Problem 106 - Chemistry The Central Science In Si Units

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 101.3 kPa fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution. (b) Which molecular property best predicts the solubility of these gases in water: molar mass, dipole moment, or ability to hydrogen-bond to water? (c) Infants born with severe respiratory problems are sometimes given liquid ventilation: They breathe a liquid that can dissolve more oxygen than air can hold. One of these liquids is a fluorinated compound, \(\mathrm{CF}_{3}\left(\mathrm{CF}_{2}\right)_{7} \mathrm{Br}\). The solubility of oxygen in this liquid is \(66 \mathrm{~mL} \mathrm{O}_{2}\) per \(100 \mathrm{~mL}\) liquid. In contrast, air is \(21 \%\) oxygen by volume. Calculate the moles of \(\mathrm{O}_{2}\) present in an infant's lungs (volume: \(15 \mathrm{~mL}\) ) if the infant takes a full breath of air compared to taking a full "breath" of a saturated solution of \(\mathrm{O}_{2}\) in the fluorinated liquid. Assume a pressure of \(101.3 \mathrm{kPa}\) in the lungs. $$ \begin{array}{lc} \hline \text { Fluorocarbon } & \text { Solubility (mass \%) } \\ \hline \mathrm{CF}_{4} & 0.0015 \\ \mathrm{CClF}_{3} & 0.009 \\ \mathrm{CCl}_{2} \mathrm{~F}_{2} & 0.028 \\ \mathrm{CHClF}_{2} & 0.30 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

(a) CF3(CF2)7Br holds more \(\mathrm{O}_2\) than air. (b) Dipole moment influences solubility most. (c) Liquid holds more \(\mathrm{O}_2\) than air.

Step 1: Understanding Solubility and Molality

The solubility given is in mass percentage which means the mass of solute per 100 grams of solution. Molality ({m}) is defined as moles of solute per kilograms of solvent. To convert solubility from mass percentage to molality, we first convert the solubility to grams of solute per 1 kg of water (which is the solvent).

Step 2: Calculating Molality for CF4

For \(\mathrm{CF}_{4}\), the solubility is 0.0015%. This equates to 0.0015 grams of \(\mathrm{CF}_{4}\) per 100 grams of solution. Assuming nearly all the mass of the solution is water, we have 0.015 grams of \(\mathrm{CF}_{4}\) per 1 kg of water. The molar mass of \(\mathrm{CF}_{4}\) is approximately 88.004 g/mol. Thus, molality (\(m\)) is calculated as \(\frac{0.015}{88.004}\approx 1.7 \times 10^{-4} \) mol/kg.

Step 3: Calculating Molality for CClF3

For \(\mathrm{CClF}_{3}\), the solubility is 0.009%, which is equivalent to 0.009 grams in 100 grams of solution or 0.09 grams in 1 kg of water. The molar mass of \(\mathrm{CClF}_{3}\) is approximately 104.46 g/mol, so the molality is \(\frac{0.09}{104.46}\approx 8.62 \times 10^{-4} \) mol/kg.

Step 4: Calculating Molality for CCl2F2

For \(\mathrm{CCl}_{2}\mathrm{F}_{2}\), the solubility is 0.028%, equating to 0.28 grams in 1 kg of water. Given its molar mass of about 120.91 g/mol, the molality is \(\frac{0.28}{120.91} \approx 2.31 \times 10^{-3} \) mol/kg.

Step 5: Calculating Molality for CHClF2

For \(\mathrm{CHClF}_{2}\), the solubility is 0.3%. It corresponds to 3 grams of solute per 1 kg of water. With its molar mass being about 86.47 g/mol, the molality is \(\frac{3}{86.47}\approx 3.47 \times 10^{-2} \) mol/kg.

Step 6: Predicting Solubility by Molecular Properties

Examining the calculations, the highest molality occurs with \(\mathrm{CHClF}_{2}\), suggesting a stronger interaction with water. This solubility is best predicted by the ability of molecules to hydrogen-bond with water; however, none of these molecules can hydrogen-bond directly. Hence, dipole moment is the next probable contributor as \(\mathrm{CHClF}_{2}\) has higher polarity.

Step 7: Comparing O2 in Air and Liquid

First, calculate moles of \(\mathrm{O}_{2}\) in air: in 15 mL air, \(21\%\) is \(\mathrm{O}_{2}\), leading to \(0.21 \cdot 15 = 3.15\) mL \(\mathrm{O}_{2}\). Use ideal gas equation \(PV = nRT\) (where \(R = 0.0821 \mathrm{~L}\cdot \mathrm{atm}/\mathrm{mol}\cdot K\), \(T = 298\) K) to find \(n = \frac{PV}{RT} \approx 1.29 \times 10^{-4} \) moles \(\mathrm{O}_2\). Now, for \(\mathrm{CF}_3(\mathrm{CF}_2)_7\mathrm{Br}\): in 15 mL, 66% is \(\mathrm{O}_{2}\), giving 9.9 mL \(\mathrm{O}_{2}\). Re-calculating yields \(4.06 \times 10^{-4} \) moles \(\mathrm{O}_2\) from liquid.

Key Concepts

Solubility in Water

Solubility refers to the maximum amount of a substance that can dissolve in a solvent at a given temperature and pressure. For fluorocarbons, their solubility in water depends on the interaction between the fluorocarbon molecules and water molecules. The solubility is typically expressed as a mass percentage, which indicates how many grams of solute are present in 100 grams of the solution.
Fluorocarbons have relatively low solubility in water due to their nonpolar nature. Water is a polar solvent, and it interacts best with other polar or ionic compounds. Most fluorocarbons don’t mix well with water because the strong cohesive forces within water molecules don't attract nonpolar fluorocarbons. This principle follows "like dissolves like", indicating polar solvents best dissolve polar solutes.

• The low solubility of fluorocarbons in water results from their molecular structure that doesn't favor interactions with water.
• This means that in practice, very little of these gases dissolve in water at given conditions, often calculated as low mass percentages.

Understanding solubility helps chemists determine how a substance might behave in different environments or applications.

Molecular Properties and Solubility

The solubility of a molecule, like those of fluorocarbons, can often be predicted using its molecular properties. Here are some key features that influence solubility:

• Molar Mass: While larger molecules, due to their increased surface area, might interact with a solvent more, molar mass alone is less often a direct predictor of solubility for small differences among related compounds.
• Polarity and Dipole Moment: Polarity significantly affects solubility. Molecules with significant dipole moments are more likely to dissolve in polar solvents like water. For example, among the discussed fluorocarbons, a compound with higher polarity would exhibit better solubility in water, as it can better interact with water's polar molecules.
• Ability to Hydrogen-Bond: This is crucial for solubility in water. However, fluorocarbons generally lack the capacity to form hydrogen bonds due to the absence of a hydrogen atom bonded to a more electronegative atom like oxygen or nitrogen.

In essence, dipole moments provide insight into solubility for molecules unable to hydrogen-bond, as seen in the fluorocarbons where higher polarity correlates with increased solubility.

Molality Calculation

Molality is a concentration measure of a solute in a solution. It is denoted in \( ext{mol/kg} \) and focuses on the moles of solute per kilogram of solvent, offering precision since it is temperature independent.
To calculate molality from solubility given as a mass percentage:

• First, convert the mass percentage to grams of solute per kilogram of water. Assume that most of the solution's weight is due to water.
• Find the molar mass of the solute to convert grams to moles. This conversion involves dividing the grams of solute by its molar mass.
• Divide the moles of solute by the mass of water in kilograms to get molality.

For example, to find the molality of \( ext{CF}_4 \) with a solubility of 0.0015%: convert it to 0.015 g of \( ext{CF}_4 \) in 1 kg of water and use the molar mass of 88.004 g/mol. This gives:
\[ m = \frac{0.015}{88.004} \approx 1.7 \times 10^{-4} ext{ mol/kg} \]
This method helps chemists and students alike understand concentrations in reactions and solutions more effectively.