Answer
Empirical formula = $C_{9}H_{12}O$
Empirical weight = 136
Work Step by Step
To find the empirical formula and weight, we first have to determine the original value of C and H in milligrams in the original sample. This value can be determined by measuring the amounts of $CO_{2} and H_{2}O$. Once this value in attained, the difference of oxygen in the 5mg sample can be determined. From this, empirical formula and weight can be deduced.
For the first part of the question,
(a) First determine the amount of $CO_{2}$ in 14.54mg sample
$14.54mg CO_{2} \times \frac{1 mmole CO_{2}}{44.01 mg CO_{2} } \times \frac{1 mmole of C}{1 mmole CO_{2} } \times \frac{ 12.01 mg C}{1 mmole C} = 3.968 mg C$
amount of Hydrogen in 3.97mg $H_{2}O$
$3.97mg H_{2}O \times \frac{1 mmole H_{2}O}{18.016 mg H_{2}O} \times \frac{2 mmoles H}{1 mmole H_{2}O} \times \frac{1.008 mg H}{1 mmole H} = 0.444mg H$
Amount of oxygen in 5.00mg sample of estradiol
$ 5.00mg estradiol - 3.968 mg C - 0.444mg H = 0.59mg O$
Hence the empirical formula -
$\frac{3.968 mg C}{12.01mg/mmole} = 0.3304 mmoles of C \div 0.037 mmoles = 8.93 \approx 9 C$
$\frac{0.444 mg H}{1.008 mg/mmole} = 0.440 mmoles H \div 0.037 mmoles = 11.9 \approx 12 H$
$\frac{0.59 mg O}{16 mg/mmole} = 0.037 mmole O \div o.o37 mmoles = 1 O$
Hence,
Empirical Formula = $C_{9}H_{12}O$
Empirical weight = $136$
