What are the relative abundances of Sb isotopes?

The stable isotopes of antimony (Sb) are 121 Sb and 123
Sb

We'll suppose that the  lighter isotope is 121 Sb.
Therefore it must be more abundant.

The atomic mass of
antimony is 121.760

121.760 = 120.904x + 122.904y
(1)

x is fractional abundance of 123
Sb

y is fractional abundance of 121
Sb

y = 1-x

We'll substitute y
= 1-x in (1):

121.760 = 120.904x +
122.904(1-x)

We'll remove the
brackets:

121.760 = 120.904x + 122.904 -
122.904x

We'll combine like terms and we'll solve for
x:

-2x = -1.144

x =
0.5720

The fractional abundance of 121 Sb is
0.5720. That means that the fractional abundance of 121 Sb is
0.4280.