Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 2 - Exercises - Page 80: 80


\[\underline{\text{85}\text{.46 amu}}\]

Work Step by Step

Calculate the abundance of Rb-85 as follows: \[\begin{align} & \text{Abundance of Rb-}85=\frac{100\,\text{percent}}{100\,\text{percent}+35\,\text{percent}}\times 100\,\text{percent} \\ & =74.07\,\text{percent} \end{align}\] Calculate the abundance of Rb-87 as follows: \[\begin{align} & \text{Abundance of Rb-}87=\frac{35\,\text{percent}}{100\,\text{percent}+35\,\text{percent}}\times 100\,\text{percent} \\ & =25.93\,\text{percent} \end{align}\] Calculate the Rb-85 fraction as follows: \[\begin{align} & \text{Fraction isotope Rb-}85=\frac{74.07}{100} \\ & =0.7407 \end{align}\] Calculate the Rb-87 fraction as follows: \[\begin{align} & \text{Fraction isotope Rb-}87=\frac{25.93}{100} \\ & =0.2593 \end{align}\] Calculate the atomic mass of the element as follows: \[\begin{align} & \text{Atomic mass}=\left( \text{fraction of Rb-}85\times \text{mass of Rb-}85 \right)+ \\ & \left( \text{fraction of Rb-}87\times \text{mass of Rb-}87 \right) \\ & =\left( 0.7407\times 85\text{ amu} \right)+\left( 0.2593\times 87\text{ amu} \right) \\ & =62.93\text{ amu}+22.53\text{ amu} \\ & =85.46\text{ amu} \end{align}\] The atomic mass of rubidium is \[\underline{\text{85}\text{.46 amu}}\].
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