Posted by Sindy on .
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= ?

Chemistry 
DrBob222,
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. 
Chemistry 
Sindy,
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 
Chemistry 
Sindy,
Okie I get 36913.6J/g* degree C but how do I change that into kJ/mol?

Chemistry 
DrBob222,
36913.6 J/g x (grams octanol/mol) = ??
In other words, look up the molar mass of octanol. 
Chemistry 
Sindy,
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 
Chemistry 
Sindy,
this is so confusing =(

Chemistry 
DrBob222,
You need to clarify.
Is benzoic 26.38 or 26.318?
Is the temperature rise 8.158, 8.518, or 8.15? 
Chemistry 
Sindy,
Benzoic is 26.38 and the temp rises 8.518

Chemistry 
Sindy,
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