Question

What is the empirical formula of a compound with the following mass percentages of carbon, hydrogen and nitrogen? C = 59.95 % H = 12.08% and N = 27.97%

Answer 1

First, in combustion analysis (experiment which gives information about the percentages), you must see if the percentages given add up to 100%. This is because the data is obtained from a combustion reaction and oxygen is consumed in the reaction accounted for and thus would not show up as a percentage.

% Oxygen= 100- (59.95+12.08+27.97)
= 0

Therefore, there is no oxygen in the compound.

Then we can divide the percentage composition of the particular molecule, by its respective molar mass:

Carbon=59.95/12.01=4.99=5

Hydrogen=12.08/1.008=11.98=12

Nitrogen=27.97/14.006=1.997=2

Therefore the emperical formula is C5H12N2.

The molar mass is 100.162 g/mol