[Aerosion]

"A 1.80g sample of octane (C₈H₁₈) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/degrees C.  The temperature of the calorimeter and its contents increased from 21.36 degrees C to 28.78 degreesw C.  What is the heat of combustion per gram of octane?  Per mole of octane?"

As soon as I get grams, I can get moles, so don't worry about that.

It's practially been hammered into my head to convert degrees Celsius to Kelvins, but since the question is applying a change in temperature, I really don't think that converting them would matter too much (as the difference would be the same, Kelvin or Celsius).

Beyond that, though, I'm not really sure.  The equation q=heat capacity * change in temberature looks tempting, but I'm not sure if the heat in that equation relates to the heat of combustion.  Please help.

[Borek]

The equation q=heat capacity * change in temberature looks tempting, but I'm not sure if the heat in that equation relates to the heat of combustion.  Please help.

You are on the right track. Where did the energy used to heat up the calorimeter came from?

[Aerosion]

The equation q=heat capacity * change in temberature looks tempting, but I'm not sure if the heat in that equation relates to the heat of combustion.  Please help.

You are on the right track. Where did the energy used to heat up the calorimeter came from?

I would say the oxygen in the air, maybe?  I do know that combustion uses oxygen and a substance to release heat (oxygen is the oxidizer and the substance is the fuel, or maybe its the other way around) but there's no oxygen in the problem, and that's where I got stuck.  Should I just assume it's there, or is the heat coming from somewhere else?

[xiankai]

assume.

energy comes from both octane and oxygen, too. otherwise octane wouldnt be needed

[Borek]

Should I just assume it's there, or is the heat coming from somewhere else?

Heat is produced in combustion. Before ignition bomb calorimeter is filled with oxygen (usually under pressure) so that reaction can proceed to the very end.

[Aerosion]

Mmhm...okay, but I'm still not seeing the answer to this question.

So oxygen is what's heating the calorimeter.  I could certainly make a little chemical equation for it now, but what am I getting the specific heat for?  The question already gives me the heat capacity of the calorimeter (11.66 kJ/C), so am I using that or the heat of oxygen (29.378 kJ/C)?  And how is this going to lead me to the heat of combustion?

I'm not asking for anyone to give me the answer or anything, but a hint to the next step would be helpful.

[Borek]

So oxygen is what's heating the calorimeter.

No! It is combustion that heats calorimeter! You have to add oxygen for the reaction to proceed, but it is reaction itself thats a source of heat. You can calculate amount of heat knowing temperature change and calorimeter heat capacity, and it will be amount of heat produced when burning 1.80g sample of octane.

[Aerosion]

Oh, okay...so, if it was an equation...

C₈H₁₈ + 13/2O₂ --> 8CO₂ + 9H₂O + heat

And then you multiply 11.66 x 7.42 to get the heat of combustion (86.5172 kJ per 1.80g octane).  You have to transfer that to however many kJ per 1 g of octane, then turn 1.80g to moles and calculate per one mol of octane, right?

EDIT:

Umm...I would think that the thermochemical equation could be used...such that...-86.52 kJ = (-202.4 kJ / 1 mol C₈H₁₈) x (1 mol C₈H₁₈ / (66g C₈H₁₈) + x

(-202.4 kJ would be from the thermochemical equation (the enthalpy from the above equation) and x would be the grams the heat of combustion produced.)

I got -28.213 for the heat.

[Borek]

Oh, okay...so, if it was an equation...

C₈H₁₈ + 13/2O₂ --> 8CO₂ + 9H₂O + heat

And then you multiply 11.66 x 7.42 to get the heat of combustion (86.5172 kJ per 1.80g octane).  You have to transfer that to however many kJ per 1 g of octane, then turn 1.80g to moles and calculate per one mol of octane, right?

OK

(-202.4 kJ would be from the thermochemical equation (the enthalpy from the above equation) and x would be the grams the heat of combustion produced.)

Completely off.

[Aerosion]

(-202.4 kJ would be from the thermochemical equation (the enthalpy from the above equation) and x would be the grams the heat of combustion produced.)

Completely off.

Yes yes, I saw that about 10 minutes after I posted the last part.  I think that was the enthalpy of the formation rather than the enthalpy of the combustion.

I think the actual one would be -5512 kJ.  Which would bring my answer to -1.04g, not -28.213 g.