Molecular Formula Chemistry Example 5

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Example 5

hard
A compound is 85.7%85.7\% C and 14.3%14.3\% H by mass, with a molar mass of 56.0 g/mol56.0\,\text{g/mol}. Determine both the empirical and molecular formulas.

Solution

  1. 1
    Assume 100100 g: C = 85.785.7 g, H = 14.314.3 g. Moles: C =85.712.01=7.14= \frac{85.7}{12.01} = 7.14, H =14.31.008=14.2= \frac{14.3}{1.008} = 14.2.
  2. 2
    Ratio: 7.147.14=1\frac{7.14}{7.14} = 1 C, 14.27.14=1.99≈2\frac{14.2}{7.14} = 1.99 \approx 2 H. Empirical formula: CH2\text{CH}_2. Empirical mass = 14.03 g/mol14.03\,\text{g/mol}.
  3. 3
    Multiplier =56.014.03=3.99≈4= \frac{56.0}{14.03} = 3.99 \approx 4. Molecular formula =C4H8= \text{C}_4\text{H}_8 (e.g., butene).

Answer

Empirical: CH2. Molecular: C4H8\text{Empirical: CH}_2\text{. Molecular: C}_4\text{H}_8
This two-step process (percent composition → empirical formula → molecular formula) is a standard method for determining the formula of an unknown compound from experimental data.

About Molecular Formula

The chemical formula showing the actual number of atoms of each element in one molecule of a compound, as opposed to the empirical formula which.

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