Chapter 5, Problem 40E

Interpretation Introduction

(a)

Interpretation:

The atomic mass of each element is to be calculated from the percentage abundances and atomic masses (u) of the natural isotopes of an element.

Concept introduction:

Elements which have same atomic number but different atomic mass are known as isotopes. The atomic mass of an element is defined as the total mass of one atom of an element. Its unit is atomic mass unit $\left(\text{a}\text{.m}\text{.u}\right)$ and is denoted by the symbol $\text{u}$.

Answer to Problem 40E

The atomic mass of the element is $\text{58}\text{.73}\text{u}$.

Explanation of Solution

The percentage abundances and atomic masses (u) of the natural isotopes of an element are given as shown below.

$\begin{array}{cc}67.88\%& 57.9353\\ 26.23\%& 59.9332\\ 1.19\%& 60.9310\\ 3.66\%& 61.9283\end{array}$

$\begin{array}{cc}1.08\%& 63.9280\end{array}$

The contribution of the isotope to the atomic mass of element is calculated by the formula as shown below.

$\text{Atomic}\text{mass}\text{of}\text{isotope}=\left(\begin{array}{l}\text{Mass}\text{of}\text{the}\text{isotope}\times \\ \text{Relative}\text{abundance}\text{of}\text{the}\text{isotope}\end{array}\right)$…(1)

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

$\begin{array}{rll}\text{Atomic}\text{mass}\text{of}\text{isotope}& =& 57.9353\text{u}\times 67.88\%\\ & =& \left(\frac{57.9353\text{u}\times 67.88}{100}\right)\text{u}\\ & =& 39.32\text{u}\end{array}$

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

$\begin{array}{rll}\text{Atomic}\text{mass}\text{of}\text{isotope}& =& 135.907\text{u}\times 26.23\%\\ & =& \left(\frac{59.9332\times 26.23}{100}\right)\text{u}\\ & =& 15.72\text{u}\end{array}$

Substitute the values of percentage abundance and atomic mass of the third isotope in equation (1).

$\begin{array}{rll}\text{Atomic}\text{mass}\text{of}\text{isotope}& =& 60.9310\text{u}\times 1.19\%\\ & =& \left(\frac{60.9310\times 1.19}{100}\right)\text{u}\\ & =& 0.725\text{u}\end{array}$

Substitute the values of percentage abundance and atomic mass of the fourth isotope in equation (1).

$\begin{array}{rll}\text{Atomic}\text{mass}\text{of}\text{isotope}& =& 61.9283\text{u}\times 3.66\%\\ & =& \left(\frac{61.9283\times 3.66}{100}\right)\text{u}\\ & =& 2.27\text{u}\end{array}$

Substitute the values of percentage abundance and atomic mass of the fifth isotope in equation (1).

$\begin{array}{rll}\text{Atomic}\text{mass}\text{of}\text{isotope}& =& 63.9280\text{u}\times 1.08\%\\ & =& \left(\frac{63.9280\times 1.08}{100}\right)\text{u}\\ & =& 0.690\text{u}\end{array}$

The atomic mass of the element is equal to the sum of atomic masses of its five isotopes.

Therefore, the atomic mass of the element is calculated as shown below.

$\begin{array}{rll}\text{Atomic}\text{mass}& =& 39.32\text{u}+15.72\text{u}+0.725\text{u}+2.27\text{u}+0.690\text{u}\\ & =& \text{58}\text{.73}\text{u}\end{array}$

Therefore, the atomic mass of the element is $\text{58}\text{.73}\text{u}$.

Conclusion

The atomic mass of the element is $\text{58}\text{.73}\text{u}$.

Interpretation Introduction

(b)

Interpretation:

The name of the element is to be identified.

Concept introduction:

Elements which have same atomic number but different atomic mass are known as isotopes. The atomic mass of an element is defined as the total mass of one atom of an element. Its unit is atomic mass unit $\left(\text{a}\text{.m}\text{.u}\right)$ and is denoted by the symbol $\text{u}$.

Answer to Problem 40E

The name of the element is cerium $\left(\text{Ni}\right)$.

Explanation of Solution

The calculated atomic mass of the element is $\text{58}\text{.73}\text{u}$.

In the periodic table, the element having atomic mass $\text{58}\text{.73}\text{u}$ is nickel.

Therefore, the name of the element is nickel $\left(\text{Ni}\right)$.

Conclusion

The element having $\text{58}\text{.73}\text{u}$ atomic mass is nickel $\left(\text{Ni}\right)$.