Really need help (1 Viewer)

[vmoore]

Hey guys,
I am really stuck on this molar heat capacity rubbish.
What i have done is an open ended investigation and i have tested the combustion of octane and octanol.

I set up with a calometre with 100mLs of water in it and i have combusted a pirit burner with octanol and octane in it 4 times.
THe average of the results are:
Octanol:
temperature change - 17.125 degrees
Fuel burnt - 0.355 g

Octane:
temperature change - 34.25 degrees
Fuel burnt - 0.9075g

I wanna know how to determine the molar heat capacity but i am not sure on how to do it!!!

this is wat i have been told (tell me if this is correct or hopelessly wrong!!)

^H = c m ^T
i was told to use 100 for m (that is the mass of the water)
so i got; ^H = 100 x 4.18 (heat capacity of water) x 34.25 (the ^T for octanol)
= 14316.5 (J or kJ ??)
then i was told to divide by the fuel burnt
1416.5 / 0.9075
= 15775.78
then finally to get the molar heat capacity i times by the molar wieght
15775.76 x (8(12) + 18 + 16)
= 2050848.8 (J or kJ ???)
then if it were J , the molar heat capacity is 2050.85jK/mol

Then id do the same for octane and compare... IS THIS RIGHT??
i am so confussed
THANK YOU IN ADVANCE!!

 

[Forbidden.]

Octanol:
Temperature change: 17.125 degrees
Fuel burnt: 0.355 g

ΔH = mCΔT
ΔH = 100 x 4.18 x 17.125 = 7158.25 J
(But that's for 0.355g)

You gotta find the value of ΔH for 1M of Octanol (C₈H₁₇OH)
i.e. (12.01 x 8) + (1.008 x 17) + 16 + 1.008 = 130.217 g mol^-1
(1M Octanol is 130.217g)

ΔH = 7158.25 x (130.217g / 0.355g) J mol^-1
ΔH = 2,625,706.592 J mol^-1
(That's too long, so we convert to kJ mol^-1, 1000 J = 1 kJ)

ΔH_c = 2,625.7 kJ mol^-1


Repeat this for your other fuels but change the required values as indicated in blue (i.e. the molecular weights for your fuels)
You may keep 100mL of water if you plan to heat 100mL of water using all fuels.