Question

What is the empirical formula for a compound that is 43.4% C, 1.2% H, 38.6% O, and 16.9% N?

Answer 1

C3HO2N

Step-by-step explanation:

C: 43.4 g x 1 mole= 3.613444679 / 1.190547056 = 3.0412.0107 g

H: 1.2 g x 1 mole= 1.190547056 / 1.190547056 = 1.00794

O : 38.6 g x 1 mole= 2.412590472 / 1.190547056 = 2.0315.9994 g

N : 16.9 g x 1 mole= 1.206561984 / 1.190547056 = 1.0114.00674 g

C=3

H=1

O=2

N=1

: C3HO2N