QuestionAntimony has isotopes ${ }^{121} \mathrm{Sb}$ (120.9038 u) and ${ }^{123} \mathrm{Sb}$ (122.9042 u). Find their percent abundances.

Studdy Solution

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 $x$ .
$x =100 - y$

STEP 10

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%.

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