29. A compound on analysis gave the following percentage composition C=54.55%, H=9.09%, O=36.36% Determine the empirical formula of the compound.30. Calculate the empirical and molecular formula of a compound containing C=76.6%,H=6.38 and rest oxygen its vapour density is 47.

29. A compound on analysis gave the following percentage composition C

1.	29. A compound on analysis gave the following percentage composition C=54.55%, H=9.09%, O=36.36% Determine the empirical formula of the compound.30. Calculate the empirical and molecular formula of a compound containing C=76.6%,H=6.38 and rest oxygen its vapour density is 47.
Answer» tion:ELEMENT [ C ] :MOLES : 54.55/12 = 4.54LEAST RATIO : 4.54/2.27 = 2ELEMENT [ H ] :MOLES : 9.09/1 = 9.09LEAST RATIO : 9.09/2.27 = 4ELEMENT [ O ] :MOLES : 36.36/16 = 2.27LEAST RATIO : 2.27/2.27 = 1From the above calculations, the empirical formula =C2H4OEmpirical formula weight =12×2+1×4+16=44 g.Vapour density (VD) =44 g.Molecular weight =2×VD=2×44=88 g.N = MOLECULAR WEIGHT / EMPIRICAL FORMULA WEIGHT = 88/44= 2Molecular Formula =2× Empirical formula= 2×(C2H4O) = C4H8O2.