Problem 111 - General Chemistry: Principles and Modern Applications

Question

A 0.168 L sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(26^{\circ} \mathrm{C}\) and a barometric pressure of \(737 \mathrm{mm} \mathrm{Hg}\). In the gas that is collected, what is the percent water vapor (a) by volume; (b) by number of molecules; (c) by mass? (Vapor pressure of water at \(26^{\circ} \mathrm{C}=25.2 \mathrm{mmHg}\).)

Step-by-Step Solution

Answer

The percentage of water vapor is around 3.4% by volume, 3.4% by number of molecules, and the percentage by mass can be computed using the expressions in Step 5

Step 1: Calculate the partial pressure of water vapor

The vapor pressure of water at \(26^{\circ} \mathrm{C}\) is given to be \(25.2 \mathrm{mmHg}\), which is the partial pressure of water vapor (since the vapor pressure of a substance is just its partial pressure in a mixture).

Step 2: Calculate the partial pressure of O2 gas

The barometric pressure is the total pressure due to O2 gas and water vapor. It's given as \(737 \mathrm{mmHg}\). So, we can subtract the partial pressure of water from the total pressure to get the partial pressure of the O2 gas: \(737-25.2 = 711.8 \mathrm{mmHg}\).

Step 3: Determine the percent by volume

The percent by volume is just the volume of one component divided by the total volume, times 100. In this instance, the volumes of the gases are proportional to their respective pressures since we're dealing with ideal gases at a constant temperature, according to Avogadro's Law. Thus, the percent volume is \( \left( \frac{25.2}{737} \right) \times 100 = 3.4\% \).

Step 4: Determine the percent by number of molecules

For ideal gases, the number of molecules is directly proportional to the pressure. So, we can calculate this in the same way as we did for percent by volume: \( \left( \frac{25.2}{737} \right) \times 100 = 3.4\% \).

Step 5: Determine the percent by mass

First, using the ideal gas law, each of the gas' moles can be determined from their partial pressures, volume, and temperature. The molar mass of O2 is \( 32 \mathrm{g/mol} \) and for H2O it is \(18 \mathrm{g/mol}\). Thus, the mass of water can be found by multiplying its moles by 18, and the mass of oxygen by multiplying its moles by 32. Add up the two masses to get the total mass. Then, calculate the percent by mass as \( \left( \frac{\mathrm{mass\,of\,water}}{\mathrm{total\,mass}} \right) \times 100 \). Calculate this to find the percentage by mass.

Key Concepts

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. In a mixture, each gas component contributes to the total pressure in proportion to its amount. In this exercise, we have a gas sample of oxygen collected over water. The total pressure is the barometric reading, which includes the pressure from both the oxygen and the water vapor. Using the vapor pressure of water at the given temperature, we subtract it from the total pressure to find the partial pressure of the oxygen.
In our scenario, the vapor pressure of water at 26°C is 25.2 mmHg. This is effectively the partial pressure of the water vapor in the collected gas mixture. To find the partial pressure of the oxygen, you subtract this vapor pressure from the total barometric pressure, which is 737 mmHg. Thus, the partial pressure of oxygen is calculated as 711.8 mmHg.

• This concept is crucial in understanding gas behavior in mixtures, such as in atmospheric studies or chemical reactions where multiple gases are involved.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form. It's a crucial concept when gases are collected over liquids, as in our exercise with oxygen over water. At any given temperature, a liquid will have a specific vapor pressure that indicates how much of it turns into gas.
For water at 26°C, the vapor pressure is 25.2 mmHg. This pressure is an inherent property of water at that temperature and tells us how much water vapor is present in the gas mixture. When collecting a gas like oxygen over water, the water's vapor pressure must be considered to accurately determine the pressure due to the collected gas.

• In real-life applications, vapor pressure is important for processes such as distillation or climatology, where vapor presence affects outcomes significantly.

Ideal Gas Law

The ideal gas law, expressed as \(PV = nRT\), connects pressure (P), volume (V), and temperature (T) to the amount of gas in moles (n), with R being the ideal gas constant. It is a key relationship in understanding gas behavior under various conditions.
In the context of this exercise, the ideal gas law can be used to determine the number of moles of oxygen and water vapor in the gas mixture. By using the known partial pressures, volume, and temperature, you can find the moles for each component. These calculations further allow the determination of percentages by mass and number of molecules in the mixture.

• Understanding the ideal gas law is fundamental for calculations involving gas reactions in chemistry, air pressure effects in physics, and more.

Avogadro's Law

Avogadro's Law states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. This principle is integral for calculations involving gaseous mixtures.
In this exercise, Avogadro's Law implies that the volume of gas is directly proportional to the number of molecules it contains. Therefore, when comparing the percent volumes or the percent number of molecules, these percentages can often be calculated in similar ways.

• The idea is that because the gases behave ideally, you can use their partial pressures as a direct indicator of their volume fractions within the mixture.