Empirical Formula
The simplest whole-number ratio of atoms of each element in a compound. Shows proportions, not actual numbers.
Molecular Formula
The actual number of atoms of each element in one molecule. Always a simple integer multiple of the empirical formula.
Worked Example
Problem
A compound contains 52.2% C, 13.0% H, and 34.8% O by mass. Its molar mass is 46 g mol⁻¹. Find the empirical and molecular formulas.
Step 1: Assume 100 g sample → masses = percentages in grams
C = 52.2 g, H = 13.0 g, O = 34.8 g
Step 2: Convert to moles (divide by \(A_r\))
C: 52.2 ÷ 12.01 = 4.35 mol
H: 13.0 ÷ 1.008 = 12.90 mol
O: 34.8 ÷ 16.00 = 2.175 mol
Step 3: Divide by the smallest (2.175)
C: 4.35 ÷ 2.175 = 2
H: 12.90 ÷ 2.175 = 6 (≈5.93, round to 6)
O: 2.175 ÷ 2.175 = 1
Empirical formula: C₂H₆O
Step 4: Compare molar masses
Empirical mass = 2(12) + 6(1) + 16 = 46
Given molar mass = 46 → multiplier = 1
Molecular formula: C₂H₆O (ethanol)
🟣 HL Extension. Combustion Analysis
For HL students, empirical formulas often come from raw combustion data rather than neat percentages. The method:
- All carbon in the sample → converted to CO₂ (weigh CO₂ to find moles of C)
- All hydrogen in the sample → converted to H₂O (weigh H₂O to find moles of H)
- Oxygen by subtraction: original mass − mass of C − mass of H
- Then proceed with the normal ratio method above
⚠️ Examiner Tip. Premature Rounding
Intermediate rounding is a fatal, cascading error. Retain at least four or five significant figures throughout. Only round at the final ratio step. Premature rounding can yield misleading ratios (e.g. 1.33 instead of 1.5), trapping you into wrong multipliers.