Q. Percentage composition of an organic compound is as follows C=10.06, H=0.84, Cl=89.10 Which of the following corresponds to its molecular formula if the vapour density is 60.0?
A) $CH_{3}Cl$
B) $CHCl_{3}$
C) $CH_{2}Cl_{2}$
D) None of these
Explanation
(b)
Atom Atomic Percentage $\frac{\mathbf{b}}{\mathbf{a}}\mathbf{= x}$ Ratio
Mass $\mathbf{(a)}$ $\mathbf{(b)}$
C 12 10.06 $\frac{10.06}{12}$ 1
H 1 0.84 $\frac{0.84}{1}$ 1
Cl 35.5 89.10 $\frac{89.10}{35.5}$ 3
Empirical formula $= CHCl_{3}$
Empirical formula mass $= 12 + 1 + 106.5 = 119.5 \approx 120$
Molecular mass $= 2 \times V.D = 2 \times 60 = 120$
$n = \frac{molar\ mass}{empirical\ formula\ mass}$
$= \frac{120}{120} = 1$
Molecular formula $= \left( CHCl_{3} \right)_{1} = CHCl_{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.