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Ch.13 - Solutions & Their Properties
All textbooksMcMurry 8th EditionCh.13 - Solutions & Their PropertiesProblem 151
Chapter 13, Problem 151

Combustion analysis of a 36.72-mg sample of the male hormone testosterone gave 106.43 mg CO2 and 32.10 mg H2O as the only combustion products. When 5.00 mg of testosterone was dissolved in 15.0 mL of a suitable solvent at 25 °C, an osmotic pressure of 21.5 mm Hg was measured. What is the molecular formula of testosterone?

Verified step by step guidance
1
Step 1: Determine the moles of carbon and hydrogen in the combustion products. Use the molar mass of CO2 (44.01 g/mol) to find the moles of carbon in 106.43 mg of CO2, and use the molar mass of H2O (18.02 g/mol) to find the moles of hydrogen in 32.10 mg of H2O.
Step 2: Calculate the mass of carbon and hydrogen in the sample. Multiply the moles of carbon by the atomic mass of carbon (12.01 g/mol) to find the mass of carbon, and multiply the moles of hydrogen by the atomic mass of hydrogen (1.008 g/mol) to find the mass of hydrogen.
Step 3: Determine the mass of oxygen in the testosterone sample. Subtract the combined mass of carbon and hydrogen from the total mass of the testosterone sample (36.72 mg) to find the mass of oxygen.
Step 4: Calculate the empirical formula of testosterone. Convert the masses of carbon, hydrogen, and oxygen to moles, then divide each by the smallest number of moles to find the simplest whole number ratio of atoms in the compound.
Step 5: Use the osmotic pressure data to find the molar mass of testosterone. Apply the formula for osmotic pressure, \( \Pi = \frac{n}{V}RT \), where \( \Pi \) is the osmotic pressure, \( n \) is the number of moles, \( V \) is the volume in liters, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. Solve for the molar mass and compare it to the empirical formula mass to determine the molecular formula.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of organic compounds. In this process, a sample is burned in excess oxygen, producing carbon dioxide and water. The masses of these products are measured, allowing for the calculation of the amounts of carbon and hydrogen in the original sample. This information is crucial for determining the empirical formula of the compound.
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Combustion of Non-Hydrocarbons

Osmotic Pressure

Osmotic pressure is the pressure required to prevent the flow of solvent into a solution through a semipermeable membrane. It is directly related to the concentration of solute particles in the solution, as described by the van 't Hoff equation. In this context, measuring the osmotic pressure of a testosterone solution helps in determining its molarity, which is essential for calculating the molecular weight and ultimately the molecular formula.
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Osmotic Pressure Formula

Molecular Formula

The molecular formula of a compound indicates the actual number of each type of atom present in a molecule. It is derived from the empirical formula, which represents the simplest whole-number ratio of the elements. By using data from combustion analysis and osmotic pressure measurements, one can calculate the molecular weight and deduce the molecular formula, providing insight into the compound's structure and properties.
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Related Practice
Textbook Question

Treatment of 1.385 g of an unknown metal M with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0 °C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and remaining HCl then gave a white crystalline compound, MClx. After dis- solving the compound in 25.0 g of water, the melting point of the resulting solution was - 3.53 °C. (e) What are the formula and molecular weight of MClx?

Textbook Question

Treatment of 1.385 g of an unknown metal M with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0 °C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and remaining HCl then gave a white crystalline compound, MClx. After dis- solving the compound in 25.0 g of water, the melting point of the resulting solution was - 3.53 °C. (f) What is the identity of the metal M?

Textbook Question

A compound that contains only C and H was burned in excess O2 to give CO2 and H2O. When 0.270 g of the com- pound was burned, the amount of CO2 formed reacted completely with 20.0 mL of 2.00 M NaOH solution according to the equation 2 OH-1aq2 + CO21g2 S CO 2- 1aq2 + H2O1l2 When 0.270 g of the compound was dissolved in 50.0 g of camphor, the resulting solution had a freezing point of 177.9 °C. [#Pure camphor freezes at 179.8 °C and has Kf = 37.7 1°C kg2>mol.] (a) What is the empirical formula of the compound?

Textbook Question
A solution prepared by dissolving 100.0 g of a mixture of sugar 1C12H22O112 and table salt (NaCl) in 500.0 g of water has a freezing point of - 2.25 °C. What is the mass of each individual solute? Assume that NaCl is completely dissociated.