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

Calculation of empirical formula

The empirical formula is `Na_2SH_(20)O_(14)`
Calculation of Molecular formula
Empirical formula mass `= (23 xx 2) + 32 + (20 xx 1 )+ (16 xx 14)`
`=322 `
`n=("Molecular mass ")/( " Empirical formula mass ")=(322)/(322)=1`
Hence molecular formula = `Na_2SH_(20)O_(14)`
Since all hydrogens are present as `H_2O` in the compound, it means 20 hydrogen atoms must have combined. It means 20 hydrogen atoms must have combined with 10 atoms of oxygen to form 10 molecules of water of crystallisation. The remaining (14 - 10 = 4) atoms of oxygen should be present with the rest of the compound.
Hence, molecular formula `= Na_2 SO_4 . 10H_2O`.