A compound on analysis was found to contain the following composition :
`Na=14.31%,S=9.97%,O=69.50 %and H=6.22 %`
Calculate the molecular formula of the compound assuming that the whole of hydrogen in the compound is present as water of crystallisation. Molecular mass of the compound is 322.

Explanation

Step 1. To calculate the percentage of oxygen. The given compound contains oxygen but is percentage is not given. This can, however, be calcuated by subtracting the sum of percentages of Na, S and H from 100 as shown below:
Sum of percentage composition of Na, S and H `=14.28+9.92+6.20=30.40`
`therefore" Percentage of oxygen"=100.00-30.40=69.60`
Step 2. To calculate the empirical formula.

Hence, the empirical formula of the compound is `Na_(2)SH_(20)O_(14)`
Step 3. To calculate the empirical formula mass. Empirical formula mass of the compound `(Na_(2)SH_(20)O_(14))`
`=2xx23.0+1xx32.0+20xx1.0+14xx16.0=322`
Step 4. To calculate the value of 'n'
`n=("Molecular mass")/("Empirical formula mass")=(322)/(322)=1`
Step 5. To calculate the molecular formula of the compound.
`"Molecular formula"=nxx"Empirical formula"=1xxNa_(2)SH_(20)O_(14)=Na_(2)SH_(20)O_(14)`
Step 6. To calculate the number of molecules of water of crystallisation. Since all the H-atoms are present in the form of `H_(2)O`, 20 atoms of hydrogen would combine with 10 oxygen atoms to produce 10 molecules of `H_(2)O`.
Thus, the number of molecules of water of crystallization present in the salt `Na_(2)SH_(20)O_(14)` is 10.
Step 7. To determine the structure of the crystalline salt. Out of the 14 oxygen atoms, 10 are present in the form of `H_(2)O`. The remaining 4 must be a part of the salt consisting of Na and S. Hence, molecular formula for the anhydrous salt is `Na_(2)SO_(4)` and that of crystalline salt will be `Na_(2)SO_(4).10H_(2)O`.