Question

In 1934, Edward A. Doisy of Washington University extracted 3000 lb of hog ovaries to isolate a few milligrams of pure estradiol, a potent female hormone. Doisy burned 5.00 mg of this precious sample in oxygen and found that 14.54 mg of CO₂ and 3.97 mg of H₂O were generated.

(a) Determine the empirical formula of estradiol.

(b) The molecular weight of estradiol was later determined to be 272. Determine the molecular formula of estradiol.

Short answer

The empirical formula of the estradiol is ${\text{C}}_{\text{9}}{\text{H}}_{\text{12}}\text{O}$.

Step-by-step solution

Empirical formula.

The empirical formula gives thesimple ratioof theelementspresent in thecompoundrather than thenumber of atomsin thecompound or molecule.

Calculate the moles of carbon dioxide and water.

Now, calculate the moles of ${\text{CO}}_{\text{2}}$,

Moles of ${\text{CO}}_{\text{2}}\text{=}\frac{{\text{massofCO}}_{\text{2}}}{{\text{molarmassofCO}}_{\text{2}}}$

$\text{=}\frac{{\text{14.54×10}}^{\text{-3}}}{\text{44.01g/mol}}$

${\text{moleofCO}}_{\text{2}}{\text{=3.3×10}}^{\text{-4}}\text{mol}$

Calculate the moles of the carbon atom,

$\begin{array}{l}{\text{molesofC=3.3×10}}^{\text{-4}}{\text{molCO}}_{\text{2}}\text{×}\frac{\text{1molC}}{{\text{1molCO}}_{\text{2}}}\\ {\text{=3.3×10}}^{\text{-4}}\text{molC}\end{array}$

Here, calculate the moles of${\text{H}}_{\text{2}}\text{O}$,

$\begin{array}{c}{\text{molesofH}}_{\text{2}}\text{O=}\frac{{\text{massofH}}_{\text{2}}\text{O}}{{\text{molarmassofH}}_{\text{2}}\text{O}}\\ \text{=}\frac{{\text{3.97×10}}^{\text{-3}}\text{g}}{\text{18.02g/mol}}\end{array}$

${\text{molesofH}}_{\text{2}}{\text{O=2.2×10}}^{\text{-4}}\text{molC}$

Calculate moles of the hydrogen atom,

$\begin{array}{c}{\text{molesofH=2.2×10}}^{\text{-4}}{\text{molH}}_{\text{2}}\text{O×}\frac{\text{2molH}}{{\text{1molH}}_{\text{2}}\text{O}}\\ {\text{=4.4×10}}^{\text{-4}}\end{array}$

Calculate the masses of carbon, hydrogen, and oxygen.

Calculate the mass of carbon (C),

${\text{massofC=3.3×10}}^{\text{-4}}\text{molC×}\frac{\text{12.01gC}}{\text{1molC}}$

$\text{=3.96mgC}$

Calculate the mass of hydrogen (H),

$\begin{array}{c}{\text{massofH=4.4×10}}^{\text{-4}}\text{molH×}\frac{\text{1.008gH}}{\text{1molH}}\\ \text{=0.44mgH}\end{array}$

Calculate mass of oxygen (O),

Calculate the moles of oxygen (O),

$\begin{array}{c}{\text{molesofO=0.60×10}}^{\text{-3}}\text{gO×}\frac{\text{1molO}}{\text{16.00gO}}\\ {\text{=0.375×10}}^{\text{-4}}\text{molO}\end{array}$

calculate the empirical formula of the estradiol

a

The mole ratio of the carbon, hydrogen, and oxygen is,

${\text{C:H:O=3.3×10}}^{\text{-4}}{\text{:4.4×10}}^{\text{-4}}{\text{:0.375×10}}^{\text{-4}}$

Now divide the mole ratio of the least mole number, and we get

$\text{C:H:O=}\frac{{\text{3.3×10}}^{\text{-4}}}{{\text{0.375×10}}^{\text{-4}}}\text{:}\frac{{\text{4.4×10}}^{\text{-4}}}{{\text{0.375×10}}^{\text{-4}}}\text{:}\frac{{\text{0.375×10}}^{\text{-4}}}{{\text{0.375×10}}^{\text{-4}}}$

$\begin{array}{l}\text{C:H:O=8.8:11.7:1}\\ \text{C:H:O=9:12:1}\end{array}$

Hence, the empirical formula is ${\text{C}}_{\text{9}}{\text{H}}_{\text{12}}\text{O}$.