A carbon compound on analysis gave the following percentage composition carbon 14.5% ; Hydrogen 1.8%; Chlorine 64.46%; Oxygen 19.24% ; Calculate the empirical formula of the compound.

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Detailed Solution

S. No	Element	%	$\frac{%}{At.Wt}$	Simple ratio
1	C	14.5	$\frac{14.5}{12} = 1.21$	$\frac{1.21}{1.20} = 1$
2	H	1.8	$\frac{1.8}{1} = 1.8$	$\frac{1.8}{1.2} = 1.5$
3	CI	64.46	$\frac{64.46}{35.5} = 1.81$	$\frac{1.81}{1.2} = 1.5$
4	O	19.24	$\frac{19.24}{16} = 1.2$	$\frac{1.2}{1.2} = 1$

multiply the simple ratio with suitable number to get integer values.
$2 \times (1 : 1.5 : 1.5 : 1) = 2 : 3 : 3 : 2$
Ratio of atoms present in the molecule
$\begin{array}{llll} C H Cl O \\ 2 3 3 2 \end{array}$
$∴$ The empirical formula of the compound is $C_{2} H_{3} {Cl}_{3} O_{2}$