Problem 15 - Chemistry: An Atoms-Focused Approach

Question

In a study of the reaction, $$ 2 \mathrm{N}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) $$ quantities of all three gases were injected into a reaction vessel. The \(\mathrm{N}_{2} \mathrm{O}\) consisted entirely of isotopically labeled \(^{15} \mathrm{N}_{2} \mathrm{O} .\) Analysis of the reaction mixture after 1 day revealed the presence of compounds with molar masses 28 \(29,30,32,44,45,\) and \(46 \mathrm{g} / \mathrm{mol} .\) Identify the compounds and account for their presence.

Step-by-Step Solution

Answer

Based on the chemical reaction, the sample of N₂O was isotopically labeled with ¹⁵N. As the reaction proceeded, the ¹⁵N₂O decomposed, generating different combinations of nitrogen and oxygen isotopes. The compounds observed had molar masses of 28, 29, 30, 32, 44, 45, and 46 g/mol, corresponding to combinations of ¹⁴N, ¹⁵N, and ¹⁶O isotopes in N₂, O₂, and N₂O molecules. The presence of these compounds indicates the interaction of isotopically labeled nitrogen gas and atmospheric nitrogen in the reaction, as well as the decomposition of ¹⁵N₂O.

Step 1: Analyzing the molar masses of each compound

Find the molar masses of the possible compounds: nitrogen (\(\mathrm{N}\)), oxygen (\(\mathrm{O}\)), and \(\mathrm{N}_{2} \mathrm{O}\). Use the periodic table to obtain the molar masses for the isotopes of nitrogen and oxygen. isotopes are: \(^{14}\mathrm{N}, ^{15}\mathrm{N}, \, \mathrm{and} \, ^{16}\mathrm{O}\).

Step 2: Identify the compounds

Using the molar masses of isotopes, determine and identify the compounds with molar masses of 28, 29, 30, 32, 44, 45, and 46 g/mol. For each mass, consider various combinations of nitrogen and oxygen isotopes in \(\mathrm{N}_{2}, \mathrm{O}_{2}\), and \(\mathrm{N}_{2}\mathrm{O}\).

Step 3: Account for the presence of compounds

Based on the reaction equation and isotopes used, explain the presence of each compound identified. Solution:

Step 1: Analyzing the molar masses of each compound

The molar masses (rounded off) of the isotopes involved are: - \(^{14}\mathrm{N}\): 14 g/mol - \(^{15}\mathrm{N}\): 15 g/mol - \(^{16}\mathrm{O}\): 16 g/mol

Step 2: Identify the compounds

For each molar mass, consider combinations of the isotopes in \(\mathrm{N}_{2}, \mathrm{O}_{2}\), and \(\mathrm{N}_{2}\mathrm{O}\): - 28 g/mol: composed of two \(^{14}\mathrm{N}\) atoms, so it is \(\mathrm{N}_{2}(^{14}\mathrm{N}\text{-}^{14}\mathrm{N})\) - 29 g/mol: composed of one \(^{14}\mathrm{N}\) atom and one \(^{15}\mathrm{N}\) atom, so it is \(\mathrm{N}_{2}(^{14}\mathrm{N}\text{-}^{15}\mathrm{N})\) - 30 g/mol: composed of two \(^{15}\mathrm{N}\) atoms, so it is \(\mathrm{N}_{2}(^{15}\mathrm{N}\text{-}^{15}\mathrm{N})\) - 32 g/mol: composed of two \(^{16}\mathrm{O}\) atoms, so it is \(\mathrm{O}_{2}(^{16}\mathrm{O}\text{-}^{16}\mathrm{O})\) - 44 g/mol: composed of two \(^{14}\mathrm{N}\) and one \(^{16}\mathrm{O}\) atoms, so it is \(\mathrm{N}_{2}\mathrm{O}(^{14}\mathrm{N}\text{-}^{14}\mathrm{N}\text{-}^{16}\mathrm{O})\) - 45 g/mol: composed of one \(^{14}\mathrm{N}\), one \(^{15}\mathrm{N}\), and one \(^{16}\mathrm{O}\) atoms, so it is \(\mathrm{N}_{2}\mathrm{O}(^{14}\mathrm{N}\text{-}^{15}\mathrm{N}\text{-}^{16}\mathrm{O})\) - 46 g/mol: composed of two \(^{15}\mathrm{N}\) and one \(^{16}\mathrm{O}\) atoms, so it is \(\mathrm{N}_{2}\mathrm{O}(^{15}\mathrm{N}\text{-}^{15}\mathrm{N}\text{-}^{16}\mathrm{O})\)

Step 3: Account for the presence of compounds

The reaction equation is: $$ 2 \, \mathrm{N}_{2} \mathrm{O}(g) \rightleftharpoons 2 \, \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) $$ Since the \(\mathrm{N}_{2}\mathrm{O}\) consisted entirely of isotopically labeled \(^{15}\mathrm{N}_{2}\mathrm{O}\), the reaction process generated various combinations of nitrogen isotopes and oxygen isotopes, resulting in \(^{14}\mathrm{N}\text{-}^{14}\mathrm{N}, \, ^{14}\mathrm{N}\text{-}^{15}\mathrm{N}, \, \text{and} \, ^{15}\mathrm{N}\text{-}^{15}\mathrm{N}\) for nitrogen gas (\(\mathrm{N}_{2}\)) and \(^{16}\mathrm{O}\text{-}^{16}\mathrm{O}\) for oxygen gas (\(\mathrm{O}_{2}\)). The compounds are present because the isotopically labeled \(\mathrm{N}_{2} \mathrm{O}\) decomposed into various nitrogen isotopes (\(^{14}\mathrm{N}\) and \(^{15}\mathrm{N}\)) and released isotopically labeled nitrogen gas (\(^{15} \mathrm{N}_{2}\)). The unlabeled atmospheric nitrogen source has also reacted with the released \(\mathrm{O}_2\), producing different nitrogen gas isotope combinations (\(^{14}\mathrm{N}\text{-}^{14}\mathrm{N}\) and \(^{14}\mathrm{N}\text{-}^{15}\mathrm{N}\)) and \(\mathrm{N}_{2}\mathrm{O}\) variants.

Key Concepts

Isotopic Labeling

Isotopic labeling involves replacing an atom in a molecule with an atom of the same element but with a different number of neutrons. This creates isotopes, which are atoms with the same number of protons but different masses. In our exercise, \(^{15}\mathrm{N}\) atoms were used to label \(\mathrm{N}_2\mathrm{O}\). By tracking these isotopic labels, scientists can follow a molecule's behavior during a chemical reaction.
This technique is helpful for understanding reaction mechanisms by revealing how reactants transform into products.

• Enables tracing the movement of specific atoms.
• Provides insight into the formation of intermediates.
• Clarifies reaction pathways.

Isotopic labeling is a crucial tool in chemistry for studying detailed molecular processes.

Molar Mass Calculation

Molar mass is the weight of one mole of an element or compound, calculated by summing the atomic masses of its constituent atoms from the periodic table. For isotopes, each has a unique molar mass.
In our example, we considered the isotopes \(^{14}\mathrm{N} (14)\), \(^{15}\mathrm{N} (15)\), and \(^{16}\mathrm{O} (16)\). By combining these, we identified possible compounds:

• \(\mathrm{N}_2 (^{14}\mathrm{N}{-}^{14}\mathrm{N})\) has 28 g/mol.
• \(\mathrm{N}_2 (^{14}\mathrm{N}{-}^{15}\mathrm{N})\) has 29 g/mol.
• \(\mathrm{N}_2\mathrm{O} (^{15}\mathrm{N}_2\text{-} ^{16}\mathrm{O})\) has 46 g/mol.

Understanding molar mass is essential for identifying chemical compounds and predicting the outcome of reactions.

Nitrous Oxide Decomposition

The decomposition of nitrous oxide (\(\mathrm{N}_2\mathrm{O}\)) is a dynamic process where it breaks down into nitrogen gas (\(\mathrm{N}_2\)) and oxygen gas (\(\mathrm{O}_2\)). This is represented by the chemical reaction:
\[ 2 \mathrm{N}_2 \mathrm{O} \rightarrow 2 \mathrm{N}_2 + \mathrm{O}_2 \]
In this reaction, isotopically labeled \(^{15} \mathrm{N}_2\mathrm{O}\) allowed us to identify different isotopic forms of nitrogen and oxygen.

• \(\mathrm{N}_2\) appears in forms like \(^{14}\mathrm{N}{-}^{15}\mathrm{N}\).
• \(\mathrm{O}_2\) appears as \(^{16}\mathrm{O}{-}^{16}\mathrm{O}\).

This decomposition helps scientists study the behavior of isotopes and their recombination during chemical transformations.

Gas Analysis

Gas analysis is key to identifying and understanding chemical reactions. It involves identifying gases based on their molar masses and isotopic compositions.
In the exercise, the analysis of reaction products revealed several compounds with distinct molar masses such as 28, 29, 30, 32, 44, and 46 g/mol. These correspond to various nitrogen and oxygen combinations.

• Understanding molar masses helps predict reaction outcomes.
• Isotopic analysis helps trace molecular changes.

By analyzing the components of a gas mixture, scientists can gain insights into the types and distributions of atoms in a reaction, leading to a deeper understanding of the underlying processes.