Q. An organic compound on analysis was found to contain 10.06% carbon, 0.84% hydrogen and 89.10% chlorine. What will be the empirical formula of the substance?
A) ${CH}_{2}{Cl}_{2}$
B) ${CHCl}_{3}$
C) ${CCl}_{4}$
D) ${CH}_{3}Cl$
Explanation
(b)
[table] Element | % age | Atomic
mass | Molar ratio | Simpler
molar ratio ; C
H
Cl | 10.06%
0.84%
89.10% | 12
1
35.5 | $\frac{10.06}{12} = 0.84$
$\frac{0.84}{1} = 0.84$
$\frac{89.10}{35.5} = 2.5$ | $\frac{0.84}{0.84} = 1$
$\frac{0.84}{0.84} = 1$
$\frac{2.5}{0.84} = 3$ [/table]
Thus, the empirical formula of the substance of $CHCl_{3}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.