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Chapter 10: Problem 105

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 \(8.102 \mathrm{~g}\). (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
To find the molar mass of the unknown gas, use the ratio of moles and the given mass increases for both argon and the unknown gas. The equation is: \[M_{unknown} = \frac{mass_{unknown} M_{Ar}}{mass_{Ar}}\] Insert the values (\(mass_{Ar} = 3.224 g\), \(M_{Ar} = 39.95 g/mol\), \(mass_{unknown} = 8.102 g\)) and solve: \[M_{unknown} = \frac{8.102 g \times 39.95 g/mol}{3.224 g} \approx 99.9 g/mol\] We assumed that both gases behave like ideal gases and the pressure, volume, and temperature conditions are the same for both gases in the flask.

Step by step solution

01

Understand the ideal gas law equation

The ideal gas law equation is given by: \[PV = nRT\] where \(P\) is the pressure, \(V\) is the volume, \(n\) is the amount of moles of gas, \(R\) is the ideal gas constant, and \(T\) is the temperature. Since we will be comparing the same flask filled with argon and the unknown gas, \(P\), \(V\) and \(T\) will be constant for both cases.
02

Find the ratio of moles

We have the mass increase when the flask is filled with argon and the unknown gas. We can use the relation between mass, moles, and molar mass to find the moles of each gas. For argon, we have: \[n_{Ar} = \frac{mass_{Ar}}{M_{Ar}}\] For the unknown gas, we have: \[n_{unknown} = \frac{mass_{unknown}}{M_{unknown}}\] In this step we can get the ratio of the moles of the unknown gas to the moles of argon gas. \[\frac{n_{unknown}}{n_{Ar}} = \frac{\frac{mass_{unknown}}{M_{unknown}}}{\frac{mass_{Ar}}{M_{Ar}}}\]
03

Calculate the molar mass of the unknown gas

Now we have: \[\frac{n_{unknown}}{n_{Ar}} = \frac{mass_{unknown} M_{Ar}}{mass_{Ar} M_{unknown}}\] Rearrange the equation to find the molar mass of the unknown gas: \[M_{unknown} = \frac{mass_{unknown} M_{Ar}}{mass_{Ar}}\] We know the mass increase of the flask due to argon gas is \(3.224 g\) and its molarity is \(39.95 g/mol\). In the case of the unknown gas, the mass increase is \(8.102 g\). Insert the values into the equation and solve for the molar mass of the unknown gas.
04

Determine the assumptions made

To arrive at the answer, we made the following assumptions: 1. Both gases behave like ideal gases. Thus, we can apply the ideal gas law. 2. The pressure, volume, and temperature conditions are the same for both gases in the flask. These assumptions are reasonable for most cases, particularly if the experiment is conducted under ordinary conditions and the unknown gas is a well-behaved gas that follows the ideal gas law.

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Key Concepts

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

Molar Mass
Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). This concept is crucial for converting between the mass of a substance and the amount in moles, allowing us to compare different gases based on their mass properties.
The formula for molar mass can be summarized as:
  • Molar mass = (mass of substance in grams) / (amount of substance in moles)
Understanding molar mass helps in calculating how much a flask of gas weighs, whether it's filled with argon or another unknown gas. Since the amount in moles gives us the number of molecules present, molar mass directly ties into identifying gases by comparing their weights under similar conditions.
This principle was used in the exercise to calculate the molar mass of the unknown gas, by leveraging the known molar mass of argon and the respective mass changes observed.
Argon Gas
Argon gas is a noble gas with the atomic number 18 and is known for its inert properties. Its molar mass is approximately 39.95 g/mol. Being colorless, odorless, and tasteless, argon does not react with other elements under normal circumstances, which makes it a perfect candidate for such experiments where reactivity needs to be minimized.
When dealing with gases like argon, we assume they behave ideally under conditions such as room temperature and atmospheric pressure. As an ideal gas, argon follows the ideal gas law formula:
  • The pressure of the gas is directly proportional to the number of moles and the temperature.
  • Any changes in the amount of gas, such as filling and evacuating a flask, can be precisely calculated using its molar mass.
This is why argon is used in the exercise as a baseline to help determine the molar mass of the unknown gas.
Evacuated Flask
An evacuated flask is a sealed container from which all the air has been removed, creating a vacuum. This is essential for accurately measuring the mass and molar mass of gases because it ensures that any mass increase is due solely to the gas introduced into the flask, without the interference of air or other gases.
By creating a vacuum, we remove all variables except for the gas of interest. This allows:
  • Precise determination of the mass of a particular gas by weighing the flask before and after filling it with the gas.
  • Clear comparison of different gases under identical conditions to deduce properties like molar mass, as showcased in the exercise where a blank flask was first filled with argon, and then with the unknown gas.
Hence, an evacuated flask is integral to experiments seeking to measure pure properties of gases without extraneous influences.

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Most popular questions from this chapter

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of \(1.25 \mathrm{~atm}\) at a temperature of \(-6^{\circ} \mathrm{C}\) (b) the absolute temperature of the gas at which \(3.33 \times 10^{-3} \mathrm{~mol}\) occupies \(478 \mathrm{~mL}\) at 750 torr; (c) the pressure, in atmospheres, if \(0.00245 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138^{\circ} \mathrm{C}\); (d) the quantity of gas, in moles, if \(126.5 \mathrm{~L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa}\)

A rigid vessel containing a \(3: 1 \mathrm{~mol}\) ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of \(2.00 \mathrm{~atm}\). If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

Indicate which of the following statements regarding the kinetic-molecular theory of gases are correct. (a) The average kinetic energy of a collection of gas molecules at a given temperature is proportional to \(\mathrm{m}^{1 / 2}\). (b) The gas molecules are assumed to exert no forces on each other. (c) All the molecules of a gas at a given temperature have the same kinetic energy. (d) The volume of the gas molecules is negligible in comparison to the total volume in which the gas is contained. (e) All gas molecules move with the same speed if they are at the same temperature.

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant. ll be produced?

Magnesium can be used as a "getter" in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of \(0.452 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$
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