Calculate empirical formula
C=41.7%,H=6.2%,N=19.3%, and O=33.1%
Thanks. any help at all will be awesome!
Well, um...make sure you gave the percentages right. I added all the percentages together and I got 100.3%. Anyway, Imma still gunna work with these numbers just so you know the process of things.
First, to make things easier, you can assume a 100 g sample.
C = 41.7% --> 41.7 g
H = 6.2 % --> 6.2 g
N = 19.3% --> 19.3 g
O = 33.1% -- > 33.1 g
Second, convert all the grams to moles of their respective substance. In other words, divide by their atomic mass. (round to about 3-4 decimal places)
C = (41.7 g) / (12.01 g/mol) = 3.472 mol
H = (6.2 g) / (1.01 g/mol) = 6.139 mol
N = (19.3 g) / (14.01 g/mol) = 1.378 mol
O = (33.1 g) / (16.00 g/mol) = 2.069 mol
Next, divide all the moles by the smallest number. In this case, divide everything by 1.378 (nitrogen).
C = 3.472 / 1.378 = 2.5
H = 6.139 / 1.378 = 4.5
N = 1.378 / 1.378 = 1
O = 2.069 / 1.378 = 1.5
Since the numbers are generally in halves (a lot of .5's), we multiply everything by 2.
C = 2.5 x 2 = 5
H = 4.5 x 2 = 9
N = 1 x 2 = 2
O = 1.5 x 2 = 3
Finally, put together the formula. Each number corresponds to the element's subscript.
C5H9N2O3