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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![### Understanding Moles in Chemistry
**Problem Statement**
The density of acetic anhydride (\(C_4H_6O_3\)) is 1.08 g/mL. Calculate the moles of acetic anhydride present in 1.69 mL of acetic anhydride.
**Solution Steps**
1. **Calculate Mass:**
- Use the density formula:
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]
- Given:
- Density = 1.08 g/mL
- Volume = 1.69 mL
- Calculate mass:
\[
\text{Mass} = \text{Density} \times \text{Volume} = 1.08 \, \text{g/mL} \times 1.69 \, \text{mL} = 1.8252 \, \text{g}
\]
2. **Calculate Moles:**
- Use the molar mass of acetic anhydride (\(C_4H_6O_3\)):
- Carbon (C): \(4 \times 12.01\) g/mol
- Hydrogen (H): \(6 \times 1.01\) g/mol
- Oxygen (O): \(3 \times 16.00\) g/mol
- Total molar mass:
\[
(4 \times 12.01) + (6 \times 1.01) + (3 \times 16.00) = 102.09 \, \text{g/mol}
\]
- Calculate moles using the formula:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.8252 \, \text{g}}{102.09 \, \text{g/mol}} \approx 0.0179 \, \text{mol}
\]
**Conclusion**
The moles of acetic anhydride present in 1.69 mL are approximately 0.0179 mol. This calculation helps in understanding the relationship between volume, density, and moles, which is fundamental in stoichiometry and various chemistry applications.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6439f801-831b-41ba-9f5d-2515f4c703aa%2F05e3a7f6-f989-444d-af9b-487f0cd78c5d%2Fkzhnszr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Moles in Chemistry
**Problem Statement**
The density of acetic anhydride (\(C_4H_6O_3\)) is 1.08 g/mL. Calculate the moles of acetic anhydride present in 1.69 mL of acetic anhydride.
**Solution Steps**
1. **Calculate Mass:**
- Use the density formula:
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\]
- Given:
- Density = 1.08 g/mL
- Volume = 1.69 mL
- Calculate mass:
\[
\text{Mass} = \text{Density} \times \text{Volume} = 1.08 \, \text{g/mL} \times 1.69 \, \text{mL} = 1.8252 \, \text{g}
\]
2. **Calculate Moles:**
- Use the molar mass of acetic anhydride (\(C_4H_6O_3\)):
- Carbon (C): \(4 \times 12.01\) g/mol
- Hydrogen (H): \(6 \times 1.01\) g/mol
- Oxygen (O): \(3 \times 16.00\) g/mol
- Total molar mass:
\[
(4 \times 12.01) + (6 \times 1.01) + (3 \times 16.00) = 102.09 \, \text{g/mol}
\]
- Calculate moles using the formula:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.8252 \, \text{g}}{102.09 \, \text{g/mol}} \approx 0.0179 \, \text{mol}
\]
**Conclusion**
The moles of acetic anhydride present in 1.69 mL are approximately 0.0179 mol. This calculation helps in understanding the relationship between volume, density, and moles, which is fundamental in stoichiometry and various chemistry applications.
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