Question

0.2 g of an organic compound contain 40.67% of carbon, 8.47% of hydrogen,37.96 ml of nitrogen are

obtained at S.T.P. from the same amount of the compound. Find out the empirical formula of the

compound?​

Answer 1

Given info : 0.2 g of an organic compound contain 40.67% of carbon, 8.47% of hydrogen,37.96 ml of nitrogen are obtained at S.T.P. from the same amount of the compound.

To find : emperical formula of the compound.

solution : mass of nitrogen gas = (molar mass of nitrogen × volume)/22400

= (28g × 37.96)/22400

= 0.04745 g

Now percentage of nitrogen in compound = 0.04745/0.2 × 100

= 23.725 %

Let's find percentage/mass ratio of each element.

for carbon ⇒40.67/12 = 3.389

For hydrogen ⇒8.47/1 = 8.47

For nitrogen ⇒23.725/14 = 1.6946

Now find simplest ratio of all gotten percentage/mass terms.

C : H : N = 3.389 : 8.47 : 1.6946

= 3.389/1.6946 : 8.47/1.6946 : 1.6946/1.6946

= 2 : 5 : 1

Therefore the emperical formula of compound is $C_2H_5N$