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

Step by step solution

01

Calculate Relative Masses of Compounds

Given the percentage of hydrogen and nitrogen in each compound, we can find the relative mass ratios: For Compound I: $$ \frac{17.75 \mathrm{g} \ \mathrm{H}}{82.25 \mathrm{g} \ \mathrm{N}} = \frac{H_{1}}{N_{1}} $$ For Compound II: $$ \frac{12.58 \mathrm{g} \ \mathrm{H}}{87.42 \mathrm{g} \ \mathrm{N}} = \frac{H_{2}}{N_{2}} $$ For Compound III: $$ \frac{2.34 \mathrm{g} \ \mathrm{H}}{97.66 \mathrm{g} \ \mathrm{N}} = \frac{H_{3}}{N_{3}} $$

02

Apply Gay-Lussac's Law

Gay-Lussac's Law states that the ratio of volumes of gases involved in a chemical reaction can be expressed in simple whole number ratios. Using the given data for each gas evolved, compare the volume ratios of hydrogen and nitrogen: For Compound I: $$ \frac{1.50 \ \mathrm{L} \ \mathrm{H}_{2}}{0.50 \ \mathrm{L} \ \mathrm{N}_{2}} = 3 $$ For Compound II: $$ \frac{2.00 \ \mathrm{L} \ \mathrm{H}_{2}}{1.00 \ \mathrm{L} \ \mathrm{N}_{2}} = 2 $$ For Compound III: $$ \frac{0.50 \ \mathrm{L} \ \mathrm{H}_{2}}{1.50 \ \mathrm{L} \ \mathrm{N}_{2}} = \frac{1}{3} $$

03

Calculate Mole Ratios

Using the volume ratios from Step 2, calculate the mole ratios of hydrogen and nitrogen in each compound given that volume is proportional to moles for gases: For Compound I: $$ \frac{3 \ \mathrm{mol} \ \mathrm{H}_{2}}{1 \ \mathrm{mol} \ \mathrm{N}_{2}} $$ For Compound II: $$ \frac{2 \ \mathrm{mol} \ \mathrm{H}_{2}}{1 \ \mathrm{mol} \ \mathrm{N}_{2}} $$ For Compound III: $$ \frac{1 \ \mathrm{mol} \ \mathrm{H}_{2}}{3 \ \mathrm{mol} \ \mathrm{N}_{2}} $$

04

Determine Molecular Formulas

Using the mole ratios found in Step 3, we can determine the molecular formulas for the three compounds: Compound I: \(NH_3\) (1 nitrogen and 3 hydrogens) Compound II: \(N_2H_4\) (2 nitrogens and 4 hydrogens) Compound III: \(N_3H\) (3 nitrogens and 1 hydrogen)

05

Determine Relative Atomic Masses

With the molecular formulas, we can now use the relative mass ratios from Step 1 along with the molecular formulas to determine the relative atomic masses: For Compound I (NH_3): $$ \frac{H_{1}}{N_{1}} = \frac{3 \times \mathrm{Mass \ of \ H}}{1 \times \mathrm{Mass \ of \ N}} $$ For Compound II (N_2H_4): $$ \frac{H_{2}}{N_{2}} = \frac{4 \times \mathrm{Mass \ of \ H}}{2 \times \mathrm{Mass \ of \ N}} $$ For Compound III (N_3H): $$ \frac{H_{3}}{N_{3}} = \frac{1 \times \mathrm{Mass \ of \ H}}{3 \times \mathrm{Mass \ of \ N}} $$ Using these equations, we can calculate the relative atomic masses: $$ \mathrm{Mass \ of \ H} = 1 $$ (By relative atomic mass convention) $$ \mathrm{Mass \ of \ N} = 14 $$ (Using the equation from Compound I) The relative atomic masses of hydrogen and nitrogen are found to be 1 and 14, respectively.

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