Can anybody help me out with this problem please. I'm not sure that I'm doing it right. A 1.841 g sample of butanoic acid, C4H8O2 (88.11 g/mol) was burned in a bomb calorimeter with excess oxygen. The temperature of the calorimeter and the water before combustion was 17.26 degrees C; after combustion the calorimeter and the water had a temperature of 24.30 degrees C.The calorimeter had a heat capacity of 697 J/K, and contained 1.372 kg of water. use these data to calculate the molar heat of combustion (in kJ) of butanoic acid. Ti = 290.26 K Tf = 297.30 K Heat capacity for calorimeter = 697/1000 = 0.697 kJ/K Heat capacity for water = 1372 g X 4.184 J/g * degrees K = 5740.448 J/g * degrees K/1000 = 5.74 kJ/g * degrees K s = heat capacity of calorimeter + heat capacity of water = 0.697 kJ/K + 5.74 kJ/g * degrees K = 6.44 kJ/g * K q = msdeltaT =(1.841 g)(6.44 kJ/g * K)(297.30 K - 290.26 K) =83.47 kJ Since qcal + qrxn = 0, qcal = -qrxn therefore, qrxn = -83.47 kJ -83.47 kJ X 88.11 g butanoic acid/1 mol butanoic acid = -7.35 x 10^3 kJ/mol can anybody confirm these results?