Question

In an organic compound of molar mass 108 g mol–1 C,H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular formula can be :

• A. C6H8N2
• B. C7H10N
• C. C5H6N3
• D. C4H18N3

Answer 1

Answer: (A)

C6H8N2

Answer 2

Molar mass = 108 gm/mole

\begin{tabular} { | c | c | c | c | c | } \hline Element & Wh. Ratio & WL ratio/Atomic mass & Simple Ratio & Simple Integer ratio \ \hline C & $9 \times$ & $\frac { 8 \pi } { 12 } = \frac { 3 \mathrm { x } } { 4 }$ & 3 & 3 \ \hline H & 1 x & x & 4 & 4 \ \hline N & $3.5 \times$ & $\frac { 3.5 \mathrm { x } } { 14 } = \frac { \mathrm { x } } { 4 }$ & 1 & 1 \ \hline \end{tabular}

$\therefore$ C3H4N

Empirical mass = 12 × 3 + 4 + 14 = 54

$\therefore$ Molecular Formula = C6H8N2