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 \(\mathrm{Br}_{2}\) 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: $$ \begin{array}{lc} \hline \text { Mass (u) } & \text { Relative Size } \\ \hline 157.836 & 0.2569 \\ 159.834 & 0.4999 \\ 161.832 & 0.2431 \\ \hline \end{array} $$ (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.

Short answer

(a) The peaks correspond to \( ^{79}\text{Br}-^{79}\text{Br}, ^{79}\text{Br}-^{81}\text{Br}, ^{81}\text{Br}-^{81}\text{Br} \). (b) Masses: 78.918 u \( ^{79}\text{Br} \), 80.916 u \( ^{81}\text{Br} \). (c) 160.268 u. (d) 80.134 u. (e) 50.67% \( ^{79}\text{Br} \), 49.33% \( ^{81}\text{Br} \).

Step-by-step solution

Define the Isotopes

Let's determine the isotopes involved. Since bromine typically has two stable isotopes, they are usually denoted as \( ^{79}\text{Br} \) and \( ^{81}\text{Br} \). Each bromine molecule \( \text{Br}_2 \) is composed of a combination of these isotopes.

Analyze Each Peak

The mass peaks in the spectrum correspond to different combinations of isotopes: - The peak at 157.836 u corresponds to \( ^{79}\text{Br} - ^{79}\text{Br} \). - The peak at 159.834 u corresponds to \( ^{79}\text{Br} - ^{81}\text{Br} \) or \( ^{81}\text{Br} - ^{79}\text{Br} \). - The peak at 161.832 u corresponds to \( ^{81}\text{Br} - ^{81}\text{Br} \).

Solve for Mass of Each Isotope

Let's calculate the mass of each isotope. Given: \( 2m_{79} = 157.836 \rightarrow m_{79} = \frac{157.836}{2} = 78.918 \text{ u} \) \( m_{79} + m_{81} = 159.834 \rightarrow m_{81} = 159.834 - m_{79} = 80.916 \text{ u} \)

Calculate Average Molecular Mass

To determine the average molecular mass of \( \text{Br}_2 \), use the formula:\[ \text{Average Molecular Mass} = 157.836 \times 0.2569 + 159.834 \times 0.4999 + 161.832 \times 0.2431 \]Calculating this gives us:\[ \text{Average Molecular Mass} = 160.268 \text{ u} \]

Calculate Average Atomic Mass

The average atomic mass of bromine is half of the average molecular mass of \( \text{Br}_2 \), given that a \( \text{Br}_2 \) molecule has two bromine atoms:\[ \text{Average Atomic Mass} = \frac{160.268}{2} = 80.134 \text{ u} \]

Determine Isotopic Abundance

Let the abundance of \( ^{79}\text{Br} \) be \( x \), thus \( ^{81}\text{Br} \) becomes \( 1-x \). Since the peak at 159.834 u involves both isotopes:\( 2x(1-x) = 0.4999 \) (mixed pairs appear twice) Solve: \[ 2x -2x^2 = 0.4999 \] Solving this quadratic equation gives \( x = 0.5067 \) and hence \( 1-x = 0.4933 \), indicating the abundances as ~50.67% for \( ^{79}\text{Br} \) and ~49.33% for \( ^{81}\text{Br} \).

Key concepts

mass spectrum

In chemistry, a mass spectrum represents a graph showing the distribution of ions based on their mass-to-charge ratio. This essential tool permits scientists to analyze the isotopic composition of elements. By examining the relative intensity of signals, we gain insights into the abundance and masses of different isotopes.
This kind of analysis is crucial for understanding isotopes, which are variants of elements differing in neutron counts. Mass spectrometry assists in identifying these isotopic combinations in molecules, such as Br₂ in bromine. Bromine's mass spectrum will show peaks corresponding to molecular weights from distinct isotope pairings.

• The mass of 157.836 u corresponds to a molecule with two identical isotopes of bromine, such as \(^{79}\text{Br} - ^{79}\text{Br}\).
• The mass of 159.834 u represents a combination of two different isotopes, like \(^{79}\text{Br} - ^{81}\text{Br}\) and vice versa.
• Lastly, 161.832 u arises from \(^{81}\text{Br} - ^{81}\text{Br}\).

Each peak on the mass spectrum offers a piece of the puzzle in determining bromine's isotopic presence.

average atomic mass

The concept of average atomic mass is critical for understanding isotopic mixtures. It demonstrates the weighted mean of an element's isotopes, factoring in their relative abundances. This average gives us an idea of the "average" weight of an atom of that element found in nature.
For bromine, the average atomic mass considers the presence of both \(^{79}\text{Br}\) and \(^{81}\text{Br}\) isotopes. First, we calculate the molecular mass of \(\text{Br}_2\). Then, since each molecule contains two atoms, the average atomic mass can be determined by dividing the molecular mass by two.
Bromine's average atomic mass calculation becomes:\[\frac{160.268\text{ u}}{2} = 80.134\text{ u}\]This indicates that, although an atom doesn't possess this mass individually, it represents the element's average atomic weight found on the periodic table.

bromine isotopes

Bromine exists naturally as two isotopes: \(^{79}\text{Br}\) and \(^{81}\text{Br}\). These isotopes differ in the number of neutrons, offering unique characteristics in scientific analysis. While they have identical chemical behavior, their mass differences hold the key to various calculations and identifications.
Determining the isotopes' exact masses and their abundance can be essential for experiments or industrial applications. Calculating them from given mass spectrum peaks requires solving simultaneous equations derived from these combinations.

• Given \(2m_{79} = 157.836\), each \(^{79}\text{Br}\) isotope has a mass of 78.918 u.
• The equation \(m_{79} + m_{81} = 159.834\) leads to finding \(^{81}\text{Br}\), with a mass of 80.916 u.

Recognizing this subtle difference allows precise work in finer scientific measurements and analyses.

molecular mass calculation

In chemistry, calculating the molecular mass of a compound, like \(\text{Br}_2\), involves summing the atomic masses of its constituent atoms. Here, the goal is to determine the average molecular weight by leveraging isotopic data.
For any element with isotopes, like bromine, the mass spectrum provides the various masses from isotope combinations. By associating these with their relative intensities in the spectrum, an average can be calculated:
\[\text{Average Molecular Mass} = 157.836 \times 0.2569 + 159.834 \times 0.4999 + 161.832 \times 0.2431\]Computing this equation, we get an average molecular mass of 160.268 u.
Such calculations help determine this average mass, offering insights into chemical reactions and behaviors by reflecting the real-world mix of isotopic combinations encountered in natural bromine.