Question

An organic compound with a vapour density 56.5 has the following percentage composition carbon 53.1%,nitrogen 12.4% oxygen 28.3%,hydrogen 6.2% . the M.F of the compound is ?

Answer 1

Answer:

53.5

Explanation:

53.1

+12.4

+28.3

+6.2

100.0

100.0

-56.5

53.5

Answer 2

Given :

• C = 53.1 %
• N = 12.4 %

• H = 6.2 %

• O = 28.3 %

• Vapor density = 56.5

➢ Relative no of atoms

Divide the % composition of the elements with their atomic masses

➜  C = 53.1 / 12 = 4.42

➜   N = 12.4 / 14 = 0.88

➜  H = 6.2 / 1 = 6.2

➜  O = 28.3 / 16 = 1.76

➢ Simple ratio

Now divide all the values obtained above with the least possible value (i.e. 0.88 in this case )

➜ C = 4.42 / 0.88 = 5

➜ N = 0.88 / 0.88 = 1  

➜ H = 6.2 / 0.88 = 7

➜ O = 1.76 / 0.88 = 2

Empirical formula = C5 N H7 O2

➜ Empirical mass = 5 x C +N +  7 x H +2 x  O

➜ Empirical mass =  60 + 14 + 7 + 32

➜ Empirical mass = 113 u

Molecular mass =  2 x Vapor density

Molecular mass = 2 x 56.5

Molecular mass = 113 u

We know that ,

➜ Molecular mass = n x empirical mass

➜ n = 113  / 113

➜ n = 1

Molecular formula =  C5 N H7 O2