Question

A You have a \(550 .\) -mL. tank of gas with a pressure of 1.56 atm at \(24^{\circ} \mathrm{C} .\) You thought the gas was pure carbon monoxide gas, CO, but you later found it was contaminated by small quantities of gaseous \(\mathrm{CO}_{2}\) and \(\mathrm{O}_{2}\). Analysis shows that the tank pressure is 1.34 atm (at \(24^{\circ} \mathrm{C}\) ) if the \(\mathrm{CO}_{2}\) is removed. Another experiment shows that \(0.0870 \mathrm{g}\) of \(\mathrm{O}_{2}\) can be removed chemically. What are the masses of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) in the tank, and what is the partial pressure of each of the three gases at \(25^{\circ} \mathrm{C} ?\)

Short answer

Mass of CO is 0.756g, CO2 is 0.220g; partial pressures at 25°C: CO is 1.206 atm, CO2 is 0.22 atm, O2 is 0.134 atm.

Step-by-step solution

Identify Known Values

We are given: 1. Total volume of gas in the tank: 550 mL 2. Total initial pressure: 1.56 atm when all gases are present. 3. Pressure with CO removed: 1.34 atm. 4. Mass of O2 removed: 0.0870 g. 5. Temperature: 24°C (or 297 K when converted to Kelvin). We need to find: The masses of CO and CO2 in the tank and the partial pressures of CO, CO2, and O2 at 25°C.

Calculate Moles of Oxygen

We first calculate the number of moles of O2 using the formula: \[\text{moles of } O_2 = \frac{\text{mass}}{\text{molar mass}}\]Given the mass of O2 is 0.0870 g and its molar mass is 32.00 g/mol, we have:\[\text{moles of } O_2 = \frac{0.0870 \, \text{g}}{32.00 \, \text{g/mol}} = 0.00272 \, \text{mol}\]

Use Ideal Gas Law for Partial Pressure of O2

The partial pressure of the remaining O2 can be calculated using the ideal gas law:\[\text{PV} = nRT\]Using \(R = 0.0821 \, \text{L atm/mol K}\) and remembering to convert volume to liters (0.550 L), we find:\[P(O_2) = \frac{nRT}{V} = \frac{(0.00272 \, \text{mol})(0.0821 \, \text{L atm/mol K})(297 \, \text{K})}{0.550 \, \text{L}}\]Calculating gives \(P(O_2) = 0.134 \, \text{atm}\).

Calculate Partial Pressures of CO and CO2

The partial pressure of CO is the pressure after removing CO2, minus the partial pressure of O2:\[P(CO) = 1.34 \, \text{atm} - 0.134 \, \text{atm} = 1.206 \, \text{atm}\]Similarly, the partial pressure of CO2 can be found from total initial pressure:\[P(CO_2) = 1.56 \, \text{atm} - 1.34 \, \text{atm} = 0.22 \, \text{atm}\]

Calculate Moles and Masses of CO and CO2

Now we use the ideal gas law to find the moles of CO and CO2. Using the volume of \(0.550 \, \text{L}\) and temperature \(297 \, \text{K}\):- For CO: \[\text{moles of } CO = \frac{P(CO) \cdot V}{RT} = \frac{1.206 \, \text{atm} \cdot 0.55 \, \text{L}}{0.0821 \, \text{L atm/mol K} \cdot 297 \, \text{K}} \approx 0.027 \, \text{mol}\]- For CO2:\[\text{moles of } CO_2 = \frac{P(CO_2) \cdot V}{RT} = \frac{0.22 \, \text{atm} \cdot 0.55 \, \text{L}}{0.0821 \, \text{L atm/mol K} \cdot 297 \, \text{K}} \approx 0.005 \, \text{mol}\]Convert these moles back to mass:- Mass of CO = \(0.027 \, \text{mol} \times 28.01 \, \text{g/mol} = 0.756 \, \text{g}\)- Mass of CO2 = \(0.005 \, \text{mol} \times 44.01 \, \text{g/mol} = 0.220 \, \text{g}\).

Key concepts

Partial Pressure

Partial pressure is a crucial concept in understanding gas mixtures. It refers to the pressure that each gas in a mixture would exert if it were alone in the volume of the container. This is essential because in a mixture, each gas acts independently, contributing to the total pressure.

In the context of this problem, we used Dalton's Law of Partial Pressures to figure out the individual contributions of CO, CO2, and O2 to the total pressure. Dalton's Law states that the total pressure of a gas mixture is the sum of the partial pressures of each individual component.

• Partial Pressure of O2: 0.134 atm, found using the ideal gas law with known moles of O2.
• Partial Pressure of CO: Calculated by subtracting the partial pressure of O2 from the pressure after CO2 is removed.
• Partial Pressure of CO2: Determined by subtracting the pressure after CO is removed from the total initial pressure.

Molar Mass

Molar mass is the mass of one mole of a given substance, usually expressed in grams per mole. It is derived from the atomic weights of the elements present in a compound.

In chemical calculations, molar mass is a pivotal figure. For example, knowing the molar mass of O2 (32.00 g/mol), CO (28.01 g/mol), and CO2 (44.01 g/mol) allows us to convert between mass and moles for these gases.

• For oxygen, the molar mass allowed us to compute the moles from the given mass using the formula: \[\text{moles of } O_2 = \frac{0.0870 \, \text{g}}{32.00 \, \text{g/mol}} \approx 0.00272 \, \text{mol}\]
• Once moles of CO and CO2 were calculated using the ideal gas law, their respective molar masses were used to find their masses by multiplying by the molar mass.

Moles

The concept of moles is fundamental in chemistry, representing a quantity of substance that contains Avogadro's number of molecules, atoms, or ions. It allows chemists to count particles by weighing them.

In our problem, moles were calculated both for pure oxygen removed from the mixture and for the components CO and CO2 that remained.

• We converted the given mass of O2 into moles. This was crucial since the calculation of partial pressures requires the number of moles as an input to the ideal gas law.
• Moles of CO and CO2 were determined using their respective partial pressures, volume of the container, and the universal gas constant in the ideal gas equation, \(PV = nRT\). This gave us insights into their respective quantities in the gas mixture.

Chemical Analysis

Chemical analysis involves the study of the composition of a material, which in our exercise means identifying and quantifying different gaseous components present in a tank.

In the given scenario, chemical analysis was used to assess the contents of the gas tank. It included removing components, such as O2 and CO2, to measure the resulting pressures and masses of individual gases.

• The initial identification of CO as the main gas was revised post-analysis when removing CO2 showed a change in pressure.
• The weight of O2 allowed calculating the moles present, further helping in determining its contribution to the overall gas pressure when compared to the other gases.
• Understanding chemical analysis helped realize each gas's contribution to pressure and mass, strengthening descriptive understanding through quantitative evaluation.