The aluminum in a 0.767 g sample of an unknown material was precipitated as aluminum hydroxide, Al(OH)3, which was then converted to Al2O3 by heating strongly. If 0.143 g of Al2O3 is obtained from the 0.767 g sample, what is the mass percent of aluminum in the sample? %
The aluminum in a 0.767 g sample of an unknown material was precipitated as aluminum hydroxide, Al(OH)3, which was then converted to Al2O3 by heating strongly. If 0.143 g of Al2O3 is obtained from the 0.767 g sample, what is the mass percent of aluminum in the sample? %
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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![### Problem Statement:
The aluminum in a 0.767 g sample of an unknown material was precipitated as aluminum hydroxide, \( \text{Al(OH)}_3 \), which was then converted to \( \text{Al}_2\text{O}_3 \) by heating strongly. If 0.143 g of \( \text{Al}_2\text{O}_3 \) is obtained from the 0.767 g sample, what is the mass percent of aluminum in the sample?
\[ \boxed{ \ \ } \% \]
### Steps for Solving the Problem:
1. **Calculate the molar masses** of \( \text{Al}_2\text{O}_3 \) and elemental aluminum (Al):
- Molar mass of \( \text{Al} \) = 26.98 g/mol
- Molar mass of \( \text{Al}_2\text{O}_3 \):
\[
2 \times \text{Al} + 3 \times \text{O} = 2 \times 26.98 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol} = 101.96 \, \text{g/mol}
\]
2. **Find the number of moles of \( \text{Al}_2\text{O}_3 \)**:
\[
\text{moles of } \text{Al}_2\text{O}_3 = \frac{0.143 \, \text{g}}{101.96 \, \text{g/mol}} \approx 0.001402 \, \text{mol}
\]
3. **Determine the moles of aluminum in** \( \text{Al}_2\text{O}_3 \):
\[
\text{Each mole of } \text{Al}_2\text{O}_3 \text{ contains 2 moles of Al}.
\]
\[
\text{moles of Al} = 2 \times 0.001402 \, \text{mol} = 0.002804 \, \text{mol}
\]
4. **Calculate the mass of aluminum** in the sample:
\[
\text{mass of Al} = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a354f1c-cdfd-41e5-a863-9a86a504af1c%2F54d08e21-01fd-43eb-ac20-500cb88035f1%2F4g3gw0t_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
The aluminum in a 0.767 g sample of an unknown material was precipitated as aluminum hydroxide, \( \text{Al(OH)}_3 \), which was then converted to \( \text{Al}_2\text{O}_3 \) by heating strongly. If 0.143 g of \( \text{Al}_2\text{O}_3 \) is obtained from the 0.767 g sample, what is the mass percent of aluminum in the sample?
\[ \boxed{ \ \ } \% \]
### Steps for Solving the Problem:
1. **Calculate the molar masses** of \( \text{Al}_2\text{O}_3 \) and elemental aluminum (Al):
- Molar mass of \( \text{Al} \) = 26.98 g/mol
- Molar mass of \( \text{Al}_2\text{O}_3 \):
\[
2 \times \text{Al} + 3 \times \text{O} = 2 \times 26.98 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol} = 101.96 \, \text{g/mol}
\]
2. **Find the number of moles of \( \text{Al}_2\text{O}_3 \)**:
\[
\text{moles of } \text{Al}_2\text{O}_3 = \frac{0.143 \, \text{g}}{101.96 \, \text{g/mol}} \approx 0.001402 \, \text{mol}
\]
3. **Determine the moles of aluminum in** \( \text{Al}_2\text{O}_3 \):
\[
\text{Each mole of } \text{Al}_2\text{O}_3 \text{ contains 2 moles of Al}.
\]
\[
\text{moles of Al} = 2 \times 0.001402 \, \text{mol} = 0.002804 \, \text{mol}
\]
4. **Calculate the mass of aluminum** in the sample:
\[
\text{mass of Al} = 0.
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