Q. An organic compound has carbon and hydrogen percentage in the ratio 6:1 and carbon and oxygen percentages in the ratio 3:4. The compound has the empirical formula
A) $C_{2}H_{6}O$
B) $CHO_{2}$
C) $CH_{4}O$
D) $CH_{2}O$
Explanation
(d)
$C:H:O = 6:1:8$
$= \frac{6}{15} \times 100:\frac{1}{15} \times 100:\frac{8}{15} \times 100$
$40:6.67:53.3$
$= \frac{40}{12}:\frac{6.67}{1}:\frac{53.3}{16}$
$= 1:2:1\ \ ie,\ \ CH_{2}O$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.