Respuesta :
Answer:
The relative atomic mass of [tex]{\rm Br}[/tex] is [tex]79.904[/tex].
The natural abundance of [tex]{\rm Br}\text{-}79[/tex] would be approximately [tex]49.31\%[/tex].
The mass of [tex]{\rm Br}\text{-}79[/tex] would be approximately [tex]78.919[/tex].
Explanation:
The natural abundance of all naturally occurring isotopes of an element (e.g., bromine) should add up to [tex]1[/tex]. Since the relative abundance of [tex]{\rm Br}\text{-}81[/tex] is [tex]50.69\%[/tex] ([tex]0.5069[/tex],) the relative abundance of [tex]{\rm Br}\text{-}79[/tex] (the only other naturally occurring isotope of bromine) should be approximately [tex]1 - 0.5069 = 0.4931[/tex].
The relative atomic mass of an element is the average of the mass of all its isotopes, weighted by the relative abundance of each isotope. It is given that the atomic mass of [tex]{\rm Br}\text{-}81[/tex] is [tex]80.9163[/tex]. If the atomic mass of [tex]{\rm Br}\text{-}79[/tex] is [tex]x[/tex], the relative atomic mass of [tex]{\rm Br}[/tex] would be:
[tex](0.5069)\, x + (1 - 0.5069)\, (80.9163)[/tex].
The expression above should be equal to the relative atomic mass of [tex]{\rm Br}[/tex] on the periodic table of elements, [tex]79.904[/tex]. Thus:
[tex]0.5069\, x + (1 - 0.5069)\, (80.9163) = 79.904[/tex].
Solving for [tex]x[/tex] gives:
[tex]\begin{aligned} x &= \frac{79.094 - (1 - 0.5069)\, (80.9163)}{0.5069} \approx 78.919 \end{aligned}[/tex].
Hence, the relative atomic mass of [tex]{\rm Br}\text{-}79[/tex] would be approximately [tex]78.919[/tex].