A mixture of two hydrocarbons, C8H18 (octane) and C7H8 (toluene), has a mass of 167 g.

Question

A mixture of two hydrocarbons, C8H18C8H18 (octane) and C7H8C7H8 (toluene), has a mass of 167 g.167 g. The hydrocarbon mixture is burned in excess oxygen to form a mixture of carbon dioxide and water that contains 1.491.49 times as many moles of carbon dioxide as water. Find the masses of C8H18C8H18 and C7H8C7H8 in the mixture.

Answer 1

The key to a solution is realize that the mixture of gases is variable. To get the relative amounts of octane and toluene on a molar basis you need to have the ratio of moles of CO2 / moles H2O be equal to 1.49.

I couldn’t see an easy way to mathematically model this so I substituted different numbers of moles of reactants into the reaction to get the proper ratio. The one that works best is:

1C8H18 + 5C7H8 + 115/2O2 —> 43CO2 + 29H2O

the ratio 43/29= 1.483 which is close enough.

Now we can calculate the mole fraction and solve:

C8H18 is 1/(1+5) x 167g = 27.8g

C7H8 is 5/(5+1) x 167g = 139.2g