Isotope Chemistry Example 2

Solution

1. 1
   Atomic mass is the weighted average of isotope masses. Use: $\bar{m} = \sum(m_i \times f_i)$, where $m_i$ is isotope mass and $f_i$ is fractional abundance.
2. 2
   Compute each term: $10 \times 0.199 = 1.99\,\text{amu}$ (from ${}^{10}\text{B}$) and $11 \times 0.801 = 8.811\,\text{amu}$ (from ${}^{11}\text{B}$).
3. 3
   Sum the contributions: $$\bar{m} = 1.99 + 8.811 = 10.80\,\text{amu}$$