Question

The element bromine is \(\mathrm{Br}_{2}\), so the mass of a \(\mathrm{Br}_{2}\) molecule is the sum of the mass of its two atoms. Bromine has two isotopes. The mass spectrum of \(\mathrm{Br}_{2}\) produces three peaks with relative masses of \(157.836,159.834,\) and 161.832 , and relative heights of \(6.337,12.499,\) and 6.164, respectively. (a) What isotopes of bromine are present in each of the three peaks? (b) What is the mass of each bromine isotope? (c) What is the average atomic mass of bromine? (d) What is the abundance of each of the two bromine isotopes?

Short answer

Isotopes: ^79Br and ^81Br; Masses: 78.918 u and 80.916 u; Average mass: 159.81 u; Abundance: Equal.

Step-by-step solution

Identify isotopes using peak masses

The three peaks correspond to different combinations of the bromine isotopes. The peaks at \(157.836\), \(159.834\), and \(161.832\) suggest that the two isotopes of bromine each differ by \(2\) units in their atomic mass. The isotopes that are possible for bromine based on these masses are \(^{79}\text{Br}\) and \(^{81}\text{Br}\). Assuming \(\text{Br}_{2}\) is composed of two isotopes, the peaks represent \(^{79}\text{Br}-^{79}\text{Br}\), \(^{79}\text{Br}-^{81}\text{Br}\), and \(^{81}\text{Br}-^{81}\text{Br}\).

Determine isotope masses from mass differences

Using the peak masses for the molecules \(^{79}\text{Br}-^{79}\text{Br}\) (\(157.836\)), \(^{79}\text{Br}-^{81}\text{Br}\) (\(159.834\)), and \(^{81}\text{Br}-^{81}\text{Br}\) (\(161.832\)), the difference in mass for one bromine atom can be calculated as follows: \[\text{Difference in masses} = 161.832 - 159.834 = 1.998 \, \text{or approximately } 2 \, \text{u}\]This indicates the isotopes are \(^{79}\text{Br}\) and \(^{81}\text{Br}\).

Calculate mass of each bromine isotope

Using the given peak masses, the calculated individual isotope masses are:- \(^{79}\text{Br} = \frac{157.836}{2} = 78.918\)- \(^{81}\text{Br} = \frac{161.832}{2} = 80.916\)

Calculate average atomic mass

To find the average atomic mass of bromine, use the relative heights as weights for the respective isotope pairs:\[\text{Average mass} = \frac{(6.337 \times 157.836) + (12.499 \times 159.834) + (6.164 \times 161.832)}{6.337 + 12.499 + 6.164} = 159.81 \, \text{u}\]

Determine isotope abundance

Due to the relative heights of the peaks, we form equations based on abundance:- Let \(x\) be the abundance of \(^{79}\text{Br}\) and \(y\) be the abundance of \(^{81}\text{Br}\), then \(x + y = 1\).- Given the peak intensity ratios, convert to percentages: \(\frac{x^2}{y^2} = \frac{6.337}{6.164}\), solve for \(x = 0.5\) and \(y = 0.5\).- Thus, the isotopes are approximately evenly abundant.

Key concepts

Mass Spectrum

A mass spectrum is a graph that displays the distribution of ions by their mass-to-charge ratio. In the case of bromine, it helps us determine the presence and arrangement of isotopes in a bromine molecule. When you look at a mass spectrum, it shows peaks that correspond to different molecular masses. For bromine, a diatomic molecule (\( \text{Br}_2 \)), these peaks indicate the possible combinations of its isotopes. In the given exercise, the peaks were observed at 157.836, 159.834, and 161.832, which are derived from the isotopic combinations of bromine atoms such as:

• \( ^{79}\text{Br}-^{79}\text{Br} \) corresponding to 157.836
• \( ^{79}\text{Br}-^{81}\text{Br} \) corresponding to 159.834
• \( ^{81}\text{Br}-^{81}\text{Br} \) corresponding to 161.832

Mass spectrum analysis is crucial as it specifically highlights the isotopes present and enables us to determine their individual masses and abundance, by analyzing the relative heights of these peaks.

Average Atomic Mass

The average atomic mass of an element is not simply the average of its isotopic masses because it must also take into account the abundance of each isotope. For bromine, calculating the average atomic mass involves the use of the relative abundances of the isotopes which contribute to the bromine molecule.
Using the peak heights from the mass spectrum, we apply these as weights in a weighted average calculation. Given:

• The peak at 157.836 with height 6.337 reflects the possibility of both bromine atoms being \( ^{79}\text{Br} \)
• The peak at 159.834 with height 12.499 is for a mix of \( ^{79}\text{Br} \) and \( ^{81}\text{Br} \)
• The peak at 161.832 with height 6.164 shows both bromine atoms as \( ^{81}\text{Br} \)

The average atomic mass is computed using the relationship:\[\text{Average atomic mass} = \frac{(6.337 \times 157.836) + (12.499 \times 159.834) + (6.164 \times 161.832)}{6.337 + 12.499 + 6.164} = 159.81 \, \text{u}\]This value reflects the combined influence of all isotopic masses and their respective abundance.

Isotope Abundance

Understanding isotope abundance is essential to grasp the natural distribution of isotopes in an element. In bromine, we often find two naturally occurring isotopes, \( ^{79}\text{Br} \) and \( ^{81}\text{Br} \). To determine their abundance, we analyze the relative intensities of the peaks on the mass spectrum.
The problem involves a system of equations derived from the peak intensities ratios since these ratios reflect the relative quantities of the isotopes present. Suppose:

• \( x \) is the proportion (or abundance) of \( ^{79}\text{Br} \)
• \( y \) is for \( ^{81}\text{Br} \)

Assume \( x + y = 1 \) since the isotopes account for the whole bromine sample.
Additionally, with data from the peaks, use equation:\[\frac{x^2}{y^2} = \frac{6.337}{6.164}\] This helps solve for \( x \approx 0.5 \) and \( y \approx 0.5 \), indicating that both isotopes have a roughly equal presence in natural bromine samples. Hence, both \( ^{79}\text{Br} \) and \( ^{81}\text{Br} \) typically exist in nearly equal amounts in naturally occurring bromine.