Question

A gaseous organic compound contains 3.6g of carbon and 0.8 g of hydrogen. The vapour density of this compound is 22. (1) calculate the emperical formula. (2) calculate the molecular formula of the compound.​

Answer 1

Explanation:

Mole of carbon atoms = 3.6/12 = 0.3

Mole of hydrogen atoms = 0.8/1= 0.8

ratio ; C:H :: 3:8

Empirical formula = C3H8

Empirical watertight= 3×12 + 8×1= 44

since, molecularly weight= 2× VD= 2× 22 = 44

factor,n = 44/44= 1

so, molecular formula= C3H8

Answer 2

The correct answer to the questions is as follows:

1. The Empirical formula of the compound is  $C_{0.3}H_{0.8}$.

2. The Molecular formula of the compound is C₃H₈.

Given:

A gaseous organic compound contains;

The mass of carbon is  = 3.6 grams.

The mass of hydrogen is = 0.8 grams.

The vapor density of this compound is = 22.

To Find:

(1) calculate the Empirical formula =?

(2) calculate the Molecular formula of the compound. =?

Solution:

We know the relationship between the vapor density of a compound and its molecular weight.

And the density of a particular gas or vapor relative to that of hydrogen at the same pressure and temperature is called as vapor density of that particular gas.

Now,

The Molecular weight of the compound  = 2 × Vapor Density of the compound.

∴ The molecular weight of our compound = 2 × 22

i.e., The molecular weight of the compound is 44 grams.

Now, the method to find out the empirical formula of any compound is as follows;

Element             Mass(Given)    Molar mass       No. of moles      

1. Carbon             3.6 grams           12 grams              $\frac{3.6}{12}$    =   0.3

2. Hydrogen        0.8 grams           1 gram                  $\frac{0.8}{1}$    =  0.8

Thus,

We get the empirical formula of the compound as follows;

Empirical Formula - $C_{0.3}H_{0.8}$.

Now, Since we know that the molecular formula is always an integral multiple of the empirical formula and also we know the molar mass of the compound, that integral constant is 10.

i.e., The molecular formula of the compound is - C₃H₈.

This is how we can calculate the Molecular and the Empirical formula of any compound.

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