Calculating Moles of Fuel Burned with Volume Concentration

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Discussion Overview

The discussion revolves around calculating the number of moles of a fuel burned, which is a mixture of 90% octane and 10% ethanol by volume. Participants explore the implications of the percentage composition and its effect on the calculation of moles, considering different interpretations of the concentration (mass vs. volume). The context includes both a homework problem and an experimental scenario.

Discussion Character

  • Homework-related
  • Debate/contested
  • Experimental/applied

Main Points Raised

  • One participant calculates the moles of octane and ethanol based on the given percentages and concludes that 0.018 moles of fuel were burned.
  • Another participant expresses uncertainty about the interpretation of the 90% octane and 10% ethanol, suggesting that the mass contribution may not align with the percentage composition due to differing molar masses.
  • A later reply questions the lack of clarity regarding whether the percentage is weight/weight (w/w), weight/volume (w/v), or volume/volume (v/v), and suggests assuming w/w unless stated otherwise.
  • One participant mentions conducting an experimental calculation and indicates that the percentage is likely volume concentration, prompting a request for guidance on how to calculate moles under this assumption.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the percentage composition of the fuel. There are competing views regarding whether the percentages refer to mass or volume, and the implications this has for the calculations.

Contextual Notes

The discussion highlights the ambiguity in the problem statement regarding the type of concentration used, which affects the calculations. There is also a mention of the experimental context, which may introduce additional variables not accounted for in the initial calculations.

FredericChopin
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Homework Statement


"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."

Homework Equations


Moles = Mass/Molar Mass

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?
 
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FredericChopin said:

Homework Statement


"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."


Homework Equations


Moles = Mass/Molar Mass




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?


Looks OK to me. :thumbs:
 
Pranav-Arora said:
Looks OK to me. :thumbs:

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?
 
FredericChopin said:


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?


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?
 
Pranav-Arora said:
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?

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?
 

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