Quantitative analysis of an organic compound (X) shows the following percentage composition.
C: 14.5%
Cl: 64.46%
H: 1.8%
Empirical formula mass of the compound (X) is:
Quantitative analysis of an organic compound (X) shows the following percentage composition.
C: 14.5%
Cl: 64.46%
H: 1.8%
Empirical formula mass of the compound (X) is:
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Solution and Explanation
\[ \text{Moles of C} = \frac{14.5}{12} = 1.21 \, \text{mol}, \quad \text{Moles of Cl} = \frac{64.46}{35.5} = 1.81 \, \text{mol}, \quad \text{Moles of H} = \frac{1.8}{1} = 1.8 \, \text{mol} \]
\[ \text{C}: \frac{1.21}{1.21} = 1, \quad \text{Cl}: \frac{1.81}{1.21} = 1.5, \quad \text{H}: \frac{1.8}{1.21} = 1.49 \]
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