Problem 108 - Chemistry The Science in Context

Question

Water and ethanol were removed from the \(\mathrm{O}_{2}\) samples in Problem 6.107 a. What is the volume of the dry \(\mathrm{O}_{2}\) gas sample at \(25^{\circ} \mathrm{C}\) and 1.00 atm? b. What is the volume of the dry \(\mathrm{O}_{2}\) gas sample at \(25^{\circ} \mathrm{C}\) and 1.00 atm if \(P_{\text {ethanol }}=50 \mathrm{mmHg}\) at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Answer: (a) The volume of the dry O₂ gas sample at 25°C and 1.00 atm is 1.53 L. (b) The volume of the dry O₂ gas sample at 25°C and 1.00 atm, taking into account the partial pressure of ethanol, is 1.64 L.

Step 1: Find moles of O₂

In problem 6.107, the moles of O₂ gas is not given. However, we can find the moles since we know the mass and molar mass of O₂. The mass of O₂ is given as 2.00 g and the molar mass of O₂ is 32.00 g/mol. Divide the mass by the molar mass to find the moles of O₂. n = (mass of O₂) / (molar mass of O₂) n = (2.00 g) / (32.00 g/mol) n = 0.0625 mol O₂

Step 2(a): Apply Ideal Gas Law for Part A

In part a, we want to find the volume of O₂ at 25°C and 1.00 atm. Use the ideal gas law, \(PV=nRT.\) Here, P = 1.00 atm, n = 0.0625 mol, R = 0.0821 L atm/mol K and T = 298 K (25°C = 298 K, since K = °C + 273). Solve for V: V = nRT / P V = (0.0625 mol)(0.0821 L atm/mol K)(298 K) / (1.00 atm) V = 1.53 L So, the volume of dry O₂ gas sample at 25°C and 1.00 atm is 1.53 L.

Step 2(b): Adjust pressure for ethanol in Part B

In part b, we need to consider the partial pressure of ethanol. The pressure of ethanol is given as 50 mmHg, which we can convert to atm by dividing by 760: P_ethanol = (50 mmHg) / (760 mmHg/atm) P_ethanol = 0.0658 atm Now we have to subtract the partial pressure of ethanol from the total pressure to find the pressure of O₂: P_O₂ = total pressure - P_ethanol P_O₂ = 1.00 atm - 0.0658 atm P_O₂ = 0.934 atm

Step 3: Apply Ideal Gas Law for Part B

Use the ideal gas law again with the new pressure value for O₂ in part b. Here, P = 0.934 atm, n = 0.0625 mol, R = 0.0821 L atm/mol K, and T = 298 K. Solve for V: V = nRT / P V = (0.0625 mol)(0.0821 L atm/mol K)(298 K) / (0.934 atm) V = 1.64 L So, the volume of dry O₂ gas sample at 25°C and 1.00 atm, considering the partial pressure of ethanol, is 1.64 L.

Key Concepts

Partial Pressure

When it comes to gases, understanding partial pressure is crucial. It's the pressure that a single gas component in a mixture of gases would exert if it alone occupied the whole volume. In the context of our exercise, we dealt with the partial pressure of ethanol. The key takeaway here is that in a mixture of gases, each gas acts independently and contributes to the total pressure. By measuring how much pressure ethanol contributes, we're able to calculate the remaining pressure for the oxygen gas. This is important because the oxygen's volume is affected by its pressure—according to Boyle’s Law—as seen when solving part b of the problem.

The bottom line: partial pressures allow us to understand and calculate the behavior of individual gases in a mixture, even when other gases are present.

Molar Mass

Molar mass is essentially the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It’s a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. We used molar mass to convert the given mass of oxygen to moles, which is a necessity for applying the ideal gas law. Remember, the ideal gas law relates the amount of gas in moles to its volume, temperature, and pressure. Understanding the conversion from grams to moles through molar mass is a fundamental step in gas stoichiometry exercises like ours.

Realize that the molar mass of a molecule like O₂ is simply the sum of the masses of the individual atoms, based on the periodic table. For O₂, each oxygen atom has a molar mass of around 16 g/mol, contributing to the molar mass of approximately 32 g/mol for the molecule.

Gas Stoichiometry

Gas stoichiometry involves calculations that relate the quantities of reactants and products in a chemical reaction involving gases. In our problem, we use these calculations to find out the volume of oxygen at certain conditions. The stoichiometry calculation bridges the gap between moles, volume, and the ideal gas law. In simpler terms, gas stoichiometry requires you to juggle through the concepts of moles (quantity of a substance), molar mass, volume, temperature, and pressure to predict the outcome of reactions involving gases.

The stoichiometric calculations are based on Avogadro’s principle which states that equal volumes of gases, at the same temperature and pressure, contain equal number of molecules. Hence, the volume directly relates to the amount in moles, and that is why using the ideal gas law becomes very handy in solving these problems.

Mole Concept

The mole concept is a connecting thread for the entire field of chemistry, including gas laws. One mole is defined as the amount of substance containing as many entities (atoms, molecules, ions, etc.) as there are atoms in 12 grams of carbon-12. This quantity is Avogadro's number, approximately 6.022 x 10²³. In our exercise, the mole concept is used to translate the mass of oxygen into a countable number of entities (in this case, molecules) which can then be used in the ideal gas law equation.

It’s important to note that every molecule will have its unique molar mass but regardless of what that mass is, one mole of any substance will always contain Avogadro’s number of molecules or atoms—the mole concept ensures that this is a constant fact in chemistry.