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

To calculate the molar enthalpy of octane (C8H18), you can use the following formula:

ΔH = q / n

Where:
ΔH is the molar enthalpy of octane (in kJ/mol)
q is the heat absorbed or released during the combustion (in kJ)
n is the number of moles of octane

To find the heat absorbed or released during the combustion, you can use the formula:

q = m × ΔT

Where:
q is the heat absorbed or released during the combustion (in kJ)
m is the mass of the substance burned (in grams)
ΔT is the change in temperature of the calorimeter (in degrees Celsius or Kelvin)

Now, let's calculate the heat absorbed or released during the combustion of octane (C8H18).

Given:
Mass of C8H18 (m) = 0.876g
Change in temperature of the calorimeter (ΔT) = 8.518°C = 8.518K

First, we need to convert the change in temperature from Celsius to Kelvin by adding 273.15.

ΔT = 8.518K

Next, we calculate the heat absorbed or released during the combustion using the equation:

q = m × ΔT

q = 0.876g × 8.518K = 7.229928kJ

Now that we have the heat (q), we can calculate the molar enthalpy (ΔH) using the equation:

ΔH = q / n

To calculate the number of moles (n) of C8H18, we need to use the molar mass (MM) of octane, which is 114.224 g/mol.

n = m / MM
n = 0.876g / 114.224 g/mol = 0.007671 mol

Now, let's calculate the molar enthalpy (ΔH).

ΔH = 7.229928kJ / 0.007671 mol = 942.87 kJ/mol

Therefore, the molar enthalpy of octane (C8H18) is 942.87 kJ/mol.