Answer :
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
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