Q. If a compound on analysis was found to contain $C = 18.5\%,\ H = 1.55\%,\ Cl = 55.04\%$ and $O = 24.81\%$ then its empirical formula is
A) $CH_{2}OCl$
B) $CH_{2}ClO_{2}$
C) $ClCH_{2}O$
D) $CHClO$
Explanation
(d)
$C:H:Cl:O = \frac{18.5}{12}:\frac{1.55}{1}:\frac{55.04}{35.5}:\frac{24.81}{16} = 1:1:1:1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.