## Chemistry: Molecular Approach (4th Edition)

$\underline{\text{16}\text{.00 g}}$
Calculate the mass of nitrogen as follows: \begin{align} & \text{Ratio}=\frac{\text{Mass of nitrogen}}{\text{Mass of}{{\text{ }}^{\text{12}}}\text{C}} \\ & \frac{7}{6}=\frac{\text{Mass of nitrogen}}{\text{12}\text{.00 g/mol }} \end{align} Rearrange the above expression as follows: \begin{align} & \text{Mass of nitrogen}=\frac{7\times 12.00\text{ g/mol}}{6} \\ & =14.00\text{ g/mol} \end{align} Calculate the mass of oxygen in ${{\text{N}}_{\text{2}}}\text{O}$ as follows: \begin{align} & \text{Ratio}=\frac{\text{2}\left( \text{Mass of nitrogen} \right)}{\text{Mass of oxygen}} \\ & \frac{7}{4}=\frac{2\left( 14.00\text{ g/mol} \right)}{\text{Mass of oxygen}} \end{align} Rearrange the above expression as follows: \begin{align} & \text{Mass of oxygen}=\frac{4\times 2\left( 14.00\text{ g/mol} \right)}{7} \\ & =16.00\text{ g/mol} \end{align} The mass of $1\text{ mol}$ of oxygen atoms is $\underline{\text{16}\text{.00 g}}$.