Answer
The molecular formula will be ($C_{18}H_{24}O_{2}$)
Work Step by Step
We need to get the empirical formula first:
Given that 14.54 mg CO2 and 3.97 H2O and we know that
CO2 molecular weight = 44 mg, contain 12 mg Carbon
H2O molecular weight = 18 mg, contain 2 mg Hydrogen
then by cross multiplication:
$\frac{12x14.5}{44}=3.9$ mg Carbon
$\frac{2x3.97}{18}=0.441$ mg Hydrogen
Oxygen = 5mg of estradiol - 3.9 - 0.441 = 0.6 mg Oxygen
C = $\frac{3.9}{12}=0.32 mol$, H = $\frac{0.441}{1}=0.441 mol$, O = $\frac{0.6}{16}=0.037mol$
Then,
C=$\frac{0.32}{0.037}\approx9$, H=$\frac{0.441}{0.037}\approx12$, O=$\frac{0.037}{0.037}\approx1$
Then the empirical formula is $C_{9}H_{12}O$
The empirical weight = 136
given that the molecular weight = 272, then 272/136=2
Then $C_{9}H_{12}O$ (x2)
the molecular formula : $C_{18}H_{24}O_{2}$
