Question

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

Answer 1

`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.