Questions: There are only two naturally-occuring stable isotopes of thallium, the masses of which are listed in the table below. Use whatever data you need from the ALEKS Periodic Table to calculate the natural abundance of each isotope and complete the table. Round your entry for ^203Ti to 2 significant digits and your entry for ^205Tl to 2 significant digits. isotope mass (amu) natural abundance --------- ^203Tl 202.97 % ^205Tl 204.97 %

There are only two naturally-occuring stable isotopes of thallium, the masses of which are listed in the table below. Use whatever data you need from the ALEKS Periodic Table to calculate the natural abundance of each isotope and complete the table.
Round your entry for ^203Ti to 2 significant digits and your entry for ^205Tl to 2 significant digits.

isotope  mass (amu)  natural abundance
---------
^203Tl  202.97  % 
^205Tl  204.97  %
Transcript text: There are only two naturally-occuring stable isotopes of thallium, the masses of which are listed in the table below. Use whatever data you need from the ALEKS Periodic Table to calculate the natural abundance of each isotope and complete the table. Round your entry for ${ }^{203} \mathrm{Ti}$ to 2 significant digits and your entry for ${ }^{205} \mathrm{Tl}$ to 2 significant digits. \begin{tabular}{|c|c|c|} \hline isotope & \begin{tabular}{c} mass \\ (amu) \end{tabular} & \begin{tabular}{c} natural \\ abundance \end{tabular} \\ \hline${ }^{203} \mathrm{Tl}$ & 202.97 & $\square \%$ \\ \hline${ }^{205} \mathrm{Tl}$ & 204.97 & $\square \%$ \\ \hline \end{tabular}
failed

Solution

failed
failed

Solution Steps

Step 1: Determine the Average Atomic Mass of Thallium

The average atomic mass of thallium (Tl) can be found on the periodic table. For this calculation, we will use the average atomic mass of thallium, which is approximately 204.38 amu.

Step 2: Set Up the Equations

Let \( x \) be the natural abundance of \( {}^{203}\mathrm{Tl} \) and \( y \) be the natural abundance of \( {}^{205}\mathrm{Tl} \). Since these are the only two isotopes, their abundances must add up to 100%:

\[ x + y = 100 \]

The average atomic mass can be expressed as a weighted average of the masses of the isotopes:

\[ 202.97x + 204.97y = 204.38 \times 100 \]

Step 3: Solve the System of Equations

First, express \( y \) in terms of \( x \):

\[ y = 100 - x \]

Substitute \( y \) into the weighted average equation:

\[ 202.97x + 204.97(100 - x) = 20438 \]

Simplify and solve for \( x \):

\[ 202.97x + 20497 - 204.97x = 20438 \]

\[ -2x + 20497 = 20438 \]

\[ -2x = -59 \]

\[ x = 29.5 \]

Now, solve for \( y \):

\[ y = 100 - 29.5 = 70.5 \]

Final Answer

The natural abundance of each isotope is:

\[ \boxed{{}^{203}\mathrm{Tl} = 29.5\%} \]

\[ \boxed{{}^{205}\mathrm{Tl} = 70.5\%} \]

Was this solution helpful?
failed
Unhelpful
failed
Helpful