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Chapter 5: Problem 112

The rate of effusion of a particular gas was measured and found to be \(24.0 \mathrm{~mL} / \mathrm{min}\). Under the same conditions, the rate of effusion of pure methane \(\left(\mathrm{CH}_{4}\right)\) gas is \(47.8 \mathrm{~mL} / \mathrm{min}\). What is the molar mass of the unknown gas?

Short Answer

Expert verified
The molar mass of the unknown gas is approximately \(4.039 \mathrm{g/mol}\).

Step by step solution

01

Graham's Law of Effusion

The Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this can be represented as: \[\frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\] where \(Rate_{1}\) and \(Rate_{2}\) are the rates of effusion of gas 1 and 2, respectively, and \(M_{1}\) and \(M_{2}\) are their molar masses.
02

Given Variables

We are given: Rate of effusion of unknown gas = \(24.0 \mathrm{~mL/min}\) Rate of effusion of methane gas \(\mathrm{(CH_4)}\) = \(47.8 \mathrm{~mL/min}\) We need to find the molar mass of the unknown gas, and we know the molar mass of methane gas \(\mathrm{(CH_4)}\) = \(12.01 + 4(1.008) = 16.042 \mathrm{~g/mol}\).
03

Substitute values in the formula

Using the Graham's Law of Effusion: \[\frac{24.0}{47.8} = \sqrt{\frac{M_{unknown}}{16.042}}\]
04

Calculate the molar mass of the unknown gas

Square both sides of the equation and solve for the molar mass of the unknown gas: \[\left(\frac{24.0}{47.8}\right)^2 = \frac{M_{unknown}}{16.042}\] \[M_{unknown} = 16.042 \times \left(\frac{24.0}{47.8}\right)^2\] \[M_{unknown} = 16.042 \times \left(\frac{576}{2289.64}\right)\] \[M_{unknown} = 16.042 \times 0.251725\] \[M_{unknown} = 4.039 \mathrm{g/mol}\] The molar mass of the unknown gas is approximately \(4.039 \mathrm{g/mol}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
The molar mass of a gas is essentially the mass of one mole of that gas, usually expressed in grams per mole. Calculating the molar mass is an essential step in determining the identity of an unknown gas through experiments like effusion.
Using Graham's Law to find the molar mass starts with identifying or knowing the rate of effusion for both gases involved.
Once you have their effusion rates, this formula comes into play:
  • \(\frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\)
In this exercise, we used the effusion rates of an unknown gas, measured as 24.0 mL/min, and methane gas, measured as 47.8 mL/min. By substituting these values and using the known molar mass of methane (16.042 g/mol), we can solve for the unknown's molar mass.
Effusion Rate
The effusion rate indicates how fast a gas travels through a small opening. Different gases have different commercial speeds or rates of effusion based upon their individual molar masses.
The principle behind this is elegantly captured in Graham's Law of Effusion. This law shows that lighter gases effuse more quickly than heavier gases, due to their lower molar mass.
By comparing effusion rates of two gases, it is possible to deduce the ratio of their molar masses. For example, the faster effusion of methane compared to an unknown gas indirectly reveals that methane, being lighter, has a smaller molar mass.
Methane Gas
Methane (\(CH_4\)) is a simple hydrocarbon with a known molar mass of 16.042 g/mol. It acts as a reference point in experiments involving Graham's Law of Effusion.
Understanding the properties of methane is crucial, as it often serves as a baseline in calculating the molar mass of unknown gases. Methane is desirable for such experiments because it is widely available and, importantly, its molar mass is well-established.
Since methane is lighter compared to many other gases, its fast effusion rate makes it an ideal candidate for comparing rate differences, allowing students and researchers to determine the molar mass of gases they know little about.

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Most popular questions from this chapter

Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction goes to completion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?

As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, the volume of the product gases collected is twice the volume of \(\mathrm{NH}_{3}\) reacted. Explain. As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure increases by some factor. Why the increase in pressure and by what factor does the total pressure increase when reactants are completely converted into products? How do the partial pressures of the product gases compare to each other and to the initial pressure of \(\mathrm{NH}_{3}\) ?

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a 1.0000-L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

A person accidentally swallows a drop of liquid oxygen, \(\mathrm{O}_{2}(l)\), which has a density of \(1.149 \mathrm{~g} / \mathrm{mL}\). Assuming the drop has a volume of \(0.050 \mathrm{~mL}\), what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of \(1.0 \mathrm{~atm}\) ?

Ideal gas particles are assumed to be volumeless and to neither attract nor repel each other. Why are these assumptions crucial to the validity of Dalton's law of partial pressures?
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