Math Snap

STEP 1

Assumptions1. Antimony has two naturally occurring isotopes, ${ }^{121} \mathrm{b}$ and ${ }^{123} \mathrm{b}$.
. The atomic mass of ${ }^{121} \mathrm{b}$ is $120.9038 \mathrm{u}$.
3. The atomic mass of ${ }^{123} \mathrm{b}$ is $122.904 \mathrm{u}$.
4. The average atomic mass of Antimony is $121.7601 \mathrm{u}$.
5. The sum of the percent natural abundances of the two isotopes is100%.

STEP 2

Let's denote the percent natural abundance of ${ }^{121} \mathrm{b}$ as $x$ and that of ${ }^{123} \mathrm{b}$ as $y$. From assumption5, we have the following equation$$x + y =100$$

STEP 3

The average atomic mass of Antimony is a weighted average of the atomic masses of its isotopes, with the weights being their percent natural abundances. This gives us the following equation$$120.9038x +122.9042y =121.7601 \times100$$

STEP 4

We now have a system of two equations, which we can solve to find the values of $x$ and $y$.
$$\begin{cases} x + y =100 \\120.9038x +122.9042y =121.7601 \times100 \end{cases}$$

STEP 5

Let's first solve the second equation for $x$.
$$x = \frac{121.7601 \times100 -122.9042y}{120.9038}$$

STEP 6

Now, substitute the value of $x$ from the second equation into the first equation.
$$\frac{121.7601 \times100 -122.9042y}{120.9038} + y =100$$

STEP 7

olve this equation for $y$.
$$y = \frac{121.7601 \times100 -120.903 \times100}{122.9042 -120.903}$$

STEP 8

Calculate the value of $y$.
$$y = \frac{121.7601 \times100 -120.9038 \times100}{122.9042 -120.9038} \approx57.36$$

STEP 9

Now, substitute the value of $y$ into the first equation to find the value of $x$.
$$x =100 - y$$

Calculate the value of $x$.
$$x =100 -57.36 =42.64$$The percent natural abundance of ${ }^{121} \mathrm{b}$ is approximately42.64% and that of ${ }^{123} \mathrm{b}$ is approximately57.36%.