(2.8)An element occurs as three isotopes with atomic masses 19.99 amu (abundance = 90.51%), 20.99 amu (abundance = 0.27%), and 21.99 amu (abundance 9.22%). What is the atomic mass of the element? 62.97 amu 20.18 amu 21.00 amu 21.01 amu
(2.8)An element occurs as three isotopes with atomic masses 19.99 amu (abundance = 90.51%), 20.99 amu (abundance = 0.27%), and 21.99 amu (abundance 9.22%). What is the atomic mass of the element? 62.97 amu 20.18 amu 21.00 amu 21.01 amu
Chemistry
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Problem Statement
An element occurs as three isotopes with atomic masses 19.99 amu (abundance = 90.51%), 20.99 amu (abundance = 0.27%), and 21.99 amu (abundance = 9.22%). What is the atomic mass of the element?
- [ ] 62.97 amu
- [ ] 20.18 amu
- [ ] 21.00 amu
- [ ] 21.01 amu
### Explanation
To determine the atomic mass of the element with given isotopes, we use the concept of the weighted average of the isotopic masses. The formula for calculating the atomic mass (\(M\)) is:
\[ M = (m_1 \times f_1) + (m_2 \times f_2) + (m_3 \times f_3) \]
where \(m_1, m_2, m_3\) are the atomic masses of the isotopes and \(f_1, f_2, f_3\) are their respective fractional abundances.
**Step-by-Step Calculation:**
1. Convert the percentage abundances into fractional abundances:
- Isotope 1: 90.51% = 0.9051
- Isotope 2: 0.27% = 0.0027
- Isotope 3: 9.22% = 0.0922
2. Perform the weighted average calculation:
\[ M = (19.99 \times 0.9051) + (20.99 \times 0.0027) + (21.99 \times 0.0922) \]
3. Resolve each term of the sum:
- \(19.99 \times 0.9051 = 18.0954\)
- \(20.99 \times 0.0027 = 0.0567\)
- \(21.99 \times 0.0922 = 2.0294\)
4. Summing these values gives the atomic mass:
\[ M = 18.0954 + 0.0567 + 2.0294 = 20.1815 \]
Therefore, the atomic mass of the element is approximately **20.18 amu** which matches one of the options](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc8db0c58-8b17-40ce-ad80-341585b72407%2F45d36549-6088-4374-a2f5-276bd200e6d8%2Fi7iwxof_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
An element occurs as three isotopes with atomic masses 19.99 amu (abundance = 90.51%), 20.99 amu (abundance = 0.27%), and 21.99 amu (abundance = 9.22%). What is the atomic mass of the element?
- [ ] 62.97 amu
- [ ] 20.18 amu
- [ ] 21.00 amu
- [ ] 21.01 amu
### Explanation
To determine the atomic mass of the element with given isotopes, we use the concept of the weighted average of the isotopic masses. The formula for calculating the atomic mass (\(M\)) is:
\[ M = (m_1 \times f_1) + (m_2 \times f_2) + (m_3 \times f_3) \]
where \(m_1, m_2, m_3\) are the atomic masses of the isotopes and \(f_1, f_2, f_3\) are their respective fractional abundances.
**Step-by-Step Calculation:**
1. Convert the percentage abundances into fractional abundances:
- Isotope 1: 90.51% = 0.9051
- Isotope 2: 0.27% = 0.0027
- Isotope 3: 9.22% = 0.0922
2. Perform the weighted average calculation:
\[ M = (19.99 \times 0.9051) + (20.99 \times 0.0027) + (21.99 \times 0.0922) \]
3. Resolve each term of the sum:
- \(19.99 \times 0.9051 = 18.0954\)
- \(20.99 \times 0.0027 = 0.0567\)
- \(21.99 \times 0.0922 = 2.0294\)
4. Summing these values gives the atomic mass:
\[ M = 18.0954 + 0.0567 + 2.0294 = 20.1815 \]
Therefore, the atomic mass of the element is approximately **20.18 amu** which matches one of the options
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