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Chapter 5: Problem 120

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

Short Answer

Expert verified
The molar mass of the unknown gas is approximately \(44.97 \: g/mol\), calculated using Graham's law of effusion formula and the given rates of effusion.

Step by step solution

01

Write down given values and Graham's law formula

To solve this problem, we should first identify the given values: Rate of effusion of unknown gas (R1) = 24.0 mL/min, Rate of effusion of methane gas (R2) = 47.8 mL/min, Molar mass of methane gas (M2) = 16 g/mol (since methane is \(\mathrm{CH}_{4}\) with 1 carbon and 4 hydrogen atoms, C = 12 g/mol and H = 1g/mol). Graham's law of effusion formula: \( \frac{R1}{R2} = \sqrt{\frac{M2}{M1}} \) Where R1 is the rate of effusion of the unknown gas, R2 is the rate of effusion of methane gas, M1 is the molar mass of the unknown gas, and M2 is the molar mass of methane gas.
02

Calculate the ratio of the two rates

According to Graham's law, we have to find the ratio between the rate of effusion of the unknown gas and methane gas: \( \frac{R1}{R2} = \frac{24.0}{47.8} \)
03

Solve for the molar mass of the unknown gas

We have the formula \( \frac{R1}{R2} = \sqrt{\frac{M2}{M1}} \) and calculated ratio \(\frac{R1}{R2}\) in step 2. Now, we will find the molar mass of the unknown gas (M1): \( \sqrt{\frac{M2}{M1}} = \frac{24.0}{47.8} \) Square both sides of the equation and solve for M1: \( \frac{M2}{M1} = \left(\frac{24.0}{47.8}\right)^2 \) \( M1 = \frac{M2}{(\frac{24.0}{47.8})^2} \) Use the given molar mass of methane gas (M2) = 16 g/mol to calculate M1: \( M1 = \frac{16}{(\frac{24.0}{47.8})^2} \)
04

Calculate the molar mass of the unknown gas

Now, we can calculate molar mass of the unknown gas (M1): \( M1 = \frac{16}{(\frac{24.0}{47.8})^2} = 44.97 \: g/mol \) The molar mass of the unknown gas is approximately 44.97 g/mol.

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Key Concepts

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

Molar Mass Determination
Molar mass determination is an essential concept in chemistry, especially when dealing with gases, as it helps us identify substances. It refers to the process of calculating the molar mass of a substance, which is the mass of one mole of molecules or atoms. Understanding this concept enables chemists to experiment with different substances accurately and make predictions about their behavior.
One common method to determine the molar mass is using Graham's law of effusion, which relates the rates at which different gases effuse. When provided with the rate of effusion of a known gas, like methane, and an unknown gas, you can determine the unknown gas's molar mass by solving Graham's equation. The formula is given by:\[\frac{R1}{R2} = \sqrt{\frac{M2}{M1}}\]where \(R1\) and \(R2\) are the rates of effusion of the unknown and known gases, respectively, and \(M1\) and \(M2\) are their molar masses.
This equation shows that gases with lower molar masses effuse faster than those with higher molar masses. By rearranging and solving the equation, we can find the molar mass of the unknown gas, making molar mass determination straightforward and practical.
Effusion Rates
Effusion rates are all about how fast a gas escapes through a small hole into a vacuum. This concept is crucial in understanding how different gases behave under similar conditions. Graham's law of effusion is used to compare effusion rates of two gases, providing insight into their molar masses.
According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means that lighter gases effuse more quickly than heavier gases. For instance, methane, which is relatively light, has a faster effusion rate compared to heavier gases. This concept is especially useful in identifying gases and understanding their molecular characteristics based on their observed effusion rates.
In our exercise, we compared the rate of effusion of an unknown gas with methane's rate. By observing that the unknown gas has a slower effusion rate than methane, we used the mathematical relationship provided by Graham's law to find the molar mass, confirming that the unknown gas has a higher molar mass than methane.
Methane (CH鈧�)
Methane, or \(\mathrm{CH}_{4}\), is a simple molecule consisting of one carbon atom and four hydrogen atoms. It is the main component of natural gas and is known for its light and flammable characteristics.
Because of its simple molecular structure, methane has a low molar mass of 16 g/mol. This low molar mass makes it an excellent reference gas for effusion and diffusion studies in which its effusion rate is often used as a benchmark.
In our exercise, methane's known effusion rate was key to finding the molar mass of an unknown gas. By measuring how fast methane effuses compared to the unknown gas, and using Graham's law, we were able to calculate the molar mass of the unknown gas. Methane's role in this context highlights its significance as an easy-to-use reference due to its uncomplicated molecular composition and predictable behavior.

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Most popular questions from this chapter

A \(1.0-\mathrm{L}\) sample of air is collected at \(25^{\circ} \mathrm{C}\) at sea level \((1.00 \mathrm{atm}) .\) Estimate the volume this sample of air would have at an altitude of 15 \(\mathrm{km}\) ( see Fig.5.30) .At \(15 \mathrm{km},\) the pressure is about 0.1 \(\mathrm{atm} .\)

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 \(\mathrm{NH}_{3}?\)

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 . \mathrm{L} / \mathrm{min}\) at 1.50 \(\mathrm{atm}\) and ambient temperature. Air is added to the chamber at 1.00 \(\mathrm{atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent \(\mathrm{O}_{2}\) and 79 \(\mathrm{mole}\) percent \(\mathrm{N}_{2},\) calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that 95.0\(\%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\) . The remainder was present as carbon in CO. Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\) . This resulted in a calculated atomic mass of 13.5 for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus \(1894,\) p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows: $$\begin{array}{lll}{\text { Mass }} & {0.2022 \mathrm{g}} & {0.2224 \mathrm{g}} \\ {\text { Volume }} & {22.6 \mathrm{cm}^{3}} & {26.0 \mathrm{cm}^{3}} \\ {\text { Temperature }} & {13^{\circ} \mathrm{C}} & {17^{\circ} \mathrm{C}} \\ {\text { Pressure }} & {765.2 \mathrm{mm} \mathrm{Hg}} & {764.6 \mathrm{mm}}\end{array}$$ If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.

A glass vessel contains 28 g of nitrogen gas. Assuming ideal behavior, which of the processes listed below would double the pressure exerted on the walls of the vessel? a. Adding 28 g of oxygen gas b. Raising the temperature of the container from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\) c. Adding enough mercury to fill one-half the container d. Adding 32 g of oxygen gas e. Raising the temperature of the container from \(30 .^{\circ} \mathrm{C}\) to \(60 .^{\circ} \mathrm{C}\)
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