Question

An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be?
A.) \[\text{C}{{\text{H}}_{\text{4}}}\text{O}\]
B.) \[\text{C}{{\text{H}}_{\text{3}}}\text{O}\]
C.) \[\text{C}{{\text{H}}_{\text{2}}}\text{O}\]
D.) \[\text{CHO}\]

Answer 1

Hint: Empirical formula is the method to determine the simplest whole-number ratio of atoms in a compound. Use the atomic weights of individual elements to calculate empirical formulas.

Complete step by step answer:
Empirical formula gives the “smallest whole number ratio between elements in a compound”.
Since, ratio for Oxygen is not mentioned, we can calculate it using the following
= 100 – ( 38.71 + 9.67 ) % = 51.62 %
For calculating empirical formulas, we need to calculate a simple ratio. We do this by calculating mole ratio for each element first and then dividing mole ratio of every element by the least mole ratio.
In this case, mole ratio of both carbon and oxygen is the least, i.e. 3.22. Therefore, we divide each mole ratio by 3.22 to calculate the simple ratio.
The process is given below –

Carbon
Given % composition = 38.71 %
Atomic mass of Carbon = 12 g
Mole ratio = 38.71/12 = 3.22
Simple ratio = 3.22 / 3.22 = 1

Hydrogen
Given % composition = 9.67 g
Atomic mass of Hydrogen = 1 g
Mole ratio = 9.67 / 1 = 9.67
Simple ratio = 9.67 / 3.22 = 3

Oxygen
Given % composition = 51.62 %
Atomic mass of Oxygen = 16 g
Mole ratio = 51.62 / 16 = 3.22
Simple ratio = 3.22 / 3.22 =1
From here, we can see that Carbon Hydrogen is present in the ratio = 1:3:1.
Therefore, the answer is – option (b) – \[\text{C}{{\text{H}}_{\text{3}}}\text{O}\].

Note: Molecular formula is different from empirical formula. The molecular formula for a compound gives the actual whole number ratio between elements in a compound.