Question

Given that the abundances of isotopes \[{}^{54}{\text{Fe}}\], \[{}^{56}{\text{Fe}}\] and \[{}^{57}{\text{Fe}}\] are $5\% $, $90\% $ and $5\% $ respectively, the atomic mass of ${\text{Fe}}$ is:
A. $55.85{\text{ u}}$
B. $55.95{\text{ u}}$
C. $55.75{\text{ u}}$
D. $55.05{\text{ u}}$

Answer 1

Hint: The average atomic mass of any element is the summation of masses of all the isotopes of the elements multiplied by their natural abundance. Thus, the equation to calculate average atomic mass is,
${\text{Average atomic mass}} = \dfrac{{\sum\limits_{i = 1}^n {\left( {{\text{mas}}{{\text{s}}_{\left( i \right)}}} \right)} \times \left( {{\text{percent isotopic abundance}}{{\text{e}}_{\left( i \right)}}} \right)}}{{100}}$.

Complete step by step answer:
Calculate the average atomic mass of ${\text{Fe}}$ using the equation as follows:
${\text{Average atomic mass}} = \dfrac{{\sum\limits_{i = 1}^n {\left( {{\text{mas}}{{\text{s}}_{\left( i \right)}}} \right)} \times \left( {{\text{percent isotopic abundance}}{{\text{e}}_{\left( i \right)}}} \right)}}{{100}}$
Thus,
${\text{Average atomic mass of Fe}} = \dfrac{{({\text{Mass of }}{}^{54}{\text{Fe}} \times {\text{% IA of }}{}^{54}{\text{Fe}}) + ({\text{Mass of }}{}^{56}{\text{Fe}} \times {\text{% IA of }}{}^{56}{\text{Fe)}} + ({\text{Mass of }}{}^{57}{\text{Fe}} \times {\text{% IA of }}{}^{57}{\text{Fe)}}}}{{100}}$
Substitute ${\text{54}}$ for the atomic mass of isotope \[{}^{54}{\text{Fe}}\], ${\text{56}}$ for the atomic mass of isotope \[{}^{56}{\text{Fe}}\], ${\text{57}}$ for the atomic mass of isotope \[{}^{57}{\text{Fe}}\], $5\% $ for the percent isotopic abundance of \[{}^{54}{\text{Fe}}\], $90\% $ for the percent isotopic abundance of \[{}^{56}{\text{Fe}}\], $5\% $ for the percent isotopic abundance of \[{}^{57}{\text{Fe}}\]. Thus,
${\text{Average atomic mass of Fe}} = \dfrac{{\left( {54 \times 5\% } \right) + \left( {56 \times 90\% } \right) + \left( {57 \times 5\% } \right)}}{{100}}$
$ = \dfrac{{270 + 5040 + 285}}{{100}}$
${\text{Average atomic mass of Fe}} = 55 \cdot 95{\text{ u}}$
Thus, the average atomic mass of ${\text{Fe}}$ is $55 \cdot 95{\text{ u}}$.

So, the correct answer is Option B .

Additional Information:
The species of atoms having the same atomic number but different atomic mass are known as isotopes. Every element in the periodic table has at least one isotope. Thus, the isotopes of iron $\left( {{\text{Fe}}} \right)$ are \[{}^{54}{\text{Fe}}\], \[{}^{56}{\text{Fe}}\] and \[{}^{57}{\text{Fe}}\].
Some isotopes are radioactive in nature. Radioactive elements are unstable and radiate excess energy. Examples of radioactive isotopes are: $^{\text{3}}{\text{H}}$, $^{{\text{14}}}{\text{C}}$, etc.

Note:
The atomic mass of an element is the combined mass of all the protons and neutrons. The average atomic mass of an element is the mean product of all the isotopes of the element and their percent isotopic abundances.