At constant volume, the heat of combustion of a particular compound is -3152.0 kj/mol. When 1.321 of the compound (molar mass = 116.30 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 8.223 C.
what is the heat capacity (calorimeter constant) of the calorimeter?

Harley,

This a question about calorimetry and is closely related to your other question about KBr. Here, though, we will solve for a different term in the equation ΔH = -C

_{cal}x ΔT. In the other question, we solved for ΔH. Here, we need to solve for C_{cal}.We are given the molar heat of combustion (-3152.0 kJ/mol) of a hypothetical substance, its molar mass (116.30 g/mol), and the mass of the sample (1.321 g). (Your question doesn't actually have units for this last quantity so I'm assuming it's grams.)
We are also given the temperature change (8.223 °C).

It might be tempting to just plug -3152.0 kJ/mole and 8.223 °C into the equation ΔH = -C

_{cal}x ΔT and solve for C_{cal}, but this will lead to an incorrect answer because the answer will have the units of kJ/mole -°C. The correct units for C_{cal}are kJ/K (or J/K). So how do we get the numerical and units parts to be correct so as to get the right answer?First, we will determine the heat of combustion of the 1.321 g sample from the other information given. So,

1.321 g of sample/ 116.30
g/mol of sample = 0.01136 mol of sample.

then, ΔH

_{sample}= -3152.0 kJ/mol x 0.01136 mol of sample = -35.807 kJ.Second, we are given ΔT in °C, but it needs to be in Kelvin. Since the size of a degree is the same in the Celsius and Kelvin scales, we just need to change this value to 8.223 K.

Third, we plug these values into the equation ΔH = -C

_{cal}x ΔT:-35.807 kJ = -C

_{cal}x 8.223 K. Rearranging gives-C

_{cal }= -35.807 kJ/ 8.223 K-C

_{cal}= -4.354 kJ/K or C_{cal}= 4.354 kJ/K after multiplying both sides of the equation by -1.Checking our answer, the units are in kJ/K as needed and the heat capacity of the calorimeter is a positive number which is expected.