Chapter 4, Problem 129A

Interpretation Introduction

Interpretation:

The average atomic mass of zirconium needs to be determined.

Concept introduction:

The isotope contains the different numbers of neutron and 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 129A

The average atomic mass of zirconium is $91.22\text{amu}$.

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	$95.908$	$2.8$

The main important value is the mass of isotope and their abundance, for the elemental average atomic mass.

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

$\text{Average}\text{atomic}\text{mass}\text{=}$

$\text{(m}Z{r}^{90}\text{×abandance)+(m}Z{r}^{91}\text{×}\text{abundance}\text{)+(m}Z{r}^{92}\text{×abundance)+(m}Z{r}^{94}\text{×abundance)+(m}Z{r}^{96}\text{×abundance}$

Take the weighted average.

Atomic mass of Zr

$=\left[\left(89.905\right)\left(0.5145\right)+\left(90.906\right)\left(0.1122\right)+\left(91.905\right)\left(0.1715\right)+\left(93.906\right)\left(0.1738\right)+\left(95.908\right)\left(0.0280\right)\right]\text{amu}$ Atomic mass $=91.22\text{amu}$

The value of average atomic mass of zirconium is equal to $91.22\text{amu}$.

Conclusion

The isotope atomic mass and their abundance from the given table helped forthe calculation of average atomic mass of the zirconium.