Question

A student weighs an empty flask and stopper and finds the mass to be 54.868 g. She then adds about \(5 \mathrm{mL}\) of an unknown liquid and heats the flask in a boiling water bath at \(100^{\circ} \mathrm{C}\). After all the liquid is vaporized, she removes the flask from the bath, stoppers it, and lets it cool. After it is cool, she momentarily removes the stopper, then replaces it and weighs the flask and condensed vapor, obtaining a mass of \(55.802 \mathrm{g} .\) The volume of the flask is known to be \(242.5 \mathrm{mL}\). The barometric pressure in the laboratory that day is \(747 \mathrm{mm}\) Hg. a. What was the pressure of the vapor in the flask in atm? \(P=\)_____________atm b. What was the temperature of the vapor in \(\mathrm{K}\) ? the volume of the flask in liters? \(T=\)_____________ \(\mathbf{K}\) \(\quad\) \(V=\)_____________\(\mathbf{L}\) c. What was the mass of vapor that was present in the flask? \(g=\) _____________ grams d. How many moles of vapor are present? \(n=\)_____________ moles What is the mass of one mole of vapor (Eq. 2 )? \(M M=\)_____________ g/mole

Short answer

a. The pressure of the vapor inside the flask is \(0.9829\,atm\). b. The temperature of the vapor inside the flask is \(373.15\,K\); the volume of the flask is \(0.2425\,L\). c. The mass of the vapor present in the flask is \(0.934\,grams\). d. The number of moles of vapor is \(0.00816\,moles\); the molar mass of the vapor is \(114.4\,g/mole\).

Step-by-step solution

a. Pressure of the vapor in atm

Given a barometric pressure of 747 mm Hg, we will convert this pressure into atmospheres (atm) since we'll be using the Ideal Gas Law later: 1 atm = 760 mm Hg \(P (atm) = \frac{747\,mm\,Hg}{760\,mm\,Hg/atm} = 0.9829\, atm\) The pressure of the vapor in the flask is equal to the barometric pressure in the laboratory. So, \(P = 0.9829\, atm\)

b. Temperature of the vapor in K and Volume of the flask in liters

Given the boiling water bath has a temperature of \(100^{\circ}C\), we will convert this temperature into Kelvin (K): \(T (K) = 100^{\circ}C + 273.15 = 373.15\, K\) The volume of the flask is given with 242.5 mL. To convert the volume to liters: \(V (L) = \frac{242.5\,mL}{1000\,mL/L} = 0.2425\, L\)

c. Mass of the vapor present in the flask

To find the mass of the vapor, we first find the difference in mass between the initial mass (empty flask) and the final mass (flask with vapor) and then convert it to grams: Initial mass of the empty flask = 54.868 g Final mass of the flask with condensed vapor = 55.802 g Difference in mass = 55.802 g - 54.868 g = 0.934 g So, the mass of the vapor present in the flask is 0.934 g.

d. Number of moles of vapor present and molar mass of the vapor

We will use the Ideal Gas Law formula to calculate the number of moles (n) of vapor present in the flask: \(PV = nRT\), where R is the gas constant = \(0.0821\,L atm /(mol K)\) We can solve for n: \(n = \frac{PV}{RT}\) Using our previous findings, \(P = 0.9829\, atm\), \(V = 0.2425\, L\), and \(T = 373.15\, K\): \(n = \frac{(0.9829\, atm)(0.2425\, L)}{(0.0821\, L\, atm /(mol\,K))(373.15\,K)} = 0.00816\,moles\) Now, we will find the molar mass (MM) of the vapor. To find the molar mass, we will divide the mass of the vapor (0.934 g) by the number of moles of vapor (0.00816 moles): \(MM = \frac{0.934\,g}{0.00816\,moles} = 114.4\,g/mole\) Finally, we have the following results: a. Pressure of the vapor inside the flask = \(0.9829\,atm\) b. Temperature of the vapor inside the flask = \(373.15\,K\); Volume of the flask = \(0.2425\,L\) c. Mass of the vapor present in the flask = \(0.934\,grams\) d. Number of moles of vapor = \(0.00816\,moles\); Molar mass of the vapor = \(114.4\,g/mole\)

Key concepts

conversion of units

In scientific experiments, like the one we're discussing, converting units, such as pressure, temperature, and volume, is essential. Measurement units must be consistent to ensure accurate results when applying formulas such as the Ideal Gas Law. Here we explore these conversions in-depth:

To determine the pressure of the vapor, we first converted the given barometric pressure from millimeters of mercury (mm Hg) to atmospheres (atm) because the Ideal Gas Law formula uses atm. This is achieved using the conversion factor: 1 atm = 760 mm Hg, allowing calculation of pressure in atm as:

• Pressure (atm) = \( \frac{747 \text{ mm Hg}}{760 \text{ mm Hg/atm}} \approx 0.9829 \text{ atm} \)

Similarly, temperature conversions were necessary. Since gas laws use Kelvin (K), the temperature in Celsius (°C) was converted using the relation Celsius to Kelvin:

• Temperature (K) = \( 100^{\circ}C + 273.15 = 373.15 \text{ K} \)

Lastly, to facilitate the Ideal Gas Law application, the volume of the flask was changed from milliliters (mL) to liters (L), because the law requires volume in liters:

• Volume (L) = \( \frac{242.5 \text{ mL}}{1000 \text{ mL/L}} = 0.2425 \text{ L} \)

These conversions are crucial for substituting into equations correctly, ensuring your calculations reflect reality.

molar mass calculation

Calculating the molar mass of a vapor involves using information from the experiment and the Ideal Gas Law. Molar mass is a critical property indicating the mass per mole of a substance, useful in identifying or understanding substance behavior.

Firstly, we calculated the mass of the vapor in grams by finding the difference in mass between the empty flask and the flask with the condensed vapor:

• Mass of vapor = Final mass - Initial mass
• Mass of vapor = 55.802 g - 54.868 g = 0.934 g

Then, using the Ideal Gas Law \( PV = nRT \), we determined the number of moles of vapor. We substituted the known values: pressure \( P = 0.9829 \text{ atm} \), volume \( V = 0.2425 \text{ L} \), the universal gas constant \( R = 0.0821 \text{ L atm/mole K} \), and temperature \( T = 373.15 \text{ K} \):

• \( n = \frac{(0.9829 \text{ atm})(0.2425 \text{ L})}{(0.0821 \text{ L atm/mole K})(373.15 \text{ K})} \approx 0.00816 \text{ moles} \)

Finally, the molar mass is calculated as the mass of the vapor divided by the number of moles obtained:

• Molar Mass \( MM = \frac{0.934 \text{ g}}{0.00816 \text{ moles}} \approx 114.4 \text{ g/mole} \)

Through these calculations, we deduced a critical characteristic of the vapor. This value can help identify the substance or guide further investigations.

chemical measurement techniques

Conducting experiments involving gases requires precise techniques to ensure accurate chemical measurements, which in turn reflect the reliability of calculated properties like molar mass.

In the described experiment, we saw the importance of securely weighing the flask before and after vaporization—one fundamental chemical measurement technique. This comparative weighing, to determine the vapor mass, requires precision in measurement:

• Ensure all instruments are calibrated correctly to avoid systematic errors.
• The flask must be allowed to completely cool to ambient temperature to avoid inaccuracies due to temperature-related mass changes.

Furthermore, handling the vapor in a controlled manner is key. Maintaining the vapor's pressure and temperature constant by using a boiling water bath regulates experimental conditions. This control allows the application of the Ideal Gas Law reliably because the law requires that the gas behaves in an 'ideal' manner, steady in temperature and pressure.

By following these rigorous measurement techniques, we guarantee precise calculations leading to valid conclusions. Hence, learning to properly carry out the experimental protocols enhances students' skillset in chemical laboratory methods.