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

Answer:

43%

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:

$121.757=120.904x+122.902-122.902x$

$121.757=-1.998x+122.902$

$121.757-122.902=-1.998x$

$-1.145=-1.998x$

$x=\frac{-1.145}{-1.998}$

$x=0.57$

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