Question

The element antimony has an atomic weight of 121.757 amu and only two naturally-occurring isotopes. One isotope, 121Sb, has an isotopic mass of 120.904 amu, and the other isotope 123Sb has a mass of 122.902 amu. Based on these data, what is the relative abundance of the heavier isotope?

Answer 1

43%

Step-by-step explanation:

As the problem says that the antimony has only two isotopes, lets call each isotope the following:

x=abundance of isotope 121Sb

1-x=abundance of isotope 123Sb

And

Atomic weight of antimony = (isotopic mass of 121Sb*x)+(isotopic mass of 123Sb*(1-x))

Replacing values we have:

`121.757=(120.904x)+(122.902(1-x))`

Solving for x:

`121.757=120.904x+122.902-122.902x`

`121.757=-1.998x+122.902`

`121.757-122.902=-1.998x`

`-1.145=-1.998x`

`x=(-1.145)/(-1.998)`

`x=0.57`

it means that the abundance of the isotope 121Sb is 57% and the abundance of isotope 123Sb is 43%