Question

An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C= 38.71% and H= 9.67%. What would be the empirical formula of the compound?

• A. CHO
• B. CH₂O
• C. CH₃O
• D. CH₄O

Answer 1

The correct option is C

CH₃O

Given:

Percentage composition is C=38.71 %, H=9.67 %

Percentage composition of Oxygen (O)= 100- (38.71+ 9.67)= 100- 48.38= 51.62 %

Step-1: Calculate the empirical formula

• Empirical formula: Empirical formula is the simplest formula of the compound which gives the simplest whole-number ratio of the atoms of the various elements present in one molecule of the compound. Example: Molecular formula of glucose is C₆H₁₂O₆ but the simplest whole-number ratio of atoms in glucose is CH₂O. Therefore, CH₂O is the empirical formula.

Element	% Of Element	Atomic mass (u)	Atomic ratio	Simplest ratio
Carbon (C)	38.71	12	$\frac{38.71}{12}=3.22$	$\frac{3.22}{3.22}=1$
Hydrogen (H)	9.67	1	$\frac{9.67}{1}=9.67$	$\frac{9.67}{3.22}=3$
Oxygen (O)	51. 62	16	$\frac{51.62}{16}=3.22$	$\frac{3.22}{3.22}=1$

As the simplest ratio of Carbon is 1, Hydrogen is 3 and Oxygen is 1, therefore the empirical formula comes out to be C₁H₃O₁ or CH₃O.

Hence, option (C) is correct.