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.

Short answer

The density of sulfur hexafluoride gas at 707 torr and \(21^{\circ}\) C is approximately 5.961 g/L. The molar mass of the vapor with a density of 7.135 g/L at \(12^{\circ}\) C and 743 torr is approximately 161.67 g/mol.

Step-by-step solution

Write down the given values and convert them to appropriate units

We are given: - The pressure, P = 707 torr - The temperature, T = \(21^{\circ}\) C We need to convert these values into appropriate units: 1. Convert the pressure from torr to atm: \(1 \, \mathrm{atm} = 760 \, \mathrm{torr}\) 2. Convert the temperature from Celsius to Kelvin: \(T(K) = T(^\circ C) + 273.15\)

Convert the given values to appropriate units

1. Convert pressure to atm: \[P = \frac{707 \, \mathrm{torr}}{760 \, \mathrm{torr/atm}} \approx 0.9303 \, \mathrm{atm}\] 2. Convert temperature to Kelvin: \[T = 21 + 273.15 = 294.15 \, \mathrm{K}\]

Rearrange the Ideal Gas Law equation to find density

Recall the Ideal Gas Law equation: \(PV = nRT\) Also, \(n = \frac{m}{M}\) (number of moles = mass/molar mass) Substitute the mass equation into the Ideal Gas Law equation: \[PV = \frac{m}{M}RT\] Now, we can rearrange the equation to find the density (\(\rho = \frac{m}{V}\)): \[\rho = \frac{m}{V} = \frac{M \times P}{R \times T}\]

Calculate the density of sulfur hexafluoride

The molar mass of sulfur hexafluoride (SF\(_6\)) is 146.06 g/mol, and R (gas constant) = 0.0821 atm·L/(mol·K). We can now substitute the values into the equation: \[\rho = \frac{146.06 \, \mathrm{g/mol} \times 0.9303 \, \mathrm{atm}}{0.0821 \, \mathrm{atm \cdot L/(mol \cdot K)} \times 294.15 \, \mathrm{K}} \approx 5.961 \, \mathrm{g/L}\] The density of sulfur hexafluoride gas at 707 torr and \(21^{\circ}\) C is approximately 5.961 g/L. ## Part (b) ##

Write down the given values and convert them to appropriate units

We are given: - The density, \(\rho\) = 7.135 g/L - The temperature, T = \(12^{\circ}\) C - The pressure, P = 743 torr We need to convert these values to appropriate units: 1. Convert the pressure from torr to atm: \(1 \, \mathrm{atm} = 760 \, \mathrm{torr}\) 2. Convert the temperature from Celsius to Kelvin: \(T(K) = T(^\circ C) + 273.15\)

Convert the given values to appropriate units

1. Convert pressure to atm: \[P = \frac{743 \, \mathrm{torr}}{760 \, \mathrm{torr/atm}} \approx 0.9776 \, \mathrm{atm}\] 2. Convert temperature to Kelvin: \[T = 12 + 273.15 = 285.15 \, \mathrm{K}\]

Rearrange the density equation to find molar mass

We have the equation for density in terms of molar mass from Part (a): \[\rho = \frac{M \times P}{R \times T}\] Now, we rearrange the equation to find molar mass (M) in terms of density: \[M = \frac{\rho \times R \times T}{P}\]

Calculate the molar mass of the vapor

Substitute the given values of density, pressure, and temperature along with the gas constant into the molar mass equation: \[M = \frac{7.135 \, \mathrm{g/L} \times 0.0821 \, \mathrm{atm \cdot L/(mol \cdot K)} \times 285.15 \, \mathrm{K}}{0.9776 \, \mathrm{atm}} \approx 161.67 \, \mathrm{g/mol}\] The molar mass of the vapor with given density, temperature, and pressure is approximately 161.67 g/mol.

Key concepts

Ideal Gas Law

Understanding the Ideal Gas Law is crucial for grasping concepts related to gas properties. At its core, this law combines several gas laws into one formula: \( PV = nRT \). It provides the relationship between pressure (P), volume (V), number of moles (n), temperature (T), and the ideal gas constant (R). In simple terms, this law conveys how gases tend to expand when heated or contract when cooled, and how they exert more pressure when compressed into a smaller volume.

For students, it is important to remember that before applying the Ideal Gas Law, conditions have to be expressed in their appropriate units: pressure in atmospheres (atm), volume in liters (L), temperature in Kelvin (K), and the number of moles as a dimensionless quantity. The gas constant, R, also has a fixed value which depends on the units being used, but commonly it is \(0.0821 \text{ atm}\text{·L/mol·K}\).

When dealing with exercises, always start by converting the given values into these units to ensure that the law is applied correctly. This step is essential and will set the foundation for solving for any unknown variable in the equation, be it pressure, volume, temperature, moles, or even the density or molar mass of a gas when combined with other equations.

Molar Mass

Molar mass is a fundamental concept in chemistry, defined as the mass of one mole of a substance, typically in grams per mole (g/mol). It is the bridge between the macroscopic world of grams that we can measure and the microscopic world of molecules and atoms counted in moles. Every substance has a unique molar mass, directly related to its molecular weight, which is the sum of the atomic weights of all atoms in a molecule.

For students tackling gas density problems, understanding molar mass allows them to convert between the mass of a gas and the number of moles present, using the relationship \(n = \frac{m}{M}\), where 'n' is the number of moles, 'm' is the mass, and 'M' is the molar mass. This becomes particularly useful when combined with the Ideal Gas Law to solve for unknown variables. A key point to remember is that molar mass values can be found on the periodic table as the atomic weight of the element and must be summed accordingly for molecular substances.

Unit Conversion

Unit conversion is a fundamental skill in all scientific calculations. It allows us to convert measurements from one unit to another so that they fit into the equations we use. In the context of gas laws, two common conversions are required: from degrees Celsius to Kelvin for temperature, and from torr to atmospheres for pressure.

To convert temperature from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For pressure, the conversion factor between torr and atmospheres is based on the equivalence \(1 \text{ atm} = 760 \text{ torr}\). Thus, to convert torr to atm, divide the pressure in torr by 760. These conversions are crucial because the Ideal Gas Law requires that pressure is in atm and temperature is in Kelvin.

Students should always perform these conversions first before applying the gas laws. This approach minimizes errors and ensures consistency in their calculations. Additionally, whenever performing these conversions, it's important to pay attention to significant figures and rounding to maintain the precision of measurements.

Density of Gases

The density of a gas is defined as its mass per unit volume, typically expressed in grams per liter (g/L). Unlike the density of solids and liquids, gas density is highly dependent on temperature and pressure, which makes the Ideal Gas Law an invaluable tool for calculating it.

By manipulating the Ideal Gas Law, we obtain the formula \(\rho = \frac{MP}{RT}\), where \(\rho\) represents density, M is molar mass, P is pressure, R is the gas constant, and T is temperature. This equation shows that as pressure increases or temperature decreases, the density of a gas increases, since gas molecules are either brought closer together or move more slowly, respectively.

For students, it's important to understand that despite gases often being less dense than solids or liquids, their densities still follow predictable patterns based on the Ideal Gas Law. The ability to calculate gas density not only helps in solving textbook problems but also in practical applications like determining the lift of a hot air balloon or the behavior of the atmosphere. Remember, accurate unit conversion is vital for these calculations to yield correct results.