Compound 'A' is found to contain 36.5 % Na,25.4 % S and 38.1 % O. Its molecular mass is 126. Calculate the molecular formula of the compound. The compound 'A' is easily oxidised to another compound 'B'. Calculate the percentage composition of 'B'.

Compound 'A' is found to contain 36.5 % Na,25.4 % S and 38.1 % O. Its

1.	Compound 'A' is found to contain 36.5 % Na,25.4 % S and 38.1 % O. Its molecular mass is 126. Calculate the molecular formula of the compound. The compound 'A' is easily oxidised to another compound 'B'. Calculate the percentage composition of 'B'.
Answer» SOLUTION :STEP I. Empirical formula of the COMPOUND 'A' `{:("Element","Percentage","Atomic mass","Gram toms (Moles)","Atomic ratio (Molar ratio)","SIMPLEST whole no. ratio"),("Na",36.5,23,(36.5)/(23)=1.59,(1.59)/(0.79)=2,2),("S",25.4,32,(25.4)/(32)=0.79,(0.79)/(0.79)=1,1),("O",38.1,16,(38.1)/(16)=2.38,(2.38)/(0.79)=3,3):}` Empirical formula of the compound `= Na_(2)SO_(3)` Step II. Molecular formula of compound 'A' Empirical formula mass `= 2 xx 23 + 32 + 3 xx 16 = 126 u` Molecular mass = 126 u `n=("Molecular mass")/("Empirical formula mass")=(126)/(126)=1` `:.` Molecular formula of the compound `'A' = n xx` Empirical formula `= 1 xx Na_(2)SO_(3)=Na_(2)SO_(3)` Step III. Percentage composition of compound 'B' `underset((A))(Na_(2)SO_(3))overset([O])(rarr)underset((B))(Na_(2)SO_(4))` Molecular mass of `B = 2 xx 23+32 + 4 xx 16 = 142 u` Percentage of `Na = (46)/(142) xx 100 = 32.4 %` Percentage of `S = (32)/(142) xx 100 = 22.54 %` Percentage of `O = (64)/(142) xx 1000 = 45.07 %`.