Question

An organic compound ‘X’ contains carbon, oxygen and hydrogen only. The percentage of carbon and hydrogen are 47.4% and 10.5% respectively. The relative molecular mass of ‘X’ is 76. Find the empirical formula and the molecular formula of ‘X’.

[Atomic weight: C = 12, O =16, H = 1]

Answer 1

Given: Percentage of carbon = 47.4%

Percentage of hydrogen = 10.5%

Percentage of oxygen = 100 − (47.4 + 10.5) % = 42.1%

Name of element	Percentage mass	Atomic mass	Relative no. of moles	Simplest ratio
Carbon	47.4	12	`47.4/12` = 3.95	`3.95/2.63` = 1.5
Hydrogen	10.5	1	`10.5/1` = 10.5	`10.5/2.63` = 4
Oxygen	42.1	16	`42.1/16` = 2.63	`2.63/2.63` = 1

Ratio between Carbon, hydrogen, and oxygen is:

15 : 4 : 1

To convert it into simple whole numbers, multiply all terms of the ratio by 2.

Hence, C : H : O = 3 : 8 : 2

Empirical formula of the compound = C₃H₈O₂

Empirical formula weight = 12 × 3 + 1 × 8 + 16 × 2

= 36 + 8 + 32

= 76

n = `"Molecular mass"/"Empirical formula mass"`

= `76/76`

= 1

Hence, the molecular formula of the compound:

= 1(C₃H₈O₂)

= C₃H₈O₂