## Chemistry: Molecular Approach (4th Edition)

$\underline{6.25\times {{10}^{18}}}$ excess electrons and $\underline{5.69\times {{10}^{-9}}\text{ g}}$
The electronic charge is $-1.6\times {{10}^{-19}}\text{ C}$. Calculate the number of electrons as follows: \begin{align} & \text{Number of electrons}=\frac{\left( -1.0\text{ C} \right)}{-1.6\times {{10}^{-19}}\text{ C}} \\ & =6.25\times {{10}^{18}} \end{align} The electron has a mass of $9.10\times {{10}^{-28}}\text{ g}$. Calculate the total mass as follows: \begin{align} & \text{Mass}=\left( 6.25\times {{10}^{18}} \right)\left( 9.10\times {{10}^{-28}}\text{ g} \right) \\ & =5.\text{69}\times \text{1}{{\text{0}}^{-9}}\text{ g} \end{align} The number of electrons necessary is $\underline{6.25\times {{10}^{18}}}$ and the collective mass is $\underline{5.69\times {{10}^{-9}}\text{ g}}$.