44 = 0.1 mol, so moles C = 0.1 mol [1]; moles H₂O = 1.8 ÷ 18 = 0.1 mol, so moles H = 2 × 0.1 = 0.2 mol [1]; divide by smallest (0.1): C = 1, H = 2 [1]; empirical formula = CH₂ [1]
Finding Molecular Formula from Empirical Formula
- Find the empirical formula (as above)
- Calculate the Mr of the empirical formula
- Divide the given Mr of the actual molecule by the Mr of the empirical formula
- Multiply every subscript in the empirical formula by this number
Worked Example: Molecular Formula
- Question: a compound has empirical formula CH₂O and a relative molecular mass of 180. Find its molecular formula. (Ar: C = 12, H = 1, O = 16) [3]
- Answer: Mr of CH₂O = 12 + 2 + 16 = 30 [1]; multiplier = 180 ÷ 30 = 6 [1]; molecular formula = CH₂O × 6 = C₆H₁₂O₆ (glucose) [1]
Common Mistakes
- ❌ Leaving a formula with a decimal subscript, e.g. C₁H₁.₅
- ✓ formulae must have whole-number subscripts - always multiply every value by the same factor until all are whole numbers
- ❌ Forgetting that each H₂O molecule contains 2 hydrogen atoms when using combustion data
- ✓ moles of H = 2 × moles of H₂O, not moles of H₂O directly
- ❌ Giving the empirical formula when the question asks for the molecular formula (or vice versa)
- ✓ check whether an Mr value is given in the question - if so, you likely need to go on to find the molecular formula
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