Question

$0.75$ g of an organic compound on complete combustion gave $1.6$ g of carbon dioxide and $0.47$ g of water. A given weight of the compound when heated with nitric acid and silver nitrate gave an equal weight of silver chloride. $0.75$ g of the compound gave $58.57$ ml of nitrogen at $1$ atm and $273$ K. The empirical formula of the compound is :

• A. $C_7{H}_{10}NCl$
• B. $C_6{H}_{12}NCl$
• C. $C_7{H}_{12}NCl$
• D. $C_7{H}_{14}NCl$

Answer 1

Answer: (A)

$C_7{H}_{10}NCl$

$0.75$ g of an organic compound on complete combustion gave $1.6$ g of carbon dioxide and $0.47$ g of water.
$100$ g of an organic compound will give $213.3$ g ($4.85$ mol) of carbon dioxide and $62.66$ g (or $3.48$ mol) of water.
$100$ g of the compound when heated with nitric acid and silver nitrate gave $100$ g (or $0.7$ moles) of silver chloride.
$0.75$ g of the compound gave $58.57$ ml of nitrogen at $1$ atm and $273$ K.
$100$ g of the compound will give $7.8$ L (or $0.7$ mol) of nitrogen at $1$ atm and $273$ K.
From the number of moles of different substances, we can obtain the number of moles of elements. For e.g. $4.85$ mol of carbon dioxide contain $4.85$ mol of $C$.
Thus, the number of moles of $C,H,N$ and $Cl$ present in $100$ g of sample are $4.85,7,0.7$ and $0.7$ respectively.
Thus, the whole number ratio of $C,H,N$ and $Cl$ is $7:10:1:1$.
The empirical formula of the compound is $C_7{H}_{10}NCl$.