Question

A laboratory analysis of an unknown sample yields 74.0% carbon, 7.4% hydrogen, 8.6% nitrogen, and 10.0% oxygen. What is the empirical formula of the sample? Give your answer in the form C#H#N#O# where the number following the element’s symbol corresponds to the subscript in the formula. (Don’t include a 1 subscript explicitly). For example, the formula CH 22 O would be entered as CH2O.

Answer 1

Let us assume that there is a 100g sample present. The respective mass of each element will then be:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Now, we divide each constituent's mass by its Mr to obtain the moles of each
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Dividing by the smallest number:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula is
C10H12NO