Chapter 2 - Exercises - Page 82: 117

\[\text{Li-6}\,=\underline{7.494\,\text{percent}}\] and \[\text{Li-7}=\underline{92.506\,\text{percent}}\]

Calculate the fraction of lithium-6 as follows: \[\text{Fraction of isotope lithium-}6=\frac{\text{Relative abundance of lithium-}6}{\text{100}}\] Calculate the fraction of lithium-7 as follows: \[\text{Fraction of isotope lithium-}7=\frac{\text{Relative abundance of lithium-}7}{\text{100}}\] Atomic mass is the average of the mass of isotopes. So, the atomic mass of lithium is \[6.941\text{ amu}\]. Calculate the relative abundance as follows: \[\begin{align} & \text{Atomic mass}=\left( \text{Fraction of lithium-}6\times \text{Mass of lithium-}6 \right)+ \\ & \left( \text{Fraction of lithium-}7\times \text{Mass of lithium-}7 \right) \\ & 6.941\text{ amu}=\left( \frac{\text{Relative abundance of lithium-}6}{\text{100}}\times \text{6}\text{.01512 amu} \right) \\ & +\left( \frac{\text{Relative abundance of lithium-}7}{\text{100}}\times 7.01601\text{ amu} \right) \end{align}\] Substitute relative abundance of lithium-7 as \[100-\text{relative abundance of lithium-}6\] in the above expression as follows: \[\begin{align} & 6.941\text{ amu}=\left( \frac{\text{Relative abundance of lithium-}6}{\text{100}}\times \text{6}\text{.01512 amu} \right) \\ & +\left( \frac{\text{100}-\text{Relative abundance of lithium-}6}{\text{100}}\times 7.01601\text{ amu} \right) \\ & \text{Relative abundance of lithium-}6=7.494\,\text{percent} \end{align}\] Calculate the relative abundance of lithium-7 as follows: \[\begin{align} & \text{Relative abundance of lithium-}7=100\,\text{percent}-7.494\,\text{percent} \\ & =92.506\,\text{percent} \end{align}\] The relative abundance of lithium-6 is \[\underline{7.494\,\text{percent}}\] and the relative abundance of lithium-7 is \[\underline{92.506\,\text{percent}}\].