Empirical Formula
The simplest whole-number ratio of atoms of each element present in a compound.
Molecular Formula
The actual number of atoms of each element present in a single molecule of a compound.
Worked Example
Determining Empirical Formula from Percentage Mass
Problem: A compound contains 52.2% C, 13.0% H, and 34.8% O by mass. Determine its empirical formula.
- Divide by molar mass: C = 52.2/12.01 = 4.35 mol, H = 13.0/1.01 = 12.87 mol, O = 34.8/16.00 = 2.18 mol
- Divide by smallest (2.18): C = 2.0, H = 5.9 ≈ 6, O = 1.0
- Empirical formula = C₂H₆O
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)
Combustion Analysis (HL Extension)
For Higher Level students, empirical formulas often come from combustion analysis data:
- All Carbon in the sample forms CO₂: \(n(\text{C}) = n(\text{CO}_2)\)
- All Hydrogen in the sample forms H₂O: \(n(\text{H}) = 2 \times n(\text{H}_2\text{O})\)
- Oxygen mass is found by subtraction: \(\text{mass of O} = \text{total sample mass} - \text{mass of C} - \text{mass of H}\)
Avoid Premature Rounding in Mole Ratios
Intermediate rounding is a fatal error in formula calculations! Only round if the ratio is within 0.05 of a whole number (e.g. 1.98 → 2). If you get 1.33, multiply everything by 3. If you get 1.50, multiply everything by 2. If you get 1.25, multiply by 4.
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