Question

An organic liquid having Carbon , hydrogen, oxygen and nitrogen contains

C=41.37%, H=5.75%, N=16.09% and rest is oxygen. Calculate the molecular formula of liquid if its V.D. is 43.3

Answer 1

Answer:C3H5NO2

ExplanatioC3H5NO2

12(3)+1(5)+14(1)+16(2)

36+5+14+32

=87

Answer 2

Answer:

The molecular formula of the liquid is $C_{3}H_{5}NO_{2}$.

Explanation:

We are given the percentage of each element.

We assume that 100%=100g. So,

1. C- 41.37g
2. H- 5.75g
3. N- 16.09g
4. O- 36.79g

We first find the moles of each element, then divide the moles with the smallest mole.

C                    H                    N                 O

$\frac{41.37}{12}$               $\frac{5.75}{1}$                  $\frac{16.09}{14}$            $\frac{36.79}{16}$

$3.44$               $5.75$                 $1.15$             $2.30$

The value is 1.15.

$\frac{3.44}{1.15}$                $\frac{5.75}{1.15}$                   $\frac{1.15}{1.15}$             $\frac{2.30}{1.15}$

$3$                   $5$                       $1$                 $2$

→ $C_{3}H_{5}NO_{2}$ its molar weight is 87g/mol.

The molar mass of this compound in its molecular formula-

$Vapor Density = \frac{Molar Mass}{2}$

$43.3 = \frac{Molar Mass}{2} \\= 86.6$

≈87

We see that the empirical formula and molecular formula are the same.