Question

A compound contains N = 25.94% and 0 = 74.06%. If its vapour density is 54.2, calculate its molecular formula.

• A. $NO$
• B. $N{O}_2$
• C. $N_{2}O$
• D. $N_2{O}_5$

Answer 1

The correct option is D $N_2{O}_5$

Let us assume the mass of the compound is $100g$.

Mass of $N=25.94g$

Mass of $O=74.06g$

Molar mass of $N=14g/mol$, $O=16g/mol$

$\therefore$ Moles of $N=\frac{25.94}{14}=1.85mol$,

Moles of $O=\frac{74.06}{16}=4.62mol$

Now, simplest atomic ratio:

$N:O=\frac{1.85}{1.85}:\frac{4.628}{1.85}$

$=1:2.5$

Now, whole number simplest atomic ratio is:

$N:O=(1:2.5)\times 2$

$=2:5$

$\therefore$ Emperical formula= $N_2{O}_5⟶1$

Emperical formula mass= $28g+80g=108g⟶2$

Given Vapour density of compound=$54.2$

We know, Molecular mass = $2\times \text{Vapour density}$

$\Rightarrow$ Molecular mass of given compound=$2\times 54.2=108.4g$ $⟶3$

Now, $n=\frac{Molarmass}{Empericalformula}=\frac{108.4}{108}=11$ $[\because 2$ & $3]$

$\therefore$ Molecular formula $=n\times Empericalformula$

$=1\times N_2{O}_5$

$=N_2{O}_5$

Hence, option $D$ is correct.