A compound on analaysis gave `Na =14.3% S=9.97% H=6.22%` and `O=69.5%` calaculate the molecular formula of the compound, if all the hydrogen in the compound is present in combination with oxyhen as water of crystallization. (molecular mass of the compound is 322).

Explanation

Empirical formula mass `=(23xx2)+(32xx1)+(16+4)+(10xx18)`
`=46+32+64+180=322`
` n= ("Molecular mass")/("Empirical formula mass")=(322)/(322)=1`
Molecular formula `=Na_(2)SO_(4).10H_(2)O`.