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(D) To find the empirical formula,we calculate the moles of each element by dividing the percentage by its atomic mass:$Na: \frac{18.60}{23} = 0.808 \

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$Na_2S_2O_3 \cdot 5H_2O$

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(D) To find the empirical formula,we calculate the moles of each element by dividing the percentage by its atomic mass:$Na: \frac{18.60}{23} = 0.808 \approx 0.8$$S: \frac{25.80}{32} = 0.806 \approx 0.8$$H: \frac{4.02}{1} = 4.02$$O: \frac{51.58}{16} = 3.22$Dividing by the smallest mole value $(0.8)$:$Na: \frac{0.8}{0.8} = 1$$S: \frac{0.8}{0.8} = 1$$H: \frac{4.02}{0.8} \approx 5$$O: \frac{3.22}{0.8} \approx 4$The empirical formula is $NaSH_5O_4$. Since all $H$ atoms are in water of crystallisation $(H_2O)$,$5$ atoms of $H$ correspond to $2.5$ molecules of $H_2O$. To get whole numbers,we multiply by $2$,giving $Na_2S_2H_{10}O_8$,which is $Na_2S_2O_3 \cdot 5H_2O$.

$A$ 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:

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