Question

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.

Short answer

The molecular formulas of the compounds are: Compound I: NH Compound II: N2H4 Compound III: N3H The relative atomic masses of hydrogen and nitrogen are 1 amu and 14 amu, respectively.

Step-by-step solution

Find the mass ratio of hydrogen and nitrogen in each compound

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: \(\frac{2.34\%}{97.66\%}\)

Calculate the mole ratio of hydrogen and nitrogen in each compound

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: \(\frac{0.50 \, \text{L} \, H_{2}}{1.50 \, \text{L} \, N_{2}}=\frac{1}{3}\)

Determine the molecular formulas of the compounds and find the relative atomic masses of hydrogen and nitrogen

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.

Key concepts

Binary Compounds

Binary compounds are chemical compounds consisting of only two different elements. In this exercise, the binary compounds are formed by hydrogen and nitrogen. These compounds are often characterized by their simple two-element nature, making them easier to analyze and understand.

For example, water (H₂O) is a binary compound composed of hydrogen and oxygen. In our exercise, we are specifically looking at binary compounds of hydrogen and nitrogen, which can take on various forms. Understanding the binary nature helps in deducing molecular formulas with given data.

By examining the different binary compounds (I, II, III), we help students understand how simple combinations of two elements can give rise to different compounds with unique properties and molecular formulas.

Mass Ratio

The mass ratio in a chemical compound is the proportion of the masses of each element present. It is derived from the percentage composition data given for each element in the compound, such as those provided in the tables from our exercise. This ratio denotes the relative masses of each element combined to form a compound.

In this exercise, to find the mass ratio of hydrogen to nitrogen, you divide the percentage mass of hydrogen by the percentage mass of nitrogen:

• For compound I: \(\frac{17.75}{82.25}\)
• For compound II: \(\frac{12.58}{87.42}\)
• For compound III: \(\frac{2.34}{97.66}\)

By looking at these calculations, one can determine how the components of the compound contribute to its whole mass, providing insights into compound composition and aiding in determining molecular formulas.

Mole Ratio

The mole ratio represents how many moles of one component of a compound relate to the moles of another component. This ratio helps identify the stoichiometry between elements in a compound. It is derived from the volumes of gas obtained from decomposing the compound into hydrogen and nitrogen.

Because gases at the same temperature and pressure occupy the same volume per mole, volumes directly correspond to moles:

• For compound I: \(\frac{1.50 \, ext{L} \, H_2}{0.50 \, ext{L} \, N_2} = 3\)
• For compound II: \(\frac{2.00 \, ext{L} \, H_2}{1.00 \, ext{L} \, N_2} = 2\)
• For compound III: \(\frac{0.50 \, ext{L} \, H_2}{1.50 \, ext{L} \, N_2} = \frac{1}{3}\)

This mole ratio is crucial because it helps to deduce the simplest ratio of atoms in the compound, which is necessary for determining the molecular formula.

Atomic Mass

Atomic mass is the mass of an atom, often expressed in atomic mass units (amu), which are based on carbon-12. Each element has a specific atomic mass, reflecting the mass of its nucleus. The atomic mass is crucial to understanding the composition of compounds like ours with hydrogen and nitrogen.

In solving the exercise, atomic mass is used to relate the mass ratio with the mole ratio to find molecular formulas. By using the known atomic mass of hydrogen, we can derive the atomic mass of nitrogen from the ratios calculated:

• Atomic mass ratio derived from Compound I gives us the atomic mass of nitrogen as \(14\, \text{amu}\) relative to hydrogen's known \(1\, \text{amu}\).

This understanding offers insight into the relative weights of elements in compounds and how they determine the molecular structures.