Question

14) a) The given carbon containing compound have 52.17% of carbon, 13.04% of hydrogen
and rest being oxygen. Its molecular mass = 46gmol-1. find its empirical formula and
molecular formula.(2017)​

Answer 1

Given :

• C = 52.17 %
• H = 13.04 %
• O = 100 - ( 52.17 + 13.04) = 100 - 65.21= 34.79  %
• Molecular mass = 46 g / mol

Solution:

➢ Relative no of atoms

Divide the % composition of the elements by their atomic masses

➜  C = 52.17 / 12 = 4.34

➜  H = 13.04 / 1 = 13.04

➜  O = 34.79 / 16 = 2.17

➢ Simple ratio

Now divide all the values obtained above with the least possible value (i.e. 2.17 in this case )

➜ C = 4.34 / 2.17 = 2

 

➜ H = 13.04 / 2.17   = 6

➜ O = 2.17 / 2.17 = 1

• Empirical formula = C2 H6 O

➜ Empirical mass = 2 x C +  6 x H +  O

➜ Empirical mass =  24 + 6 + 16

➜ Empirical mass = 46 u

We know that,

➜ Molecular mass = n x empirical mass

➜ n = 46  / 46

➜ n = 1

Similarly,

➜ Molecular formula = n x empirical formula

➜ Molecular formula = n x C2 H6 O

➜ Molecular formula = 1 x C2 H6 O

➜ Molecular formula =  C2 H6 O