Chapter 4, Problem 128A

Interpretation Introduction

Interpretation: The isotope to which the average atomic mass of zirconium is closest needs to be determined.

Concept introduction:

The isotope contains a different number of neutrons and the same number of protons. 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 128A

The calculated average atomic mass of zirconium $=90.93$ is closest to the isotope zirconium $-90$ .

Explanation of Solution

The given table is:

The Relative abundance of zirconium isotopes
Element	Mass (amu)	Abundance (%)
Zirconium-90	$89.905$	$51.45$
Zirconium-91	$90.906$	$11.22$
Zirconium-92	$91.905$	$17.15$
Zirconium-94	$93.906$	$17.38$
Zirconium-96	$95.908$	$2.8$

For the calculation of the average atomic mass of zirconium, the mass and abundance are taken from the given table.

Where m = mass of each isotope and the abundance value is given in percentage form so fraction value is used for the calculation.

The average atomic mass can be calculated as follows:

  $\begin{array}{c}\text{A}=\text{(m}{\text{Zr}}^{90}\text{×abandance)+(m}{\text{Zr}}^{91}\text{×}\text{abundance}\text{)+(m}{\text{Zr}}^{92}\text{×abundance)}\end{array}$

   $\begin{array}{c}{\text{+(mZr}}^{94}{\text{×abundance)+(mZr}}^{96}\text{×abundance)}\end{array}$

Putting the values,

  $\begin{array}{c}\text{A}=(89.90\times .514)+(90.90\times 0.112)+(91.90\times .171)+(93.90\times .173)+(95.90\times .028)\end{array}$

   $\begin{array}{c}=46.20+10.18+15.71+16.24+2.6\end{array}$

   $\begin{array}{c}=90.93\end{array}$

With respect to the relative abundances of each isotope, the zirconium-90 is going to be the closest.

Conclusion

The value of the average atomic mass of the zirconium is calculated from the given atomic mass and their abundance of zirconium isotopes.