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.

Answer 1

Answer:

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

Answer 2

Answer:

The empirical formula is $C_2H_4O$.

Calculation:

We shall assume 100%=100grams.

1. Carbon- 54.55g
2. Hydrogen- 9.09g
3. Oxygen- 36.6g

We find moles of each element.

$C$                       $H$                     $O$

$\frac{54.55}{12} \\4.55$                  $\frac{9.09}{1} \\9.09$                  $\frac{36.36}{16}\\ 2.27$

Divide the smallest moles by the moles of each.

$\frac{4.55}{2.27} \\2$                   $\frac{9.09}{2.27} \\4$                     $\frac{2.27}{2.27} \\1$

$C_2H_4O_$ is the empirical formula.