What is the empirical formula of a substance that is 53.5% C, 15.5% H, and 31.19% N by weight?

1 Answer
anor277
Feb 25, 2016

#C_2H_7N# is the simplest whole number ratio defining constituent atoms in a species, and is thus the empirical formula.

Explanation:

As always with these problems, we assume that there are #100# #g# of substance, and calculate the elemental composition (we divide thru by the atomic masses of each component):

#%C# #=# #(53.5*g)/(12.011*g*mol^-1)# #=# #4.45# #mol# #C#.

#%H# #=# #(15.5*g)/(1.00794*g*mol^-1)# #=# #15.38# #mol# #H#.

#%N# #=# #(31.19*g)/(14.01*g*mol^-1)# #=# #2.23# #mol# #N#.

We divide thru by the smallest quotient, that of #N# to give an EMPIRICAL formula of #C_2H_7N# (i.e. we divide each molar quantity by #2.23#!). Without details of molecular weight, this is as far as we can go.

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