What is the empirical formula of Kaolin the percentage composition of which is as under :
`Al_(2)O_(3) = 39.5 %, SiO_(2)=46.6 %,H_(2)O=13.9 %`
(Atomic masses : `Al = 27, Si= 28,O = 16,H=1`)
Hint : Radical mass of `Al_(2)O_(3) = 102,SiO_(2) = 60,H_(2)O=18`.

Answer

`Al_(2)O_(3).2SiO_(2).2H_(2)O`

Explanation

Empirical formula of Kaolin may be calculated as follows
`{:("Radical","Percentage","Radical mass","Radical ratio","Simplest ratio","Simplest whole no. ratio"),("Al"_(2)"O"_(3),39.5,102,(39.5)/(102)=0.38,(0.38)/(0.38)=1.0,1),("SiO"_(2),46.6,60,(46.6)/(60)=0.78,(0.78)/(0.38)=2.05,2),("H"_(2)"O",13.9,18,(13.9)/(18)=0.77,(0.77)/(0.38)=2.03,2):}`
Empirical formula of Kaolin `= Al_(2)O_(3).2SiO_(2).2H_(2)O`.