Question

A chemist vaporized a liquid compound and determined its density. If the density of the vapor at \(90^{\circ} \mathrm{C}\) and \(753 \mathrm{mmHg}\) is \(1.585 \mathrm{~g} / \mathrm{L},\) what is the molecular weight of the compound?

Short answer

The molecular weight of the compound is approximately 48 g/mol.

Step-by-step solution

Understand the Relationship

We need to find the molecular weight of the compound. The density is given, and it is related to the molecular weight through the ideal gas law. At given conditions, we use the formula involving ideal gas law:\[PV = nRT \rightarrow n = \frac{PV}{RT}\]where \(n\) is the number of moles, \(P\) is pressure, \(V\) is volume, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.

Convert Pressure to Proper Units

Pressure is given in mmHg, and we need to convert it to atm for using the gas constant (\(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\)). 1 atm = 760 mmHg, so: \[P = \frac{753}{760} = 0.9908 \text{ atm}\]

Convert Temperature to Kelvin

Temperature is given in Celsius and needs to be converted to Kelvin. Use the conversion formula:\[T \text{ (in K)} = 90 + 273.15 = 363.15 \text{ K}\]

Relate Density to Molar Mass

Density (\(\rho\)) of the gas can be related to the molecular weight (\(M\)) using the formula derived from the ideal gas law:\[M = \frac{\rho RT}{P}\]where \(\rho = 1.585 \text{ g/L}\), \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\), \(T = 363.15 \text{ K}\), and \(P = 0.9908 \text{ atm}\).

Calculate the Molecular Weight

Substitute the values into the equation:\[M = \frac{1.585 \times 0.0821 \times 363.15}{0.9908}\]Calculating this gives:\[M \approx 48.05 \, \text{g/mol}\]

Round Off the Answer

The molecular weight should be reported with appropriate significant figures, considering the accuracy of the given data. Let's present only two significant figures as in density. Rounding 48.05 gives us 48 g/mol.

Key concepts

Ideal Gas Law

The Ideal Gas Law is one of the fundamental equations in chemistry and helps us describe the behavior of gases under various conditions. It is represented by the formula: \[PV = nRT\]Where:

• \(P\) is the pressure of the gas, typically measured in atmospheres (atm).
• \(V\) is the volume occupied by the gas, often in liters (L).
• \(n\) represents the number of moles of the gas.
• \(R\) is the ideal gas constant, with a value of approximately 0.0821 \(\frac{L \cdot atm}{mol \cdot K}\).
• \(T\) is the temperature in Kelvin (K).

The equation shows how these properties are interrelated. Changes in one variable can affect others, which is crucial for predicting gas behavior. For instance, if pressure increases while temperature and volume remain constant, the number of moles, \(n\), must adjust accordingly. This principle is essential for understanding how gases behave when confined in a container or assessed during experiments.

Density of Gases

Density is an important concept that connects the mass of a gas to its volume, and it is measured in units such as grams per liter (g/L). For gases, density can be derived from the Ideal Gas Law, forming the equation:\[M = \frac{\rho RT}{P}\]Where:

• \(M\) is the molecular weight or molar mass of the gas.
• \(\rho\) represents the density.
• \(R\) is the ideal gas constant.
• \(T\) is the temperature in Kelvin.
• \(P\) is the gas pressure in atmospheres.

This formula arises from the relationship between these factors through the manipulation of the ideal gas equation. By knowing the density of the gas at specific conditions (pressure and temperature), we can compute the molecular weight, thus determining the identity or characteristics of the gas. This is particularly useful in experimental chemistry, where the gas's quality may be assessed by its density.

Temperature and Pressure Conversions

Converting temperature and pressure to the appropriate units is crucial for applying the Ideal Gas Law correctly. Temperature is often provided in degrees Celsius but must be converted to Kelvin for calculations involving the gas constant. This conversion is done using the formula:\[T \text{(K)} = T \text{(°C)} + 273.15\]Similarly, pressure might be measured in millimeters of mercury (mmHg) but needs conversion to atmospheres (atm), as this is the standard unit used in the Ideal Gas Law. The conversion formula is:\[P \text{(atm)} = \frac{P \text{(mmHg)}}{760}\]Understanding these conversions:

• Helps in accurately applying the Ideal Gas Law.
• Ensures consistency in units, which is crucial for the correct computation of the gas properties.
• Allows scientists to compare experimental results achieved under different conditions.

Proper conversion of units is essential in scientific calculations, ensuring that all inputs are harmonious with the constants used.