The chemical element Boron (B) exists in nature as a mixture of exactly two stable isotopes: and . I — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
The chemical element Boron (B) exists in nature as a mixture of exactly two stable isotopes: $^{10}B$ and $^{11}B$. If the highly precise average atomic mass of Boron reported on the periodic table is $10.81\text{ u}$, what is the exact percentage relative abundance of the heavier $^{11}B$ isotope?
💡 Solution & Explanation
Let the percentage abundance of the $^{11}B$ isotope be $x\%$. Consequently, the percentage abundance of the lighter $^{10}B$ isotope must be $(100 - x)\%$. The formula for average atomic mass is: $M_{\text{avg}} = \frac{\sum (\% \text{ abundance} \times \text{isotopic mass})}{100}$. $10.81 = \frac{(x \times 11) + ((100 - x) \times 10)}{100}$. $1081 = 11x + 1000 - 10x$. $1081 - 1000 = x \Rightarrow x = 81$. Therefore, the relative abundance of the $^{11}B$ isotope is exactly $81\%$.