Molecular Formula Chemistry Example 2

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Example 2

medium
A compound has an empirical formula of NO2\text{NO}_2 and a molar mass of 92 g/mol92\,\text{g/mol}. Determine the molecular formula. (N = 14.0114.01, O = 16.0016.00)

Solution

  1. 1
    Empirical formula mass of NO2=14.01+2(16.00)=46.01 g/mol\text{NO}_2 = 14.01 + 2(16.00) = 46.01\,\text{g/mol}.
  2. 2
    Find the multiplier: n=molar massempirical formula mass=9246.01=2.00n = \frac{\text{molar mass}}{\text{empirical formula mass}} = \frac{92}{46.01} = 2.00.
  3. 3
    Molecular formula =(NO2)2=N2O4= (\text{NO}_2)_2 = \text{N}_2\text{O}_4 (dinitrogen tetroxide).

Answer

N2O4\text{N}_2\text{O}_4
Determining the molecular formula requires knowing both the empirical formula and the molar mass. The molar mass can be found experimentally through techniques like mass spectrometry or gas density measurements.

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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