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.

Short answer

The molar mass of the unknown volatile substance determined by the Dumas-bulb method is approximately \(62.13 \, \text{g/mol}\).

Step-by-step solution

Write down the ideal gas law

The ideal gas law equation is \(PV=nRT\), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Convert the given temperature to Kelvin

In order to work with the ideal gas law equation, we need to have the temperature in Kelvin. To convert from Celsius to Kelvin, add 273.15. The given temperature is \(100^{\circ}\)C, so the temperature in Kelvin is: \(T = 100 + 273.15 = 373.15 K\)

Convert the pressure from torr to atm

The ideal gas law requires pressure to be in atmospheres (atm). We're given the pressure in torr, so we need to convert it to atm. We can do this by using the conversion factor: 1 atm = 760 torr So, the pressure in atm is: \(P =\frac{752 \, \text{torr}}{760 \, \text{torr/atm}} = 0.9895 \, \text{atm}\)

Convert the volume to liters

The volume should be in liters for use in the ideal gas law equation. We are given the volume in \(\text{cm}^{3}\), and we can convert it to liters using the conversion factor: 1 L = 1000 cm³ So, the volume in liters is: \(V =\frac{354\, \text{cm}^{3}}{1000 \, \text{cm}^{3}/\text{L}} = 0.354 \, \text{L}\)

Rewrite the ideal gas law equation to include molar mass

We want to find out the molar mass (M) of the unknown substance, so we need to incorporate this into the ideal gas law equation. We can do this by using the formula: \(PV=\frac{m}{M}RT\) where m is the mass of the gas.

Plug in the given values and solve for the molar mass

Now we can plug in the values we've converted for the pressure, volume, and temperature, as well as the given mass of the gas, and solve for M: \(0.9895\,\text{atm} \cdot 0.354\,\text{L} =\frac{0.846\,\text{g}}{M}\cdot 0.0821\,\frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}} \cdot 373.15\,\text{K}\) Now, let's solve for M: \(\frac{0.846\,\text{g}}{M} =\frac{0.9895\,\text{atm} \cdot 0.354\,\text{L}}{0.0821\,\frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}} \cdot 373.15\,\text{K}}\) \(M =\frac{0.846\,\text{g}}{\frac{0.9895\,\text{atm} \cdot 0.354\,\text{L}}{0.0821\,\frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}} \cdot 373.15\,\text{K}}}=62.13 \,\text{g/mol}\)

Report the molar mass of the unknown substance

The molar mass of the unknown volatile substance is approximately 62.13 g/mol.

Key concepts

Molar Mass Calculation

One core aspect of understanding chemical properties is determining the molar mass of a substance. Molar mass is essentially the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). It tells us how much one mole of particles of that substance would weigh.

When calculating molar mass in the context of gases, we use the ideal gas law equation, which relates pressure, volume, and temperature to the number of moles, expressed as:

• The ideal gas law: \( PV = nRT \)
  • P = Pressure in atmospheres (atm)
  • V = Volume in liters (L)
  • n = Moles of the gas
  • R = Ideal gas constant (0.0821 \( \text{L atm/mol K} \))
  • T = Temperature in Kelvin (K)

Using this formula, we can rearrange and solve for the molar mass \(M\) by incorporating the known mass of the gas, \(m\), as \( m = n \times M \). This allows us to find the molar mass through the relationship \( PV = \frac{m}{M}RT \). Reorganizing this will give us \( M = \frac{m}{n} \).
Hence, understanding molar mass calculation is crucial in chemical studies as it helps us relate mass to moles, a common necessity in laboratory and theoretical chemistry.

Dumas-Bulb Method

The Dumas-Bulb Method is a traditional technique used to determine the molar mass of a volatile liquid. Named after Jean-Baptiste Dumas, this method involves evaporating a liquid into a gas under known conditions of temperature and pressure. It's a practical application of the ideal gas law.

To employ the Dumas-Bulb Method:

• Evaporate a known mass of liquid in a weighed glass bulb at a known temperature and pressure.
• Allow the vapor to reach an equilibrium.
• Return the system to room temperature, condensing the vapor, and weighing the bulb again to determine the mass of the vapor.
• Measure the system's volume, equating it to the volume the gas occupied.

These steps allow us to find the molar mass using the rearranged gas equation: \( M = \frac{mRT}{PV} \).

This method relies on the ideal gas law's assumption that the gas behaves ideally, which is mostly valid for gases at low pressures and high temperatures. Beyond providing insights into molar masses, the Dumas-Bulb Method is an engaging way to employ practical lab skills and comprehensive calculations in chemistry.

Gas Constant

The Gas Constant, often denoted as \( R \), is a crucial component in the ideal gas law equation. It serves as a bridge linking the macroscopic and microscopic levels of molecular movement in chemistry. The value of \( R \) we frequently use is 0.0821 \( \text{L atm/mol K} \).

Here's why the Gas Constant is significant:

• It represents the proportionality factor that relates the energy scale in physics to thermodynamic temperature, pressure, and volume.
• \( R \)'s versatility allows it to be used in a wide array of conditions and units, aligning the equation's workings according to the units for pressure, volume, and temperature.
• Because of \( R \), we can predict the behavior of ideal gases under varying conditions. It helps in manipulating and predicting experimental setups and outcomes.

Understanding \( R \) not only enhances one's grasp of chemical equations but also provides the toolset needed for advanced theoretical predictions in ideal and real gases.