Question

There are two different isotopes of bromine atoms. Under normal conditions, elemental bromine consists of \(\mathrm{Br}_{2}\) molecules, and the mass of a Br molecule is the sum of the masses of the two atoms in the molecule. The mass spectrum of \(\mathrm{Br}_{2}\) consists of three peaks: (a) What is the origin of each peak (of what isotopes does each consist)? (b) What is the mass of each isotope? (c) Determine the average molecular mass of a \(\mathrm{Br}_{2}\) molecule. (d) Determine the average atomic mass of a bromine atom. (e) Calculate the abundances of the two isotopes.

Short answer

The origin of the three peaks in the mass spectrum are due to the combinations of the two isotopes of bromine, \(\mathrm{Br}_{x}\mathrm{Br}_{x}\), \(\mathrm{Br}_{x}\mathrm{Br}_{y}\), and \(\mathrm{Br}_{y}\mathrm{Br}_{y}\). The masses of the isotopes are \(m_{1} = 79\) amu and \(m_{2} = 81\) amu. The average molecular mass of a \(\mathrm{Br}_{2}\) molecule is calculated using the masses and relative abundances of the peaks. The average atomic mass of a bromine atom is obtained by dividing the average molecular mass of \(\mathrm{Br}_{2}\) by 2. Finally, the abundances of the two isotopes can be determined using the average atomic mass and the masses of the two isotopes.

Step-by-step solution

Identify the isotopes present in the mass spectrum peaks

Since Br exists as a diatomic molecule, the different possible combinations of isotopes will be in the form of \(\mathrm{Br}_{x}\mathrm{Br}_{x}\), \(\mathrm{Br}_{x}\mathrm{Br}_{y}\), and \(\mathrm{Br}_{y}\mathrm{Br}_{y}\), where x and y both represent different isotopes of Br. Hence, these three combinations correspond to the three peaks in the mass spectrum.

Assign masses to each peak

Let m1 and m2 be the masses of the two isotopes x and y of bromine. We can assign masses for each peak as: Peak 1 - \(m_{1} + m_{1} = 2m_{1}\) Peak 2 - \(m_{1} + m_{2}\) Peak 3 - \(m_{2} + m_{2} = 2m_{2}\)

Determine the masses of each isotope

From the mass spectrum data given, to find \(m_{1}\) and \(m_{2}\) for the two isotopes, one should use the exact masses provided in the spectrum for the most abundant combinations (\(\mathrm{Br}_{x}\mathrm{Br}_{x}\) and \(\mathrm{Br}_{y}\mathrm{Br}_{y}\)). These masses are approximately 158 and 162 amu. From this information, we can identify the bromine isotopes as having masses: \(m_{1} = 79\) amu (Isotope x) \(m_{2} = 81\) amu (Isotope y)

Calculate the average molecular mass of \(\mathrm{Br}_{2}\)

Using the masses of the peaks and their relative abundances, we can determine the average molecular mass of \(\mathrm{Br}_{2}\). Let's denote the abundances of the two isotopes as \(p_{1}\) for isotope x, and \(p_{2}\) for isotope y. Average molecular mass = \((2m_{1} \cdot p_{1}^{2}) + ((m_{1} + m_{2}) \cdot 2 \cdot p_{1} \cdot p_{2}) + (2m_{2} \cdot p_{2}^{2})\)

Calculate the average atomic mass of bromine

To find the average atomic mass of bromine, we can use the average molecular mass of the \(\mathrm{Br}_{2}\) molecule and divide it by 2. Average atomic mass of bromine = \(\frac{(\text{Average molecular mass of } \mathrm{Br}_{2})}{2}\)

Calculate the abundances of the two isotopes

Using the average atomic mass of bromine and the masses of the two isotopes, we can determine the abundance percentages \(p_{1}\) and \(p_{2}\). \(p_{1} = \frac{(\text{Average atomic mass} - m_{2})}{(m_{1} - m_{2})}\) \(p_{2} = \frac{(m_{1} - \text{Average atomic mass})}{(m_{1} - m_{2})}\) Using these equations, we'd utilize mass spectrum information to compute the corresponding abundance percentages for each bromine isotope.

Key concepts

Isotope Abundances

Isotope abundances refer to the proportions of different isotopes of an element present in a sample. Isotopes are atoms of the same element that have the same number of protons but a different number of neutrons, resulting in different atomic masses. In mass spectrometry, each isotope contributes to a distinct peak, and by analyzing these peaks, we can identify the isotopes and calculate their abundances.

Mass spectrometry provides us with a mass spectrum, which displays the mass-to-charge ratio on the x-axis and the relative abundance on the y-axis. In elemental bromine (\r\(\mathrm{Br}_2\)), there are two isotopes with masses 79 and 81 amu. When these isotopes combine to form the diatomic molecule, we observe three peaks, each representing a different combination of isotopes. From the spectrum, we can deduce the relative amounts of each isotopic molecule present, which informs us about the individual isotope abundances.

For example, from the intensity of the peaks in a bromine mass spectrum, relative abundances can be calculated for each isotope. This gives us a fuller understanding of the natural composition of bromine in the given sample.

Molecular Mass Calculation

The molecular mass calculation involves finding the mass of a molecule by summing the masses of its constituent atoms. In mass spectrometry, the molecular mass of a compound is reflected in the peaks that represent the different possible combinations of isotopes. For diatomic molecules like \r\(\mathrm{Br}_2\), finding the molecular mass is straightforward if the atomic masses of the isotopes are known.

In a mass spectrum of \r\(\mathrm{Br}_2\), we can calculate the molecular mass of the molecule by assigning masses to the peaks corresponding to the various isotope combinations. For instance, if a peak represents a molecule consisting of two isotopes with masses \r\(m_1\) and \r\(m_2\), the molecular mass for that peak is \r\(m_1 + m_2\). The average molecular mass, however, takes the isotope abundances into account, providing a weighted average based on how frequently each isotope combination occurs in the sample.

Atomic Mass Average

The average atomic mass of an element is the weighted average of the masses of the isotopes of that element, considering their natural abundance. This value is what we typically see on the periodic table. To calculate this average, we consider both the mass of each isotope and its relative abundance in nature.

Using the example of bromine, with isotopes of masses 79 amu and 81 amu, if we have determined the molecular mass of \r\(\mathrm{Br}_2\) and the individual abundances of the isotopes (\r\(p_1\) and \r\(p_2\)), we can compute the average atomic mass. It is essentially half of the average molecular mass of \r\(\mathrm{Br}_2\) because the molecule is composed of two bromine atoms. This average atomic mass is an important concept in chemistry, as it links the microscopic world of atoms to the macroscopic mass measurements we use in everyday laboratory work.

Considering the existing step-by-step solution, understanding how to determine the average atomic mass provides a foundational knowledge that can improve comprehension of the results yielded by mass spectrometry and the calculations of molecular masses and isotope abundances.