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A compound has the following composition by weight ; $\mathrm{Na}=18.60 \%, \mathrm{~S}=25.80 \%, \mathrm{H}=4.02 \%$ and $\mathrm{O}=51.58 \%$ Assuming that all the hydrogen atoms in the compound are part of water of crystallization, the correct molecular formula of the compound is
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The correct answer is:
$\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}$
\begin{array}{|c|c|c|c|}
\hline Elements & \% & moles & Simplest ratio \\
\hline \mathrm{Na} & 18.6 & \frac{18.6}{23}=0.8 & 1 \times 2 \\
\hline \mathrm{S} & 25.8 & \frac{25.8}{32}=0.8 & 1 \times 2 \\
\hline \mathrm{O} & 51.58 & \frac{51.58}{16}=3.22 & 4 \times 2 \\
\hline \mathrm{H} & 4.02 & \frac{4.02}{1}=4.02 & 5 \times 2 \\
\hline
\end{array}
$\Rightarrow$ Formula is $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}$
\hline Elements & \% & moles & Simplest ratio \\
\hline \mathrm{Na} & 18.6 & \frac{18.6}{23}=0.8 & 1 \times 2 \\
\hline \mathrm{S} & 25.8 & \frac{25.8}{32}=0.8 & 1 \times 2 \\
\hline \mathrm{O} & 51.58 & \frac{51.58}{16}=3.22 & 4 \times 2 \\
\hline \mathrm{H} & 4.02 & \frac{4.02}{1}=4.02 & 5 \times 2 \\
\hline
\end{array}
$\Rightarrow$ Formula is $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}$
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