Question

Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at \(25^{\circ} \mathrm{C},\) and their solubilities in water at \(25^{\circ} \mathrm{C}\) and 1 atm fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution. (b) Explain why the molarity of each of the solutions should be very close numerically to the molality. (c) Based on their molecular structures, account for the differences in solubility of the four fluorocarbons. (d) Calculate the Henry's law constant at \(25^{\circ} \mathrm{C}\) for CHClF \(_{2}\), and compare its magnitude to that for \(\mathrm{N}_{2}\left(6.8 \times 10^{-4} \mathrm{~mol} / \mathrm{L}-\mathrm{atm}\right) .\) Can you account for the difference in magnitude?

Short answer

Calculating the molality of the saturated solutions of the fluorocarbons, we have: For CF₄: molality = 0.000101 mol/kg For CClF₃: molality = 0.000506 mol/kg For CCl₂F₂: molality = 0.001628 mol/kg For CHClF₂: molality = 0.010815 mol/kg Analyzing the properties of fluorocarbons, we can see that the solubility of these gases in water is mainly governed by their dipole moment. All these compounds are unable to form hydrogen bonds with water molecules, and their differences in solubility cannot be explained solely based on molar mass. Calculating the moles of O₂ in an infant's lungs for a full breath of air and a full breath of fluorinated liquid: For air: n₁ = 0.00111 moles O₂ For fluorinated liquid: n₂ = 0.00348 moles O₂ Thus, a full "breath" of saturated O₂ solution in the fluorinated liquid contains more moles of O₂ than a full breath of air, which would be advantageous for the infant with severe respiratory problems.

Step-by-step solution

Calculate the molality of a saturated solution for each fluorocarbon

For a mass percentage of solubility, we need to first define the mass of water and fluorocarbon such that their solubility percentage matches the given data. We can assume 100 grams of water is present. Then, using the given mass percentages, we can calculate the mass of each fluorocarbon in a saturated solution. After finding the mass, we can find the moles of each fluorocarbon using their respective molar mass. Finally, using the formula for molality, we can calculate the molality of each saturated solution. Molality = moles of solute(kg of solvent)^{-1} Let's find the molality of each saturated solution.

Analyze the properties of fluorocarbons to determine which best predicts solubility

We need to investigate the molar mass, dipole moment, and hydrogen-bonding ability of the given fluorocarbons to determine which of these factors best explains their solubility in water.

Calculate the moles of O₂ present in an infant's lungs for a full breath of air and a full breath of fluorinated liquid

We are given that the solubility of oxygen in the fluorinated liquid is 66 mL O₂ per 100 mL liquid, and that air is 21% oxygen by volume. We are also given the volume of an infant's lungs as 15 mL. We can use these values to calculate the moles of O₂ present in an infant's lungs for each scenario using the ideal gas law equation (PV=nRT). We will assume a pressure of 101.3 kPa in the lungs, and the ideal gas constant R = 8.314 J mol^{-1} K^{-1}. By comparing the moles of O₂ in each scenario, we can determine which is more advantageous for the infant.

Key concepts

Molality Calculation

Understanding the concept of molality is essential for solving a variety of chemistry problems, including those concerned with fluorocarbon solubility. Molality, which is a measure of the concentration of a solution, is defined as the moles of solute per kilogram of solvent. This value does not change with temperature since it's based on mass, which is different from molarity that depends on volume and can vary with temperature changes.

To calculate molality, you can follow these steps: First, you need to determine the mass of the solvent, typically in kilograms. In most textbook exercises, water is used as the solvent. Then, you'll need to find out how much of the solute (in this case, a fluorocarbon) is present in the solution. If the fluorocarbon's solubility is given in mass percentage, you must calculate the mass of solute that can dissolve in a specific mass of solvent. Afterward, convert this mass to moles by using the molar mass of the solute. Finally, apply the molality formula:\[\text{Molality} = \frac{\text{moles of solute}}{\text{kg of solvent}}\].

For example, if you have a 10% mass percentage of solute in water, and you assume you have 100 grams of water, you would weigh out 10 grams of solute. Convert this mass to moles by dividing by the molar mass of the solute, and then divide the moles by the mass of water converted into kilograms to get the molality of the solution.

Henry's Law Constant

Henry's Law is a crucial concept when understanding the behavior of gases in liquids, useful when considering fluorocarbons in water. Put simply, Henry's Law states that the concentration of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The constant of proportionality in this relationship is known as Henry's Law constant (often denoted as \( k_H \)).

The solubility of a fluorocarbon can be related to its partial pressure by the formula:\[ C = k_H \times P \], where \( C \) is the solubility of the gas (in mol/L), \( k_H \) is Henry's Law constant, and \( P \) is the partial pressure of the gas.

When you're calculating the Henry's Law constant for a fluorocarbon, you'll take into account the solubility in mol/L and the pressure in atmospheres. The greater the constant, the more soluble the gas is at a given pressure. For comparison, if the Henry's Law constant for \( \mathrm{N}_2 \) is \( 6.8 \times 10^{-4} \mathrm{~mol/L\cdot atm} \), a fluorocarbon with a higher constant would be more soluble in water than nitrogen at the same conditions. This property is particularly important when considering the efficacy of different substances in applications like refrigeration.

Molecular Structure and Solubility

The solubility of a substance is intimately tied to its molecular structure. Fluorocarbons, for instance, show varied solubility in water based on their structure. There are several key structural features that influence a molecule's solubility in a particular solvent. The first is the presence of polar or nonpolar bonds within the molecule; polar molecules are more likely to dissolve in polar solvents like water. Fluorocarbons are typically nonpolar and thus have low solubility in polar solvents.

However, characteristics like molecular size and shape, as well as the type and strength of intermolecular forces, also play central roles. Small molecules or those with compact shapes can often dissolve more easily, whereas larger or more irregularly shaped molecules may have more difficulty. Furthermore, intermolecular forces such as hydrogen bonding, dipole-dipole interactions, and London dispersion forces directly affect the solubility. Substances with strong intermolecular forces are more likely to be soluble in solvents that can engage in similar types of interactions.

For instance, a fluorocarbon with a high degree of fluorination, having a strong carbon-fluorine bond, will be less likely to interact favorably with water molecules and therefore less soluble in water. This is why, when comparing fluorocarbons, their individual molecular structures offer a valuable perspective on their expected behaviors and solubility profiles in various environments.