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Chapter 3: Problem 139

Anabolic steroids are performance enhancement drugs whose use has been banned from most major sporting activities. One anabolic steroid is fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right)\) . Calculate the percent composition by mass of fluoxymesterone.

Short Answer

Expert verified
The percent composition by mass of fluoxymesterone is approximately 71.31% carbon, 8.69% hydrogen, 5.65% fluorine, and 14.35% oxygen.

Step by step solution

01

Determine the molar mass of each element

Using the periodic table, we can find the molar mass of each element: Molar mass of carbon (C): 12.01 g/mol Molar mass of hydrogen (H): 1.01 g/mol Molar mass of fluorine (F): 19.00 g/mol Molar mass of oxygen (O): 16.00 g/mol
02

Calculate the molar mass of fluoxymesterone

To calculate the molar mass of fluoxymesterone, we multiply the molar mass of each element by the number of its atoms in the molecule and add them up. Molar mass of fluoxymesterone = (20 × molar mass of C) + (29 × molar mass of H) + (1 × molar mass of F) + (3 × molar mass of O) Molar mass of fluoxymesterone = (20 × 12.01) + (29 × 1.01) + (1 × 19.00) + (3 × 16.00) Molar mass of fluoxymesterone = 240.2 + 29.29 + 19.00 + 48.00 Molar mass of fluoxymesterone = 336.49 g/mol
03

Calculate the percent composition by mass of each element

To calculate the percent composition of each element in fluoxymesterone, we divide the molar mass of the element (multiplied by its number of atoms) by the total molar mass of the compound, and then multiply by 100. Percent of Carbon (C) = \(\frac{20 \times 12.01}{336.49}\) × 100 = 71.31% Percent of Hydrogen (H) = \(\frac{29 \times 1.01}{336.49}\) × 100 = 8.69% Percent of Fluorine (F) = \(\frac{1 \times 19.00}{336.49}\) × 100 = 5.65% Percent of Oxygen (O) = \(\frac{3 \times 16.00}{336.49}\) × 100 = 14.35% The percent composition by mass of fluoxymesterone is approximately 71.31% carbon, 8.69% hydrogen, 5.65% fluorine, and 14.35% oxygen.

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Most popular questions from this chapter

An element X forms both a dichloride \(\left(\mathrm{XCl}_{2}\right)\) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right) .\) Treatment of 10.00 $\mathrm{g} \mathrm{XCl}_{2}\( with excess chlorine forms 12.55 \)\mathrm{g} \mathrm{XCl}_{4}\( . Calculate the atomic mass of \)\mathrm{X},$ and identify \(\mathrm{X}\) .

Acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. $$ 2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If $15.0 \mathrm{g} \mathrm{C}_{3} \mathrm{H}_{6}, 10.0 \mathrm{g} \mathrm{O}_{2},\( and 5.00 \)\mathrm{g} \mathrm{NH}_{3}$ are reacted, what mass of acrylonitrile can be produced, assuming 100\(\%\) yield?

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