What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?

For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:

\(\begin{align}{\rm{0}}{\rm{.130 g nitrogen}} \times \frac{{{\rm{mol of nitrogen}}}}{{14.007}} &= 0.00928{\rm{ mol of nitrogen}}\\{\rm{0}}{\rm{.370 g oxygen }} \times \frac{{{\rm{mol of oxygen}}}}{{16.00}} &= 0.02318{\rm{ mol of oxygen}}\end{align}\)

Next, derive the nitrogen-to-oxygen molar ratio by dividing by the lesser number of moles:

\(\begin{align}\frac{{{\rm{ }}0.00928}}{{0.00928}} &= 1{\rm{ mol of N}}\\\frac{{0.02318}}{{0.00928}} &= 2.50{\rm{ mol of oxygen}}\end{align}\)

Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct nitrogen-to-oxygen ratio:\(2\left( {{{\rm{N}}_1}{{\rm{O}}_{2.5}}} \right) = {{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}\)

Hence, the empirical formula will beN₂O₅