Answered

Below lists the natural abundances of the bromine isotopes. The average atomic mass of bromine [tex](Br)[/tex] is 79.90 amu. Given this information, what is the mass contribution of [tex]Br-79[/tex]?

\begin{tabular}{|l|l|l|}
\hline Isotope & Relative Atomic Mass & Natural (Fractional) Abundance (\%) \\
\hline Bromine-79 & 79.918 & 50.69 \\
\hline Bromine-81 & 80.916 & 49.31 \\
\hline
\end{tabular}

A. [tex]40.51 \, \text{amu}[/tex]
B. [tex]39.90 \, \text{amu}[/tex]
C. [tex]39.41 \, \text{amu}[/tex]
D. [tex]41.02 \, \text{amu}[/tex]



Answer :

GinnyAnswer
To determine the mass contribution of [tex]\(Br-79\)[/tex], we can use the information given about its relative atomic mass and its natural abundance. Here's the step-by-step solution:

1. Identify the relative atomic mass of [tex]\(Br-79\)[/tex]:
The relative atomic mass of [tex]\(Br-79\)[/tex] is [tex]\(79.918\)[/tex] amu.

2. Identify the natural (fractional) abundance of [tex]\(Br-79\)[/tex]:
The natural abundance of [tex]\(Br-79\)[/tex] is given as [tex]\(50.69\%\)[/tex]. To use this value in calculations, we need to convert it into a fraction by dividing by 100.
[tex]\[ \text{Fractional abundance of } Br-79 = \frac{50.69}{100} = 0.5069 \][/tex]

3. Calculate the mass contribution of [tex]\(Br-79\)[/tex]:
The mass contribution of an isotope can be calculated by multiplying the relative atomic mass by its fractional abundance.
[tex]\[ \text{Mass contribution of } Br-79 = \text{Relative atomic mass of } Br-79 \times \text{Fractional abundance of } Br-79 \][/tex]
[tex]\[ \text{Mass contribution of } Br-79 = 79.918 \times 0.5069 \][/tex]

4. Determine the numerical result:
Performing the multiplication:
[tex]\[ \text{Mass contribution of } Br-79 \approx 79.918 \times 0.5069 \approx 40.51 \text{ amu} \][/tex]

Therefore, the mass contribution of [tex]\(Br-79\)[/tex] is approximately [tex]\(40.51\)[/tex] amu.

Given the options:
- [tex]\(40.51 \text{ amu}\)[/tex] (Correct Answer)
- [tex]\(39.90 \text{ amu}\)[/tex]
- [tex]\(39.41 \text{ amu}\)[/tex]
- [tex]\(41.02 \text{ amu}\)[/tex]

The correct answer is [tex]\(40.51 \text{ amu}\)[/tex].