Problem 30 - Chemistry: An Atoms-Focused Approach

Question

Suppose the same mass of each of these components of natural gas was completely combusted: (a) \(\mathrm{CH}_{4},\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{8},\) and (d) \(\mathrm{C}_{4} \mathrm{H}_{10} .\) Which one would produce the greatest mass of \(\mathrm{CO}_{2} ?\)

Step-by-Step Solution

Answer

Answer: Butane (C4H10) produces the greatest mass of carbon dioxide when completely combusted, given the same initial mass for each component.

Step 1: Write the balanced chemical equations for the combustion of natural gas components

The balanced chemical equations for the combustion of methane (CH4), ethane (C2H6), propane (C3H8), and butane (C4H10) are as follows: a) CH4 + 2O2 → CO2 + 2H2O (Methane) b) 2C2H6 + 7O2 → 4CO2 + 6H2O (Ethane) c) C3H8 + 5O2 → 3CO2 + 4H2O (Propane) d) 2C4H10 + 13O2 → 8CO2 + 10H2O (Butane)

Step 2: Calculate molar mass and mass ratios of CO2 produced to the mass of initial fuel

Calculate the molar mass of the components (Methane, Ethane, Propane, and Butane) and CO2: Methane (CH4) Molar Mass: 1 C = 12 g/mol 4 H = 4 * 1 g/mol = 4 g/mol M(CH4) = 12 + 4 = 16 g/mol Ethane (C2H6) Molar Mass: 2 C = 2 * 12 g/mol = 24 g/mol 6 H = 6 * 1 g/mol = 6 g/mol M(C2H6) = 24 + 6 = 30 g/mol Propane (C3H8) Molar Mass: 3 C = 3 * 12 g/mol = 36 g/mol 8 H = 8 * 1 g/mol = 8 g/mol M(C3H8) = 36 + 8 = 44 g/mol Butane (C4H10) Molar Mass: 4 C = 4 * 12 g/mol = 48 g/mol 10 H = 10 * 1 g/mol = 10 g/mol M(C4H10) = 48 + 10 = 58 g/mol Carbon dioxide (CO2) Molar Mass: 1 C = 12 g/mol 2 O = 2 * 16 g/mol = 32 g/mol M(CO2) = 12 + 32 = 44 g/mol Now, we will calculate the mass ratios: Methane (CH4): 1 mol CH4 produces 1 mol CO2 Mass ratio, R (CH4) = 44/16 = 2.75 Ethane (C2H6): 1 mol C2H6 produces 2 mol CO2 Mass ratio, R(C2H6) = (2 * 44)/30 = 2.93 Propane (C3H8): 1 mol C3H8 produces 3 mol CO2 Mass ratio, R(C3H8) = (3 * 44)/44 = 3 Butane (C4H10): 1 mol C4H10 produces 4 mol CO2 Mass ratio, R(C4H10) = (4 * 44)/58 = 3.03

Step 3: Determine which component produces the highest mass ratio

Comparing the calculated mass ratios: R(Methane) = 2.75 R(Ethane) = 2.93 R(Propane) = 3 R(Butane) = 3.03 The highest mass ratio is for Butane (C4H10), which is 3.03. This indicates that, given an equal mass of the four components, Butane will produce the greatest amount of CO2 when completely combusted.

Key Concepts

Molar Mass Calculation

Molar mass is the mass of one mole of a given substance. It is expressed in grams per mole (g/mol). Understanding how to calculate molar mass is crucial, especially when working with chemical reactions such as combustion. To calculate the molar mass, simply add up the atomic masses of all the atoms in the molecular formula.

• For methane (\(\text{CH}_4\): 1 carbon (\(\text{C}\)) + 4 hydrogens (\(\text{H}\)), the molar mass equals 16 g/mol.
• For ethane (\(\text{C}_2\text{H}_6\)), calculate: 2 carbons and 6 hydrogens, resulting in a molar mass of 30 g/mol.
• Propane (\(\text{C}_3\text{H}_8\)) consists of 3 carbons and 8 hydrogens, yielding 44 g/mol.
• Butane (\(\text{C}_4\text{H}_{10}\)): has a molar mass of 58 g/mol, with 4 carbons and 10 hydrogens.

Each component's molar mass is needed to compute how much CO2 will result from its complete combustion.

Carbon Dioxide Production

When hydrocarbons undergo combustion in the presence of oxygen, they produce carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). The amount of carbon dioxide generated depends on the chemical composition of the hydrocarbon. Let's dive into some examples:

• Methane combustion: Releases one mole of CO2 for every mole of methane combusted.
• Ethane Reaction: Yields two moles of CO2 for each mole of ethane combusted. \(2\text{C}_2\text{H}_6 + 7\text{O}_2 → 4\text{CO}_2 + 6\text{H}_2\text{O}\)
• Burning propane forms three moles of CO2 from each mole of propane.\(\text{C}_3\text{H}_8 + 5\text{O}_2 → 3\text{CO}_2 + 4\text{H}_2\text{O}\)
• For butane, combustion results in 4 moles of CO2 per mole of butane.\(2\text{C}_4\text{H}_{10} + 13\text{O}_2 → 8\text{CO}_2 + 10\text{H}_2\text{O}\)

Therefore, among these hydrocarbons, butane yields the most CO2 per unit mass.

Natural Gas Components

Natural gas is a mixture of light hydrocarbons, primarily methane, ethane, propane, and butane. Each component has distinct combustion characteristics. Here's a quick breakdown:

• Methane (\(\text{CH}_4\)): It is the simplest hydrocarbon and the primary component of natural gas.
• Ethane (\(\text{C}_2\text{H}_6\)): Less abundant than methane but significant in petrochemical industries.
• Propane (\(\text{C}_3\text{H}_8\)) and butane (\(\text{C}_4\text{H}_{10}\)): Used as liquefied petroleum gas (LPG) for heating and cooking.

Understanding these components is key to determining which hydrocarbon produces the most carbon dioxide, a critical factor in assessing environmental impacts.