Q12-1P - Organic Chemistry (Mcmurry)

Question

The male sex hormone testosterone contains only C, H, and O and has a mass of 288.2089 amu, as determined by high-resolution mass spectrometry. What is the likely molecular formula of testosterone?

Step-by-Step Solution

Answer

The likely molecular formula of testosterone is ${\mathrm{C}}_{19}\mathrm{H}{}_{28}\mathrm{O}_2$.

Step 1: Molecular mass

The molecular mass of a substance is the sum of the atomic masses of all its atoms in a molecule.

Step 2: Assumption

• Let's assume that in this molecule, only hydrogen and oxygen atoms are present. 
• The molecular mass of carbon is 12, and the molecular mass of hydrogen is 1.
• The sum of the molecular mass of C and H is $12+1=13$
• Now divide the given molecular mass by 13.
• The quotient indicates the number of carbons, and the number of hydrogens is equal to the quotient plus the remainder. Given by: $\frac{288}{13}=22\frac{2}{13}$
• This suggests that there are 22 carbon and 24 hydrogens.
• So, the formula is  ${\mathrm{C}}_{22}\mathrm{H}{}_{24}$

Step 3: Actual molecules possible:

As oxygen is also part of the molecule, so all possible combinations are:

  $\begin{array}{l}{\mathrm{C}}_{20}\mathrm{H}{}_{32}\mathrm{O}\\ {\mathrm{C}}_{19}\mathrm{H}{}_{28}\mathrm{O}_2\\ {\mathrm{C}}_{18}\mathrm{H}{}_{24}\mathrm{O}_3\end{array}$

The molecular mass of oxygen is 16.

Step 4: Mass of each predicted molecule:

• The precise molecular mass of carbon is 12.0g/mol, the mass of hydrogen is 1.008g/mol, and the mass of oxygen is 15.994g/mol.
• The molecular mass of each molecule:

$\begin{array}{c}{\mathrm{C}}_{20}\mathrm{H}{}_{32}\mathrm{O}=(20\times \mathrm{molar}\mathrm{mass}\text{ of C})+(1\times \mathrm{molar}\mathrm{mass}\text{ of O})+(32\times \mathrm{molar}\mathrm{mass}\text{ of H})\\ =(20\times 12)+(1\times 15.994)+(32\times 1.008)\\ =(240+15.994+32.256)\\ =288.25\mathrm{g}\end{array}$

$\begin{array}{c}{\mathrm{C}}_{19}\mathrm{H}{}_{28}\mathrm{O}_2=(19\times \mathrm{molar}\mathrm{mass}\text{ of C})+(2\times \mathrm{molar}\mathrm{mass}\text{ of O})+(28\times \mathrm{molar}\mathrm{mass}\text{ of H})\\ =(19\times 12)+(2\times 15.994)+(28\times 1.008)\\ =(228+31.988+28.224)\\ =288.212\mathrm{g}\end{array}$

$\begin{array}{c}{\mathrm{C}}_{18}\mathrm{H}{}_{24}\mathrm{O}_3=(18\times \mathrm{molar}\mathrm{mass}\text{ of C})+(3\times \mathrm{molar}\mathrm{mass}\text{ of O})+(24\times \mathrm{molar}\mathrm{mass}\text{ of H})\\ =(18\times 12)+(3\times 15.994)+(24\times 1.008)\\ =(216+47.982+24.192)\\ =288.174\mathrm{g}\end{array}$

Step 5: Comparison:

On comparing the calculated molecular masses with the given molecular mass of  288.2089 amu, it can be seen that  ${\mathrm{C}}_{19}\mathrm{H}{}_{28}\mathrm{O}_2$ has the same value as given on rounding off.

So, the correct molecular formula is ${\mathrm{C}}_{19}\mathrm{H}{}_{28}\mathrm{O}_2$.