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A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09%, O = 36.36 %. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.I am preparing for ftre so that i need

  1. A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09%, O = 36.36 %. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.I am preparing for ftre so that i need

Grade:9

1 Answers

Abishek arun
153 Points
6 years ago
Element
Percentage
Relative number of moles
Simple ratio moles
Simplest whole number ratio
C
54.54
  54.54
------- -  =  4.53
    12
  4.53
--------  = 2 
  2.27
2
H
9.09
   9.09
--------    =  9.09
     1
  9.09
--------  = 4
  2.27
4
O
36.36
  36.36
----------  =  2.27
    16
  2.27
---------  =  1
  2.27
1
 
Calculation of Molecular formula
Empirical formula mass = 12 x 2 + 1 x 4 + 16 x 1 = 44
Molecular mass =  2 x   Vapour density
 = 2 x 44 = 88
             Molecular mass                       88
n = ------------------------------------  = ---------   = 2
        Empirical Formula mass              44
 
= C2H4O        x   2
= C4H8O2
Element
Percentage
Relative number of moles
Simple ratio moles
Simplest whole number ratio
C
54.54
  54.54
------- -  =  4.53
    12
  4.53
--------  = 2 
  2.27
2
H
9.09
   9.09
--------    =  9.09
     1
  9.09
--------  = 4
  2.27
4
O
36.36
  36.36
----------  =  2.27
    16
  2.27
---------  =  1
  2.27
1
 
Calculation of Molecular formula
Empirical formula mass = 12 x 2 + 1 x 4 + 16 x 1 = 44
Molecular mass =  2 x   Vapour density
 = 2 x 44 = 88
             Molecular mass                       88
n = ------------------------------------  = ---------   = 2
        Empirical Formula mass              44
 
= C2H4O        x   2
= C4H8O2

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