Chapter 4, Problem 116A

Interpretation Introduction

Interpretation:

The mass number, X, of the third isotope needs to be determined.

Concept introduction:

The isotope contains different numbers of neutron and same number of proton. So, the atomic mass of the isotope is different. The calculation of the atomic mass of elements is termed as the average atomic mass.

$\text{Average}\text{atomic}\text{mass}\text{=}\text{(}\text{mass}\text{of}\text{isotope×relative}\text{abundance)}$

Answer to Problem 116A

The mass number for the third isotope of magnesium is $26\text{amu}$.

Explanation of Solution

Given information:

The abundances of the natural occurring isotopes are $79.0\%,10\%$ and $11\%$ for $Mg$, $Mg$, $Mg$ respectively.

$\begin{array}{l}\text{Average}\text{atomic}\text{mass}\text{=}\\ \text{(mass}\text{of}\text{M}{\text{g}}^{\text{24}}\text{×relative}\text{abundance)+(mass}\text{of}\text{M}{\text{g}}^{\text{25}}\text{×relative}\text{abundance}\text{)+(mass}\text{of}\text{M}{\text{g}}^{\text{X}}\text{×relative}\text{abundance)}\end{array}$

$24.305=\left(24\right)\left(0.790\right)+\left(25\right)\left(0.10\right)+X\left(0.11\right)$

$24.305=18.96+2.5+X\left(0.11\right)$

$24.305–18.96–2.5=X\left(0.11\right)$

$2.845=X\left(0.11\right)$

$X=26\text{ amu}$

Conclusion

The value of $X=26\text{ amu}$