Question

antimony has two naturally occuring isotopes, (^{121}\text{sb}) and (^{123}\text{sb}). (^{121}\text{sb}) has an atomic mass of 120.9038 u, and (^{123}\text{sb}) has an atomic mass of 122.9042 u. antimony has an average atomic mass of 121.7601 u. what is the percent natural abundance of each isotope?
(^{121}\text{sb}): (\boxed{quad}) %
(^{123}\text{sb}): (\boxed{quad}) %

Explanation:

Step1: Define variables

Let \( x \) be the percent abundance of \( ^{121}\text{Sb} \) (in decimal form), so the percent abundance of \( ^{123}\text{Sb} \) is \( 1 - x \) (in decimal form).

Step2: Set up the average atomic mass equation

The average atomic mass is the sum of the product of each isotope's atomic mass and its percent abundance (in decimal). So:
\[
120.9038x + 122.9042(1 - x) = 121.7601
\]

Step3: Expand and simplify the equation

\[
120.9038x + 122.9042 - 122.9042x = 121.7601
\]
\[
120.9038x - 122.9042x = 121.7601 - 122.9042
\]
\[
-2.0004x = -1.1441
\]

Step4: Solve for \( x \)

\[
x = \frac{-1.1441}{-2.0004} \approx 0.5719
\]

Step5: Convert to percent and find the other abundance

Percent abundance of \( ^{121}\text{Sb} \): \( 0.5719 \times 100 = 57.19\% \)
Percent abundance of \( ^{123}\text{Sb} \): \( 100 - 57.19 = 42.81\% \)

Answer:

\( ^{121}\text{Sb} \): \( \boldsymbol{57.19\%} \)
\( ^{123}\text{Sb} \): \( \boldsymbol{42.81\%} \)