Question

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and gasoline (assumed to be all octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\) ) are both used as automobile fuel. If gasoline is selling for \(\$ 2.20 / \mathrm{gal},\) what would the price of ethanol have to be in order to provide the same amount of heat per dollar? The density and \(\Delta H_{\mathrm{f}}^{\circ}\) of octane are \(0.7025 \mathrm{~g} / \mathrm{mL}\) and \(-249.9 \mathrm{~kJ} / \mathrm{mol}\), respectively, and of ethanol are \(0.7894 \mathrm{~g} / \mathrm{mL}\) and \(-277.0 \mathrm{~kJ} / \mathrm{mol}\) respectively \((1 \mathrm{gal}=3.785 \mathrm{~L})\).

Short answer

Ethanol should be priced at approximately $15.00 per gallon.

Step-by-step solution

Calculate the Mass of Gasoline in 1 Gallon

First, calculate the mass of gasoline in 1 gallon. We know that 1 gallon is 3.785 liters and the density of octane is 0.7025 g/mL.\\Convert the volume to milliliters: 1 gallon = 3785 mL.Calculate the mass of gasoline:Mass = Density \( \times \) Volume = 0.7025 g/mL \( \times \) 3785 mL = 2658.9625 g.

Convert Mass of Octane to Moles

Next, find the number of moles of octane in the mass we calculated.The molar mass of octane (\(\mathrm{C}_{8} \mathrm{H}_{18}\)) is: \\[ 8(12.01) + 18(1.01) = 114.22 \, \mathrm{g/mol} \].Now, use the mass to find the moles:\[ \text{Moles of octane} = \frac{2658.9625 \text{ g}}{114.22 \text{ g/mol}} \approx 23.28 \text{ mol}. \]

Calculate Heat Provided by Gasoline

The enthalpy change \(\Delta H_{\text{f}}^{\circ}\) for octane is \(-249.9 \text{ kJ/mol}\). Calculate the total heat produced by the gasoline:\[ \text{Total heat} = 23.28 \text{ mol} \times (-249.9 \text{ kJ/mol}) = -5814.372 \text{ kJ}. \]

Calculate Heat per Dollar for Gasoline

Gasoline sells for $2.20 per gallon. Calculate the heat provided per dollar:\[ \frac{-5814.372 \, \text{kJ}}{2.20 \, \text{ extdollar}} = -2642.8964 \, \text{kJ/ extdollar}. \]

Calculate Mass of Ethanol to Provide Equal Heat

We want ethanol to provide heat at \( -2642.8964 \text{ kJ/ extdollar} \). Let the mass of ethanol be \(m\), with the molar mass = 46.08 g/mol. The enthalpy \(\Delta H_{\text{f}}^{\circ}\) is \(-277.0 \text{ kJ/mol}\). Model: \[ \frac{-277.0 \text{ kJ/mol} \times \frac{m}{46.08 \text{ g/mol}}}{\text{ extdollar} } = -2642.8964 \text{ kJ/ extdollar} \]. Solve for \(m\): \[ m = \frac{(-2642.8964 \times 46.08)}{-277.0} \approx 438.90 \text{ g}. \]

Convert Ethanol Mass to Volume

Use the density of ethanol to find the volume:Density is 0.7894 g/mL.Calculate the volume required:\[ \text{Volume} = \frac{438.90 \text{ g}}{0.7894 \text{ g/mL}} \approx 556.08 \text{ mL}. \]Convert to liters if necessary: 556.08 mL = 0.55608 L.

Calculate Ethanol Price per Gallon

We know that 0.55608 liters results in \$2.20 worth of ethanol.Convert to gallons (1 gallon = 3.785 L):\[ \text{Ethanol Price/gallon} = \frac{2.20 \, \text{ extdollar} \times 3.785 \, \text{L}}{0.55608 \, \text{L}} \approx 15.00 \, \text{ extdollar/gallon}. \]

Key concepts

Enthalpy Change

Understanding enthalpy change is crucial when comparing fuels like ethanol and gasoline. Enthalpy change, denoted as \( \Delta H_{\text{f}}^{\circ} \), refers to the amount of heat absorbed or released during a chemical reaction at a constant pressure. For this exercise, we have two substances: ethanol and octane, which is a major component of gasoline.

The given enthalpy changes for octane and ethanol are \(-249.9 \, \text{kJ/mol}\) and \(-277.0 \, \text{kJ/mol}\) respectively. A negative value indicates that the reaction is exothermic, releasing heat. When you're comparing fuels, enthalpy change helps you assess how much energy per mole is released during combustion. Knowing these measurements, we can determine how much energy each fuel provides, impacting decisions like cost-efficiency for fuel choices. Calculating this allows you to quantify energy production per liter and understand the heat output that different fuels provide. Understanding enthalpy is fundamental to making informed fuel decisions across various applications.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance and it's measured in grams per mole (g/mol). Calculating molar mass is essential when converting grams to moles or vice versa, which are necessary steps to fully analyze the energy content from a chemical substance.

For octane (\(\mathrm{C}_{8} \mathrm{H}_{18}\)), you calculate the molar mass by adding the atomic weights of carbon (C) and hydrogen (H) found in one molecule of octane:\[ 8 \times 12.01 \, \text{(C)} + 18 \times 1.01 \, \text{(H)} = 114.22 \, \text{g/mol}. \]
Likewise, for ethanol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)) the calculation is:\[ 2 \times 12.01 \, \text{(C)} + 6 \times 1.01 \, \text{(H)} + 16.00 \, \text{(O)} = 46.08 \, \text{g/mol}. \]
These calculations allow you to convert between the weight of a substance and the number of moles—essential for finding how much heat will be released when combustion takes place.

Density and Volume Conversion

In fuel chemistry, density is used to relate mass and volume of liquid substances. It is expressed in g/mL or g/L, and it allows you to convert between the volume of the liquid and its mass.

For octane, with a density of \(0.7025 \, \text{g/mL}\), and ethanol with a density of \(0.7894 \, \text{g/mL}\), these figures give ways to calculate how much of each substance you'll have based on given volumes.

• You first convert gallons to milliliters, knowing \(1 \, \text{gallon} = 3785 \, \text{mL}\).
• Multiply by the density to find the mass: e.g., \(\text{mass} = \text{density} \times \text{volume}\).

Important conversions serve multiple steps in determining cost-efficiency and equivalent fuel comparison while considering the volume that is practically available for use.

Heat per Dollar Calculation

When comparing how much energy different fuels provide for their price, you must calculate heat per dollar. This calculation helps consumers figure out the most cost-efficient fuel.

Using the formula to determine heat per dollar involves knowing total heat produced per standard volume of fuel and its cost. In this case, gasoline provides \(-5814.372 \, \text{kJ}\) for \$2.20. Therefore, the heat per dollar is computed using:
\[ \frac{-5814.372 \, \text{kJ}}{2.20 \, \text{dollar}} = -2642.8964 \, \text{kJ/dollar}. \]
The comparable steps for ethanol involve setting them to match this heat output for its cost, solving for the required mass and volume based on density, finally leading to an equivalent dollar cost per comparable gallon. Calculating heat per dollar effectively guides your decision-making to obtain the best value for vehicle fuel.