Question

Anabolic steroids are performance enhancement drugs whose use has been banned from most major sporting activities. One anabolic steroid is fluoxymesterone $\left(C_{30}{H}_{29}{\mathrm{FO}}_3\right)$. Calculate the percent composition by mass of fluoxymesterone.

Short answer

Themass percentage composition of fluoxymesterone is 100%.

Step-by-step solution

Definition of fluoxymesterone

The molar calculation of fluoxymesterone is found using the mass percent composition of each element.

Calculation of each element

Now, $\left({\mathrm{C}}_{30}{\mathrm{H}}_{29}{\mathrm{FO}}_3\right)$ themolar mass is:

$$\begin{gathered} =20\times 12.01\mathrm{g}/\mathrm{mol}+29\times 1.008\mathrm{g}/\mathrm{mol}+19.00\mathrm{g}/\mathrm{mol}+3\times 16.00\mathrm{g}/\mathrm{mol} \\ =\left(240.2+29.232+19.00+48.00\right)\mathrm{g}/\mathrm{mol} \\ =336.43\mathrm{g}/\mathrm{mol} \\ \mathrm{Hence},\mathrm{the}\mathrm{mass}\mathrm{of}\left({\mathrm{C}}_{30}{\mathrm{H}}_{29}{\mathrm{FO}}_3\right)\mathrm{is}336.43\mathrm{g}/\mathrm{mol} \end{gathered}$$

Calculation of mass

$$\begin{gathered} \mathrm{C}\mathrm{mass}\mathrm{in}1\mathrm{mol}=20\mathrm{mol}\times 12.01\mathrm{g}/\mathrm{mol}=240.2\mathrm{g} \\ \mathrm{H}\mathrm{mass}\mathrm{in}1\mathrm{mol}=29\mathrm{mol}\times 1.008\mathrm{g}/\mathrm{mol}=29.23\mathrm{g} \\ \mathrm{F}\mathrm{mass}\mathrm{in}1\mathrm{mol}=1\mathrm{mol}\times 19.00\mathrm{g}/\mathrm{mol}=19.00\mathrm{g} \\ \mathrm{O}\mathrm{mass}\mathrm{in}1\mathrm{mol}=3\mathrm{mol}\times 16.00\mathrm{g}/\mathrm{mol}=48.00\mathrm{g} \end{gathered}$$

Calculation of percent composition

$$\begin{gathered} \mathrm{Percent}\mathrm{composition}=\frac{\mathrm{Element}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}}\times 100 \\ \mathrm{Now}, \\ \mathrm{C}\mathrm{mass}\mathrm{percent}: \\ =\frac{240.2\mathrm{g}}{336.43\mathrm{g}}\times 100 \\ =71.40\% \\ \mathrm{H}\mathrm{mass}\mathrm{percent} \\ =\frac{29.23\mathrm{g}}{336.43\mathrm{g}}\times 100 \\ =8.69\% \\ \mathrm{F}\mathrm{mass}\mathrm{percent} \\ =\frac{19.00\mathrm{g}}{336.43\mathrm{g}}\times 100 \\ =5.65\% \\ \mathrm{O}\mathrm{mass}\mathrm{percent} \\ =\frac{48.00\mathrm{g}}{336.43\mathrm{g}}\times 100 \\ =14.27\% \end{gathered}$$

Hence, the overall percent composition is 100%.