There are two different isotopes of bromine atoms. Under normal conditions, elemental bromine consists of \(\mathrm{Br}_{2}\) molecules, and the mass of a \(\mathrm{Br}_{2}\) 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: $$ \begin{array}{cc} \hline \text { Mass (amu) } & \text { Relative Size } \\ \hline 157.836 & 0.2569 \\ 159.834 & 0.4999 \\ 161.832 & 0.2431 \\ \hline \end{array} $$ (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.

Step by step solution

01

Identify the possible isotopes in the mass spectrum

Since we know that the mass spectrum of \(\mathrm{Br}_2\) consists of three peaks, we can assume that there are two isotopes of bromine, which we will call \(Br_1\) and \(Br_2\). The values in the mass spectrum represent the masses of one \(\mathrm{Br}_2\) molecule, composed of two atoms of either \(Br_1\), \(Br_2\) or a combination of both. The possible combinations are: 1. \(Br_1-Br_1\) 2. \(Br_1-Br_2\) 3. \(Br_2-Br_2\) The mass of each combination corresponds to one of the peaks in the mass spectrum table.

02

Assign the mass values to the isotopes

Now we will assign the mass values from the mass spectrum to each combination. The lowest mass peak, 157.836 amu, corresponds to the \(Br_1-Br_1\) combination. The highest mass peak, 161.832 amu, corresponds to the \(Br_2-Br_2\) combination. Finally, the middle mass peak, 159.834 amu, corresponds to the \(Br_1-Br_2\) combination.

03

Calculate the mass of each isotope

From step 2, we have the mass of a \(Br_1-Br_1\) molecule (157.836 amu) and a \(Br_2-Br_2\) molecule (161.832 amu). Since there are two atoms of the same isotope in each molecule, we can find the mass of each isotope by dividing the mass of the molecule by 2: Mass of \(Br_1\) = \(\frac{157.836}{2}\) = 78.918 amu Mass of \(Br_2\) = \(\frac{161.832}{2}\) = 80.916 amu

04

Determine the average molecular mass of \(\mathrm{Br}_2\)

To find the average molecular mass of a \(\mathrm{Br}_2\) molecule, we multiply the mass of each type of molecule by its corresponding relative size and sum up the results: Average molecular mass = \((157.836 \times 0.2569)\) + \((159.834 \times 0.4999)\) + \((161.832 \times 0.2431)\) = 40.59410 + 79.91417 + 39.32410 = 159.83237 \approx 159.832$ amu

05

Determine the average atomic mass of a bromine atom

To determine the average atomic mass of a bromine atom, we simply divide the average molecular mass of \(\mathrm{Br}_2\) by 2: Average atomic mass = \(\frac{159.83237}{2} = 79.91619 \approx 79.916\) amu

06

Calculate the abundances of the two isotopes

To find the abundances of the two isotopes, we can use the relative size values from the mass spectrum. We know that the combined abundances of \(Br_1-Br_1\) and \(Br_1-Br_2\) give the total abundance of \(Br_1\), and the combined abundances of \(Br_2-Br_2\) and \(Br_1-Br_2\) give the total abundance of \(Br_2\). We can represent this with the following equations: Abundance of \(Br_1\) = \(x_1\) = \(0.2569 + 0.4999 = 0.7568\) Abundance of \(Br_2\) = \(x_2\) = \(0.2431 + 0.4999 = 0.7430\) Thus, the abundances of the two isotopes are approximately 75.68% for \(Br_1\) and 74.30% for \(Br_2\).

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