Problem: Determine the empirical formulas of the compounds with the following compositions by mass.62.1% C, 5.21% H, 12.1% N, and the remainder O

Step 1: 62.1% C, 5.21% H, 12.1% N, and the remainder O by mass. Recall that mass percent is given by:

$$\begin{gathered} \mathbf{100}\mathbf{\%}\mathbf{=}\mathbf{\%}\mathbf{C}\mathbf{+}\mathbf{\%}\mathbf{H}\mathbf{+}\mathbf{\%}\mathbf{N}\mathbf{+}\mathbf{\%}\mathbf{O} \\ \mathbf{100}\mathbf{\%}\mathbf{=}\mathbf{62}\mathbf{.}\mathbf{1}\mathbf{\%}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{21}\mathbf{\%}\mathbf{+}\mathbf{12}\mathbf{.}\mathbf{1}\mathbf{\%}\mathbf{+}\mathbf{\%}\mathbf{O} \\ \mathbf{\%}\mathbf{O}\mathbf{=}\mathbf{100}\mathbf{\%}\mathbf{-}(62.1\%+5.21\%+12.1\%) \\ \mathbf{\%}\mathbf{O}\mathbf{=}\mathbf{100}\mathbf{\%}\mathbf{-}\mathbf{79}\mathbf{.}\mathbf{41}\mathbf{\%} \end{gathered}$$

 % O = 20.59%

$\boxed{\mathbf{\%}\mathbf{ }\mathbf{mass}\mathbf{=}\frac{\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{X}}{\mathbf{total}\mathbf{ }\mathbf{mass}}\mathbf{\times }\mathbf{100}}$

Assuming we have 100 g of the compound, this means we have 62.1 g C, 5.21 g H, 12.1 g N, and 20.59 g O.