The element antimony has an atomic weight of 121.757 u and only two naturally occurring isotopes. One isotope has an abundance of 57.3% and an isotopic mass of 120.904 u. What is the mass of the other isotope?

Md Ariful Islam

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Md Ariful Islam

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The mass of antimony's second naturally occurring is $"122.902 u"$

The thing to know when doing isotope abundance problems is that the abundances of the two must add up to 100%. This implies that the abundance of antimony's second naturally occurring isotope must be

$"abundance isotope 2" = "100%" - "57.3%" = "42.7%"$

The average of an element can be written as the sum of the isotopic weights of its isotopes, each multiplied by its respective abundance.

It's useful to write percent abundances as decimal abundances; in this case, the two decimal abundances must add up to 1 and are determined by dividing the percent abundances by 100.

So, the equation will be

$"0.427" * "x" + "0.573" * "120.904" = "121.757"$

where $"x"$ is the isotopic mass of the second isotope. Solving for $"x"$ will give you

$"x" = (121.757 - 0.573 * 120.904)/0.427 = color(green)("122.902 u")$

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