To check the principle of multiple proportions, a series of pure binary compounds ( $P_{m} Q_{n}$ )

Question

To check the principle of multiple proportions, a series of pure binary compounds ( $P_{m} Q_{n}$ ) were analyzed and their composition is tabulated below. The correct option(s) is (are) $$\begin{tabular}{|c|c|c|} \hline Compound & Weight $\%$ of $P$ & Weight $\%$ of $\boldsymbol{Q}$ \ \hline 1 & 50 & 50 \ \hline 2 & 44.4 & 55.6 \ \hline 3 & 40 & 60 \ \hline \end{tabular}$$

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(b, c) : Option (a) $ightarrow$ For compound 3, Let the mass of compound be 100 g .\ Mass of $P=40 \mathrm{~g}$, Atomic mass of $P=A_{p}$\ Mass of $Q=60 \mathrm{~g}$, Atomic mass of $Q=A_{Q}$\ Molar ratio of $P$ and $Q$ is\ $\frac{40 / A_{P}}{60 / A_{Q}}=\frac{3}{4} \Rightarrow \frac{40}{60} 	imes \frac{A_{Q}}{A_{P}}=\frac{3}{4}$\ $\frac{2}{3} 	imes \frac{A_{Q}}{A_{P}}=\frac{3}{4} \Rightarrow 9 A_{P}=8 A_{Q}$\ For compound 2, molar ratio of $P$ and $Q$ is $P: Q$.\ $\frac{44.4}{A_{P}}: \frac{55.6}{A_{Q}} \Rightarrow \frac{44.4}{\frac{8}{9} A_{Q}}: \frac{55.6}{A_{Q}}=9: 10$\ So, the empirical formula for compound 2 is $P_{9} Q_{10}$.\ Hence, option (a) is incorrect.\ Let atomic weight of $P$ is $A_{p}$ and atomic weight of $Q$ is $A_{Q}$.\ For compound 3,\ $\frac{40}{A_{p}}: \frac{60}{A_{Q}}=\frac{3}{2} \Rightarrow \frac{2}{3} 	imes \frac{A_{Q}}{A_{P}}=\frac{3}{2}$\ $\Rightarrow 4 A_{Q}=9 A_{P}$\ $\because \quad$ Atomic weight of $P$ is 20 and atomic weight of $Q$ is 45 .\ Then, $4 A_{Q}=9 A_{P}$\ $4 	imes 45=9 	imes 20$\ $180=180$\ Hence, the option (b) is correct.\ Molar ratio of $P$ and $Q$ in compound 2 is\ $\frac{44.4}{A_{P}}: \frac{55.6}{A_{Q}}=\frac{1}{1} \Rightarrow \frac{A_{P}}{A_{Q}}=\frac{44.4}{55.6}$\ Molar ratio of atom $P$ and $Q$ in compound 1 is\ $\frac{50}{A_{p}}: \frac{50}{A_{Q}} \Rightarrow \frac{A_{Q}}{A_{p}}=\frac{55.6 A_{Q}}{44.4 A_{Q}}=\frac{5}{4}$\ $A_{Q}: A_{P}=5: 4$\ Hence, empirical formula is $P_{5} Q_{4}$.\ So, the option (c) is correct.\ For compound 1, molar ratio of atom $P$ and $Q$ is $P: Q$.\ Let the mass of compound be 100 g .\ Mass of $P=50 \mathrm{~g}$\ Mass of $\mathrm{Q}=50 \mathrm{~g}$ $$\begin{tabular}{|c|c|c|} \hline & Moles & Simple ratio \ \hline $P$ & $\frac{50}{70}=0.714$ & 1 \ \hline $Q$ & $\frac{50}{35}=1.4$ & 2 \ \hline \end{tabular}$$ The empirical formula is $P Q_{2}$.\ Hence, the option (d) is incorrect.\