Problem 52 - Chemistry Central Science

Question

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise \(10.51 .\) The unknown vapor had a mass of \(0.846 \mathrm{~g} ;\) the volume of the bulb was \(354 \mathrm{~cm}^{3}\), pressure 752 torr, and temperature \(100^{\circ} \mathrm{C}\). Calculate the molar mass of the unknown vapor.

Step-by-Step Solution

Answer

The molar mass of the unknown vapor is approximately \(24.1 \frac{g}{mol}\).

Step 1: Convert Units

First, we need to convert the given units into the SI Units. Convert temperature from Celsius to Kelvin, pressure from torr to Pascals, and volume from cm³ to m³. Temperature: \[T = 100^{\circ} C + 273.15 = 373.15 K\] Pressure: \[P = 752 torr \times \frac{101325 Pa}{760 torr} = 101225 Pa\] Volume: \[V = 354 cm^3 \times \frac{1 m^3}{10^6 cm^3} = 3.54 \times 10^{-4} m^3\] Mass: \[m = 0.846 g\]

Step 2: Apply the Ideal Gas Law

The Ideal Gas Law formula is given as \(PV = nRT\). We will rearrange this formula to solve for the number of moles (n), by dividing both sides by RT: \[n = \frac{PV}{RT}\] Substitute the given values and solve for n: \[n = \frac{(101225 Pa)(3.54 \times 10^{-4} m^3)}{(8.314 J/mol \cdot K)(373.15 K)}\]

Step 3: Calculate the Number of Moles (n)

By calculating the number of moles (n): \[n = 0.03503 mol\]

Step 4: Calculate the Molar Mass of the Unknown Vapor

Now that we have the number of moles (n) and the mass (m) of the substance, we can calculate the molar mass (M) using the formula: \[M = \frac{m}{n}\] Substitute the given values: \[M = \frac{0.846 g}{0.03503 mol}\]

Step 5: Determine the Molar Mass of the Unknown Vapor

By determining the molar mass of the unknown vapor: \[M = 24.1 \frac{g}{mol}\] The molar mass of the unknown vapor is approximately \(24.1 \frac{g}{mol}\).

Key Concepts

Molar Mass Calculation

The concept of molar mass is central to understanding the composition of substances. It tells us how much one mole of a substance weighs. The Dumas-bulb method helps in determining the molar mass of volatile substances by using the mass, temperature, pressure, and volume of their vapors. To calculate molar mass, the formula is \[M = \frac{m}{n}\]where:

• \(M\) is the molar mass.
• \(m\) is the mass of the substance (in grams).
• \(n\) is the number of moles.

In the given exercise, we calculated the molar mass after finding the number of moles using the Ideal Gas Law. This direct approach offers a simple yet powerful way to understand the molar mass in practical scenarios.
The calculated molar mass was found to be approximately \(24.1 \frac{g}{mol}\). This value helps in identifying the unknown vapor.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that relates pressure, volume, temperature, and number of moles of a gas. It is expressed as:\[PV = nRT\]where:

• \(P\) is the pressure of the gas in Pascals (Pa).
• \(V\) is the volume in cubic meters (m³).
• \(n\) is the number of moles.
• \(R\) is the ideal gas constant, \(8.314 \frac{J}{mol \cdot K}\).
• \(T\) is the temperature in Kelvin (K).

By rearranging this equation to \(n = \frac{PV}{RT}\), we can solve for the number of moles. This is precisely what was done in the exercise, using the conditions of the vapor to determine the amount of substance present. The Ideal Gas Law provides a comprehensive way to analyze gases under various conditions, giving insights into their behavior.

Unit Conversion

Unit conversion is an essential skill in chemistry, especially when using formulas like the Ideal Gas Law, which require consistent units. In many exercises, like the one at hand, converting units ensures accuracy.
Here's a quick rundown of key conversions used in this exercise:

• **Temperature:** Celsius to Kelvin by adding 273.15.
• **Pressure:** Torr to Pascals using the conversion factor \(\frac{101325 Pa}{760 torr}\).
• **Volume:** Cubic centimeters to cubic meters using \(\frac{1 m^3}{10^6 cm^3}\).

Accurate conversions create a seamless application of the Ideal Gas Law, ensuring that measurements correlate properly. This harmonization across measurements makes subsequent calculations trustworthy and meaningful.

Chemical Vapor Analysis

Chemical vapor analysis is an intriguing method used to understand unknown substances, leveraging techniques like the Dumas-bulb method. By analyzing the vapor phase of a substance, we can deduce its molar mass, composition, and often identity.
The Dumas-bulb method specifically measures the vapor mass, volume, temperature, and pressure to explore the properties of volatile substances. This approach is valued for its simplicity and effectiveness, especially useful when dealing with substances that are difficult to weigh directly due to their volatility.
In this exercise, chemical vapor analysis was crucial in calculating the molar mass and identifying the unknown vapor. Such analyses are a cornerstone in laboratory techniques, assisting in the discovery and understanding of new materials and compounds.