Naturally occurring elemental Boron consists entirely of exactly two stable isotopes, and . Mass spe — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
Naturally occurring elemental Boron consists entirely of exactly two stable isotopes, $^{10}B$ and $^{11}B$. Mass spectrometry reveals the standard average relative atomic mass of Boron to be precisely $10.8\text{ u}$. Which of the following statements are mathematically and logically correct deductions from this data?
💡 Solution & Explanation
Step 1: Set up the abundance equation. Let the percentage abundance of $^{11}B$ be $x\%$. Consequently, $^{10}B$ is $(100 - x)\%$. Step 2: Apply the average atomic mass formula. Average Mass = $\frac{(11 \times x) + 10 \times (100 - x)}{100} = 10.8$. Step 3: Solve for $x$. $11x + 1000 - 10x = 1080 \Rightarrow x = 80$. Thus, $^{11}B$ is $80\%$ (Option A is true) and $^{10}B$ is $20\%$ (Option B is true). Step 4: Evaluate theoretical implications. The average is a weighted mean, naturally sitting closer to the more abundant isotope (Option C is true). Because $80\%$ of all atoms are $^{11}B$, a macroscopic 1 mole sample ($N_A$ atoms) will contain exactly $0.80$ moles of $^{11}B$ atoms (Option D is true).