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?

Short answer

The mass of the oxygen in the container is approximately \(4.03 \mathrm{~g}\).

Step-by-step solution

Write down given information and formulas

Given information: - Mass of helium: \(1.42 \mathrm{~g}\) - Partial pressure of helium: \(42.5 \mathrm{~torr}\) - Partial pressure of oxygen: \(158 \mathrm{~torr}\) Formulas: - Ideal Gas Law: \(PV = nRT\) - Dalton's Law of Partial Pressures: \(P_{total} = P_{He} + P_{O_2}\)

Calculate the moles of helium

First, we need to find the moles of helium using its mass and molar mass. The molar mass of helium is \(4.00 \mathrm{\frac{g}{mol}}\). Moles of helium: \(n_{He} = \frac{mass_{He}}{molar~mass_{He}}\) \(n_{He} = \frac{1.42 \mathrm{~g}}{4.00 \mathrm{~\frac{g}{mol}}}\) \(n_{He} = 0.355 \mathrm{~mol}\)

Calculate the total pressure in the flask

Using Dalton's Law of Partial Pressures, calculate the total pressure in the flask. \(P_{total} = P_{He} + P_{O_2}\) \(P_{total} = 42.5 \mathrm{~torr} + 158 \mathrm{~torr}\) \(P_{total} = 200.5 \mathrm{~torr}\)

Use the Ideal Gas Law to determine the volume of the flask

We can use the Ideal Gas Law to find the volume of the flask containing helium, assuming that both gases occupy the same volume. Room temperature is assumed to be \(298 \mathrm{~K}\). We also need to convert the pressure from torr to atm using the conversion factor: \(1 \mathrm{~atm} = 760 \mathrm{~torr}\). \(PV = nRT\) Solve for \(V\): \(V = \frac{nRT}{P}\) Plug in the helium values and constants: \(V = \frac{0.355 \mathrm{~mol} \times 0.0821 \mathrm{~\frac{L \cdot atm}{mol \cdot K}} \times 298 \mathrm{~K}}{42.5 \mathrm{~torr} \times \frac{1 \mathrm{~atm}}{760 \mathrm{~torr}}}\) The volume of the flask is: \(V = 2.46 \mathrm{~L}\)

Calculate the moles of oxygen

Using the Ideal Gas Law, we can now find the moles of oxygen in the flask. \(PV = nRT\) Solve for \(n\): \(n = \frac{PV}{RT}\) Insert the oxygen values: \(n_{O_2} = \frac{158 \mathrm{~torr} \times \frac{1 \mathrm{~atm}}{760 \mathrm{~torr}} \times 2.46 \mathrm{~L}}{0.0821 \mathrm{~\frac{L \cdot atm}{mol \cdot K}} \times 298 \mathrm{~K}}\) \(n_{O_2} = 0.126 \mathrm{~mol}\)

Calculate the mass of oxygen

Now that we have the moles of oxygen, we can calculate its mass by multiplying the moles by the molar mass of oxygen. The molar mass of oxygen is \(32.00 \mathrm{\frac{g}{mol}}\). Mass of oxygen: \(mass_{O_2} = n_{O_2} \times molar~mass_{O_2}\) \(mass_{O_2} = 0.126 \mathrm{~mol} \times 32.00 \mathrm{~\frac{g}{mol}}\) \(mass_{O_2} = 4.03 \mathrm{~g}\) The mass of the oxygen in the container is approximately \(4.03 \mathrm{~g}\).