Answer to Question #124706 in Organic Chemistry for Kabir Bashir

Question

4 b). An organic compound contains 47.3% Carbon, 10.59% Hydrogen and 42.11% Oxygen. Calculate the empirical formula of the compound. If the vapour density of the compound is 22, find
the molecular formula of the compound.

Answer 1

Percentage of carbon = 47.3

Mass of carbon= 47.3 g

Molecular Mass of carbon= 12 g

No of mole of carbon= 47.3/12

= 3.94 mole

Percentage of hydrogen =10.59

Mass of hydrogen= 10.59g

Molecular Mass oh hydrogen= 1 g

Number of mole of hydrogen= 10.59/1

= 10.59 mole

Percentage of oxygen= 42.11

Mass of oxygen=42.11g

Molecular Mass of oxygen=16g

Number of mole of oxygen=42.11/16

=2.63

C:H:O=3.94:10.59:2.63

Simplefy the ratio

C= 3.94/2.63=1.5 ×2 = 3

H=10.59/2.63=4 ×2=8

O=2.63/2.63=1×2=2

Empirical formula= C3H8O2

Empirical formula Mass =76

Vapour density of the compound is =38

Molecular Mass of the compound=76

n=76/76=1

Molecular formula=

(Empirical formula)n

Molecular formula= C3H8O2