Problem 149 - General Chemistry

Question

Acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), forms stable pairs of molecules held together by two hydrogen bonds. Such molecules - themselves formed by the association of two simpler molecules-are called dimers. The vapor over liquid acetic acid consists of a mixture of monomers (single acetic acid molecules) and dimers. At \(100.6^{\circ} \mathrm{C}\) the total pressure of vapor over liquid acetic acid is \(436 \mathrm{mmHg}\). If the vapor consists of \(0.630\) mole fraction of the dimer, what are the masses of monomer and dimer in \(1.000 \mathrm{~L}\) of the vapor? What is the density of the vapor?

Step-by-Step Solution

Answer

The masses of monomer and dimer are approximately 0.418 g and 1.419 g, respectively. The density of the vapor is 1.837 g/L.

Step 1: Define the Problem Parameters

We are given that the vapor over liquid acetic acid at \(100.6^{\circ} \mathrm{C}\) has a total pressure of \(436 \mathrm{mmHg}\). The vapor consists of \(63\%\) (or \(0.630\) mole fraction) dimers. We need to find the masses of monomer and dimer in \(1.000\ \mathrm{L}\) of the vapor, and then calculate the vapor's density.

Step 2: Calculate Moles of Dimer and Monomer

Assume \(1\) mole of the total vapor mixture. Let \(x\) be the number of moles of dimer and \(1-x\) be the moles of monomer. Given \(x = 0.630\), hence \((1-x) = 0.370\). Since each dimer consists of 2 monomers, the moles of monomer in terms of dimers is \(2x = 2(0.630) = 1.260\). The total moles of monomer considering both the free monomers and those in dimers is \(\text{total monomers} = 0.370 + 1.260 = 1.630\).

Step 3: Calculate Molar Masses

Calculate the molar mass of monomer \(\mathrm{CH}_{3} \mathrm{COOH} = 2(12.01) + 4(1.01) + 2(16.00)\) which results in \(60.05\, \mathrm{g/mol}\). The molar mass of the dimer \((\mathrm{CH}_{3} \mathrm{COOH})_2\) is twice that, \(120.10\, \mathrm{g/mol}\).

Step 4: Find Masses of Dimer and Monomer

For \(436 \mathrm{mmHg}\) total pressure, convert to atm: \(436 \div 760 = 0.5737\, \mathrm{atm}\). Use the ideal gas law to find total moles of gas in \(1.000\, \mathrm{L}\): \(PV = nRT\), with \(P = 0.5737\, \mathrm{atm}\), \(V = 1.000\, \mathrm{L}\), \(R = 0.0821\, \mathrm{L} \cdot \mathrm{atm} / (\mathrm{mol} \cdot \mathrm{K})\), \(T = 373.75\, \mathrm{K}\). Calculate \(n = \frac{PV}{RT} = \frac{0.5737 \times 1}{0.0821 \times 373.75} = 0.01882\, \mathrm{mol}\). Then, masses: \(\text{mass of monomer} = 0.01882 \times 0.37 \times 60.05\) and \(\text{mass of dimer} = 0.01882 \times 0.63 \times 120.10\).

Step 5: Calculate the Density of the Vapor

Total mass of the vapor = \(\text{mass of monomer} + \text{mass of dimer}\). Density \(= \frac{\text{total mass}}{\text{volume}} = \frac{\text{total mass in g}}{1.000 \, \mathrm{L}}\). Calculate based on computed masses.

Key Concepts

Hydrogen Bonding

Acetic acid molecules are excellent examples of hydrogen bonding in action. When these molecules pair up to form dimers, they are held together by two hydrogen bonds. A hydrogen bond occurs when a hydrogen atom is attracted to a highly electronegative atom, such as oxygen or nitrogen. In acetic acid, these bonds stabilize the dimer structure, making it distinct from individual acetic acid molecules.

The hydrogen bond is one of the strongest types of intermolecular forces, although it is still much weaker than covalent or ionic bonds. This weaker bond strength allows the dimers to form and break fairly easily.

• Hydrogen bonds are important for the unique properties of water and biological molecules.
• They contribute to the high boiling point of acetic acid compared to other molecules of similar mass without hydrogen bonds.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase. In the case of acetic acid, the vapor phase consists of both monomers and dimers. The total vapor pressure at a given temperature (100.6°C in this problem) is the sum of the partial pressures exerted by both monomers and dimers.

The concept of vapor pressure is crucial because it indicates how volatile a substance is. Substances with high vapor pressures evaporate more readily, while those with low vapor pressures are less volatile.

• Total pressure of 436 mmHg represents the combined pressure of acetic acid molecules in dimer and monomer forms.
• Vapor pressure is a key factor in the boiling point of a substance.

Ideal Gas Law

The ideal gas law is expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature.

In this scenario, we use the ideal gas law to calculate the number of moles of acetic acid in the gas phase. Despite the presence of dimers and monomers, the vapor obeys the ideal gas law under the conditions given. It's important to adjust for pressure in atm and temperature in Kelvin for accurate calculations.

• The total pressure in atm is used to find the total number of moles in the vapor mixture.
• This law helps to link macroscopic quantities like pressure and volume to the microscopic quantity of moles.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is the ratio of the number of moles of one component to the total number of moles in the mixture.

For the vapor over liquid acetic acid, the mole fraction of the dimer is 0.630, indicating that 63% of the gas phase consists of dimers. Understanding mole fraction is key in determining the relative amounts of components in a mixture.

• For calculations, assume 1 mole of total vapor to simplify the math.
• Mole fraction helps in determining the proportion of substances which affects mass and density calculations.