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 \( 63.67 \mathrm{g/mol} \).

Step by step solution

01

Write down the given information

The rate of effusion of the unknown gas is 24.0 mL/min, and the rate of effusion of methane (CH4) gas is 47.8 mL/min.
02

Write down the Graham's law of effusion formula

According to Graham's law, the ratio of the rates of effusion of two gases is the inverse of the square root of the ratio of their molar masses. Mathematically, this can be written as: \( \frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \) In this problem, we will consider the unknown gas as gas 1 and methane as gas 2.
03

Plug in the given values into the formula

Now, we will plug in the rates of effusion of the unknown gas and methane into the Graham's law formula: \( \frac{24.0}{47.8} = \sqrt{\frac{M_{CH_4}}{M_{unknown}}} \)
04

Find the molar mass of methane

Methane (CH4) has 1 carbon atom and 4 hydrogen atoms, so its molar mass would be: \( M_{CH_4} = 1 \times 12.01 + 4 \times 1.008 = 16.042 \mathrm{g/mol} \)
05

Solve for the molar mass of the unknown gas

Now, we have the following equation to solve for the molar mass of the unknown gas: \( \frac{24.0}{47.8} = \sqrt{\frac{16.042}{M_{unknown}}} \) First, square both sides of the equation: \( \left(\frac{24.0}{47.8}\right)^2 = \frac{16.042}{M_{unknown}} \) Next, calculate the left side of the equation: \( \left(\frac{24.0}{47.8}\right)^2 \approx 0.2521 \) Now, rearrange the equation to isolate the molar mass of the unknown gas: \( M_{unknown} = \frac{16.042}{0.2521} \) Finally, solve for the molar mass of the unknown gas: \( M_{unknown} \approx 63.67 \mathrm{g/mol} \)
06

State the answer

The molar mass of the unknown gas is approximately 63.67 g/mol.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Rate of Effusion
The rate of effusion is a term used in chemistry to describe how fast a gas escapes through a tiny hole into a vacuum. It's an important concept in gas laws and helps to understand the behavior of gases under various conditions. According to Graham's Law of Effusion, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. In simpler terms, lighter gases effuse faster than heavier ones.

When comparing the effusion rates of two gases, we can set up a ratio that allows us to calculate one if we know the other. This is particularly useful when trying to identify an unknown gas, as demonstrated in the exercise provided. We can see how the effusion rate of the unknown gas relates to that of a known gas (methane, in this case), and this relationship can lead us to determine the molar mass of the unknown gas. Understanding the relationship between effusion rates is a key step in solving problems related to gas properties and behaviors.
Molar Mass Calculation
The molar mass of a substance is the weight of one mole of that substance. It's a fundamental concept in chemistry because it relates the microscopic world of atoms and molecules to the macroscopic world we can measure. When calculations require converting between grams and moles, the molar mass serves as the conversion factor.

In the context of the exercise, once we know the quantitative relationship between the rates of effusion, we use Graham's Law to set up a proportion. To solve for the molar mass of the unknown gas, we first need the molar mass of the known gas, which is methane in this case. Calculating molar mass typically involves summing the atomic masses of all atoms in the molecule, taken from the periodic table. Methane's molar mass calculation shows it’s the sum of the atomic masses of one carbon atom and four hydrogen atoms. Concerning the unknown gas, once we have the proportion set up, it's just a matter of simple algebra to solve for its molar mass.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and the steps by which these reactions occur, known as the reaction mechanism. It involves understanding how various factors like temperature, concentration, and the presence of catalysts affect the speed of a chemical reaction. While the problem at hand doesn't directly involve reaction rates, the principles of kinetics – particularly the factors that influence how rapidly molecules move – do inform our understanding of effusion.

Graham's Law ties into kinetics by describing the movement of gas particles through a small opening. This can be considered a physical process akin to reactions in that the particles’ rates of movement are governed by their kinetic energies. Essentially, lighter gas molecules, having higher average speeds at a given temperature (as described by kinetic molecular theory), will effuse more rapidly. Thus, mastering the basics of chemical kinetics not only helps in conceptualizing reactions but also in understanding phenomena such as effusion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A steel cylinder contains \(5.00\) moles of graphite (pure carbon) and \(5.00\) moles of \(\mathrm{O}_{2}\). The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by \(17.0 \% .\) Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in the final gaseous mixture.

An organic compound contains \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O} .\) Combustion of \(0.1023 \mathrm{g}\) of the compound in excess oxygen yielded \(0.2766 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0991 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) A sample of \(0.4831 \mathrm{g}\) of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129 ). At \(\mathrm{STP}, 27.6 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) was obtained. In a third experiment, the density of the compound as a gas was found to be \(4.02 \mathrm{g} / \mathrm{L}\) at \(127^{\circ} \mathrm{C}\) and \(256\) torr. What are the empirical and molecular formulas of the compound?

A compressed gas cylinder contains \(1.00 \times 10^{3} \mathrm{g}\) argon gas. The pressure inside the cylinder is \(2050 .\) psi (pounds per square inch) at a temperature of \(18^{\circ} \mathrm{C}\). How much gas remains in the cylinder if the pressure is decreased to \(650 .\) psi at a temperature of \(26^{\circ} \mathrm{C} ?\)

Trace organic compounds in the atmosphere are first concentrated and then measured by gas chromatography. In the concentration step, several liters of air are pumped through a tube containing a porous substance that traps organic compounds. The tube is then connected to a gas chromatograph and heated to release the trapped compounds. The organic compounds are separated in the column and the amounts are measured. In an analysis for benzene and toluene in air, a \(3.00-\mathrm{L}\) sample of air at \(748\) torr and \(23^{\circ} \mathrm{C}\) was passed through the trap. The gas chromatography analysis showed that this air sample contained \(89.6\) ng benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) and \(153 \mathrm{ng}\) toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right) .\) Calculate the mixing ratio (see Exercise 121 ) and number of molecules per cubic centimeter for both benzene and toluene.

Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to \(400^{\circ} \mathrm{C}\) in a nickel reaction vessel. A \(100.0\) -\(\mathrm{mL}\) nickel container is filled with xenon and fluorine, giving partial pressures of \(1.24\) atm and \(10.10\) atm, respectively, at a temperature of \(25^{\circ} \mathrm{C}\). The reaction vessel is heated to \(400^{\circ} \mathrm{C}\) to cause a reaction to occur and then cooled to a temperature at which \(\mathrm{F}_{2}\) is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining \(\mathrm{F}_{2}\) gas is transferred to another A \(100.0\) -\(\mathrm{mL}\) nickel container, where the pressure of \(\mathrm{F}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(7.62\) atm. Assuming all of the xenon has reacted, what is the formula of the product?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free