Molecular Formula Chemistry Example 3

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Example 3

medium
A compound is 40% C, 6.7% H, 53.3% O by mass. Its molar mass is 180 g/mol. Find the molecular formula.

Solution

  1. 1
    Assume 100 g of the compound. Moles of each element: C:4012.01=3.33,H:6.71.008=6.65,O:53.316.00=3.33\text{C}: \frac{40}{12.01} = 3.33,\quad \text{H}: \frac{6.7}{1.008} = 6.65,\quad \text{O}: \frac{53.3}{16.00} = 3.33
  2. 2
    Divide by the smallest (3.33) to get the mole ratio: C : H : O = 1 : 2 : 1. The empirical formula is CH2O\text{CH}_2\text{O}.
  3. 3
    Empirical formula mass =12.01+2(1.008)+16.00=30.03 g/mol= 12.01 + 2(1.008) + 16.00 = 30.03\,\text{g/mol}. Multiplier =18030.03=5.99≈6= \frac{180}{30.03} = 5.99 \approx 6. Molecular formula =(CH2O)6=C6H12O6= (\text{CH}_2\text{O})_6 = \text{C}_6\text{H}_{12}\text{O}_6 (glucose).

Answer

The molecular formula is C6H12O6\text{C}_6\text{H}_{12}\text{O}_6 (glucose).
To find a molecular formula from percent composition: (1) convert percentages to moles, (2) find the simplest whole-number ratio (empirical formula), (3) compare the empirical formula mass to the given molar mass to find the multiplier.

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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