Chemistry
posted by Sindy .
The combustion of benzoic acid releases 26.38kJ/g and is often used to calibrate a bomb calorimete. The combustion of 1.045g of benzoic acid caused the temperature of the calorimeter to rise by 5.985 degreeC.Using the same calorimeter, a sample of 0.876g of C8H18 was burned. the temp increased by 8.518 degreeC. what is the molar enthalpy of octane?
I don't know how to go about doing this question. What formula should I used I listed all the given:
Benzoic acid
deltaH= 26.318kJ/g
m= 1.045g
delta T= 5.985= 278.98K
C8H18
m= 0.876
mm= 114.224
delta T= 8.15= 281.52K
delta H= ?

First you need to address the data. I see 26.318 kJ/g for delta H in the table but 26.38 in the question. Be sure to take care of that first. To calculate the calorimeter constant, that is
mass x specific heat x delta T.
q = delta H, 26,380 J (check that number) = 1.045 x Cp x 5.985. Solve for Cp.
For the octanol,
[mass octanol x specific heat octanol x delta T] + [Cp x delta T] = 0
Solve for specific heat octanol which will be in units of J/g*C.
Then change that to kJ/mol.
Check my thinking. Check my work. 
Okie i get Cp= 42178.8J/g delta C
and for the octanol
im using q= m*Cs*T
q= 0.876*Cs*8.158
q= 7.1395g degreeC * Cs
In your work i don't understand thisequation:
[mass octanol x specific heat octanol x delta T] + [Cp x delta T] = 0 
Okie I get 36913.6J/g* degree C but how do I change that into kJ/mol?

36913.6 J/g x (grams octanol/mol) = ??
In other words, look up the molar mass of octanol. 
36913.6 J/g * 114g/mol= 4208150.4J/mol and to convert that to kJ= 4208150400 kJ/mol.
the correct answer is 5116 kJ/mol 
this is so confusing =(

You need to clarify.
Is benzoic 26.38 or 26.318?
Is the temperature rise 8.158, 8.518, or 8.15? 
Benzoic is 26.38 and the temp rises 8.518

I did is again:
Benzoic acid: q= m*Cs*T
Cs = 4.217kJ/g°C
q1 + q2 = 0
[m*Cs*T] + [26.38kJ] = 0
[.876g*Cs*8.518°C]+ [26.38kJ] = 0
7.462g°C * Cs = 26.38kJ
Cs= 3.535kJ/g°C
to get the molar enthalpy:
3.535kJ/g°C * 114g/mol= 403.03kJ/mol* °C