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Antimony, Sb, has two stable isotopes with determined masses of 120.904 amu(Sb¹²¹) and 122.904 amu(Sb¹²³). What are the relative abundance of these Isotopes?

1 answer

Let X = fraction with 120.904
Then 1-X = fraction with 122.904
My periodic table shows 121.775 for atomic mass Sb but you should use the number on your chart or class notes.
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120.904X + (122/904-X) = 121.775
Solve for X = fraction and multiply by 100 to convert to percent.
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