Question

Lithium has an atomic mass of 6.941 amu. lithium has two common isotopes. the one iotope has a mass of 6.015 amu and a relative abundance of 7.49%
A. what is the abundance of the other isotope?
B. what is the mass of the other isotope?
make the answer make sense and clear?

Answer 1

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**.