Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The molar mass of a Volatile Liquid using The Dumas Method
Transcribed Image Text:**Data Analysis**
Calculate the molar mass of your unknown for each trial, and report the average. If your instructor provides you with the accepted molar mass for your unknown, calculate the percent error of your average molar mass.
**For Your Consideration**
- Why is it not necessary to know the mass of the volatile liquid initially added to the flask?
- What assumption is being made in determining each of the following?
- The mass of the vapor.
- The volume of the vapor.
- The temperature of the vapor.
- The pressure of the vapor.
- How will the unknown’s reported molar mass be affected in each of the following cases?
- The atmospheric pressure is recorded too high.
- A small piece of the foil cap was lost before the final mass was measured.
- Condensed water vapor from the bath was trapped underneath the foil cap.
- The vapor’s volume is assumed to be 125 mL (marked on the side of the flask) instead of the measured volume of the flask.
- Heating the flask too long resulted in air outside the flask entering and displacing some of the vapor inside the flask.
- The thermometer is miscalibrated and reads a constant amount too low over its entire temperature range.
![# Molar Mass of a Volatile Liquid
## Trial 1
- **Temperature:** \(374.15 \, \text{K}\)
- **Pressure:**
\[
766 \, \text{mmHg} \left(\frac{1 \, \text{atm}}{760 \, \text{mmHg}}\right) = 1.0078 \, \text{atm}
\]
- **Volume:** \(0.158 \, \text{L}\)
- **Moles of vapor (\(n_{\text{vapor}}\)):**
\[
0.00518 \, \text{mol} \, (5.18 \times 10^{-3} \, \text{mol})
\]
- **Mass of Liquid (MM):**
\[
0.156 \, \text{g} = \frac{30.1 \, \text{g/mol}}
{5.18 \times 10^{-3} \, \text{mol}}
\]
## Trial 2
- **Temperature:** \(374.15 \, \text{K}\)
- **Pressure:**
\[
1.0078 \, \text{atm}
\]
- **Volume:** \(0.160 \, \text{L}\)
- **Moles of vapor (\(n_{\text{vapor}}\)):**
\[
0.00525 \, \text{mol} \, (5.25 \times 10^{-3} \, \text{mol})
\]
- **Mass of Liquid (MM):**
\[
42.8 \, \text{g/mol}
\]
## Average Molar Mass
\[
\text{Avg} = 36.45 \, \text{g/mol}
\]
## Percent Error Calculation
\[
\%\text{ error} = \frac{88.29 \, \text{g/mol} - 36.45 \, \text{g/mol}}{88.2 \, \text{g/mol}} \times 100
\]
\[
\%\text{ error} = 58.7\%
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F668a52c6-cc0e-4197-a3c4-5560835d26b4%2Fcb6c88fa-e494-4d74-b413-19753e2f30a8%2Fanfo44h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:# Molar Mass of a Volatile Liquid
## Trial 1
- **Temperature:** \(374.15 \, \text{K}\)
- **Pressure:**
\[
766 \, \text{mmHg} \left(\frac{1 \, \text{atm}}{760 \, \text{mmHg}}\right) = 1.0078 \, \text{atm}
\]
- **Volume:** \(0.158 \, \text{L}\)
- **Moles of vapor (\(n_{\text{vapor}}\)):**
\[
0.00518 \, \text{mol} \, (5.18 \times 10^{-3} \, \text{mol})
\]
- **Mass of Liquid (MM):**
\[
0.156 \, \text{g} = \frac{30.1 \, \text{g/mol}}
{5.18 \times 10^{-3} \, \text{mol}}
\]
## Trial 2
- **Temperature:** \(374.15 \, \text{K}\)
- **Pressure:**
\[
1.0078 \, \text{atm}
\]
- **Volume:** \(0.160 \, \text{L}\)
- **Moles of vapor (\(n_{\text{vapor}}\)):**
\[
0.00525 \, \text{mol} \, (5.25 \times 10^{-3} \, \text{mol})
\]
- **Mass of Liquid (MM):**
\[
42.8 \, \text{g/mol}
\]
## Average Molar Mass
\[
\text{Avg} = 36.45 \, \text{g/mol}
\]
## Percent Error Calculation
\[
\%\text{ error} = \frac{88.29 \, \text{g/mol} - 36.45 \, \text{g/mol}}{88.2 \, \text{g/mol}} \times 100
\]
\[
\%\text{ error} = 58.7\%
\]
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