Problem 16 - Chemistry: An Atoms-Focused Approach

Question

A mixture of \(^{13} \mathrm{CO},^{12} \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in a scaled reaction vessel was used to follow the reaction $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) $$ Analysis of the reaction mixture after 1 day revealed the presence of compounds with molar masses 28,29,32,44 and \(45 \mathrm{g} / \mathrm{mol}\). Identify the compounds and account for their presence.

Step-by-Step Solution

Answer

Question: Identify the compounds with molar masses 28, 29, 32, 44, and 45 g/mol, and explain their presence in a reaction mixture of \(^{13} \mathrm{CO},^{12} \mathrm{CO}_{2}\), and \(\mathrm{O}_{2}\) after 1 day of reaction. Answer: The compounds and their molar masses are: 1. 28 g/mol: \(^{12}\mathrm{CO}\) - formed from \(^{12}C\) carbon isotope and oxygen 2. 29 g/mol: \(^{13}\mathrm{CO}\) - formed from \(^{13}C\) carbon isotope and oxygen 3. 32 g/mol: \(\mathrm{O_2}\) - unreacted oxygen 4. 44 g/mol: \(^{12}\mathrm{CO}_2\) - formed from reaction between \(^{12}C\) carbon isotope and oxygen 5. 45 g/mol: \(^{13}\mathrm{CO}_2\) - formed from reaction between \(^{13}C\) carbon isotope and oxygen These compounds are present in the reaction mixture because the given reaction can involve both carbon isotopes (\(^{12}C\) and \(^{13}C\)) forming different compounds with varying molar masses depending on the isotopes involved.

Step 1: Identify the given molar masses

We are given reaction mixture contains compounds with molar masses 28, 29, 32, 44 and 45 g/mol.

Step 2: Recall the molar masses of the reactants and isotopes

From the periodic table, recall the molar masses of elements involved: - Carbon: \(^{12}C = 12 \thinspace g/mol\), \(^{13}C = 13 \thinspace g/mol\) - Oxygen: \(^{16}O = 16 \thinspace g/mol\).

Step 3: Match the given molar masses with potential compound combinations

Combine the isotopes of carbon and oxygen to create compounds and find their molar masses: 1. \(^{12} \mathrm{C} + ^{16} \mathrm{O} = \mathrm{CO} = 12 + 16 = 28 \thinspace g/mol\) 2. \(^{13} \mathrm{C} + ^{16} \mathrm{O} = \mathrm{^{13}CO} = 13 + 16 = 29 \thinspace g/mol\) 3. \(2 \times ^{16} \mathrm{O} = \mathrm{O_2} = 2 \times 16 = 32 \thinspace g/mol\) 4. \(^{12} \mathrm{C} + 2 \times ^{16} \mathrm{O} = \mathrm{CO_2} = 12 + 2 \times 16 = 44 \thinspace g/mol\) 5. \(^{13} \mathrm{C} + 2 \times ^{16} \mathrm{O} = \mathrm{^{13}CO_2} = 13 + 2 \times 16 = 45 \thinspace g/mol\)

Step 4: Identify the compounds and account for their presence in the reaction

Based on the known molar masses of reactants and their isotopes, we are able to identify the compounds in the reaction mixture after 1 day: 1. 28 g/mol: \(^{12}\mathrm{CO}\) - formed from \(^{12}C\) carbon isotope and oxygen 2. 29 g/mol: \(^{13}\mathrm{CO}\) - formed from \(^{13}C\) carbon isotope and oxygen 3. 32 g/mol: \(\mathrm{O_2}\) - unreacted oxygen 4. 44 g/mol: \(^{12}\mathrm{CO}_2\) - formed from reaction between \(^{12}C\) carbon isotope and oxygen 5. 45 g/mol: \(^{13}\mathrm{CO}_2\) - formed from reaction between \(^{13}C\) carbon isotope and oxygen These compounds are present in the reaction mixture because the given reaction can involve both carbon isotopes (\(^{12}C\) and \(^{13}C\)) forming different compounds with varying molar masses depending on the isotopes involved.

Key Concepts

Carbon Isotopes

Carbon isotopes play a crucial role in chemical reactions, especially when tracing reactions involving carbon compounds. There are two stable carbon isotopes:

• \(^{12}C\): This is the most abundant carbon isotope, accounting for about 98.9% of carbon in nature. It has 6 protons and 6 neutrons, making it a common component in organic and inorganic compounds.
• \(^{13}C\): Though less common, it makes up about 1.1% of carbon isotopes. It contains 6 protons and 7 neutrons, slightly altering the physical properties of molecules containing it, such as increased mass.

The difference in mass between these isotopes allows scientists to distinguish between molecules using tools like mass spectrometry. In the chemical reaction considered, the presence of both isotopes contributes significantly to the diversity of products formed, allowing researchers to understand isotope effects on reaction dynamics.

Molar Mass Calculation

Calculating molar mass is fundamental in chemistry for identifying and quantifying substances in a reaction. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Here's how to calculate it:

• Start by identifying the elemental composition of a compound. For instance, carbon monoxide (CO) consists of one carbon atom and one oxygen atom.
• Use periodic table values to find the atomic masses. For CO, the atomic mass of carbon (C) can be 12 g/mol or 13 g/mol for different isotopes, while oxygen (O) is 16 g/mol.
• Add these values together. For example, CO with \(^{12}C\) has a molar mass of 28 g/mol, while CO with \(^{13}C\) has 29 g/mol.

This accurate calculation helps to determine unknown compounds in a mixture by comparing with the given or measured molar masses, as seen in separating isotopes or identifying products in a reaction.

Reaction Equilibrium

Reaction equilibrium is a state in a chemical reaction where the forward and reverse reaction rates are equal. This means that the concentrations of reactants and products remain constant over time. In the given reaction:
\[ 2 \mathrm{CO}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) \]Equilibrium is reached when the rate of carbon monoxide (\(\mathrm{CO}\)) and oxygen (\(\mathrm{O}_2\)) forming carbon dioxide (\(\mathrm{CO}_2\)) equals the rate of its decomposition back into reactants.
When dealing with isotopes, like in the given scenario, these equilibrium states include different isotopic variations, producing a mix of compounds like \(^{12}\mathrm{CO}_2\) and \(^{13}\mathrm{CO}_2\). Understanding equilibrium helps predict product formation rates and the extent of reactions, crucial in optimizing conditions for desired reactions.

Gas Mixture Analysis

Analyzing gas mixtures involving different isotopes and molecules provides key insights into reaction dynamics. The method typically involves:

• Sampling a known volume of the reaction mixture and using analytical techniques, such as gas chromatography or mass spectrometry, to separate and quantify components based on their molar masses.
• Comparison of measured molar masses with calculated or known values to identify substances present in the mixture, such as distinguishing between \(^{12}\mathrm{CO}\) and \(^{13}\mathrm{CO}\) based on their subtle mass differences.
• Understanding the formation rates and contributions of different isotopic molecules, which helps researchers deduce the behavior and pathway of reactions under various conditions.

In the reaction analyzed, different compounds like \(^{12}\mathrm{CO}_2\) and \(^{13}\mathrm{CO}_2\) were observed, each presenting a distinct molar mass, showcasing the importance of isotope analysis in studying gas-phase reaction dynamics.