To solve the problem, we will go step by step:
A. What is the abundance of the other isotope?
Lithium has two common isotopes, and the total abundance must equal 100%. Since one isotope has a relative abundance of 7.49%, the abundance of the other isotope can be found by subtracting this value from 100%:
\[ \text{Abundance of the other isotope} = 100% - 7.49% = 92.51% \]
B. What is the mass of the other isotope?
To find the mass of the other isotope, we can use the weighted average formula, which applies to atomic masses of isotopes:
\[ \text{Average atomic mass} = ( \text{mass of isotope 1} \times \text{abundance of isotope 1}) + ( \text{mass of isotope 2} \times \text{abundance of isotope 2}) \]
We know:
- Average atomic mass of lithium = 6.941 amu
- Mass of the first isotope (Isotope 1) = 6.015 amu
- Abundance of Isotope 1 = 7.49% = 0.0749 (as a decimal)
- Abundance of Isotope 2 = 92.51% = 0.9251 (as a decimal)
Substituting the values we know into the equation:
\[ 6.941 = (6.015 \times 0.0749) + (m_2 \times 0.9251) \]
Where \( m_2 \) is the mass of the other isotope. First calculate the contribution of the first isotope:
\[ 6.015 \times 0.0749 = 0.4500 , \text{amu} , (approximately) \]
Now we can substitute this value back into the equation:
\[ 6.941 = 0.4500 + (m_2 \times 0.9251) \]
Now isolate \( m_2 \):
\[ 6.941 - 0.4500 = m_2 \times 0.9251 \]
\[ 6.491 = m_2 \times 0.9251 \]
Now, divide both sides by 0.9251 to find \( m_2 \):
\[ m_2 = \frac{6.491}{0.9251} \approx 7.01 , \text{amu} \]
Summary:
A. The abundance of the other isotope is 92.51%.
B. The mass of the other isotope is approximately 7.01 amu.