91Ó°ÊÓ

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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Formulas

Molecular formulas represent the specific number of atoms in a molecule. They show the exact count of each type of atom present in a compound. To determine the molecular formulas for compounds, we look at the ratios derived from the data.
In the provided exercise, we calculated the mole ratios for each compound to identify the correct molecular formula.

• For Compound I, the mole ratio of 3 hydrogen atoms to 1 nitrogen atom gives us the formula \(NH_3\).
• Compound II displayed a 2 hydrogen to 1 nitrogen mole ratio resulting in \(N_2H_4\).
• Lastly, Compound III showed an inverse ratio of 1 hydrogen to 3 nitrogen atoms, forming \(N_3H\).

These formulas are based on the combinations that abide by the whole number ratios according to the mole concept, showing how different atoms come together to form a molecule.

Gay-Lussac's Law

Gay-Lussac's Law indicates that the volumes of gases participating in a chemical reaction maintain a simple integer ratio under constant temperature and pressure. This is a key concept in identifying chemical formulas and understanding gaseous relationships.
In the exercise, we used Gay-Lussac’s Law to establish the ratio of the gaseous volumes of hydrogen (\(H_2\)) and nitrogen (\(N_2\)) released on decomposition:

• Compound I showed a volume of \(1.50 ext{ L } ext{ H}_2\) to \(0.50 ext{ L } ext{ N}_2\), simplifying to 3:1.
• For Compound II, the volume relation was \(2.00 ext{ L } ext{ H}_2\) to \(1.00 ext{ L } ext{ N}_2\), equating to a 2:1 ratio.
• Compound III showed \(0.50 ext{ L } ext{ H}_2\) to \(1.50 ext{ L } ext{ N}_2\), which simplifies to 1:3.

Understanding these volume ratios helps us deduce the mole ratios needed to find molecular formulas, making Gay-Lussac's Law very useful in chemical stoichiometry and reaction analysis.

Relative Atomic Masses

Relative atomic masses are used to compare the average mass of elements. They're central for calculating molecular weights and stoichiometry in chemistry.
In this exercise, we've found relative atomic masses by examining the percentage mass data and combining it with molecular formula insights.

• The data indicated that the atomic mass of hydrogen is 1, aligning with its light nature and standard relative mass.
• For nitrogen, we computed a relative atomic mass of 14, derived from the known distribution of nitrogen in different oxides and compounds.

By understanding relative atomic masses, you can balance reaction equations and predict reactant-product quantities efficiently. These calculations are crucial for any quantitative chemistry exploration, allowing chemists to scale reactions correctly.