Question

The vapor pressure of acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right)\) at \(19^{\circ} \mathrm{C}\) is 5.33 kPa. A 3.50-g sample of acetone is placed in a \(2.00 \mathrm{~L}\) container at \(22^{\circ} \mathrm{C}\). If the container is closed and the acetone is allowed to reach equilibrium with its vapor, how many grams of liquid acetone remain? (Assume that the vapor behaves like an ideal gas.)

Short answer

The short answer: First, we convert the given mass of acetone (3.50 g) into moles using its molar mass (58.08 g/mol). Second, we use the Ideal Gas Law (PV=nRT) to find the number of moles of acetone vapor at equilibrium using the volume of the container (2.00 L), the temperature (22°C converted to 295.15 K), and the vapor pressure at 22°C (approximately 5.76 kPa). Next, we convert the moles of acetone vapor back to mass using the molar mass of acetone. Finally, we subtract the mass of acetone vapor from the initial mass of acetone (3.50 g) to find the mass of liquid acetone remaining in the container.

Step-by-step solution

Convert the given mass of acetone into moles

Using the molar mass of acetone (58.08 g/mol), we can convert the given mass of acetone (3.50 g) into moles: moles of acetone = (mass of acetone) / (molar mass of acetone) = (3.50 g) / (58.08 g/mol)

Calculate the volume of the gas in equilibrium

Since the volume of the container is given as 2.00 L, we have the volume where the liquid acetone and the vapor are in equilibrium.

Find the number of moles of acetone vapor at equilibrium

Using the Ideal Gas Law, PV=nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin. First, convert the temperature from Celsius to Kelvin: T = 22°C + 273.15 = 295.15 K Then, rearrange the Ideal Gas Law to solve for the number of moles (n): n = PV/RT The vapor pressure at 19°C is given as 5.33 kPa, but we need the vapor pressure at 22°C. Assuming that the relationship is linear, for an increase of 1°C, the vapor pressure rises approximately 1% (this can vary, a better approach is to find and use the relationship between vapor pressure and temperature from a table or empirical equation). With this assumption, the vapor pressure at 22°C is: P = 5.33 kPa * (1.03 ^ (22-19)) ≈ 5.76 kPa Now we can find the number of moles of acetone vapor in equilibrium: n = (5.76 kPa)(2.00 L) / (8.314 J/mol·K)(295.15 K)

Find the mass of acetone vapor and subtract from the initial mass of acetone

Convert the moles of acetone vapor back to mass using the molar mass: mass of acetone vapor = (moles of acetone vapor) * (molar mass of acetone) Finally, find the mass of liquid acetone remaining: mass of liquid acetone = initial mass of acetone - mass of acetone vapor With these steps, we should now be able to find the mass of liquid acetone remaining in the container after reaching equilibrium with its vapor.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that provides a useful relationship between pressure, volume, temperature, and the amount of gas in moles. The law is expressed as \( PV = nRT \), where

• \( P \) is the pressure of the gas,
• \( V \) is the volume it occupies,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant, with a value of 8.314 J/mol·K,
• \( T \) is the temperature in Kelvin.

This equation assumes that the gas behaves ideally, which means that the interactions between gas molecules are negligible, and the volume occupied by gas molecules themselves is very small compared to the container's volume.
In our exercise, the Ideal Gas Law helps to determine the number of moles of acetone vapor present when a closed system reaches equilibrium.

Equilibrium

Equilibrium in chemistry refers to the state in which the rate of the forward reaction equals the rate of the reverse reaction. In the context of vapor pressure within a closed container, equilibrium is achieved when the rate of evaporation of the liquid equals the rate of condensation of the vapor.
This means that the amount of liquid acetone vaporizing and the vapor condensing back into liquid is balanced, and thus the overall amount of acetone in either state remains constant. At equilibrium, the vapor pressure becomes stable and can be used to calculate the amount of substance in each phase.
In the container from the exercise, achieving equilibrium means that the acetone vapor pressure equals the calculated pressure at 22°C, allowing us to use further steps to find out how much liquid acetone remains.

Molar Mass

Molar mass is the mass of one mole of a substance. It serves as a bridge between the atomic scale and the macroscopic scale, as it allows us to convert between the mass of a compound and the amount in moles.
Molar mass is expressed in grams per mole (g/mol). For acetone, \(\( \mathrm{CH_3COCH_3} \),\) the molar mass is 58.08 g/mol. This value is calculated by adding up the atomic masses of each element within the molecule: three hydrogens, one carbon, one oxygen, and again three hydrogens.
In our exercise, the molar mass is essential for converting the mass of the acetone sample (3.50 grams) into moles, which is a necessary step before applying the Ideal Gas Law to find the behavior of acetone vapor in equilibrium.

Temperature Conversion

Temperature conversion is a fundamental skill in scientific calculations, necessary for utilizing equations such as the Ideal Gas Law that require temperature inputs in Kelvin rather than Celsius.
To convert Celsius to Kelvin, a straightforward method is adding 273.15 to the Celsius temperature. So, for instance, converting 22°C to Kelvin involves simply calculating 22 + 273.15, resulting in 295.15 Kelvin.
This conversion ensures that temperature-based calculations remain consistent with the absolute temperature scale required in thermodynamics, where calculations can be applied universally regardless of the process at hand. In our exercise, converting the given temperature to Kelvin facilitates proper use of the Ideal Gas Law to calculate the number of moles in the acetone vapor.