Question

\(\Theta\) In the "Chemistry in Focus" segment Cars of the Future, the claim is made that the combustion of gasoline for some cars causes about 11 b of \(\mathrm{CO}_{2}\) to be produced for each mile traveled. Fstimate the gas mileage of a car that produces about \(11 \mathrm{~b}\) of \(\mathrm{CO}_{2}\) per mile traveled. Assume gasoline has a density of \(0.75 \mathrm{~g} / \mathrm{mL}\) and is \(100 \%\) octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right) .\) While this last part is not true, it is close enough for an estimation. The reaction can be represented by the following unbalanced chemical equation: $$ \mathrm{C}_{8} \mathrm{H}_{18}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$

Short answer

The estimated gas mileage of the car that produces about 11 b of CO₂ per mile traveled is approximately 7.95 miles per gallon.

Step-by-step solution

Balance the chemical equation

To balance the chemical equation, we need to find the correct coefficients for each chemical species involved. The balanced equation is: \[ \mathrm{C}_{8} \mathrm{H}_{18}+12.5\mathrm{O}_{2} \rightarrow 8\mathrm{CO}_{2}+9\mathrm{H}_{2} \mathrm{O} \]

Calculate the molar mass of octane (𝐶_8𝐻_18)

To determine the molar mass, we need to sum the atomic masses of all the atoms in the chemical formula: Molar mass of 𝐶_8𝐻_18 = (8 × (12.01 g/mol)) + (18 × (1.01 g/mol)) = 96.08 g/mol + 18.18 g/mol = 114.26 g/mol

Calculate the mass of CO₂ produced

We know that 11 b (bornes) of CO₂ is produced per mile traveled. We need to get this mass in grams. Since 1 b is equivalent to 100 grams, 11 b of CO₂ is equal to 11×100 g = 1100 g. Now we can determine the moles of CO₂ produced using the molar mass of CO₂: Moles of CO₂ produced = 1100 g / (44.01 g/mol) = 24.99 mol From the balanced equation, 1 mol of C_8H_18 produces 8 mol of CO₂. So we can determine the moles of C_8H_18 consumed: Moles of C_8H_18 = moles of CO₂ produced / 8 = 24.99 mol / 8 = 3.124 mol Now we can calculate the mass of C_8H_18 consumed using its molar mass: Mass of C_8H_18 = moles of C_8H_18 × molar mass of C_8H_18 = 3.124 mol × 114.26 g/mol = 356.9 g

Estimate the gas mileage

We have found the mass of the gasoline (C_8H_18) consumed for one mile traveled (356.9 g). We also know the density of gasoline is 0.75 g/mL. Now, we will find the volume of gasoline consumed: Volume of gasoline = mass of gasoline / density = 356.9 g / (0.75 g/mL) = 475.9 mL Since we have found out the amount of gasoline consumed per mile (475.9 mL), now we can express it as gas mileage: Gas Mileage = 1 mile / volume of gasoline = 1 mile / 475.9 mL ≈ 0.0021 miles/mL To convert to miles per gallon (mpg), we need to remember there are 3,785.41 mL in a gallon: Gas Mileage (mpg) = 0.0021 miles/mL × 3,785.41 mL/gallon ≈ 7.95 mpg So, the estimated gas mileage of the car is approximately 7.95 miles per gallon.

Key concepts

Stoichiometry

Stoichiometry is the area of chemistry that involves calculating the quantities, often in moles, of reactants and products in a chemical reaction. This process involves using balanced chemical equations to ensure the law of conservation of mass is upheld—meaning atoms are neither created nor destroyed during a reaction.

For students tackling stoichiometric problems, a helpful tip is to always start with a balanced chemical equation. This establishes the correct molar ratios between the different chemical species involved. As showcased in the gasoline combustion example, once the equation is balanced, it's possible to determine how many moles of a product (such as carbon dioxide, CO₂) are produced from a specific amount of reactants (like octane, C₈H₁₈).

For clarity and to reduce possible errors, students should:

• Write down the balanced chemical equation
• Work systematically, calculating one substance at a time
• Confirm units are consistent (convert if necessary)
• Double-check molar ratios to avoid miscalculating proportions

Molar Mass Calculation

Understanding how to calculate molar mass is crucial for stoichiometry, as it allows conversion between mass in grams and amount in moles. The molar mass is determined by summing up the atomic masses of every atom in a compound—taken from the periodic table—multiplied by the number of each type of atom present in the molecule.

In solving problems like the one provided—estimating gas mileage based on CO₂ production from gasoline combustion—accurately calculating the molar mass of octane, C₈H₁₈, is key. This involves multiplying the atomic mass of carbon (12.01 g/mol) by 8, and of hydrogen (1.01 g/mol) by 18, then summing these products.

Students often benefit from breaking this process into steps:

Identify

• Identify the molecular formula of the substance. In the example, C₈H₁₈ for octane.

Calculate

• Calculate each element's contribution separately, then add them together for total molar mass.

Confirm

• Always cross-check the calculated molar mass with reliable sources or the periodic table for accuracy.

Gasoline Combustion

Gasoline combustion is a chemical reaction where gasoline reacts with oxygen to produce water (H₂O), carbon dioxide (CO₂), and energy. This type of reaction is highly exothermic, powering the engines of many vehicles. By approximating gasoline as octane (C₈H₁₈), the balanced equation for its combustion in oxygen shows the stoichiometric ratios of the reactants and products.

In the context of the provided exercise, the stoichiometry and balanced equation led to an understanding of the environmental impact—indicating how much CO₂ is produced—and economic aspects, reflecting gas mileage. To improve the estimation of gas mileage, consider:

• Including the energy content per mole of gasoline, which can show the efficiency of the vehicle's engine
• Examining different environmental conditions or operating scenarios, which might alter the combustion efficiency

The practical implications of gasoline combustion in automotive engineering and environmental science are significant and multifaceted. Engaging with exercises that involve gasoline combustion allows students to better grasp the importance of such combustion reactions in our daily lives as well as the necessity to accurately understand and calculate their outcomes.