Problem 53 - Chemistry The Central Science In Si Units

Question

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below \(100^{\circ} \mathrm{C}\) in a boiling-water bath and determine the mass of vapor required to fill the bulb. From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, \(1.012 \mathrm{~g} ;\) volume of bulb, \(354 \mathrm{~cm}^{3}\); pressure, \(98.93 \mathrm{kPa} ;\) temperature, \(99^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The molar mass of the unknown liquid is approximately 88.96 g/mol.

Step 1: Convert Units as Necessary

First, convert the given temperature to Kelvin. Since the temperature is given as \(99^{\circ}\ \mathrm{C}\), we need to convert it to Kelvin by adding 273.15. This gives us \(T = 99^{\circ}\ \mathrm{C} + 273.15 = 372.15\ \mathrm{K}\). Next, convert the volume from \(\mathrm{cm}^{3}\) to liters by dividing by 1000. So, \(V = \frac{354}{1000} = 0.354\ \mathrm{L}\).

Step 2: Use the Ideal Gas Law

The ideal gas law equation is \(PV = nRT\), where \(P\) is the pressure in \(\mathrm{kPa}\), \(V\) is the volume in liters, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant (which is \(8.314\ \mathrm{L}\cdot\mathrm{kPa}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1}\)), and \(T\) is temperature in Kelvin. Rearrange the ideal gas law to solve for \(n\): \[ n = \frac{PV}{RT} \].

Step 3: Calculate Moles of Gas

Substitute the converted values into the ideal gas law equation: \[ n = \frac{98.93\ \mathrm{kPa} \times 0.354\ \mathrm{L}}{8.314\ \mathrm{L}\cdot\mathrm{kPa}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1} \times 372.15\ \mathrm{K}} \]. This yields \(n \approx 0.011377\ \mathrm{mol}\).

Step 4: Calculate Molar Mass

The molar mass \(M\) is calculated using the formula \(M = \frac{m}{n}\), where \(m\) is the mass of the vapor (1.012 g) and \(n\) is the number of moles. Substitute the values to find \[ M = \frac{1.012\ \mathrm{g}}{0.011377\ \mathrm{mol}} \approx 88.96\ \mathrm{g/mol} \].

Key Concepts

Molar Mass Determination

Determining the molar mass of a substance allows us to understand how heavy one mole of that substance is. In the Dumas-bulb technique, we weigh the vaporized sample when it fills a bulb. This measurement is crucial because it reveals the mass of gas occupying a known volume under specified conditions of temperature and pressure. When we know both the mass of the vapor and the number of moles calculated using the ideal gas law, we can find the molar mass. The formula used is \[ M = \frac{m}{n} \]where - \(M\) is the molar mass, - \(m\) is the mass of the vapor, and - \(n\) is the number of moles.
By dividing the mass of the vapor by how many moles we have calculated, we receive the molar mass in grams per mole \(\mathrm{g/mol}\). This technique is accurate and commonly used for liquids that vaporize easily at lower boiling points.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas. The equation is written as \[ PV = nRT \]where - \(P\) is the pressure, - \(V\) is the volume, - \(n\) is the number of moles, - \(R\) is the ideal gas constant \(8.314 \ \mathrm{L \cdot kPa \cdot mol^{-1} \cdot K^{-1}}\), and - \(T\) is the temperature in Kelvin.
This equation allows you to solve for any one of these variables as long as the others are known. For our problem, we rearranged the equation to find the number of moles \(n\):\[ n = \frac{PV}{RT} \]. This calculation is pivotal because it allows us to connect the real and measurable properties (pressure, volume, and temperature) with the mole concept, which is crucial for determining molar mass.

Unit Conversion

In chemistry, correct and consistent units are essential for ensuring accurate calculations. Before using the ideal gas law, it's crucial to convert all measurements to the right units.
Temperature needs to be in Kelvin, which is done by adding 273.15 to Celsius measurements. For example, \(99^{\circ}\ \mathrm{C}\) becomes \(372.15\ \mathrm{K}\).- Volume should be in liters. Since you often deal with milliliters or cubic centimeters, divide by 1000 to convert \(\mathrm{cm}^3\) to liters. In our case, \(354\ \mathrm{cm}^3\) converts to \(0.354\ \mathrm{L}\). - Pressure is often given in \(\mathrm{kPa}\), which matches the units in our gas constant \(R\).
These initial conversions set the stage for accurate calculations, avoiding errors that could arise from mismatched units.

Vaporization Process

Vaporization is the phase change process in which a liquid turns into a gas. This process happens when a liquid gains enough energy to overcome intermolecular forces and enter the gas phase. This is particularly important in the Dumas-bulb method where the liquid vaporizes to fill up a bulb.
The vaporization process is significant because this state change is necessary for determining vapor mass at a controllable volume and temperature, making the calculation of molar mass possible. - As the liquid is heated in a boiling-water bath, it boils below \(100^{\circ}\ \mathrm{C}\), making it safe and convenient to handle.
Ensure the entire sample vaporizes and fills the bulb completely without any remaining liquid or air. A measured increase in the vapor weight indicates the mass of the gas that would fill the space available in the bulb.