Chemistry / Unit 5: Chemical Names and Formulas
5.6 Empirical and Molecular Formulas
Notes
The empirical formula is the simplest whole-number ratio of atoms. The molecular formula is the actual number of atoms in a molecule.
- Glucose: molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$, empirical formula $\text{CH}_2\text{O}$.
- Finding an empirical formula: (1) assume 100 g so percents become grams, (2) convert grams to moles, (3) divide each by the smallest number of moles, (4) multiply to get whole numbers if needed (for example, 1.5 becomes 3 when you multiply by 2).
- Finding a molecular formula: $\displaystyle n = \frac{\text{molar mass}}{\text{empirical formula mass}}$, then multiply all subscripts by $n$.
- Example: 40.0% C, 6.7% H, 53.3% O gives $\displaystyle \frac{40.0}{12.01} = 3.33$, $\displaystyle \frac{6.7}{1.01} = 6.63$, $\displaystyle \frac{53.3}{16.00} = 3.33$; ratio 1 : 2 : 1, so $\text{CH}_2\text{O}$.
Practice
Which is the empirical formula of $\text{C}_6\text{H}_6$ (benzene)?
Divide both subscripts by 6.
A compound is 30.4% N and 69.6% O by mass. What is its empirical formula? (N = 14.01, O = 16.00)
N: $\displaystyle \frac{30.4}{14.01} = 2.17$ mol. O: $\displaystyle \frac{69.6}{16.00} = 4.35$ mol. Ratio $4.35 \div 2.17 = 2.00$, so $\text{NO}_2$.
A compound has the empirical formula $\text{CH}_2$ and a molar mass of 56.12 g/mol. What is its molecular formula?
$\text{CH}_2 = 12.01 + 2(1.01) = 14.03$ g/mol. $\displaystyle \frac{56.12}{14.03} = 4$, so multiply by 4.
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