Question

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise 10.53. 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 vapor can be calculated using the ideal gas law and the given mass, volume, pressure, and temperature. First, convert all given quantities to SI units. Then, apply the ideal gas law (PV=nRT) to solve for the number of moles (n). Finally, calculate the molar mass by dividing the mass of the vapor by the number of moles. After following these steps, we find the molar mass of the unknown vapor to be \(\frac{0.846\,\text{g}}{n}\), where n is the calculated number of moles.

Step-by-step solution

Convert the given quantities to SI units

First, we need to convert all the given quantities to the SI units. - Mass of the unknown vapor = \(0.846\,\text{g}\) - Volume of the bulb = \(354\, \text{cm}^3\) = \(354 \times 10^{-6}\, \text{m}^3\) - Pressure = \(752\, \text{torr}\) = \(\frac{752}{760}\, \text{atm}\) = \(0.9895\, \text{atm}\). (We know that 1 atm = 760 torr) - Temperature = \(100^{\circ}\, \text{C}\) = \(100 + 273.15\) = \(373.15\, \text{K}\)

Apply the ideal gas law

We can now use the ideal gas law to find the number of moles (n) of the unknown vapor: PV = nRT where: - P is the pressure in atm - V is the volume in m³ - n is the number of moles - R is the ideal gas constant = 0.0821 L·atm/mol·K - T is the temperature in K First, we should change the value of R to be in m³·atm/mol·K, so the units are consistent. R = \(0.0821 \frac{L \cdot atm}{mol \cdot K} = 0.0821 \times 10^{-3} \frac{m^3 \cdot atm}{mol \cdot K} = 8.21 \times 10^{-5} \frac{m^3 \cdot atm}{mol \cdot K}\)

Solve for the number of moles

We can rearrange the ideal gas law equation to solve for the number of moles: n = \(\frac{PV}{RT}\) Plug in the values: n = \(\frac{(0.9895 \, \mathrm{atm})(354 \times 10^{-6}\, \mathrm{m^3})}{(8.21 \times 10^{-5} \frac{\mathrm{m^3 \cdot atm}}{\mathrm{mol \cdot K}})(373.15\, \mathrm{K})}\)

Calculate the molar mass of the unknown vapor

To find the molar mass of the unknown vapor, we will divide the mass of the vapor by the number of moles: Molar mass = \(\frac{Mass\,of\,vapor}{Number\,of\,moles}\) Using the given mass and the calculated moles, we get: Molar mass = \(\frac{0.846\,\text{g}}{n}\) After calculating the number of moles, plug the value into the equation and find the molar mass of the unknown vapor.

Key concepts

Molar Mass Calculation

When learning about molar mass calculation, we're essentially learning how to find the mass of one mole of a substance. A mole is a fundamental unit in chemistry, representing Avogadro's number (\(6.022 \times 10^{23}\) particles). To calculate molar mass, we use the relation:
\[ \text{Molar Mass} = \frac{\text{Mass of sample}}{\text{Number of moles}} \]
In the context of the Dumas-bulb method used in the exercise, molar mass helps us determine the quantity of a volatile substance through its mass and the number of moles calculated from the ideal gas law. Correct measurement of the vapor's mass is critical for an accurate molar mass calculation. This calculation can be thrown off by any residual substance or error in measurement, so precision is key.

Ideal Gas Law

The ideal gas law is a cornerstone in understanding gas behavior under various conditions of pressure, volume, and temperature. It ties all these variables to the quantity of gas present in moles, captured in the equation:
\[ PV = nRT \]
The challenge with using the ideal gas law in calculations like these is ensuring all measurements are in proper units that align with the gas constant, R. Once the pressure (P), volume (V), and temperature (T) are known, and with R as a constant, one can solve for the number of moles (n) of the gas. It's crucial to remember that the ideal gas law assumes no interactions between gas molecules and that the molecules do not occupy space. While these assumptions aren't perfectly true, they're close enough for many practical calculations, like molar mass determination for gases at high temperature and low pressure.

Unit Conversion

Unit conversion is the bridge between different measurement systems—an essential process for solving problems in science and chemistry. Without proper unit conversion, your results may end up being meaningless because you’ve mixed incompatible units.
In the solution steps, we see an example of this: volume was converted from cubic centimeters to cubic meters, pressure from torr to atmospheres, and temperature from degrees Celsius to Kelvin. Here’s the rationale:

• Cubic centimeters to cubic meters: SI units are standard in scientific calculations, and volume should be in cubic meters to match the units of 'R' in the ideal gas law.
• Torr to atmospheres: Since 'R' is given in terms of atmospheres, not torr, conversion is necessary to ensure consistency.
• Degrees Celsius to Kelvin: Temperature in the ideal gas law must be in Kelvin because it is an absolute scale, reflecting actual kinetic energy levels of particles.

Always use consistent units within your formulas to accurately calculate physical quantities.

Vapor Density Determination

Understanding vapor density determination is vital when working with gases. Vapor density refers to the density of a vapor in relation to the density of an ideal gas under the same conditions. It gives us insight into the relative heaviness of a gas compared to something we're familiar with, like dry air.
To find vapor density, we can compare the molar mass of the gas to the molar mass of air (roughly 29 g/mol). Vapor density is a practical way to identify substances in a gaseous state when paired with Molar Mass Calculations. The Dumas-bulb method, as used in our exercise, facilitates this by capturing the vapor in a known volume and measuring its mass under known conditions. Once you know the molar mass, determining vapor density becomes a matter of simple division.
Using the ideal gas law to find the number of moles and knowing the mass of the vapor allows students to calculate both the molar mass and vapor density. Industries that deal with gases, like pharmaceuticals and petrochemicals, often use these calculations for quality control and product specification purposes.