Problem 100 - 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, \(^{79} \mathrm{Br}(50.7 \%\) abundance) and \(^{81} \mathrm{Br}(49.3 \%\) 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

The peaks correspond to \(^{79}\mathrm{Br}-^{79}\mathrm{Br}\), \(^{79}\mathrm{Br}-^{81}\mathrm{Br}\), and \(^{81}\mathrm{Br}-^{81}\mathrm{Br}\). The ratios align due to isotopic abundances.

Step 1: Identify molecular species for each peak

The three mass spectrum peaks are at \(m/Z\) values of 158, 160, and 162. These correspond to the molecular species \(^{79}\mathrm{Br}-^{79}\mathrm{Br}\), \(^{79}\mathrm{Br}-^{81}\mathrm{Br}\), and \(^{81}\mathrm{Br}-^{81}\mathrm{Br}\) respectively. This is because the sum of the isotopic masses gives: 79+79=158, 79+81=160, and 81+81=162.

Step 2: Calculate probabilities for each molecular species

For \(m/Z=158\) (\(^{79}\mathrm{Br}-^{79}\mathrm{Br}\)), the probability is \((0.507)^2\). For \(m/Z=160\) (\(^{79}\mathrm{Br}-^{81}\mathrm{Br}\)), the probability is \(2 \times 0.507 \times 0.493\).For \(m/Z=162\) (\(^{81}\mathrm{Br}-^{81}\mathrm{Br}\)), the probability is \((0.493)^2\).

Step 3: Calculate the ratio of the probabilities

Calculate the numerical probabilities:- \((0.507)^2 = 0.257\)- \(2 \times 0.507 \times 0.493 = 0.5\)- \((0.493)^2 = 0.243\)From these probabilities, the ratio is approximately 0.257 : 0.5 : 0.243.

Step 4: Simplify the probability ratio

Approximating to the simplest ratio: - Convert to nearest simple integers: approximately 1 : 2 : 1, which aligns with the given intensity ratio of the peaks in the mass spectrum.

Key Concepts

Isotopes

Isotopes are different forms of the same element, where each isotope has the same number of protons but a different number of neutrons. This means isotopes of an element have different atomic masses. Isotopes are located in the same place on the periodic table, hence the name."iso"(same) and "topos" (place) in Greek.

Bromine has two naturally occurring isotopes:

• \(^{79}\text{Br}\) which has 44 neutrons and an atomic mass of 79.
• \(^{81}\text{Br}\) which has 46 neutrons and an atomic mass of 81.

The isotopic composition of an element determines many of its properties, including how it behaves in mass spectrometry. In the case of bromine, the nearly equal abundances of its two isotopes \(^{79}\text{Br}\) and \(^{81}\text{Br}\) lead to unique patterns in mass spectra. Understanding isotopes is crucial for interpreting spectrometry results, especially for diatomic molecules like \(\text{Br}_2\) where isotopic combinations yield different spectral lines.

Probability in Chemistry

Probability in chemistry often involves determining the likelihood of various outcomes, such as the formation of molecules with different isotopic compositions. It can guide us in predicting the ratios at which different molecular species are present.

In the case of bromine's mass spectrometry peaks:

• For peak at \(m/Z = 158\) from \(^{79}\text{Br}-^{79}\text{Br}\), the probability is computed as \((0.507)^2 = 0.257\), since both bromine atoms are \(^{79}\text{Br}\).
• For peak at \(m/Z = 160\) from \(^{79}\text{Br}-^{81}\text{Br}\), the probability is \(2 \times 0.507 \times 0.493 = 0.5\). The factor of 2 arises because either bromine can be \(^{79}\text{Br}\) or \(^{81}\text{Br}\).
• For peak at \(m/Z = 162\) from \(^{81}\text{Br}-^{81}\text{Br}\), the probability calculation is \((0.493)^2 = 0.243\).

Calculations like these are fundamental to determining expected intensities and understanding why peaks in mass spectra appear at certain ratios.

Bromine

Bromine is a chemical element with the symbol Br and atomic number 35. It is a halogen and is known for its reddish-brown liquid form at room temperature. Unlike other elements, bromine commonly forms diatomic molecules, \(\text{Br}_2\), in its natural form.

The isotopic nature of bromine, with almost equal amounts of \(^{79}\text{Br}\) and \(^{81}\text{Br}\), contributes to its interesting behavior in mass spectrometry. When diatomic molecules, like \(\text{Br}_2\), enter a mass spectrometer, they yield a mass spectrum with peaks that reflect the sum of the atomic masses of the isotopes present in each molecule.

In a mass spectrum of bromine, you will observe peaks corresponding to different combinations of \(^{79}\text{Br}\) and \(^{81}\text{Br}\), such as 158 (from two \(^{79}\text{Br}\) atoms), 160 (from one \(^{79}\text{Br}\) and one \(^{81}\text{Br}\) atoms), and 162 (from two \(^{81}\text{Br}\) atoms). This makes bromine a particularly interesting case study for understanding isotopic patterns in mass spectrometry.