1.

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.

Answer» Determination of empirical formula of the compound
`{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Na",14.31,23,(14.31)/(23)=0.622,(0.622)/(0.311)=2,2),("S",9.97,32,(9.97)/(32)=0.311,(0.311)/(0.311)=1,1),("H",6.22,1,(6.22)/(1)=6.22,(6.22)/(0.311)=20,20),("O",69.50,16,(69.50)/(16)=4.34,(4.34)/(0.311)=14,14):}`
The empirical of the compound `= Na_(2)SH_(20)O_(14)`
Step II. Determination of molecular formula of the compound
Empirical formula mass `= 2xx 23+32+20xx1+14xx16`
`=46+32+20+224=332`
Molecular mass 322 (Given)
`:. n=("Molecular mass")/("Empirical formula mass")=(322)/(322)=1`
Molecular formula `= n xx` Empirical formula `= 1 xx Na_(2)SH_(20)O_(14)=Na_(2)SH_(20)O_(14)`
The whole of hydrogen is present in combination with oxygen as molecules of water of crystallisation
`:.` The number of `H_(2)O` molecules = 10
No. of oxygen atoms not involved in the formation of `H_(2)O` molecules = 4
`:.` Molecular formula of the compound `= Na_(2)SO_(4).10 H_(2)O`


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