A mixture of two hydrocarbons, C8H18 (octane) and C7H8 (toluene), has a mass of 201 g.

Question

A mixture of two hydrocarbons, C8H18 (octane) and C7H8 (toluene), has a mass of 201 g. The hydrocarbon mixture is burned in excess oxygen to form a mixture of carbon dioxide and water that contains 1.46 times as many moles of carbon dioxide as water. Find the masses of C8H18 and C7H8 in the mixture.

mass of C8H18=

mass of C7H8=

Answer 1

Starting with a balanced chemical reaction is correct, as the reaction will indeed form products in a stoichiometric ratio. This is the correct balanced reaction:

2C8H18 + 2C7H8 + 43O2 ==> 30CO2 + 26H2O

From this point on this becomes a math problem involving stoichiometric ratios. The starting mass of the hydrocarbon mixture is given as 201 g. Thus we know that for any X grams of Octane we must have (201-X) grams of Toluene. All of these calculations are conveniently carried out in a spreadsheet.

(Xg / 114.2 g/mole Octane) = moles Octane

((201-X)/92.13 g/mole Toluene) = moles Toluene

moles of CO2 produced = 8(moles Octane) + 7(moles Toluene)

moles H2O produced = 18/2 (moles Octane) + 8/2(moles Toluene)

Using these formulas in a spreadsheet one can easily find the value of X to produce the desired molar ratio of products. You are welcome to use my spreadsheet here: https://docs.google.com/spreadsheets/d/1R94TZqQ6ur4_G9xQgLmRjuDCGw_fI5LZp1xMo89KrcQ/edit?usp=sharing

For this problem using the given conditions, 43.93 g of Octane and the corresponding 157.07g of Toluene result in a product molar ratio = 1.46.

Answer 2

The first step is writing the balanced equation to where these are the conditions outlined in the problem:

2C8H18 + 2C7H8 + 43O2 ==> 30CO2 + 26H2O

Now you have to balance the equation accounting for all of the Hydrogen atoms, carbon atoms and Oxygen atoms on each side, while also keeping in mind that the Carbon dioxide molecule on the product side should have at least 1.46 times as many moles as the water molecule when it comes to moles.

You can start off with your Carbons. Look to see how many you will have on each side. On the reactant side you have a total of 15. But then watch your Hydrogens as well especially on the product side. What number would you need to put in front of H20 to get the Hydrogens to balance on both sides?

2C8H18 + 2C7H8 + 43O2 ==> 30CO2 + 26H2O

This up above is a balanced equation for the reaction but we would have to look at the ratio between the CO2 and H20 is CO2 1.46 times greater than H2O as is in terms of moles?

30/26 = 1.1538

So no.

What you can now do is estimate what this ratio would need to look like.

X/26 = 1.46

And solve for X. For this ratio to be 1.46 or so, X would need to be about

X= 26*1.46= 38 moles of the CO2

Looking back at our equation, if we put a 3 in front of our toluene molecule and keep the 2 in front of octane we would be a lot closer to that ratio of 1.46.

Reducing it more we could do:

X/13= 1.46. Which means 19 carbon moles, to 13 water moles.

How could we simplify and write this?

1.5C8H18 + C7H8 + 25.5O2 ==> 19CO2 + 13H2O+ 4.5H2

If you put a 1.5 in front of octane you will have 19 carbons on both sides and the right amount of moles of water to create that ratio of 1.46.

1.5C8H18 + C7H8 + 25.5O2 ==> 19CO2 + 13H2O+ 4.5H2

Now leading with the above equation you would have the same amount of atoms on each side of the equation of Carbons, Hydrogens and even Oxygens but you would have some of the Hydrogen going off to diatomic hydrogen gas, but this would still allow you to keep this ratio of 1.46:

19 Carbons on both sides

35 Hydrogens on both sides

51 Oxygens on both sides

Now to get rid of half moles we can multiply the entire equation by 2 where moles are:

3C8H18 + 2C7H8 + 51O2 ==> 38CO2 + 26H2O+ 9H2

Now with the proper balanced equation you will use the atomic weight of both to figure out what portion is of the moles then do a mole ratio, find octane. The remaining will then be toluene.

At weight of octane: 114.26g/mol

At weight of toluene: 92.15g/mol

201g* 1mol/206.41g,. notice how we combined both atomic weights because we were told this is the mass of the mixture now we can find total moles of the mixture:

Grams cancel top to bottom and you will have moles: 201/206.41= 0.97379 moles of mixture of hydrocarbons.

We know that 3 moles of the octane react with 2 moles of the toluene. So we can set up fractions and ratios to figure out the masses of each.

3+2= 5 parts total of the mixture.

Octane's part is 3/5 for the moles. 0.6* 201g= 120.6g of Octane.

The remainder is 201-120.6= 80.4g of toluene, which works out to be the 2/5 parts as well: 0.4* 201g= 80.4g of toluene.