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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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