To check the principle of multiple proportions, a series of pure...

(A) If empirical formula of $3$ is ${P}_{3}{Q}_{4}$, then molecular formula is ${\left({P}_{3}{Q}_{4}\right)}_{n}$

$⇒ \dfrac{3{M}_{P}}{3{M}_{P} + 4{M}_{Q}} = \dfrac{40}{100} = \dfrac{2}{5}$

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

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

$∴$ For ${P}_{3}{Q}_{5},$ of $P = \left(\dfrac{3{M}_{P}}{3{M}_{P} + 5{M}_{Q}}\right) \times 100 = \dfrac{8{M}_{Q}}{23{M}_{Q}} \times 100 = 34 . 78$

(B) If empirical formula of compound $3$ is ${P}_{3}{Q}_{2}$, we have

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

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

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

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

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

$∴ \dfrac{{M}_{P}}{{M}_{P} + {M}_{Q}} = \dfrac{4}{9} = \dfrac{44 . 44}{100}$

$9Mp = 5Mp + 4MQ$

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

If empirical formula is assumed as ${P}_{5}{Q}_{4}$,

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

Hence ${P}_{5}{Q}_{4}$ is the empirical formula of compound

(D) If empirical formula of $I$ is ${P}_{2}Q$, we have

$$

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

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

Hence, atomic weight of $P$ and $Q$ cannot be $70$ and $35$ respectively.