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Chapter 5: Problem 128

Boron consists of two isotopes, \(^{10} \mathbf{B}\) and \(^{11} \mathbf{B}\). Chlorine also has two isotopes, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\) How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?

Short Answer

Expert verified
The mass spectrum of BCl₃ will have 6 peaks, corresponding to the approximate masses of 115, 116, 117, 118, 119, and 120 amu. This results from the unique combinations of isotopes in each BCl₃ molecular ion: B¹⁰Cl³⁵Cl³⁵Cl³⁵, B¹⁰Cl³⁵Cl³⁵Cl³⁷, B¹⁰Cl³⁵Cl³⁷Cl³⁷, B¹¹Cl³⁵Cl³⁵Cl³⁵, B¹¹Cl³⁵Cl³⁵Cl³⁷, and B¹¹Cl³⁵Cl³⁷Cl³⁷.

Step by step solution

01

Isotope combinations

First, we must identify all the possible combinations of isotopes in BCl3. Since we have two isotopes each of Boron and Chlorine, the following combinations can be formed: 1. B¹⁰Cl³⁵Cl³⁵Cl³⁵ 2. B¹⁰Cl³⁵Cl³⁵Cl³⁷ 3. B¹⁰Cl³⁵Cl³⁷Cl³⁷ 4. B¹¹Cl³⁵Cl³⁵Cl³⁵ 5. B¹¹Cl³⁵Cl³⁵Cl³⁷ 6. B¹¹Cl³⁵Cl³⁷Cl³⁷
02

Mass calculation

Now that we have identified the unique combinations of isotopes in BCl3, we can sum the mass numbers in each combination to find the approximate mass of each peak: 1. B¹⁰Cl³⁵Cl³⁵Cl³⁵: \( 10 + 35 + 35 + 35 = 115 \) amu 2. B¹⁰Cl³⁵Cl³⁵Cl³⁷: \( 10 + 35 + 35 + 37 = 117 \) amu 3. B¹⁰Cl³⁵Cl³⁷Cl³⁷: \( 10 + 35 + 37 + 37 = 119 \) amu 4. B¹¹Cl³⁵Cl³⁵Cl³⁵: \( 11 + 35 + 35 + 35 = 116 \) amu 5. B¹¹Cl³⁵Cl³⁵Cl³⁷: \( 11 + 35 + 35 + 37 = 118 \) amu 6. B¹¹Cl³⁵Cl³⁷Cl³⁷: \( 11 + 35 + 37 + 37 = 120 \) amu
03

Counting peaks and summarizing results

Based on the calculations, each unique combination gives us a distinct peak, resulting in a total of 6 peaks. Therefore, the mass spectrum of BCl3 will have 6 peaks with approximate masses of 115, 116, 117, 118, 119, and 120 amu.

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Most popular questions from this chapter

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