Chemistry - Mole Concept Question with Solution | TestHub

(A) If empirical formula of $3$ is ${\mathrm{P}}_3{\mathrm{Q}}_4$, then molecular formula is ${\left({\mathrm{P}}_3{\mathrm{Q}}_4\right)}_{\mathrm{n}}$

$\Rightarrow \frac{3{\mathrm{M}}_{\mathrm{P}}}{3{\mathrm{M}}_{\mathrm{P}}+4{\mathrm{M}}_{\mathrm{Q}}}=\frac{40}{100}=\frac{2}{5}$

$15{\mathrm{M}}_{\mathrm{p}}=6{\mathrm{M}}_{\mathrm{p}}+8{\mathrm{M}}_{\mathrm{Q}}$

$9{\mathrm{M}}_{\mathrm{P}}=8{\mathrm{M}}_{\mathrm{Q}}$

$\therefore$ For ${\mathrm{P}}_3{\mathrm{Q}}_5,\%$ of $\mathrm{P}=\left(\frac{3{\mathrm{M}}_{\mathrm{P}}}{3{\mathrm{M}}_{\mathrm{P}}+5{\mathrm{M}}_{\mathrm{Q}}}\right)\times 100=\frac{8{\mathrm{M}}_{\mathrm{Q}}}{23{\mathrm{M}}_{\mathrm{Q}}}\times 100=34.78\%$

(B) If empirical formula of compound $3$ is ${\mathrm{P}}_3{\mathrm{Q}}_2$, we have

$\frac{3{\mathrm{M}}_{\mathrm{P}}}{3{\mathrm{M}}_{\mathrm{P}}+2{\mathrm{M}}_{\mathrm{Q}}}=\frac{2}{5}$

$15{\mathrm{M}}_{\mathrm{P}}=6{\mathrm{M}}_{\mathrm{P}}+4{\mathrm{M}}_{\mathrm{Q}}$

$9\mathrm{Mp}=4{\mathrm{M}}_{\mathrm{Q}}$

If ${\mathrm{M}}_{\mathrm{p}}=20$

${\mathrm{M}}_{\mathrm{Q}}=\frac{180}{4}=45$
(C) If empirical formula of $2$ is $\mathrm{PQ}$

$\therefore \frac{{\mathrm{M}}_{\mathrm{P}}}{{\mathrm{M}}_{\mathrm{P}}+{\mathrm{M}}_{\mathrm{Q}}}=\frac{4}{9}=\frac{44.44}{100}$

$9\mathrm{Mp}=5\mathrm{Mp}+4\mathrm{MQ}$

$5{\mathrm{M}}_{\mathrm{P}}=4{\mathrm{M}}_{\mathrm{Q}}$

If empirical formula is assumed as ${\mathrm{P}}_5{\mathrm{Q}}_4$,

$\%$ of $\mathrm{P}=\left(\frac{5{\mathrm{M}}_{\mathrm{P}}}{5{\mathrm{M}}_{\mathrm{P}}+4{\mathrm{M}}_{\mathrm{Q}}}\right)\times 100=50$

Hence ${\mathrm{P}}_5{\mathrm{Q}}_4$ is the empirical formula of compound

(D) If empirical formula of $I$ is ${\mathrm{P}}_2\mathrm{Q}$, we have

$\% \mathrm{of} \mathrm{P}=\left(\frac{2{\mathrm{M}}_{\mathrm{P}}}{2{\mathrm{M}}_{\mathrm{P}}+{\mathrm{M}}_{\mathrm{Q}}}\right)\times 100=50$

$4{\mathrm{M}}_{\mathrm{p}}=2{\mathrm{M}}_{\mathrm{p}}+{\mathrm{M}}_{\mathrm{Q}}$

$2{\mathrm{M}}_{\mathrm{p}}={\mathrm{M}}_{\mathrm{Q}}$

Hence, atomic weight of $\mathrm{P}$ and $\mathrm{Q}$ cannot be $70$ and $35$ respectively.