Question

In an organic compound, the percentage of carbon and hydrogen are 40 and 6.7, respectively. If the vapor density of compound is 30, what is the molecular formula of the compound.

Answer 1

Step-1: Calculate the empirical formula

• The empirical formula weight is obtained by the addition of the atomic weight of the various atoms present in 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.

Percentage of Carbon= 40

Percentage of Hydrogen= 6.7

The percentage of two elements combined together is less than 100.

Percentage of Oxygen = 100-(40+6.7)= 53.3 %.

Element	% Of Element	Atomic mass (u)	Atomic ratio	Simplest ratio
Carbon (C)	40	12	$\frac{40}{12}=3.33$	$\frac{3.33}{3.33}=1$
Hydrogen (H)	6.7	1	$\frac{6.7}{1}=6.7$	$\frac{6.7}{3.33}=2$
Oxygen (O)	53.3	16	$\frac{53.3}{16}=3.33$	$\frac{3.33}{3.33}=1$

As the simplest ratio of Carbon =1, Hydrogen =2 and Oxygen = 1, therefore the empirical formula comes out to be CH₂O_.

Step-2: Calculate the molecular formula

• Molecular formula: Molecular formula is the actual formula of the compound which gives the actual number of atoms present in the molecule of the compound. For example, the molecular formula of glucose is C₆H₁₂O₆.
• The relationship between the empirical formula and the molecular formula is given as follows: Molecular formula = n× Empirical formula, where n is an integer such as 1,2,3…..etc.

$n=\frac{Molecularweightofthecompound}{Empiricalformulamass}$

Given, Vapor density = 30

Molecular weight= 2 x Vapor density

Molecular weight = 2x 30= 60 u

The empirical formula weight

Atomic weight of Carbon= 12 u

Atomic weight of Hydrogen= 1 u

Atomic weight of Oxygen= 16 u

Empirical formula weight of CH₂O= 12 +2 +16= 30 u

$n=\frac{Molecularweightofthecompound}{Empiricalformulamass}$

$n=\frac{60}{30}=2$

As, Molecular formula = n× Empirical formula, therefore molecular formula of CH₂O is (CH₂O)₂ i.e.; C₂H₄O_2.