Chapter 5, Problem 57E

The element lanthanum has two stable isotopes, lanthanum $-138$ with an atomic mass of $\text{137}\text{.9071}\text{u}$ and, lanthanum $-139$ with an atomic mass of $\text{138}\text{.9063}\text{u}$. From atomic mass of $\text{La}$, $\text{138}\text{.9}\text{u}$ what conclusion can you make about the relative percentage abundance of the isotopes?

Interpretation Introduction

Interpretation:

The relative percentage abundance of two isotopes of lanthanum is to be calculated.

Concept introduction:

Every atom of particular element has the same number of protons. This number of proton is termed as atomic number of the element. It represented by the symbol $\text{Z}$. Atoms of the same elements that have different masses and different number of neutrons are known as isotopes.

Answer to Problem 57E

The percentage abundance of lanthanum $-138$ and lanthanum $-139$ are $0.63\%$, and $\text{99}\text{.37}\text{\%}$ respectively.

Explanation of Solution

The relation between atomic mass of the atom and atomic mass of their isotopes is shown by an equation given below.

$\text{Atomic}\text{mass}=\frac{\left(\begin{array}{c}\left(\text{\%}\text{of}\text{the}\text{first}\text{isotope}\times \text{atomic}\text{mass}\text{of}\text{the}\text{first}\text{isotope}\right)\\ \text{+}\left(\begin{array}{c}\text{\%}\text{of}\text{the}\text{second}\text{isotope}\times \text{atomic}\text{mass}\text{of}\text{the}\text{second}\\ \text{isotope}\end{array}\right)\end{array}\right)}{100}$…(1)

It is given that the atomic mass of lanthanum is $\text{138}\text{.9}\text{u}$.

The atomic mass of lanthanum $-138$ is given as $\text{137}\text{.9071}\text{u}$.

The atomic mass of lanthanum $-139$ is given as $\text{138}\text{.9063}\text{u}$.

The percentage abundance of the lanthanum $-138$ is taken as $\text{x}$.

Then the percentage abundance of the lanthanum $-139$ is calculated by the formula given below.

$\text{\%}\text{abundance}\text{of}\text{lanthanum}-138+\%\text{abundance}\text{of}\text{lanthanum}-139=100$…(2)

Rearrange the equation (2).

$\%\text{abundance}\text{of}\text{lanthanum}-139=100-\text{\%}\text{abundance}\text{of}\text{lanthanum}-138$…(3)

Substitute the value of percentage of first isotope in equation (3).

$\%\text{abundance}\text{of}\text{lanthanum}-139=100-\text{x}$…(4)

Substitute the values of atomic mass, percentage of the first isotope, atomic mass of the first isotope and percentage of the second isotope in equation (1).

$\begin{array}{rll}\text{Atomic}\text{mass}\text{of}\text{lanthanum}& =& \frac{\left(\begin{array}{l}\left(\begin{array}{c}\text{\%}\text{abundance}\text{of}\text{lanthanum}-138\times \\ \text{atomic}\text{mass}\text{of}\text{lanthanum}-138\end{array}\right)\text{+}\\ \left(\begin{array}{l}\text{\%}\text{abundance}\text{of}\text{lanthanum}-139\times \\ \text{atomic}\text{mass}\text{lanthanum}-139\end{array}\right)\end{array}\right)}{100}\\ & =& \frac{\left(\begin{array}{l}\text{x}\times \text{137}\text{.9071}\text{u}+\left(100-\text{x}\right)\\ \times \text{138}\text{.9063}\text{u}\end{array}\right)}{100}\\ & =& \frac{\left(\begin{array}{l}\text{x}\times \text{137}\text{.9071}\text{u}+100\times \text{138}\text{.9063}\text{u}\\ -\text{x}\times \text{138}\text{.9063}\text{u}\end{array}\right)}{100}\\ & =& \frac{\left(\text{13890}\text{.63}-0.9992\text{x}\right)\text{u}}{100}\end{array}$…(5)

Substitute the value of atomic mass of lanthanum and rearrange the equation (5) in terms of $\text{x}$.

$\begin{array}{rll}\text{138}\text{.9}\text{u}& =& \frac{\left(\text{13890}\text{.63}-0.9992\text{x}\right)\text{u}}{100}\\ 0.9992\text{x}& =& \text{13890}\text{.63}-\text{13890}\\ & =& \text{0}\text{.63}\\ \text{x}& =& \frac{\text{0}\text{.63}}{0.9992}\end{array}$

$\text{x}=0.63\%$

Substitute the value of $\text{x}$ in equation (4).

$\begin{array}{rll}\%\text{abundance}\text{of}\text{lanthanum}-139& =& 100-\text{0}\text{.63}\\ & =& \text{99}\text{.37}\text{\%}\end{array}$

Therefore, the percentage abundance of lanthanum $-138$ and lanthanum $-139$ are $0.63\%$, and $\text{99}\text{.37}\text{\%}$ respectively.

Conclusion

The percentage abundance of lanthanum $-138$ and lanthanum $-139$ are calculated as $0.63\%$ and $\text{99}\text{.37}\text{\%}$ respectively.