Question

The mass spectrum of bromine $\left({\mathrm{Br}}_2\right)$consists of three peaks with the following relative sizes:

Mass(u)	Relative size
157.84	0.2534
159.84	0.5000
161.84	0.2466

How do you interpret these data, and what is the relative abundance of the bromine isotopes?

Short answer

$$\begin{gathered} \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}78.92\mathrm{u}=\frac{\mathrm{mass}\mathrm{of}\mathrm{isotope}}{\mathrm{Total}\mathrm{mass}\mathrm{of}\mathrm{all}\mathrm{isotopes}}\times 100\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}78.92\mathrm{u}=\frac{78.92\mathrm{u}}{159.84\mathrm{u}}\times 100\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}78.92\mathrm{u}=49.37\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}80.92\mathrm{u}=\frac{\mathrm{mass}\mathrm{of}\mathrm{isotope}}{\mathrm{Total}\mathrm{mass}\mathrm{of}\mathrm{all}\mathrm{isotopes}}\times 100\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}80.92\mathrm{u}=\frac{80.92\mathrm{u}}{159.84\mathrm{u}}\times 100\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}80.92\mathrm{u}=50.63\% \\ \mathrm{The}\mathbf{isotope}\mathbf{}\mathbf{percent}\mathbf{}\mathbf{abundance}\mathrm{is}49.37\%\mathrm{and}50.63\% \end{gathered}$$

Step-by-step solution

Definition of mass spectrum

In the mass spectrum, the relative size or height of the peak indicates % abundance. The mass spectrum of bromine ${\mathbf{Br}}_{\mathbf{2}}$consists of three peaks.

Calculation of mass

The mass and their corresponding peaks of ${\mathbf{Br}}_{\mathbf{2}}$are as follows:

Mass(u)	Relative size
157.84	0.2534
159.84	0.5000
161.84	0.2466

The size of the peak corresponding to 159.84u is twice the size of peaks corresponding to 157.84u and 161.84u.

The lighter isotope of bromine gives the peak at a lower mass. The heavier isotope gives the peak at a higher mass. The peak at 157.84 u is due to the combination of two bromine atoms, and the mass of each bromine atom is half of 157.84u, that is,

$\frac{157.84}{2}=78.92\mathrm{U}$

Mass of bromine isotopes

The mass of bromine atoms that gives the peak at 161.84u is.

$\frac{161.84}{2}=80.92u$

The mass 159.84u is the average of the remaining two masses (157.84u+161.84u). So, the peak at 159.84u corresponds to the bromine molecule that contains the bromine atoms of mass 78.92u and 80.92u.

  Calculation of average molecular mass

$$\begin{gathered} \mathrm{Average}\mathrm{molecular}\mathrm{mass}=\underset{\mathrm{i}=1}{\overset{\mathrm{n}}{\sum \left({\mathrm{mass}}_{\mathrm{i}}\right)}}\left(\mathrm{relative}{\mathrm{abundance}}_{\mathrm{i}}\right) \\ \mathrm{Average}\mathrm{molecular}\mathrm{mass}=\left(157.84\mathrm{u}\times 0.2534\right)+\left(159.84\mathrm{u}\times 0.5000\right)+\left(161.84\mathrm{u}\times 0.2466\right) \\ \mathrm{Average}\mathrm{molecular}\mathrm{mass}=159.83\mathrm{u} \\ \mathrm{Average}\mathrm{molecular}\mathrm{mass}=\frac{159.83\mathrm{u}}{2} \\ \mathrm{Average}\mathrm{molecular}\mathrm{mass}=79.91\mathrm{u} \end{gathered}$$

 Calculation of percent abundance

$$\begin{gathered} \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}78.92\mathrm{u}=\frac{\mathrm{mass}\mathrm{of}\mathrm{isotope}}{\mathrm{Total}\mathrm{mass}\mathrm{of}\mathrm{isotopes}}\times 100\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}78.92\mathrm{u}=\frac{78.92\mathrm{u}}{159.84\mathrm{u}}\times 100\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}78.92\mathrm{u}=49.37\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}80.92\mathrm{u}=\frac{\mathrm{mass}\mathrm{of}\mathrm{isotope}}{\mathrm{Total}\mathrm{mass}\mathrm{of}\mathrm{isotopes}}\times 100\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}80.92\mathrm{u}=\frac{80.92\mathrm{u}}{159.84\mathrm{u}}\times 100\% \\ \mathrm{Percent}\mathrm{abundance}\mathrm{of}\mathrm{Br}\mathrm{with}\mathrm{mass}80.92\mathrm{u}=50.63\% \end{gathered}$$

Interpretation

The mass spectrum of each peak differs by two units, and bromine is two isotopes that differ by units. The two isotopes masses are 78.92 u and 80.92 u. Then, the average molecular masses are${\mathrm{Br}}_2=159.83\mathrm{u},\mathrm{Br}=79.91$