Chapter 2, Problem 2.39QP

Interpretation Introduction

Interpretation:

The atomic mass of ${}^{\text{24}}\text{Mg}$ should be calculated.

Concept introduction:

Atomic mass is the total material present in one atom of an element. Namely,

1. (1) Protons
2. (2) Electrons
3. (3) Neutrons

The Average atomic mass of an element can be calculated by using given formula,

$\text{Average atomic mass= }\sum \left(\text{Mass of isotope × Percent abundance}\right)$

To find: The atomic mass of ${}^{\text{24}}\text{Mg}$.

Answer to Problem 2.39QP

Solution

The atomic mass of ${}^{\text{24}}\text{Mg}$ $\text{= 23}\text{.98}\text{amu }$

Explanation of Solution

The atomic mass of ${}^{\text{24}}\text{Mg}$ is given below,

$\text{Average atomic mass= }\sum \left(\text{Mass of isotope × Percent abundance}\right)$

Multiplying the mass of an isotope by its fractional abundance will give the contribution to the average atomic mass of that particular isotope.

The atomic masses of ${}^{\text{25}}\text{Mg =}24.9858\text{ amu}\text{=}\text{10\% }$

The atomic masses of ${}^{\text{26}}\text{Mg =}25.9825\text{ amu}\text{=}\text{11}\text{.01\% }$

The atomic masses of ${}^{\text{24}}\text{Mg =}78.99\text{\% }$

The atomic mass of ${}^{\text{24}}\text{Mg}$, x, given the contribution to the average atomic mass of ${}^{\text{25}}\text{Mg}$ and ${}^{\text{26}}\text{Mg}$, and the average atomic mass of magnesium.

$\begin{array}{l}{\text{The fractional abundance of }}^{\text{25}}\text{Mg =}\text{10\% }\text{=}\frac{\text{10}\text{.0}}{\text{100}}\\ \text{= 0}\text{.1}\end{array}$

$\begin{array}{l}{\text{The fractional abundance of }}^{\text{26}}\text{Mg =}\text{11}\text{.01\% }\text{=}\frac{\text{11}\text{.01}}{\text{100}}\\ \text{= 0}\text{.1101}\end{array}$

$\begin{array}{l}\text{Atomic mass calculation is given below,}\\ \left(\text{0}\text{.1000}\right)\left(\text{24}{\text{.9858 amu }}^{\text{25}}\text{Mg}\right)\text{+}\left(\text{0}\text{.1101}\right)\left(\text{25}{\text{.9825 amu }}^{\text{26}}\text{Mg}\right)\text{+}\left(\text{0}\text{.7899}\right)\left({\text{x amu}}^{\text{24}}\text{Mg}\right)\text{=}\text{24}\text{.3050}\\ \text{ 2}\text{.4985}\text{+}\text{2}\text{.8606}\text{+}\text{0}\text{.7899}\text{(x)}\text{= 24}\text{.3050}\\ \text{0}\text{.7899}\text{(x)}\text{=}\text{24}\text{.3050}\text{-}5.3591\\ \text{0}\text{.7899}\text{(x)}\text{=}\text{18}\text{.9459}\\ \text{(x)}\text{=}\frac{\text{18}\text{.9459}}{\text{0}\text{.7899}}\\ \text{(x)}\text{=}\text{23}\text{.98}\text{amu}\\ \end{array}$

Conclusion

Conclusion

The atomic mass of ${}^{\text{24}}\text{Mg}$ is calculated.