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

`A = Na_(2)SO_(3),B=Na_(2)SO_(4)` ; In `'B' ; Na = 32.4 % ; S = 22.54 % and O = 45.07 %`

Explanation

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 %`.