Question

A compound has a chemical composition of .62 g carbon, .10 g hydrogen, .28 g oxygen. What is the empirical formula?

Answer 1

Answer: Hopefully is correct :)

C6H12O2

Step-by-step explanation:

0.62 C/12= 0.052

0.10H/ 1= 0.1

0.28/16= 0.018

C= 0.052/0.018= 2.8

H= 0.1/0.018= 5.5

O=1

C= 2.8 x 2= C6

H= 5.5 x 2= H12

O= 1 x 2= O2

Answer: C6H12O2

Answer 2

First we divide the masses by molar masses:
0.62 g carbon / (12.01 g/mol carbon) = 0.0516 mol C
0.10 g hydrogen / (1.01 g/mol H) = 0.099 mol H
0.28 g oxygen / (14 g/mol O) = 0.02 mol O
Since O has the smallest amount, we divide the other values by it:
0.0516 mol C / 0.02 mol O = 2.58 ~ 2.5
0.099 mol H / 0.02 mol O = 4.95 ~ 5
Therefore we have a ratio of C2.5, H5, and O. Since an empirical formula must only contain integers, we multiply everything by two to get:
C5H10O2