Chapter 2, Problem 2.38QP

The element rubidium has two naturally occurring isotopes. The atomic mass of ⁸⁵Rb (72.17 percent abundant) is 84.911794 amu. Determine the atomic mass of ⁸⁷Rb (27.83 percent abundant). The average atomic mass of Rb is 85.4678 amu.

Interpretation Introduction

Interpretation:

The atomic mass of ${}^{\text{87}}\text{Rb}$ 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 ${}^{\text{87}}\text{Rb}$.

Answer to Problem 2.38QP

The atomic mass of ${}^{\text{87}}\text{Rb}$ is $\text{86}\text{.91}\text{amu}$

Explanation of Solution

The atomic mass of ${}^{\text{87}}\text{Rb}$ is needs to know.

The atomic mass of ${}^{\text{87}}\text{Rb}$ 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 ${}^{\text{85}}\text{Rb =}\text{84}\text{.911 amu }$.

The atomic mass of ${}^{\text{87}}\text{Rb}$, x, given the contribution to the average atomic mass of ${}^{\text{85}}\text{Rb}$ and the average atomic mass of rubidium.

$\begin{array}{l}{\text{The fractional abundance of }}^{\text{87}}\text{Rb =}72.17\%\text{=}\frac{\text{72}\text{.17}}{\text{100}}\\ \text{= 0}\text{.7217}\end{array}$

$\begin{array}{l}{\text{The fractional abundance of }}^{\text{85}}\text{Rb =}27.83\%\text{=}\frac{27.83}{\text{100}}\\ \text{= 0}\text{.2783}\end{array}$

$\begin{array}{l}\text{Atomic mass calculation is given below,}\\ \left(\text{0}\text{.7217}\right)\left(\text{84}{\text{.911 amu }}^{\text{85}}\text{Rb}\right)\text{+}\left(\text{0}\text{.2783}\right)\left(\text{x}\right)\text{=}\text{85}\text{.4678}\\ \text{ 61}\text{.280}\text{+}\text{0}\text{.2783}\text{(x)}\text{= 85}\text{.4678}\\ \text{0}\text{.2783}\text{(x)}\text{=}\text{85}\text{.4678}\text{-}\text{ 61}\text{.280}\\ \text{0}\text{.2783}\text{(x)}\text{=}\text{24}\text{.1878}\\ \text{(x)}\text{=}\frac{\text{24}\text{.1878}}{\text{0}\text{.2783}}\\ \text{(x)}\text{=}\text{86}\text{.91}\text{amu}\end{array}$

Conclusion

Conclusion

The atomic mass of ${}^{\text{87}}\text{Rb}$ is calculated.