Problem 160 - Chemistry and Chemical Reactivity

Question

The highest mass peaks in the mass spectrum of \(\mathrm{Br}_{2}\) occur at \(m / Z 158,160,\) and \(162 .\) The ratio of intensities of these peaks is approximately \(1: 2: 1 .\) Bromine has two stable isotopes, \(^{79} \mathrm{Br}(50.7 \% \text { abundance })\) and \(^{81} \mathrm{Br}\) \((49.3 \% \text { abundance })\) (a) What molecular species gives rise to each of these peaks? (b) Explain the relative intensities of these peaks. (Hint: Consider the probabilities of each atom combination.)

Step-by-Step Solution

Answer

(a) \(^{79}\text{Br}-^{79}\text{Br}\), \(^{79}\text{Br}-^{81}\text{Br}\) / \(^{81}\text{Br}-^{79}\text{Br}\), \(^{81}\text{Br}-^{81}\text{Br}\); (b) Reflects isotopic abundance, resulting in 1:2:1 intensity ratio.

Step 1: Identify the Isotopic Composition of Bromine

Bromine has two stable isotopes: \(^{79}\mathrm{Br}\) and \(^{81}\mathrm{Br}\), with abundances of 50.7% and 49.3% respectively. We need to consider the diatomic molecule \(\mathrm{Br}_2\), which means combining these isotopes as possible molecular species.

Step 2: Determine Possible Molecular Species

The molecular species formed can be \(^{79}\text{Br}-^{79}\text{Br}\), \(^{79}\text{Br}-^{81}\text{Br}\), and \(^{81}\text{Br}-^{81}\text{Br}\). These combinations correspond to the molecular masses of 158, 160, and 162 respectively.

Step 3: Calculate the Probability of Each Molecular Species

The probability of \(^{79}\text{Br}-^{79}\text{Br}\) is \((0.507)^2 = 0.257049\). The probability of \(^{79}\text{Br}-^{81}\text{Br}\) or \(^{81}\text{Br}-^{79}\text{Br}\) is \(2 \times (0.507 \times 0.493) = 0.499722\). The probability of \(^{81}\text{Br}-^{81}\text{Br}\) is \((0.493)^2 = 0.243049\). The probabilities confirm the relative intensities in the 1:2:1 ratio for peaks at mass 158, 160, and 162 respectively.

Key Concepts

Isotopic Composition

Isotopic composition refers to the presence and proportion of different isotopes within an element. Isotopes are variants of a chemical element that have the same number of protons but differ in the number of neutrons. This difference results in each isotope having a distinct atomic mass. In the case of bromine, there are two stable isotopes: \(^{79}\mathrm{Br}\) and \(^{81}\mathrm{Br}\). These isotopes have a natural abundance of 50.7% and 49.3% respectively.
Understanding the isotopic composition is crucial in mass spectrometry as it helps to predict the possible molecular species and their relative masses that will appear in the spectra. When analyzing a diatomic molecule like \(\mathrm{Br}_2\), the isotopes can combine in different ways, resulting in molecular species with different masses.

Molecular Species

Molecular species are different combinations of isotopes that form a molecule. For a diatomic molecule such as \(\mathrm{Br}_2\), the possible molecular species are determined by the combination of its isotopes. Bromine's isotopes \(^{79}\mathrm{Br}\) and \(^{81}\mathrm{Br}\) can combine in three distinct ways to form \(\mathrm{Br}_2\):

• \(^{79}\text{Br}-^{79}\text{Br}\)
• \(^{79}\text{Br}-^{81}\text{Br}\)
• \(^{81}\text{Br}-^{81}\text{Br}\)

The masses of these molecular species are 158, 160, and 162 amu respectively.
Each combination corresponds to a specific peak in the mass spectrometry graph. By examining these peaks, chemists can determine the composition of the molecule and gain insights into the isotopic composition of the elements involved.

Relative Intensities

The relative intensity of a peak in a mass spectrometry graph indicates the abundance or probability of a particular molecular ion. For \(\mathrm{Br}_2\), we observe three prominent peaks with mass-to-charge ratios (m/z) of 158, 160, and 162, appearing in an approximate ratio of 1:2:1.
This pattern is a result of the combination probabilities of . The probability of each molecular species is calculated by multiplying the abundances of the isotopes involved:

• The probability of \(^{79}\mathrm{Br}-^{79}\mathrm{Br}\) is \(0.507\times0.507 = 0.257\).
• The probability of either \(^{79}\mathrm{Br}-^{81}\mathrm{Br}\) or \(^{81}\mathrm{Br}-^{79}\mathrm{Br}\) is \(2\times(0.507\times0.493) = 0.500\).
• The probability of \(^{81}\mathrm{Br}-^{81}\mathrm{Br}\) is \(0.493\times0.493 = 0.243\).

These probabilities align with the observed 1:2:1 ratio, showcasing how the molecular abundance directly influences peak intensity in mass spectrometry.

Bromine Isotopes

Bromine isotopes play a vital role in chemical analysis, particularly in mass spectrometry. With isotopes \(^{79}\mathrm{Br}\) and \(^{81}\mathrm{Br}\), bromine is an excellent example to demonstrate isotopic effects in mass spectra due to the similar natural abundances of these isotopes.
Bromine's unique isotopic distribution makes it possible for peaks to display clearly defined ratios, as seen in the \(\mathrm{Br}_2\) mass spectrum. The two bromine isotopes allow for the distinct molecular species \(^{79}\mathrm{Br}-^{79}\mathrm{Br}\), \(^{79}\mathrm{Br}-^{81}\mathrm{Br}\), and \(^{81}\mathrm{Br}-^{81}\mathrm{Br}\), making bromine compounds perfect candidates for illustrating how isotopes affect molecular mass and peak intensities.
The subtle differences in mass and abundance between \(^{79}\mathrm{Br}\) and \(^{81}\mathrm{Br}\) help chemists analyze molecules based on the observed peaks, facilitating a deeper understanding of isotopic variation and its effects on chemical behavior.