Question

A compound having mol., mass $=78$ contains $C=92.31$ and $H=7.69$ Its molecular formula is

• A. $C_5{H}_{12}$
• B. $C_5{H}_{18}$
• C. $C_4{H}_{30}$
• D. $C_6{H}_6$

Answer 1

Answer: (D)

The empirical formula of compound can be calculated as

Element	Percentage	Percentage/At.wt.	Ratio
C	92.31	92.31/12=7.7	1
H	7.69	7.69/1 = 7.69	1

Thus the empirical formula is $CH$
The empirical formula wt $=12+1=13$
Value of ' $n^{\prime }=\frac{\text{ Molecular wt }}{\text{ Empirical formula wt }}$
$=\frac{78}{13}=6$
Molecular formula $=$ (Empirical formula) $\times n$
$=(CH)\times 6=C_6{H}_6$