Question

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
Molecular formula of a compound is​

Answer 1

Answer:

Empirical formula = C2H4ORead more on Sarthaks.com - https://www.sarthaks.com/914817/a-compound-on-analysis-gave-the-following-percentage-composition-c-54-55-h-9-09-o-36-36

Answer 2

Explanation:

C= 54.55% or 54.55gm

H= 9.09 % or 9.09 gm

O = 36.36% or 36.36 gm

find mole of all

C = 54.55\12 = [ c =12 gm ]

H = 9.09 \ 1 = [4 = 1 gm]

O = 36.36 \16 = [ 0.16 gm]

{ C= 4.54 \ 2.27 =2 }

{4 = 9.09 \ 2.07= 4.004 }

{o= 2.27\ 2.27 = 1 }

empirical formula is C 2 H 4 O