Question

In his 1989 essay "The End of Nature," William McKibben states that "... a clean burning automobile engine will emit \(5.5 \mathrm{lb}\) of carbon in the form of carbon dioxide for every gallon of gasoline it consumes." Assume that the gasoline is \(\mathrm{C}_{8} \mathrm{H}_{18}\) and its density is \(0.703 \mathrm{~g} / \mathrm{mL}\) (a) Write a balanced equation for the complete combustion of gasoline. (b) Is McKibben's assertion correct? Support your answer with calculations.

Short answer

McKibben's assertion is incorrect; calculations show 18.09 lb of CO2 per gallon, not 5.5 lb.

Step-by-step solution

Write the Combustion Reaction

The complete combustion of gasoline, assumed as octane (\(\text{C}_8\text{H}_{18}\)), with oxygen forms carbon dioxide and water. The balanced chemical equation is:\[\text{C}_8\text{H}_{18} + 12.5\,\text{O}_2 \rightarrow 8\,\text{CO}_2 + 9\,\text{H}_2\text{O}\]

Calculate Molar Mass of Reactants and Products

Calculate the molar mass of octane and carbon dioxide based on their molecular formulas:\[\text{Molar mass of } \text{C}_8\text{H}_{18} = (8 \times 12.01 + 18 \times 1.01)\, \text{g/mol} = 114.22\, \text{g/mol}\]\[\text{Molar mass of } \text{CO}_2 = (1 \times 12.01 + 2 \times 16.00)\, \text{g/mol} = 44.01\, \text{g/mol} \]

Moles of Octane per Gallon

First, convert 1 gallon to mL, knowing that 1 gallon = 3785 mL:\[ \text{Volume of 1 gallon} = 3785\, \text{mL} \]Calculate the mass of gasoline using its density:\[ \text{Mass of 1 gallon of } \text{C}_8\text{H}_{18} = 3785\, \text{mL} \times 0.703\, \text{g/mL} = 2661.55\, \text{g} \]Calculate the moles of octane:\[ \text{Moles of } \text{C}_8\text{H}_{18} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2661.55\, \text{g}}{114.22\, \text{g/mol}} \approx 23.31\, \text{moles} \]

Calculate Moles of CO2 Produced

From the balanced equation, every mole of octane produces 8 moles of \(\text{CO}_2\). Thus, for 23.31 moles of \(\text{C}_8\text{H}_{18}\):\[ \text{Moles of } \text{CO}_2 = 23.31\, \text{moles } \text{C}_8\text{H}_{18} \times 8 = 186.48\, \text{moles } \text{CO}_2 \]

Convert CO2 to Pounds

Convert the moles of \(\text{CO}_2\) to grams, then to pounds (1 lb = 453.592 g):\[ \text{Mass of } \text{CO}_2 = 186.48\, \text{moles} \times 44.01\, \text{g/mol} = 8204.9448\, \text{g} \]Convert to pounds:\[ \text{Mass in pounds} = \frac{8204.9448\, \text{g}}{453.592\, \text{g/lb}} \approx 18.09\, \text{lb} \]

Conclusion: Compare to McKibben's Assertion

William McKibben claims the emissions are 5.5 lb of carbon per gallon of gasoline. Our calculation shows the emissions as approximately 18.09 lb of \(\text{CO}_2\) per gallon, which is significantly higher. Thus, McKibben's assertion is not supported by these calculations, as they pertain to the emission of carbon dioxide rather than carbon alone.

Key concepts

Balanced Chemical Equation

In a balanced chemical equation, the number of atoms for each element must be equal on both sides of the reaction. This ensures the law of conservation of mass is satisfied.

When gasoline, represented as octane (\( ext{C}_8 ext{H}_{18}\)), combusts completely, it reacts with oxygen (\( ext{O}_2\)) to produce carbon dioxide (\( ext{CO}_2\)) and water (\( ext{H}_2 ext{O}\)).

The balanced equation for the combustion is:\[\text{C}_8 ext{H}_{18} + 12.5\, \text{O}_2 \rightarrow 8\, \text{CO}_2 + 9\, \text{H}_2 ext{O}\]

• 1 molecule of octane reacts with 12.5 molecules of oxygen.
• This forms 8 molecules of carbon dioxide and 9 molecules of water.

It's important to balance equations to accurately predict the products and their quantities.

Molar Mass Calculation

Molar mass helps us convert between grams and moles, an essential step in chemical calculations. Each element's atomic mass contributes to the total molar mass of a compound.

For octane (\( ext{C}_8 ext{H}_{18}\)), the molar mass is calculated by adding:

• 8 carbon atoms: \(8 \times 12.01 \text{ g/mol} = 96.08\text{ g/mol}\)
• 18 hydrogen atoms: \(18 \times 1.01 \text{ g/mol} = 18.18 \text{ g/mol}\)
• Total for octane: \(114.22 \text{ g/mol}\)

Similarly, for carbon dioxide (\( ext{CO}_2\)), the calculation is:

• 1 carbon atom: \(1 \times 12.01 \text{ g/mol} = 12.01 \text{ g/mol}\)
• 2 oxygen atoms: \(2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol}\)
• Total for carbon dioxide: \(44.01 \text{ g/mol}\)

Molar masses aid in converting between mass and the amount of substance.

Carbon Dioxide Emissions

Carbon dioxide emissions are a significant concern associated with the combustion of fossil fuels. This process releases \(\text{CO}_2\) into the atmosphere, contributing to climate change.

We calculate emissions by understanding the relation:

• 1 mole of octane produces 8 moles of \(\text{CO}_2\).
• From the equation, 23.31 moles of octane yields 186.48 moles of \(\text{CO}_2\), corresponding to approximately 8204.94 grams.
• This converts to roughly 18.09 pounds of \(\text{CO}_2\) per gallon.

William McKibben's statement refers to carbon, while our calculation addresses total \(\text{CO}_2\) emissions.

Understanding these emissions is crucial to evaluating the environmental impact of fuel consumption.