Consider the following data for three binary compounds of hydrogen and nitrogen: $$\begin{array}{|lcc|}\hline & \% \text { H (by Mass) } & \% \text { N (by Mass) } \\\\\hline \text { I } & 17.75 & 82.25 \\\\\text { II } & 12.58 & 87.42 \\\\\text { III } & 2.34 & 97.66 \\\\\hline\end{array}$$ When 1.00 L of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$\begin{array}{|lcc|}\hline & \mathrm{H}_{2}(\mathrm{L}) & \mathrm{N}_{2}(\mathrm{L}) \\\\\hline \\\\\mathrm{I} & 1.50 & 0.50 \\\\\mathrm{II} & 2.00 & 1.00 \\\\\mathrm{III} & 0.50 & 1.50 \\\\\hline\end{array}$$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

To determine the mass ratio of hydrogen and nitrogen, we can use the percentage composition by mass provided in the table. Divide the percentage of hydrogen by the percentage of nitrogen in each compound. For Compound I: \(\frac{17.75\%}{82.25\%}\) For Compound II: \(\frac{12.58\%}{87.42\%}\) For Compound III: \(\frac{2.34\%}{97.66\%}\)

Now, we will use the volumes of H2 and N2 obtained after decomposition to determine the mole ratio of hydrogen and nitrogen in each compound. Since 1 mole of any gas occupies the same volume under the same conditions, the volume ratios are equal to the mole ratios. For Compound I: \(\frac{1.50 \, \text{L} \, H_{2}}{0.50 \, \text{L} \, N_{2}}=3\) For Compound II: \(\frac{2.00 \, \text{L} \, H_{2}}{1.00 \, \text{L} \, N_{2}}=2\) For Compound III: \(\frac{0.50 \, \text{L} \, H_{2}}{1.50 \, \text{L} \, N_{2}}=\frac{1}{3}\)

To find the molecular formulas, we will use the mass ratios and mole ratios obtained in the previous steps. Divide the mass ratio by the mole ratio to find the atomic mass ratio of hydrogen and nitrogen in each compound. Since we know the atomic mass of hydrogen, we can find the atomic mass of nitrogen relative to hydrogen. For Compound I: \(\frac{17.75\%}{82.25\%} = \frac{3x}{x}\) For Compound II: \(\frac{12.58\%}{87.42\%} = \frac{2x}{x}\) For Compound III: \(\frac{2.34\%}{97.66\%} = \frac{x}{3x}\) Solving these equations, we get: For Compound I: \(x = 1\) (i.e., 1 hydrogen atom and 1 nitrogen atom), so the molecular formula is NH. For Compound II: \(x = 2\) (i.e., 2 hydrogen atoms and 1 nitrogen atom), so the molecular formula is N2H4. For Compound III: \(x = 3\) (i.e., 1 hydrogen atom and 3 nitrogen atoms), so the molecular formula is N3H. Now we can find the atomic mass of nitrogen relative to hydrogen: Using Compound I: \(1 \, H \equiv 17.75 \Rightarrow 1 \, N \equiv 82.25\) Atomic mass ratio of hydrogen to nitrogen \(=1:14\) Since the atomic mass of hydrogen is 1 amu, the relative atomic mass of nitrogen is 14 amu.