There were 8.35 x 10^{23} molecules of carbon dioxide produced when a sample of octane was burned in air. If the combustion process is known to produce an 86.7% yield, how many grams of octane were in the same?

The equation I have is:

2C_{8}H_{18} + 25O_{2} 16CO_{2} + 18H_{2}O

So far so good.

I converted molecules of CO_{2} to moles then to grams and came out with 6.10 g of CO_{2}.

This step is not required and is wrong as are all the steps after this

Since it's in a 1:8 ratio, I divided 6.10g by 8 and got 0.7625.

As DeraDevil has pointed out the 1:8 ratio is moles not mass so you can not simply divide the weight of carbon dioxide to get the answer.

Then I took 0.7625 x 86.7% and got 0.661 g of C_{8}H_{18}.

This step is wrong. If the yield is 86.7% would more or less octane be required to give an ammount of CO

_{2}?

2C_{8}H_{18} + 25O_{2} 16CO_{2} + 18H_{2}O

There were 8.35 x 10^{23} molecules of carbon dioxide produced

1) Is 8.35 x 10

^{23} more or less than Avogadro's number?

2) So how many moles of carbon dioxide were formed?

If for example 8 moles of CO

_{2} were formed from the

**complete** combustion that would mean 1 mole of octane was burnt.

3) Now using your numbers how many moles of Octane buring completely would give the CO

_{2} formed and calculated in 2?

As the yield is 86.7% would that mean more or less octane is need to give this amount of CO

_{2}?

4) So how many moles of octane were needed to produce the amount of CO2 formed when the yield was 86.7%?

5) Now convert the number of moles in part 4 to a weight using the molecular weight of octane.