Problem 50 - Chemistry: The Central Science

Question

(a) Calculate the density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr.

Step-by-Step Solution

Answer

\(a)\) The density of sulfur hexafluoride (SF6) gas at 707 torr and 21°C is 5.88 g/L. \(b)\) The molar mass of the unknown vapor with a density of 7.135 g/L at 12°C and 743 torr is 172.76 g/mol.

Step 1: Solve part (a) - Calculate the density of SF6

First, we need to calculate the number of moles of SF6 present per unit volume. We rearrange the Ideal Gas Law equation to solve for n/V: \[\frac{n}{V} = \frac{P}{RT}\] Given pressure, P = 707 torr. We will convert this to atm: \(P = \frac{707}{760} = 0.93 \, \mathrm{atm}\). Temperature, T = 21°C. We will convert this to Kelvin: \(T = 21 + 273.15 = 294.15 \, \mathrm{K}\). R = 0.0821 \(\mathrm{L~atm~K^{-1}~mol^{-1}}\). Plugging in these values, we get: \[\frac{n}{V} = \frac{0.93}{0.0821 \times 294.15} = 0.0403 \, \mathrm{mol~L^{-1}}\] To calculate the density of SF6, we multiply the moles per unit volume by its molar mass (M). The molar mass of SF6 (1 sulfur atom = 32.07 g/mol and 6 fluorine atoms = 6 × 19.00 g/mol) is: \[M_\mathrm{SF6} = 32.07 + 6 \times 19.00 = 146.07 \, \mathrm{g/mol}\] Finally, we get the density of SF6: \[\rho_\mathrm{SF6} = 0.0403 \times 146.07 = 5.88 \, \mathrm{g/L}\]

Step 2: Solve part (b) - Calculate the molar mass of the unknown vapor

To find the molar mass of an unknown vapor, we will first find the number of moles per unit volume using the Ideal Gas Law equation as before: \[\frac{n}{V} = \frac{P}{RT}\] Given pressure, P = 743 torr. We will convert this to atm: \(P = \frac{743}{760} = 0.977 \, \mathrm{atm}\). Temperature, T = 12°C. We will convert this to Kelvin: \(T = 12 + 273.15 = 285.15 \, \mathrm{K}\). Plugging in these values, we get: \[\frac{n}{V} = \frac{0.977}{0.0821 \times 285.15} = 0.0413 \, \mathrm{mol~L^{-1}}\] Using given density, \(\rho = 7.135 \, \mathrm{g/L}\). Now, we use the relationship between density, molar mass (M), and the number of moles (n): \[\rho = \frac{nM}{V}\] \[M = \frac{\rho V}{n} = \frac{\rho}{\frac{n}{V}}\] Plugging in the values, we get: \[M = \frac{7.135}{0.0413} = 172.76 \, \mathrm{g/mol}\] The molar mass of the unknown vapor is 172.76 g/mol.

Key Concepts

Density Calculation

Understanding gas density involves knowing how much mass is packed into a certain volume. For a gas, the density \( \rho \) can be calculated using the formula: \( \rho = \frac{PM}{RT} \), where:

• \( P \) is the pressure.
• \( M \) is the molar mass.
• \( R \) is the ideal gas constant.
• \( T \) is the temperature in Kelvin.

By plugging in these values, you can find the gas's density in grams per liter. It's similar to weighing the air! This formula is particularly useful when working with gases like sulfur hexafluoride.

Molar Mass

Molar mass, represented as \( M \), is the mass of one mole of a substance, typically in grams per mole. It bridges the gap between how much an element weighs on its own and how it contributes to a compound. To find the molar mass, you sum the atomic masses of all atoms in a molecule.
For instance, for sulfur hexafluoride (SF6), you add the atomic mass of one sulfur (32.07 g/mol) and six fluorine atoms (6 x 19.00 g/mol), giving you 146.07 g/mol. This value is crucial when calculating gas densities using the ideal gas law.

Sulfur Hexafluoride (SF6)

Sulfur hexafluoride (SF6) is a heavy, inert gas. It's often used as an insulating gas in electrical equipment, thanks to its non-reactive nature and high density. SF6 is composed of one sulfur atom and six fluorine atoms, leading to its chemical formula.
The molecular structure leads to specific properties like a high molar mass of 146.07 g/mol, which affects how we calculate its density under given conditions. Understanding these properties will aid in effectively using the ideal gas law to find its behavior under different pressures and temperatures.

Pressure Conversion

Pressure conversion is an essential step when using the ideal gas law. Pressure can be given in various units, but when dealing with equations like \( PV = nRT \), pressure is often required in atmospheres (atm). To convert from torr (another common unit) to atm, you divide the pressure in torr by 760 (since 1 atm = 760 torr).

• Example: For 707 torr, we convert to 0.93 atm.

This conversion ensures the units are consistent with the gas constant \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \). Consistent units make the math much simpler.

Temperature Conversion

When working with gases, temperature must always be in Kelvin for calculations. Conversion from Celsius to Kelvin is straightforward: add 273.15 to the Celsius temperature.

• Example: 21°C converts to 294.15 K.

Using Kelvin is important because the gas laws rely on absolute temperature measurements. It ensures that all calculations are accurate and based on the absolute scale, which prevents negative temperatures or zero from causing errors in the equations.