Final answer:
The percent abundance of antimony isotopes is found by setting up an equation using the known average atomic mass and individual masses of isotopes, which results in approximately 57.21% for 121Sb and 42.79% for 123Sb.
Step-by-step explanation:
To calculate the percent abundance of antimony isotopes, we can use the average atomic mass of antimony (121.760 amu) and the masses of the two isotopes: 121Sb (120.904 amu) and 123Sb (122.904 amu). Let the abundance of 121Sb be x and consequently, the abundance of 123Sb will be (1-x) since the total abundance must equal 100%.
The average atomic mass is calculated with the equation:
Average atomic mass = (x)(mass of 121Sb) + ((1-x))(mass of 123Sb)
Inserting the given values, we get:
121.760 amu = (x)(120.904 amu) + ((1-x))(122.904 amu)
By solving this equation for x, we can find the percent abundance of isotope 121Sb, and by subtracting x from one, we find the percent abundance of isotope 123Sb.
After solving, we find that the abundance of 121Sb is approximately 57.21% and the abundance of 123Sb is approximately 42.79%.
Therefore the percent abundances of the isotopes are:
- 121Sb: 57.21%
- 123Sb: 42.79%