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 \(\mathrm{CHClF}_{2}\), 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 dif- ference in magnitude?

Short answer

In summary, the molalities of the given fluorocarbons are: \(1.7 \times 10^{-5} ~mol/kg\) (CF4), \(1.4 \times 10^{-4} ~mol/kg\) (CClF3), \(4.6 \times 10^{-4} ~mol/kg\) (CCl2F2), and \(7.3 \times 10^{-3} ~mol/kg\) (CHClF2). The molarities are numerically close to molalities since the density of most solutions is close to 1 g/mL or 1 kg/L. The differences in solubility can be attributed to the varying polarizability of the fluorocarbons, which impacts their ability to dissolve in water, a highly polar solvent. Lastly, the Henry's law constant for CHClF2 is found to be \(7.3 \times 10^{-3} ~mol/L-atm\), which is higher than that of N2 due to the presence of more polarizable atoms in its molecular structure.

Step-by-step solution

(a) Calculate the molality of a saturated solution.

First, we must convert the mass percentage solubilities to molalities. To do this, we divide the mass percentage by the molar mass of each fluorocarbon, solving for moles of solute. Then, we will divide by the mass of solvent, in this case, water (in kg). The molality (m) is given by: \(m = \frac{moles~ of ~solute}{mass~of~solvent(kg)}\) For the four fluorocarbons: 1. CF4 Molar mass of CF4 = \(12.01 + 4 \times 19 = 88.01 ~g/mol\) Percentage mass = 0.0015% Mass of solute in 100 g of solution = \((100 g) \times (0.0015/100) = 0.0015 g\) Number of moles = \(0.0015 / 88.01\) So, the molality = \({moles}/{0.1} =\) \(1.7 \times 10^{-5} ~mol/kg\) Repeat the same steps for the other three fluorocarbons to find their molality. 2. CClF3 Molality = \(1.4 \times 10^{-4} ~mol/kg\) 3. CCl2F2 Molality = \(4.6 \times 10^{-4} ~mol/kg\) 4. CHClF2 Molality = \(7.3 \times 10^{-3} ~mol/kg\)

(b) Explain why molarity is close numerically to molality.

Molarity and molality are related but slightly different concepts. Molarity is the number of moles of solute per liter of solution, while molality is the number of moles of solute per kilogram of solvent. Since the densities of most solutions are close to 1 g/mL or 1 kg/L, the volume of solvent is roughly equal to its mass. That is, 1 kg of solvent (water) has a volume close to 1 L. Therefore, the values of molarity and molality are very close numerically for most solutions.

(c) Account for differences in solubility based on molecular structure.

The solubilities of the fluorocarbons vary due to their different molecular structures. Generally speaking, solubility is determined by the fundamental principle of "like dissolves like." In other words, polar solutes tend to dissolve in polar solvents, while nonpolar solutes dissolve in nonpolar solvents. Water is a highly polar solvent, while the fluorocarbons are nonpolar molecules. As more polarizable atoms (fluorine and chlorine) are added to the molecular structure, the molecule becomes more polar and thus more soluble in water. Comparing the given compounds from least to most soluble, we find that CF4 (tetrafluoromethane) has all polarizable fluorine atoms around the central carbon atom, and it is therefore less soluble in water. In contrast, CHClF2 has a mix of fluorine and chlorine atoms, along with a hydrogen atom, making it relatively more polar and its solubility in water higher.

(d) Calculate Henry's law constant and explain the difference.

Henry's law constant (K_H) can be calculated using the formula: \(K_{H} = \frac{c}{P}\) where c is the concentration (mol/L) and P is the partial pressure (atm). For CHClF2, we must convert the molality to molarity. As we explained earlier, they are approximately the same for most solutions. So, we can use the molality value as an approximate molarity value: \({7.3 \times 10^{-3} mol/kg}\) ≈ \(7.3 \times 10^{-3} mol/L\) Since the partial pressure is given as 1 atm, we can calculate K_H: \(K_{H} = \frac{7.3 \times 10^{-3}}{1} = 7.3 \times 10^{-3} ~mol/L-atm\) Comparing the value of CHClF2 with the given value of N2: \(K_{H, CHClF2} = 7.3 \times 10^{-3} mol/L-atm\) \(K_{H, N2} = 6.8 \times 10^{-4} mol/L-atm\) CHClF2 has a significantly higher Henry's law constant compared to N2. This difference can be explained by their molecular structures. CHClF2 has more polarizable atoms (fluorine and chlorine) than N2, which is a nonpolar diatomic molecule. The greater polarizability of CHClF2 increases its interaction with water molecules, leading to a higher solubility and a higher Henry's law constant.

Key concepts

Molarity vs Molality

When discussing concentrations of solutions, molarity and molality are two important terms to understand. Molarity is defined as the number of moles of solute per liter of solution. It is usually represented by the symbol \( M \). On the other hand, molality refers to the number of moles of solute per kilogram of solvent, and it is denoted by the symbol \( m \). These metrics help chemists understand the composition of a solution.

One of the primary differences between the two is how they are measured. Molarity is dependent on the volume of the solution, which can change with temperature since liquids expand and contract with temperature fluctuations. This means molarity might vary with temperature changes. Contrastingly, molality is based on the mass of the solvent and remains unaffected by changes in temperature.

Because many solutions have densities around 1 g/mL, in certain situations, such as the given exercise with fluorocarbons and water, molarity and molality can be almost numerically equal. This is particularly true when working with water where 1 kg of water approximately equals 1 L. Therefore, for dilute aqueous solutions at room temperature, these two units often yield similar values, making them somewhat interchangeable in practice for many calculations.

Henry's Law Constant

Henry's Law is a critical concept for understanding how gases dissolve in liquids. It states that the concentration of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. The proportionality constant in this relationship is known as Henry's Law constant, denoted as \( K_H \).

The formula is:\[ K_H = \frac{c}{P} \]
where \( c \) is the concentration of the dissolved gas (in mol/L), and \( P \) is the partial pressure of the gas (in atm).

In the exercise, CHClF\(_2\) was discussed regarding its higher Henry's Law constant compared to nitrogen (\( N_2 \)). This indicates that CHClF\(_2\) will dissolve more readily in water than nitrogen at the same pressure. The higher \( K_H \) value for CHClF\(_2\) is attributed to its molecular structure which includes more polarizable atoms such as chlorine and fluorine, enhancing its interaction with water molecules and therefore increasing solubility. In contrast, nitrogen is nonpolar and has a lower \( K_H \), which is why it’s less soluble in water.

Molecular Structure Influence on Solubility

The molecular structure of a compound plays a pivotal role in determining its solubility. This is primarily based on the "like dissolves like" principle, where polar substances tend to dissolve in polar solvents, and nonpolar substances dissolve in nonpolar solvents.

In the context of fluorocarbon solubility in water, a highly polar solvent, the variations in solubility can be explained by examining each molecule's structure. CF\(_4\) (tetrafluoromethane) is a completely nonpolar molecule because of its symmetrical structure, making it less soluble in water. As we introduce more elements like chlorine into the structure, as seen in CClF\(_3\), CCl\(_2\)F\(_2\), and CHClF\(_2\), the molecules become increasingly polar. The presence of chlorine atoms increases the intermolecular forces with water via dipole interactions leading to enhanced solubility.

Hence, the solubility increases from CF\(_4\) to CHClF\(_2\) because the molecular structure becomes more polar, which aligns better with the polar nature of water. Understanding these interactions helps in predicting and explaining the solubility behavior of various compounds in different solvents.