Question

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 1

Explanation:

ELEMENT [ C ] :

MOLES : 54.55/12 = 4.54

LEAST RATIO : 4.54/2.27 = 2

ELEMENT [ H ] :

MOLES : 9.09/1 = 9.09

LEAST RATIO : 9.09/2.27 = 4

ELEMENT [ O ] :

MOLES : 36.36/16 = 2.27

LEAST RATIO : 2.27/2.27 = 1

From the above calculations, the empirical formula =C2H4O

Empirical 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

= 2

Molecular Formula =2× Empirical formula

= 2×(C2H4O)

= C4H8O2.