Can someone please help me with this Chemistry question

Question

Octane, C₈H₁₈, a constituent of gasoline burns according to the equation:

C₈H₁₈(l) + 12.5 O₂(g) ---> 8CO₂(g) + 9H₂O(l)

0.1111g of C₈H₁₈ was burned in the presence of excess O₂ in a bomb calorimeter. The heat capacity of the calorimeter was 1.726 x 10³ J/degree C. The temperature of the calorimeter and 1.200 x 10³ grams of water rose from 21.22 degrees Celsius to 23.05 degrees Celsius. Calculate the heat of combustion per gram of octane.

Answer 1

Usually, this question is answered like this:

q=heat change of calorimeter

mc_waterΔt=heat change of water

C_calorimeterΔt=heat change of calorimeter, so

q=m_waterc_waterΔt_water+C_calorimeterΔt_calorimeter

q=(1200g)(4.184 J/gC)(23.05-21.22 C)+(1726 J/C)(23.05-21.22 C)

q=12346.644 J

Since the calorimeter got warmer, it's endothermic with respect to the calorimeter and exothermic with respect to the reaction. same number, opposite signs:

|+q_calorimeter|=|-q_rxn|

q_rxn= -12346.644 J

Or, in kJ

q_rxn= -12.346 kJ

Calculate the heat of combustion per gram of octane.

This should, again, be a negative value since it's exothermic .Divide the energy by the grams.

-12346.644 J / 0.1111g C₈H₁₈

= -111130.9 J/g

OR

-12.346644 kJ / 0.1111g C₈H₁₈

= -111.1309 kJ/g

Answer 2

You are looking for the heat of combustion per gram (it is usually per mole). The heat released by the reaction cause the temperature of the calorimeter and the water to increase:

-mΔH_Comb = m_wc_wΔΤ + C_CalΔT (If it was the heat of combustion per mole, the LHS would be -nΔH. The negative sign is there because exothermic reactions have negative enthalpies. If you just want to give the the heat of combustion as a positive number, don't include the sign)

You know that c_w is 4.284 J/gK

Calculate the RHS and divide by the mass of octane in order to find the heat of combustion. Units all work out, but you should always make sure that they cancel appropriately.