This is one of the problems i have to do...COMPLETELY LOST on it. I dont even know where to start or what to do. Please, please, please could someone help!?!?

One of the serious problems that we have not yet solved is the use of fossil fuels, especially gasoline, to power our automobiles. Companies around the world are constantly trying to find a suitable replacement for gasoline. If we use octane (C8H18) as the representative of what is contained in gasoline, and compare it to methanol (CH3OH), currently a promising substitute for gasoline, what will be the energy difference on a per gram basis for the two. For the reaction of octane with oxygen to produce carbon dioxide and water, 48.1 kJ of heat are evolved per gram of octane burned. For the reaction ofmethanol, CH3OH, with oxygen to produce CO2 and gaseous water, 638.6 kJ/mole methanol. On a per gram basis, from which one will more energy be produced. The density of octane is 0.7025 g/cm3 and the density of methanol is 0.7914 g/cm3. If burning octane in an automobile is 30% efficient (i.e., 30% of the energy produced is used to move the car) and burning methanol is 40% efficient, which fuel will have the larger tank to have the same mileage capability?

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a good place to start is molecular weight

MW octane = 8*12+18 = 114
MW methanol = 12+3+16+1 = 32

For octane, the energy produced is 48.1 KJ/g (given)

For methanol, the energy produced is 638.6 KJ/mol (given). To convert this to KJ/g, divide by MW of methanol. Then, 638.6/32 = 19.96 KJ/g

for octane:
48.1 KJ/g * 0.7025 g/cm^3 = 33.79 KJ/cm^3 * (0.3) = 10.137 KJ/cm^3 utilized energy

for methanol:
19.96 KJ/g * 0.7914 g/cm^3 = 15.8 KJ/cm^3 * (0.4) = 6.32 KJ/cm^3 utilized energy

More energy is stored in octane per unit volume than in methanol, thus methanol will require a larger tank.

Andy Resnick