Question

Bromine has two naturally occurring isotopes (Br79(\mathrm{Br}-79 and Br-8181) and an atomic mass of 79.90479.904 amu.

(a) If the natural abundance of Br79\mathrm{Br}-79 is 50.69%50.69 \%, what is the natural abundance of Br-8181?

(b) If the mass of Br-8181 is 80.9163amu80.9163 \mathrm{amu}, what is the mass of Br-7979?

Solution

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a.\textbf{a.} The sum of all relative abundances of all isotopes of an element has to be equal to 100%.

100%=i=1nXi\begin{equation} 100 \% = \sum^{n}_{i=1}{X_{i}} \end{equation}

Where nn is the number of all naturally occurring isotopes and XiX_{i} is the relative abundance of a particular isotope.

So for this example. There are two naturally occurring isotopes (n=2n=2). If we know the abundance of one of them, we can calculate the abundance of the other:

1=XBr79+XBr81=0,5069+XBr81\begin{equation} 1 = X_{Br-79} + X_{Br-81} = 0,5069 + X_{Br-81} \end{equation}

Using simple algebra we can calculate XBr81X_{Br-81}

XBr81=10,5069=0,4931=49,31%\begin{equation} X_{Br-81} = 1-0,5069 = 0,4931 = 49,31\% \end{equation}

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