A compound on analysis gave the following percentage composition :
Na= 14.31%, S = 9.97%, H = 6.22%, O = 69.50%.
Calculate the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. Molecular mass of the compound is 322.
(At. wt. of Na = 23, S = 32, H = 1, O = 16)

Explanation

Calculation of empirical formula:

`therefore` The empirical formula of the given compound is `Na_(2)SH_(20)O_(14)`.
Calculation of molecular formula : Empirical formula mass = `(2 xx 22.99) + 32.06 + (20 xx 1.008) + (14 xx 16.0) = 322.2`
Molecular mass (given) = 322
`n=("Molecular mass")/("Empirical formula mass")`
`=322/(322.2) = 0.99938=1`
`therefore` Molecular formula = `1 xx` Empirical formula
`=1 xx Na_(2)SH_(20)O_(14)`
`=Na_(2)SH_(20)O_(14)`
Since, as per question, all the 20 hydrogen atoms in combination with oxygen are present as water of crystallisation , one molecule of the given compound must contain 10 molecules of `H_2O`. This is because 20 atoms of hydrogen present in the molecular formula obtained above will combine with 10 atoms of oxygen to form 10 molecules of water. Now, only 14 - 10 = 4 atoms of oxygen are left. They must be present in the main molecule.
Hence, the molecular formula of the given compound can be written as `Na_(2)SO_(4).10H_(2)O`