A molecular formula gives the actual number of each type of atom in one molecule. It is related to the empirical formula by an integer factor \(n\): \[ \text{molecular formula} = n \cdot (\text{empirical formula}) \]
Problem. A compound has empirical formula CH and molar mass \(78.0\ \text{g/mol}\). Determine its molecular formula.
Step 1: Compute the empirical-formula mass
Use standard atomic masses (rounded to typical introductory-chemistry precision): \(M(\text{C}) \approx 12.01\ \text{g/mol}\), \(M(\text{H}) \approx 1.008\ \text{g/mol}\).
| Atom in empirical unit | Count | Atomic mass (g/mol) | Contribution (g/mol) |
|---|---|---|---|
| C | 1 | \(12.01\) | \(1 \cdot 12.01 = 12.01\) |
| H | 1 | \(1.008\) | \(1 \cdot 1.008 = 1.008\) |
| Total empirical-formula mass | \(\,M(\text{CH}) = 12.01 + 1.008 = 13.018\ \text{g/mol}\) | ||
Step 2: Find the integer multiple \(n\)
The factor \(n\) is the ratio of the molar mass to the empirical-formula mass: \[ n = \frac{78.0\ \text{g/mol}}{13.018\ \text{g/mol}} \approx 5.99 \approx 6 \] The result rounds to the nearest whole number because molecular formulas must have whole-number subscripts.
Step 3: Multiply subscripts in the empirical formula by \(n\)
- Empirical formula: \(\text{C}_1\text{H}_1\)
- Multiply each subscript by \(6\): \(\text{C}_{1\cdot 6}\text{H}_{1\cdot 6} = \text{C}_6\text{H}_6\)
Result. The molecular formula is \(\text{C}_6\text{H}_6\).
Visualization: empirical unit repeated to form the molecular formula
Quick checklist for any molecular formula problem
- Compute empirical-formula mass (sum of atomic masses with subscripts).
- Compute \(n = \dfrac{\text{molar mass}}{\text{empirical-formula mass}}\) and round to the nearest whole number.
- Multiply every subscript in the empirical formula by \(n\).