Question

Combustion analysis of \(1.00 \mathrm{~g}\) of the male sex hormone, testosterone, yields \(2.90 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.875 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) What are the mass percents of carbon, hydrogen, and oxygen in testosterone?

Short answer

Answer: The mass percents of carbon, hydrogen, and oxygen in testosterone are 79.1%, 4.91%, and 16.0%, respectively.

Step-by-step solution

Calculate the moles of carbon and hydrogen in the combustion products

First, we'll calculate the number of moles of carbon and hydrogen in \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\), using their respective molar masses. 1 mole of \(\mathrm{CO}_{2}\) has a molar mass of \(12.01 \mathrm{~g(C)} + 2 \times 16.00 \mathrm{~g(O)} = 44.01 \mathrm{~g(CO_{2})}\) 1 mole of \(\mathrm{H}_{2}\mathrm{O}\) has a molar mass of \(2 \times 1.01 \mathrm{~g(H)} + 16.00 \mathrm{~g(O)} = 18.02 \mathrm{~g(H_{2}O)}\) Now we can calculate the moles of carbon and hydrogen in the products: Moles of Carbon in \(\mathrm{CO}_{2}: \frac{2.90 \mathrm{~g}}{44.01 \mathrm{~g/mol}} = 0.0659 \mathrm{~mol}\) Moles of Hydrogen in \(\mathrm{H}_{2}\mathrm{O}: \frac{0.875 \mathrm{~g}}{18.02 \mathrm{~g/mol}} = 0.0486 \mathrm{~mol}\)

Calculate the mass of carbon and hydrogen in testosterone

We'll now convert the moles of carbon and hydrogen to mass using their molar masses. Mass of Carbon in testosterone: \(0.0659 \mathrm{~mol} \times 12.01 \mathrm{~g/mol} = 0.791 \mathrm{~g(C)}\) Mass of Hydrogen in testosterone: \(0.0486 \mathrm{~mol} \times 1.01 \mathrm{~g/mol} = 0.0491 \mathrm{~g(H)}\)

Calculate the mass of oxygen in testosterone

The mass of oxygen can be calculated by subtracting the mass of carbon and hydrogen from the total mass of testosterone. Mass of Oxygen in testosterone: \(1.00 \mathrm{~g} - 0.791 \mathrm{~g(C)} - 0.0491 \mathrm{~g(H)} = 0.160 \mathrm{~g(O)}\)

Calculate the mass percents of carbon, hydrogen, and oxygen in testosterone

Lastly, we can calculate the mass percents of each element using the following formula: Mass Percent of Element = \(\frac{\text{Mass of Element}}{\text{Total Mass}} \times 100\) Mass Percent of Carbon: \(\frac{0.791 \mathrm{~g(C)}}{1.00 \mathrm{~g}} \times 100 = 79.1 \%\) Mass Percent of Hydrogen: \(\frac{0.0491 \mathrm{~g(H)}}{1.00 \mathrm{~g}} \times 100 = 4.91 \%\) Mass Percent of Oxygen: \(\frac{0.160 \mathrm{~g(O)}}{1.00 \mathrm{~g}} \times 100 = 16.0 \%\) So, the mass percents of carbon, hydrogen, and oxygen in testosterone are 79.1%, 4.91%, and 16.0%, respectively.

Key concepts

Mass Percent Composition

Understanding the mass percent composition is crucial when studying the elemental makeup of a compound. In essence, it is a measure of the relative masses of individual elements compared to the total mass of the compound. To calculate the mass percent of an element, you take the mass of the element in the sample and divide it by the total mass of the sample, then multiply the result by 100 to get a percentage.

For example, if we have 1.00 g of a compound and find that it contains 0.791 g of carbon, the mass percent of carbon would be calculated using the formula: \[ \text{Mass Percent of Carbon} = \left( \frac{0.791 \, \text{g}}{1.00 \, \text{g}} \right) \times 100 = 79.1\% \].

This method is not only applicable to carbon but to other elements as well. In the context of our exercise, after conducting a combustion analysis, the mass percent composition is determined to find out how much carbon, hydrogen, and oxygen are present in testosterone. It provides valuable insights into the compound's empirical formula, which is fundamental in many fields of chemistry.

Molar Mass

Molar mass is another fundamental concept in chemistry, and it plays a vital role in completing our understanding of mass percent composition and stoichiometry. The molar mass can be considered the bridge between the mass of a substance and the number of moles contained in that mass. It is expressed as grams per mole (g/mol) and can be found by summing up the atomic masses of elements present in a molecule according to the periodic table.

Look at how the concept is applied in the exercise: The molar mass of \( \text{CO}_2 \) is calculated by adding the atomic masses of one carbon atom (12.01 g/mol) and two oxygen atoms (16.00 g/mol each), resulting in \( 12.01 \, \text{g(C)} + 2 \times 16.00 \, \text{g(O)} = 44.01 \, \text{g/mol} \). This allows us to convert the mass of carbon dioxide into moles of carbon, a step necessary for later calculations in the exercise.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In the context of our combustion analysis problem, stoichiometry involves using balanced chemical equations and molar relationships to draw conclusions about the composition of the original substance.

In the given exercise, stoichiometry allows us to understand how the masses of products formed from the combustion (carbon dioxide and water) relate to the amounts of carbon and hydrogen in the original testosterone molecule. Calculations based on stoichiometric coefficients from a balanced combustion equation facilitate the conversion of mass to moles, and then back to mass - as we initially calculate moles of carbon and hydrogen from the masses of \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) produced, and ultimately determine the mass of these elements present in the original compound. This step-by-step approach is central in stoichiometry, as it reveals the composition of testosterone.