# Chemistry Moles Calculation

1. Nov 1, 2013

### FredericChopin

1. The problem statement, all variables and given/known data
"1.79 g of a fuel was burned in the presence of oxygen. The fuel is a composite of 90% octane and 10% ethanol. Calculate the number of moles of this fuel that was burned."

2. Relevant equations
Moles = Mass/Molar Mass

3. The attempt at a solution
90% of 1.79 g is 1.61 g. This was the mass of octane burned.
10% of 1.79 g is 0.18 g. This was the mass of ethanol burned.

Moles of octane = Mass of octane burned/Molar mass of octane
Moles of octane = 1.61/(8*C + 18*H)
Moles of octane = 1.61/(8*12 + 18*1)
Moles of octane = 1.61/(96 + 18)
Moles of octane = 1.61/114
Moles of octane = 0.014 mol

Moles of ethanol = Mass of ethanol burned/Molar mass of ethanol
Moles of ethanol = 0.18/(2*C + 6*H + O)
Moles of ethanol = 0.18/(2*12 + 6*1 + 16)
Moles of ethanol = 0.18/(24 + 6 + 16)
Moles of ethanol = 0.18/46
Moles of ethanol = 0.0039 mol

Total moles burned = Moles of octane burned + Moles of ethanol burned
Total moles burned = 0.014 + 0.0039
Total moles burned = 0.018 mol

Therefore 0.018 moles of the fuel was burned.

Have I done this correctly?

2. Nov 2, 2013

### Saitama

Looks OK to me. :thumbs:

3. Nov 2, 2013

### FredericChopin

I still have the feeling I've done this wrong because saying the fuel is made of 90% octane doesn't mean that 90% of its mass comes from octane.

It's like looking at a compound made of (for the sake of argument) hydrogen and lead, and that compound was 90% hydrogen and 10% lead. Does that mean 90% of its mass comes from hydrogen? Definitely not! Lead is way heavier than hydrogen, even if it makes up 10% of its content.

What do you think?

4. Nov 2, 2013

### Saitama

Well, I am no expert at all in this but usually, the problem should mention what kind of percentage is that? Is it w/w, w/v or v/v? I thought you are asked to assume that percentage is w/w. Does the book state this?

About the example you quote, I would say why not. If the percentage is w/w, then in a 100g of substance, 90g is hydrogen and 10g of lead. Did someone restrict you from using only 10g of lead?

5. Nov 2, 2013

### FredericChopin

Actually, this is an experimental calculation I'm doing and it's not from a textbook of any kind. What I was doing was testing the efficiency of fuels by looking at the molar enthalpy of combustion as the percentage octane and ethanol changed.

It says on the spirit burner "90% Octane 10% Ethanol", so I'm guessing that's volume concentration and not mass fraction or concentration (v/v). What would I do to calculate the number of moles of fuel burned in this case?