Answer to Question #136989 in General Chemistry for mica

Question

What is the percent composition of aconitine, C34H47NO11?



% C
% H
% N
% O

Answer 1

Given molicular formula = C34H47NO11

In this compound the ratio of number of moles of constituent atoms 

     C : H : N : O = 34 : 47 : 1 : 11

So, number fraction of moles of C atoms = 34/(34+47+1+11)

       =( 34/93 )

 So, percent of composition of C atom = (34/93)×100

                       =36.559%

     (upto three nearest decimal place)

Similarly, number fraction of moles of H atoms = 47/(34+47+1+11)

       =( 47/93 )

        So, percent of composition of C atom = (47/93)×100

                       =50.538%

      (upto three nearest decimal place)

So, number fraction of moles of N atoms = 1/(34+47+1+11)

       =( 1/93 )

 So, percent of composition of C atom = (1/93)×100

                       =1.075%

     (upto three nearest decimal place)

So, number fraction of moles of O atoms = 11/(34+47+1+11)

       =( 11/93 )

 So, percent of composition of O atom = (11/93)×100

                       =11.828%

       (upto three nearest decimal place)

Hence, 

the percent composition of C34H47NO11 is......

     

     36.559%  C

      

     50.538%  H

     1.075%   N

  

    11.828%  O