Question

A compound has the following composition by weight $:Na=18.60\%,S=25.80\%,H=4.02\%$ and $O=51.58\%.$ Assuming that all the hydrogen atoms in the compound are part of water of crystallisation, the correct molecular formula of the compound is

• A. $N{a}_2{S}_2{O}_3\cdot 3H_{2}O$
• B. $N{a}_{2}S{O}_4\cdot 5H_{2}O$
• C. $N{a}_{2}S{O}_4\cdot 10H_{2}O$
• D. $N{a}_2{S}_2{O}_3\cdot 5H_{2}O$

Answer 1

Answer: (D)

Elements	$\%$ of Elements	At mass of element	Moles of element	Simplest molar ratio	Simplest whole no
$Na$	$18.6$	$23$	$\frac{18.6}{23}=0.8$	$\frac{0.8}{0.8}=1$	$1\times 2=2$
$S$	$25.8$	$32$	$\frac{25.8}{32}=0.8$	$\frac{0.8}{0.8}=1$	$1\times 2=2$
$O$	$51.58$	$16$	$\frac{51.58}{16}=3.22$	$\frac{3.22}{0.8}=4$	$4\times 2=8$
$H$	$4.02$	$1$	$\frac{4.02}{1}=4.02$	$\frac{4.02}{0.8}=5$	$5\times 2=10$

Thus, the empirical formula of compound is $N{a}_2{S}_2{H}_{10}{O}_8$. As it is given the all the hydrogen atoms in the compound are part of water of crystallisation, therefore molecular formula will be $N{a}_2{S}_2{O}_3\cdot 5H_{2}O$