A compound on analysis gave the following percentage composition: Na=14.31% S = 9.97%, H = 6.22%, O = 69.5%, calcualte 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 [Na = 23, S = 32, H = 1, 0 = 16].

Explanation

All H combines with 10 oxygen atoms to form as `10H_(2)O`
So the empirical formula is `Na_(2)SO_(4).10H_(2)O`
Empirical formula mass = `(23xx2) + (32 xx 1) + (16xx4) + (10xx 18)`
= 46 + 32 + 64 + 180 = 322
`n=("Molecular mass")/("Emperical formula mass")=322/322=1`
Molecular formula=`Na_(2)SO_(4).10H_(2)O`