# 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

Abishek arun
153 Points
9 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