Problem 69 - General Chemistry: Principles and Modern Applications

Question

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(24^{\circ} \mathrm{C}\) The volume of gas is 1.16 L. In a subsequent experiment, it is determined that the mass of \(\mathrm{O}_{2}\) present is 1.46 g. What must have been the barometric pressure at the time the gas was collected? (Vapor pressure of water \(=22.4 \text { Torr. })\)

Step-by-Step Solution

Answer

The barometric pressure at the time the gas was collected was 1.15 atm.

Step 1: Identification of Moles of the Oxygen Gas

Use the molar mass of oxygen to identify the number of moles in the given mass. The molar mass of \(O_2\) is \(32.00 \, g/mol\). Thus, the number of moles \(n\) of \(O_2\) is given by the formula: \(n = \frac{mass}{molar \,mass} = \frac{1.46 \, g}{32.00 \, g/mol} = 0.046 \, mol\).

Step 2: Calculation of Total Pressure using the Ideal Gas Law

We can use the Ideal Gas law \(PV = nRT\), where \(P\) is the total pressure, \(V\) is the volume in liters, \(n\) is the number of moles, \(R\) is the gas constant (\(0.0821 \, L \cdot atm/(mol \cdot K)\)), and \(T\) is the temperature in Kelvin. First, convert the temperature from Celsius to Kelvin using the formula \(K = °C + 273.15\) which gives \(T = 24 + 273.15 = 297.15 \, K\). Then, substitute the appropriate values to find \(P\): \(P = \frac{nRT}{V} = \frac{0.046 \, mol \cdot 0.0821 \, L \cdot atm/(mol \cdot K) \cdot 297.15 \, K}{1.16 \, L} = 1.12 \, atm\).

Step 3: Applying Dalton's Law of Partial Pressure

Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. In this case, that means the total pressure \(P_{total}\) is equal to the pressure of oxygen \(P_{O2}\) plus the vapor pressure of water \(P_{H2O}\). Since we're looking for the barometric pressure, which is the same as the total pressure, we rearrange the equation to be: \(P_{total} = P_{O2} + P_{H2O}\) or \(P_{Barometric} = P_{O2} + P_{H2O}\). The given water pressure is \(22.4 \, Torr = 0.0295 \, atm\). Therefore, the barometric pressure is \(1.12 \, atm + 0.0295 \, atm = 1.15 \, atm\).

Key Concepts

Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures is an important principle used to understand gas mixtures. It tells us that the total pressure in a container is the sum of the partial pressures of each gas present.
In simpler terms, each gas in a mixture behaves independently and contributes to the total pressure as if it were the only gas there. To express it mathematically, you can write:

• \( P_{\text{total}} = P_{1} + P_{2} + P_{3} + \ldots \)

where each \( P \) represents the partial pressure of a different gas.### Application to the Problem In the exercise, you needed to find the barometric pressure when oxygen was collected over water. According to Dalton's Law, the total pressure equals the partial pressure of oxygen plus the vapor pressure of water. The exercise helps demonstrate how Dalton’s Law allows you to calculate barometric pressure by combining these individual pressures.

Vapor Pressure

Vapor pressure is another concept important to our understanding of gas behavior, especially when gases are collected over a liquid. Vapor pressure is simply the pressure exerted by a vapor in equilibrium with its solid or liquid phase. Think of it as the evaporation "force" of the liquid. Every liquid at a given temperature has its own characteristic vapor pressure. For water at 24°C, it has a known vapor pressure of 22.4 Torr, as mentioned in the exercise.
The significance of vapor pressure in this exercise is that it must be taken into account when determining the total pressure of a gas collected over a liquid, such as in the collection of oxygen over water. ### Considering Vapor Pressure When you collect a gas over water, water vapor contributes to the total pressure. Thus, to find the pressure of a gas itself, like oxygen in this exercise, you first need to subtract the vapor pressure of water from the total atmospheric pressure. The inclusion of vapor pressure in calculations reflects real-world conditions more accurately.

Molar Mass of Oxygen

The molar mass is a measure of the mass of one mole of a substance. For diatomic oxygen, \( O_2 \), the molar mass is calculated from the atomic mass of oxygen (16 g/mol) multiplied by two, equaling 32 g/mol. This value is fundamental for converting between the mass of a substance and the amount in moles.### Calculating Moles of OxygenIn the problem, you used the mass of the collected oxygen gas, 1.46 g, to determine the number of moles.

• The formula used is: \( n = \frac{\text{mass}}{\text{molar mass}} \)

By substituting 32 g/mol for the molar mass, you could find the moles of oxygen to be \( 0.046 \, mol \).
This step is critical as it relates directly to using the Ideal Gas Law for further calculations.The molar mass and mole concept bridge the gap between macroscopic measurements and the microscopic understanding of substances. The calculation of moles allows us to use tools like the Ideal Gas Law, connecting physical mass with observable gas behavior.