A compound contains 48.98% C, 2.72% Hand 48.32% Cl by mass. Its empirical formula is

A compound contains 48.98% C, 2.72% Hand 48.32% Cl by mass. Its empirical formula is

A
$C_{13} H_{8} {Cl}_{5}$
B
$C_{3} H_{2} Cl$
C
$C_{12} H_{10} {Cl}_{5}$
D
$C_{15} H_{12} {Cl}_{6}$

Solution:

Ratio of atoms $C : H : Cl :: \frac{48.98}{12} : \frac{2.72}{1} : \frac{48.32}{35.5} :: 4.08 : 2.72 : 1.36 :: 3 : 2 : 1$

Empirical formula = $C_{3} H_{2} Cl$

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