Question

Suppose you are given two 1 -L flasks and told that one contains a gas of molar mass \(30,\) the other a gas of molar mass 60 , both at the same temperature. The pressure in flask \(\mathrm{A}\) is \(\mathrm{X}\) atm, and the mass of gas in the flask is \(1.2 \mathrm{~g} .\) The pressure in flask \(\mathrm{B}\) is \(0.5 \mathrm{X}\) atm, and the mass of gas in that flask is \(1.2 \mathrm{~g} .\) Which flask contains the gas of molar mass \(30,\) and which contains the gas of molar mass \(60 ?\)

Short answer

The gas in Flask A has a molar mass of 30, while the gas in Flask B has a molar mass of 60.

Step-by-step solution

Recognize the given information

The given information is: 1. Flask A has a gas with pressure X atm and mass 1.2 g. 2. Flask B has a gas with pressure 0.5X atm and mass 1.2 g. 3. Both flasks have the same volume (1 L) and temperature.

Calculate the number of moles in each flask using the Ideal Gas Law

We can rewrite the Ideal Gas Law equation to express the number of moles (n) as follows: n = \(\frac{PV}{RT}\) Since the problem does not explicitly provide the temperature and ideal gas constant, we will use the ratios of moles instead of their actual values. For Flask A (nA): nA = \(\frac{P_A \times V_A}{R \times T}\) For Flask B (nB): nB = \(\frac{P_B \times V_B}{R \times T}\) Now we can find the ratio of moles between the two flasks: \(\frac{n_A}{n_B}\) = \(\frac{P_A \times V_A}{P_B \times V_B}\)

Substitute the given values into the ratio equation

Using the given pressure and volume values, plug them into the ratio equation: \(\frac{n_A}{n_B}\) = \(\frac{X \times 1~L}{0.5X \times 1~L}\) = \(\frac{X}{0.5X}\) = 2 This tells us that there are twice as many moles of gas in Flask A compared to Flask B.

Calculate and compare molar masses

We can determine the molar mass (MM) of each gas using the formula: Molar Mass (MM) = \(\frac{Mass}{Moles}\) For Flask A: MM_A = \(\frac{1.2~g}{2n_B}\) For Flask B: MM_B = \(\frac{1.2~g}{n_B}\) Comparing both Molar Masses: \(\frac{MM_A}{MM_B}\) = \(\frac{1.2~g / 2n_B}{1.2~g / n_B}\) = 0.5 This tells us that the molar mass of the gas in Flask A is half the molar mass of the gas in Flask B.

Determine which flask contains which gas

Based on the results from Step 4, we can conclude that Flask A contains the gas with a lower molar mass (30) and Flask B contains the gas with a higher molar mass (60).

Key concepts

Molar Mass Calculation

Molar mass calculation is the process of determining the mass of one mole of a substance, typically expressed in grams per mole (g/mol). The molar mass of a substance is numerically equivalent to its atomic or molecular weight in unified atomic mass units (u).
To calculate the molar mass, you take the mass of a sample and divide it by the number of moles of the substance in that sample. This can be represented by the equation:\[ Molar\,Mass(MM) = \frac{Mass}{Moles} \]When dealing with gases, the molar mass becomes particularly useful because it allows us to use the ideal gas law to relate the physical properties of the gas – such as pressure, volume, and temperature – to the amount of substance.
In the case of the exercise involving two flasks containing gases of different molar masses, we used the ideal gas law to infer the number of moles from pressures. With a fixed mass of gas in both flasks, the gas with fewer moles must have a larger molar mass, allowing us to determine which flask contains which gas. This aligns with the core concept that the molar mass is a property intrinsic to a specific chemical species and independent of its physical state. Moreover, understanding molar mass calculations enables students to fundamentally grasp the stoichiometric relationships in chemical reactions.

Practical Tip:

When solving problems involving molar mass, always ensure that you're working with consistent units (e.g., convert milligrams to grams if necessary) and check your substance's molecular formula to account for all constituent atoms correctly.

Gas Pressure Comparison

Gas pressure comparison is an important concept when investigating gases in different conditions, as pressure can directly affect the behavior and properties of a gas within a container.
Pressure (\( P \)), according to the ideal gas law, is directly proportional to the number of moles (\( n \)) of a gas, when temperature (\( T \)) and volume (\( V \)) are held constant.In the provided exercise, the pressure in one flask is double that in the other (\( X \) to \( 0.5X \) atm). This tells us that, if temperature and volume remain constant, the number of moles in the higher-pressure flask (Flask A) must also be higher. By comparing gas pressures, we can gain insights into the relative amounts of substance in each flask.

Exercise Application:

In applying this to our flask problem, when we determined there were twice as many moles in Flask A due to its pressure being double that of Flask B's, it allowed us to compare molar masses and deduce which gas was in which flask.When comparing gas pressures, it's also important to understand units and conversions (e.g., converting between atmospheres, pascals, and torr) to accurately communicate and compare different systems.

Stoichiometry

Stoichiometry is a division of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction.
This branch of chemistry allows us to predict the amounts of substances consumed and produced in reactions, based on the principle of conservation of mass and the known stoichiometric coefficients from the balanced chemical equation.In practice, stoichiometry requires a solid understanding of molar relationships, because reactions occur on a mole-to-mole basis. To solve problems involving stoichiometry, one typically follows these steps:

• Write and balance the chemical equation.
• Convert known masses to moles using molar mass.
• Use the balanced equation to set up mole ratios.
• Convert moles back to grams if the problem asks for masses of products or reactants.

While the textbook exercise did not directly involve a chemical reaction, the stoichiometric principles applied in determining the molar mass reflect the fundamental stoichiometric concept of using mole ratios. By understanding the stoichiometry involved in gas law calculations, students can enhance their ability to predict and quantify the outcomes of chemical reactions.