A 99.99% pure, 0.4903 g sample containing only carbon, hydrogen, and nitrogen is subjected to combustion analysis, resulting in the formation of 1.373 g CO2, 0.2791 g H2O, and 0.1860 g NO. What is the empirical formula of the sample?

Answer 1

Here is what I would do.
The weight of the sample is 0.4903 x 0.9999 = 0.49025.

mols C = 1.373/44.01 = ?
mols H = 0.2791 x 2.016/18.02 = ?
mols N = 0.1860/14.007 = ?

Now find the ratio of each with no number being less than 1.00. The easy way to do that is to divide the smallest number by itself. Then divide the other numbers by the same small number. Round to whole numbers to find the empirical formula. I came up with C5H5N
but I wouldn't be satisfied with that (that's pyridine). Why? Because g carbon + g nitrogen + g H don't add to 0.4903 but to a larger number. I would redo the elemental analysis. If I calculate these grams I get 0.3747 g C, 0.03122 g H and 0.0868 g N and that adds to 0.4927g. In my book that's too far from 0.4903 g. I wouldn't sleep right at night.