## Chemistry: Molecular Approach (4th Edition)

$\underline{9.4\times {{10}^{13}}}$ excess electrons and $\underline{8.5\times {{10}^{-17}}\text{ kg}}$
The electronic charge is $-1.60218\times {{10}^{-19}}\text{ C}$. Calculate the number of electrons as follows: \begin{align} & \text{Number of electrons}=\left( -15\,\text{ }\!\!\mu\!\!\text{ C} \right)\times \frac{{{10}^{-6}}}{1\,\text{ }\!\!\mu\!\!\text{ }}\times \left( \frac{1\text{ electron}}{-1.60218\times {{10}^{-19}}\,\text{C}} \right) \\ & =9.4\times {{10}^{13}}\,\text{electrons} \end{align} The electron has a mass of $0.00091\times {{10}^{-27}}\text{ kg}$. Calculate the total mass as follows: \begin{align} & \text{Mass}=\frac{0.00091\times {{10}^{-27}}\text{ kg}}{1\,\text{electron}}\times 9.4\times {{10}^{13}}\,\text{electrons} \\ & =\text{8}\text{.5}\times \text{1}{{\text{0}}^{-17}}\,\text{kg} \end{align} The number of excess electrons acquired is $\underline{9.4\times {{10}^{13}}}$ and the collective mass is $\underline{8.5\times {{10}^{-17}}\text{ kg}}$.