Question

You vaporize a liquid substance at \(100^{\circ} \mathrm{C}\) and \(755 \mathrm{mmHg}\). The volume of \(0.548 \mathrm{~g}\) of vapor is \(237 \mathrm{~mL}\). What is the molecular weight of the substance?

Short answer

The molecular weight of the substance is approximately 44.01 g/mol.

Step-by-step solution

Identify the Given Variables

We are given the mass of the substance, which is 0.548 g, the volume of the vapor is 237 mL, the temperature is 100°C, and the pressure is 755 mmHg.

Convert Units Where Necessary

Convert the volume from mL to L since we usually need it in liters for gas calculations: \[ V = 237 \, \text{mL} = 0.237 \, \text{L} \]Convert pressure from mmHg to atm, knowing that 1 atm = 760 mmHg: \[ P = \frac{755 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.993 \text{atm} \]

Apply the Ideal Gas Law

Use the Ideal Gas Law formula \( PV = nRT \), where: - \( P \) is the pressure in atm,- \( V \) is the volume in liters,- \( n \) is the number of moles,- \( R \) is the ideal gas constant (0.0821 L·atm/mol·K),- \( T \) is the temperature in Kelvin.Convert the temperature to Kelvin:\[ T = 100^{\circ}C + 273.15 = 373.15 \, K \].

Solve for Number of Moles

Rearrange the Ideal Gas Law to solve for n: \[ n = \frac{PV}{RT} \].Substitute in the known values:\[ n = \frac{(0.993 \, \text{atm})(0.237 \, \text{L})}{(0.0821 \, \text{L}·\text{atm/mol}·\text{K})(373.15 \, K)} \].Calculate \( n \) from this expression.

Calculate the Molecular Weight

Calculate the molecular weight (M) using the formula \( M = \frac{\text{mass}}{n} \), where the mass is 0.548 g.Substitute \( n \) from the previous step into the expression:\[ M = \frac{0.548 \, \text{g}}{n} \].Calculate the molecular weight.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, and temperature of a gas to the number of moles of the gas present. It's given by the formula \( PV = nRT \), where:

• \( P \) is the pressure of the gas in atmospheres (atm),
• \( V \) is the volume of the gas in liters (L),
• \( n \) is the number of moles of the gas,
• \( R \) is the ideal gas constant, typically \( 0.0821 \, \text{L} \cdot \text{atm/mol} \cdot \text{K} \),
• \( T \) is the temperature of the gas in Kelvin (K).

This law assumes that gases behave ideally, meaning there are no interactions between the molecules, and the size of the molecules themselves is negligible. While not perfectly accurate, this approximation works well for many gases under common conditions. By rearranging the equation to get \( n = \frac{PV}{RT} \), we can solve for the number of moles in a gas sample, as shown in the exercise.

Unit Conversion

Unit conversion is a key step in solving problems involving the Ideal Gas Law and other scientific calculations. Accurate conversions ensure that all measurements align correctly within a formula.
Let's look at the necessary conversions in the exercise:

• Volume: Convert from milliliters (mL) to liters (L) because the Ideal Gas Law requires volume in liters. Simply divide the volume in mL by 1000. So, \( 237 \, \text{mL} = 0.237 \, \text{L} \).
• Pressure: Convert from millimeters of mercury (mmHg) to atmospheres (atm). As 1 atm is equivalent to 760 mmHg, use the conversion \( P = \frac{755 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.993 \, \text{atm} \).
• Temperature: Convert temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature. For example, \( T = 100^{\circ}C + 273.15 = 373.15 \, K \).

Correct unit conversion is crucial to apply the Ideal Gas Law accurately.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. In calculating molecular weight using the Ideal Gas Law, vapor pressure might seem unrelated, but understanding pressure is key.
In the context of the problem statement, the vapor pressure tied to "755 mmHg" refers to the atmospheric pressure under which the vaporization occurred. Knowing this pressure allows us to convert it into atm and use it directly in the Ideal Gas Law.
Vapor pressure helps in determining how volatile a substance is, and it can vary with temperature. No additional vapor pressure adjustments were necessary for this scenario because it directly influenced the Ideal Gas Law calculation under given conditions.
When dealing with gas laws, remember that the ambient pressure combines with any vapor pressure from a substance still transitioning between phases, influencing how pressure is measured and used.

Mole Calculation

Mole calculation is essential for determining the molecular weight of a substance using the Ideal Gas Law. The number of moles links mass and volume to a definitive molecular count, impacting molecular weight determination.
Using the Ideal Gas Law, we first calculated the number of moles (\( n \)) through rearranging to \( n = \frac{PV}{RT} \). By substituting the pressure (0.993 atm), volume (0.237 L), temperature (373.15 K), and the gas constant (0.0821 L·atm/mol·K), we find the mole amount in the vapor sample.
Here's how molecular weight computations connect:

• Mass of the substance is known, given as 0.548 g.
• Mole calculation of the gas gives us a clue about the quantity in terms of fundamental entities (moles).
• Finally, the molecular weight \( M \) is calculated using \( M = \frac{\text{mass}}{n} \).

This result gives an insight into the substance's molar mass, derived from observations and computations connecting macroscopic measurements to molecular structure.