Answer
Empirical Formula: $C_{13}H_{18}O_2$
Molecular Formula: $C_{13}H_{18}O_2$
Work Step by Step
1. Divide the percentages by the molar mass of the atoms:
C: 75.69% $\div$ 12 = 6.3075
H: 8.8 $\div$ 1 = 8.8
O: 15.51 $\div$ 16 = 0.969
These are the proportions:
$C_{6.3075}H_{8.8}O_{0.969}$
2. Now we need to transform these numbers to integers numbers:
*We can divide all numbers by the minor of them:
C : 6.3075 / 0.969 = 6.5
H : 8.8 / 0.969 = 9
O: 0.969 / 0.969 = 1
* To get rid of the .5, we can multiply all numbers by 2
C: 6.5 * 2 = 13
H: 9 * 2 = 18
O: 1 * 2 = 2
The Empirical Formula is:
$C_{13}H_{18}O_2$
To find the Molecular Formula, let's calculate the molecular mass of the empirical, and compare:
$C_{13}H_{18}O_2$ : 13*12 + 18*1 + 2*16 = 206g/mol
Since the molar mass of the empirical is equal to the molar mass given in the problem, we don't need to change anything.
Molecular Formula:
$C_{13}H_{18}O_2$
