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Chapter 10: Problem 104

When a large evacuated flask is filled with argon gas, its mass increases by 3.224 \(\mathrm{g}\) . When the same flask is again evacuated and then filled with a gas of unknown molar mass, the mass increase is 8.102 g. (a) Based on the molar mass of argon, estimate the molar mass of the unknown gas. (b) What assumptions did you make in arriving at your answer?

Short Answer

Expert verified
The molar mass of the unknown gas is estimated to be 100.4 g/mol. We made the following assumptions: the volume, temperature, and pressure of the flask remain the same for both gases; the unknown gas behaves as an ideal gas; and Avogadro's hypothesis holds true for both gases.

Step by step solution

01

Identify the given information and the variables

We are given the mass increase for the argon gas and the unknown gas. We can denote the mass increase of the argon gas as m鈧�=3.224 g and the mass increase of the unknown gas as m鈧�=8.102 g. We also know the molar mass of argon gas M鈧�=39.95 g/mol. We need to calculate the molar mass of the unknown gas, which we can denote as M鈧�.
02

Calculate number of moles of argon gas

We can calculate the number of moles of argon gas using the mass of argon gas and its molar mass. The formula to find the number of moles (n) is: n = mass / molar mass For argon gas, we can write: n鈧� = m鈧� / M鈧� n鈧� = 3.224 g/ 39.95 g/mol n鈧� = 0.0807 mol
03

Apply Avogadro's hypothesis

Avogadro's hypothesis states that equal volumes of gases at the same temperature and pressure have the same number of moles. As we are assuming that the volume, temperature, and pressure of the flask remain the same for both gases, we can apply Avogadro's hypothesis. Since the number of moles is the same for both gases, we can write: n鈧� = n鈧� 0.0807 mol = n鈧�
04

Calculate the molar mass of the unknown gas

Now that we know the number of moles of the unknown gas, we can find its molar mass using the mass of the unknown gas. We can use the formula: M鈧� = m鈧� / n鈧� M鈧�= 8.102 g / 0.0807 mol M鈧� = 100.4 g/mol The estimated molar mass of the unknown gas is 100.4 g/mol.
05

Identify assumptions made in the calculations

The following assumptions were made during the calculations: 1. The volume, temperature, and pressure of the flask remain the same for both gases. 2. The unknown gas behaves as an ideal gas. 3. Avogadro's hypothesis holds true for both gases. Finally, the molar mass of the unknown gas is estimated to be 100.4 g/mol.

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Key Concepts

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

Avogadro's Hypothesis
Avogadro's Hypothesis is an important concept in gas chemistry. It states that equal volumes of gases at the same temperature and pressure contain the same number of particles, or moles. This is true regardless of the kind of gas we're dealing with.

In our exercise, we dealt with two gases in a flask. Since we assume the conditions of temperature, pressure, and volume remain constant for both gases, Avogadro's Hypothesis allows us to equate the number of moles of argon to that of the unknown gas.

Therefore, when we calculated the number of moles of argon, we could directly set it equal to the moles of the unknown gas. This simplifies the solution significantly, since without this principle, we would need additional information about the unknown gas, which we don鈥檛 have.
Ideal Gas Law
The Ideal Gas Law is a fundamental tool in understanding gas behavior. It鈥檚 expressed as PV = nRT, where:
  • P is pressure.
  • V is volume.
  • n is the number of moles.
  • R is the gas constant.
  • T is temperature.
Even though we did not directly use the Ideal Gas Law formula in solving our exercise, the steps rely on its principles.

For both gases in the flask, the conditions鈥攑ressure, volume, and temperature鈥攁re assumed to be constant. This aligns with the assumptions derived from the Ideal Gas Law. By maintaining these conditions, it allows us to assert that both gases behave ideally.

This assumption simplifies calculations and helps us use the molar masses to determine the properties of the gases, such as mass change, similar to how we substitute in the gas law.
Moles Calculation
Calculating the number of moles is crucial for solving gas-related problems. Moles represent the quantity of particles in a given mass of substance. The formula used is:
  • n = \( \frac{\text{mass}}{\text{molar mass}} \)
In our case, this formula was used to find the number of moles of argon gas in the flask.

Given that argon has a known molar mass of 39.95 g/mol, and the mass increase observed was 3.224 g, we simply divided these values to get the moles of argon. This step is essential because it then allows us to determine the moles of the unknown gas using Avogadro's Hypothesis.

Ultimately, understanding moles calculations helps bridge the gap between mass and molar mass, giving us a complete understanding of the gas properties inside the flask.

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Most popular questions from this chapter

Suppose you have two 1 -L flasks, one containing \(\mathrm{N}_{2}\) at STP, the other containing \(\mathrm{CH}_{4}\) at STP. How do these systems compare with respect to (a) number of molecules, (b) density, (c) average kinetic energy of the molecules, (d) rate of effusion through a pinhole leak?

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8},\) liquefies under modest pressure, allowing a large amount to be stored in a container. (a) Calculate the number of moles of propane gas in a \(110-\) L container at 3.00 atm and \(27^{\circ} \mathrm{C}\) (b) Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is 0.590 \(\mathrm{g} / \mathrm{mL}\) . (c) Calculate the ratio of the number of moles of liquid to moles of gas. Discuss this ratio in light of the kinetic-molecular theory of gases.

In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In 1 hr the average cockroach running at 0.08 \(\mathrm{km} / \mathrm{hr}\) consumed 0.8 \(\mathrm{mL}\) of \(\mathrm{O}_{2}\) at 1 atm pressure and \(24^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in 1 hr by a 5.2 -g cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 1 -qt fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, will the cockroach consume more than 20\(\%\) of the available \(\mathrm{O}_{2}\) in a 48 -hr period? (Air is 21 \(\mathrm{mol} \% \mathrm{O}_{2}\) . \()\)

A 4.00 -g sample of a mixture of \(\mathrm{CaO}\) and \(\mathrm{BaO}\) is placed in a 1.00 -L vessel containing \(\mathrm{CO}_{2}\) gas at a pressure of 730 torr and a temperature of \(25^{\circ} \mathrm{C}\) . The \(\mathrm{CO}_{2}\) reacts with the CaO and \(\mathrm{BaO},\) forming \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3} .\) When the reaction is complete, the pressure of the remaining \(\mathrm{CO}_{2}\) is 150 torr. (a) Calculate the number of moles of \(\mathrm{CO}_{2}\) that have reacted. (b) Calculate the mass percentage of CaO in the mixture.

Consider the combustion reaction between 25.0 \(\mathrm{mL}\) of liquid methanol (density \(=0.850 \mathrm{g} / \mathrm{mL} )\) and 12.5 \(\mathrm{L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.
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