Question

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:

Answer 1

Correct Answer: 1655

Element	% Mass	Molar Ratio	Simplest Ratio
C	14.5	$ \frac{14.5}{12} = 1.2 $	$ \frac{1.2}{1.2} = 1 $
Cl	64.46	$ \frac{64.46}{35.5} \approx 1.8 $	$ \frac{1.8}{1.2} = 1.5 $
H	1.8	$ \frac{1.8}{1} = 1.8 $	$ \frac{1.8}{1.2} = 1.5 $
O	19.24	$ \frac{19.24}{16} \approx 1.2 $	$ \frac{1.2}{1.2} = 1 $

Steps:

1. Divide % composition by atomic mass to get molar ratio.
2. Divide each ratio by the smallest value to get simplest ratio.
3. Multiply all ratios by a factor to convert to whole numbers if necessary.

Empirical Formula: $ C_2H_3Cl_3O_2 $

Molar Mass: $ 165.5\,\text{g/mol} $ (or $ 1655 \times 10^{-1}\,\text{g/mol} $)