Question

The automobile fuel called E85 consists of \(85 \%\) ethanol and \(15 \%\) gasoline. E85 can be used in the so-called flex-fuel vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ), that the average heat of combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) is \(5400 \mathrm{~kJ} / \mathrm{mol}\), and that gasoline has an average

Short answer

The heat of combustion of gasoline is approximately \(47.3 \mathrm{~kJ/g}\) or \(34.9 \mathrm{~kJ/mL}\). Comparing this to ethanol's heat of combustion (\(29.8 \mathrm{~kJ/g}\) or \(21.1 \mathrm{~kJ/mL}\)), we find that gasoline provides more energy per unit of mass or volume than ethanol.

Step-by-step solution

Determine the molar mass of octanes C8H18

To determine the heat of combustion of gasoline, we need to find the molar mass of C8H18. The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. So, the molar mass of C8H18 is: Molar mass = (8 x 12.01) + (18 x 1.01) \[ \approx 114.23 \mathrm{~g/mol} \] 2.

Calculate the energy per gram of gasoline

Now, we will calculate the energy per gram of gasoline using the heat of combustion of octanes (C8H18), which is \(5400 \mathrm{~kJ/mol}\). To do this, we will divide the given energy per mole by the molar mass: Energy per gram = \(\dfrac{5400 \mathrm{~kJ/mol}}{114.23 \mathrm{~g/mol}}\) ≈ \[47.3 \mathrm{~kJ/g} \] 3.

Calculate the energy per milliliter of gasoline

Next, we will calculate the energy per milliliter of gasoline. We will use the average density of gasoline, which is 0.739 g/mL, to convert the energy per gram of gasoline to energy per milliliter: Energy per milliliter = Energy per gram * Density = \(47.3 \mathrm{~kJ/g} * 0.739 \mathrm{~g/mL}\) ≈ \[34.9 \mathrm{~kJ/mL} \] 4.

Compare the heat of combustion of gasoline and ethanol

Finally, let's compare the heat of combustion of gasoline and ethanol: For gasoline: - Heat of combustion: \(47.3 \mathrm{~kJ/g}\) or \(34.9 \mathrm{~kJ/mL}\) For ethanol: - Heat of combustion: \(29.8 \mathrm{~kJ/g}\) or \(21.1 \mathrm{~kJ/mL}\) From the comparison, we can see that the heat of combustion of gasoline is higher than that of ethanol, both on a per-gram and per-milliliter basis. This means that gasoline provides more energy per unit of mass or volume than ethanol, which is why gasoline is usually preferred over ethanol for use in internal combustion engines.

Key concepts

Energy Per Gram of Fuel

Understanding the energy content of fuels is crucial when assessing their efficiency and impact on engine performance. The heat of combustion is a measure of the total energy released when a fuel is burned completely. It is often expressed in energy per gram (kJ/g), which shows the energy value of a fuel in terms of its mass. To calculate this, you divide the heat of combustion in kilojoules per mole by the molar mass of the fuel. The resulting figure provides a benchmark for comparing the efficiency of different fuels.

For example, if a fuel like gasoline has a higher energy per gram compared to another fuel like ethanol, it means that gasoline can release more energy for the same amount of mass. This is an essential concept for students to understand, as it affects decisions regarding fuel selection for different applications, such as in automotive engines or power generation.

Molar Mass Calculation

In chemistry, molar mass is a fundamental concept that represents the mass of one mole of a substance (elements or compounds). It is usually expressed in grams per mole (g/mol) and can be calculated by summing the atomic masses of all atoms in a molecular formula. For students, grasping how to calculate molar mass is key to understanding many aspects of chemical reactions and processes.

The molar mass calculation aids in the evaluation of the heat of combustion on a per gram basis since the energy content of a fuel is typically given per mole of substance. As illustrated in the solution, by knowing the molar mass of octane (C8H18) in gasoline, which includes combining the atomic weights of carbon and hydrogen, students can then determine the energy content per gram of the fuel – a practical skill for analyzing fuel efficiency.

Fuel Comparison

Comparing fuels is an important practice in both ecological and economic contexts. It helps consumers and industries to choose the most suitable fuel for their needs based on specific parameters like cost, availability, and energy content. Heat of combustion figures, energy per gram, and energy per milliliter are all vital for these comparisons.

From the solution provided, we learn that gasoline has a higher heat of combustion than ethanol. Such comparisons inform decisions about fuel efficiency. When considering energy per gram, a higher value indicates that a smaller amount of the fuel is required to produce the same amount of energy, which impacts fuel economy. It is important for students to understand these comparisons to make informed choices about fuels, especially given the current emphasis on sustainability and the move towards alternative energy sources.

Flex-Fuel Vehicles

Flex-fuel vehicles (FFVs) are designed to operate on more than one type of fuel, usually gasoline blended with either ethanol or methanol. These vehicles have an internal combustion engine and a fuel system that can handle different fuel mixtures. The most common type of flexible-fuel used in FFVs is E85, a blend of 85% ethanol and 15% gasoline.

FFVs provide the advantage of versatility, allowing drivers to select fuels based on availability, price, or environmental concerns. Understanding the chemistry behind the various fuel options, particularly how their heat of combustion translates into actual vehicle performance, is crucial. For students studying environmental science or automotive engineering, knowledge about FFVs is directly applicable to discussions about reducing greenhouse gas emissions and our reliance on non-renewable energy resources.