Question

A compound of empirical formula AB has vapour density three times it's empirical formula weight. Find the molecular formula

Answer 1

Answer:

     Molecular Formula ⇒ A₆B₆

Explanation:

Molecular Formula = {Empirical formula}ₙ

n = Molecular Mass / Empirical Mass = MM / EM

n = MM / EM ----{i}

Vapor density = Molecular mass / 2

Molecular mass = 2.Vapor density ----{ii}

From question - Vapor density = 3 × Empirical Mass ----{iii}

from {ii} and {iii}

Molecular mass = 2 × 3 × Empirical Mass

Molecular mass = 6 × Empirical Mass

6 = Molecular mass / Empirical Mass ----{iv}

from {i} and {iv}

n = 6

Molecular Formula - {Empirical formula}ₙ

                                = {AB}ₙ

                                = {AB}₆

Molecular Formula = A₆B₆

Answer 2

Given:

Empirical formula of the compound $=AB$

Vapour Density $\[ = 3 \times Empirical\,formula\,weight\]$

To Find: Molecular formula of the compound

Solution:

The relation between the molecular formula and the empirical formula for the given compound can be given as:

$\[Molecular\,formula = {(Empirircal\,formula)_n}\]$

Therefore,

$\[Molecular\,formula = {(AB)_n}\]$        ... (A)

Also, we have

$\[n = \frac{{Molecular\,mass}}{{Empirical\,mass}}\]$                         ... (i)

The relation between molecular mass and vapour density is

$\[Molecular\,mass = 2 \times Vapour\,density\]$       ... (ii)

According to the question,

$\[Vapour\,density = 3 \times Empirical\,mass\]$        ... (iii)

From the equations (ii) and (iii), we have

$\[Molecular\,mass = 2 \times 3 \times Empirical\,mass\]$

$\[Molecular\,mass = 6 \times Empirical\,mass\]$

Therefore, the value of 'n' is 6.

Now, from equation (A), we have

$\[Molecular\,formula = {(AB)_n}\]$

$\[Molecular\,formula = {(AB)_6}\]$

$\[Molecular\,formula = {A_6}{B_6}\]$

Hence, the molecular formula of the compound is $\[{A_6}{B_6}\]$.