Q. A compound contains 54.55% carbon, 9.09 % hydrogen, 36.36% oxygen. The empirical formula of this compound is
A) $C_{3}H_{5}O$
B) $C_{4}H_{8}O_{2}$
C) $C_{2}H_{4}O_{2}$
D) $C_{2}H_{4}O$
Explanation
(d)
Given, % of C=54.55%
% of H=9.09%
%of O=36.36%
[table] Element | % | At. no. | Ratio of atoms | Simplest ration ; C
H
O | 54.55
9.09
36.06 | 12
1
16 | 54.55/12=4.54
9.09/1=9.09
36.16/16=2.27 | 4.54/2.27=2
9.09/2.27=4
2.27/2.27=1 [/table]
$\therefore$ Empirical formula is $C_{2}H_{4}O$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.