Problem 101 - Chemistry: The Central Science

Question

A sample of \(1.42 \mathrm{~g}\) of helium and an unweighed quantity of \(\mathrm{O}_{2}\) are mixed in a flask at room temperature. The partial pressure of helium in the flask is \(42.5\) torr, and the partial pressure of oxygen is 158 torr. What is the mass of the oxygen in the container?

Step-by-Step Solution

Answer

The mass of oxygen in the container is approximately \(2.849 \, \mathrm{g}\).

Step 1: List the given information

We have the following information given: - Mass of helium (He): \(1.42 \mathrm{~g}\) - Partial pressure of helium (P(He)): \(42.5 \mathrm{~torr}\) - Partial pressure of oxygen (P(O_2)): \(158 \mathrm{~torr}\) - Room temperature: assumed to be \(25^\circ \mathrm{C}\) (or \(298 \mathrm{~K}\)) Additionally, we need to consider the following constants: - Gas constant (R): \(0.0821\, \dfrac{\mathrm{L. atm}}{\mathrm{K. mol}}\) - Molecular weight of helium: \(4.00 \dfrac{\mathrm{g}}{\mathrm{mol}}\) - Molecular weight of oxygen: \(32.0 \dfrac{\mathrm{g}}{\mathrm{mol}}\)

Step 2: Convert pressures to atm

First, let's convert the given pressures from torr to atm, as the constant R is in atm. We can do this using the conversion factor of \(1 \mathrm{~atm} = 760 \mathrm{~torr}\). P(He) = \(\frac{42.5 \mathrm{~torr}}{760 \mathrm{~torr/atm}} = 0.05592 \mathrm{~atm}\) P(O_2) = \(\frac{158 \mathrm{~torr}}{760 \mathrm{~torr/atm}} = 0.2079 \mathrm{~atm}\)

Step 3: Calculate the number of moles using ideal gas law formula

Now, let's find the total moles of helium, using the ideal gas law formula (PV = nRT), and molecular weight of helium. Rearrange the formula to solve for n: \(n_{He} = \frac{P_{He}V}{RT}\) We need the volume (V) as well. To find V, rearrange the formula again, then plug in the given values: \(PV = nRT \Rightarrow V = \frac{nRT}{P}\) For helium: \(V_{He} = \frac{(1.42 \mathrm{g})/(4.00 \: \mathrm{g/mol})(0.0821 \, \mathrm{L \cdot atm/K \cdot mol})(298 \, \mathrm{K})}{0.05592 \, \mathrm{atm}}\) \(V_{He} = 2.287 \mathrm{~L}\) As both helium and oxygen occupy the same volume, V remains the same. Thus, the number of moles of helium is: \(n_{He}=\frac{1.42 \, \mathrm{g}}{4.00 \, \mathrm{g/mol}}=0.355 \mathrm{~mol}\) Now, we can find the number of moles of oxygen using the ideal gas law: \(n_{O_2} = \frac{P_{O_2}V}{RT}\) \(n_{O_2} = \frac{(0.2079 \mathrm{~atm})(2.287 \mathrm{~L})}{(0.0821 \, \mathrm{L \cdot atm/K \cdot mol})(298 \, \mathrm{K})}\) \(n_{O_2} = 0.08903 \, \mathrm{mol}\)

Step 4: Calculate the mass of oxygen

Finally, we can calculate the mass of oxygen using the molecular weight of oxygen and the number of moles calculated: Mass of \(O_2 = n_{O_2} \times\) molecular weight of \(O_2\) Mass of \(O_2 = (0.08903 \, \mathrm{mol})(32.0 \, \mathrm{g/mol})\) Mass of \(O_2 = 2.849 \, \mathrm{g}\) The mass of oxygen in the container is approximately \(2.849 \, \mathrm{g}\).

Key Concepts

Partial Pressure

In a mixture of gases, each gas contributes to the total pressure. This is known as partial pressure. Partial pressure is the pressure a gas would exert if it alone occupied the entire volume. It's a crucial concept in understanding how gases behave in a mixture. In our problem, helium and oxygen both have their own partial pressures. Helium's partial pressure is given as 42.5 torr, while oxygen's is 158 torr.

To find the total pressure in the flask, you can add these two partial pressures together. This is derived from Dalton's Law of Partial Pressures, which states that the total pressure is the sum of the individual partial pressures in a mixture of non-reacting gases. Partial pressure helps you understand the individual contribution of each gas in a mixture.

Molecular Weight

Molecular weight is the mass of a given molecule. It is measured in grams per mole (g/mol) and helps convert between mass and moles of a substance. In this problem, the molecular weight of helium is given as 4.00 g/mol, and oxygen as 32.0 g/mol.

Molecular weight allows us to connect the macroscopic (grams) with the microscopic (moles) world. For example, using the molecular weight of helium, you can find the number of moles from a known mass and vice versa. It's a fundamental concept for calculating the amount of substance present in a chemical sample.

Moles Calculation

Calculating moles is a key step in using the Ideal Gas Law, which is \[ PV = nRT \] where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

In this exercise, knowing the partial pressure of helium and using the room temperature, we can calculate its moles by rearranging the formula to solve for \( n \):

\[ n = \frac{PV}{RT} \]

Once you know the volume from helium, you use it to find the moles of oxygen. Calculating moles helps us translate information about pressure, volume, and temperature into a tangible amount of substance, which can then be used to find mass.

Conversion of Units

Unit conversion is fundamental for applying scientific concepts to real-world problems. In this exercise, pressure is initially given in torr, but we need to convert it to atmospheres (atm) because the gas constant \( R \) is in \( \text{L atm/K mol} \). The conversion factor is \(1 \text{ atm} = 760 \text{ torr}\).

By applying this conversion, helium's pressure becomes 0.05592 atm and oxygen's pressure becomes 0.2079 atm. This allows us to use them correctly in the Ideal Gas Law.

Conversions ensure consistency in units across calculations, making them crucial for accurate scientific work. Remember to always check your units to avoid errors and make calculations seamless.