Rubidium is a soft, silvery-white metal that has two common isotopes, $""^85Rb$ and $""^87Rb$. If the abundance of $""^85Rb$ is 72.2% and the abundance of $""^87Rb$ is 27.8%, what is the average atomic mass of rubidium?

Md Ariful Islam

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Md Ariful Islam

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As you know, the average of an element is calculated by taking the weighted average of the atomic masses of its stable .

More specifically, the average atomic mass of an element is calculated by taking into account the atomic masses of its stable and their decimal abundance, which is imply the percent abundance divided by $100$.

$color(blue)("avg. atomic mass" = sum_i ("isotope"_i xx "abundance"_i))$

So, you know that rubidium has two stable isotopes, $""^85"Rb"$ and $""^87"Rb"$, each with its respective percent abundance.

When the problem doesn't provide the atomic masses of the stable isotopes, you can use their . As you know, an isotope's tells you how many protons and neutrons it contains in its nucleus.

In this case, $""^85"Rb"$ has a mass number equal to $85$ and $""^87"Rb"$ has a mass number equal to $87$. This means that you can approximate the atomic masses of the two isotopes to be $"85 u"$ and $"87 u"$, respectively.

The average atomic mass of rubidium will thus be

$"avg. atomic mass" = "85 u" xx 0.722 + "87 u" xx 0.278$

$"avg. atomic mass " = " 85.556 u"$

Rounded to three , the number of sig figs you have for the percent abundances of the two isotopes, the answer will be

$"avg. atomic mass " = color(green)(" 85.6 u")$

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