Question

A 0.54 g sample of fructose (MW = 180. g/mol) is burned in a bomb calorimeter that has a heat capacity of 2.69 kJ/^oC. The temperature of the calorimeter increases by 3.16^oC. Calculate the molar heat of combustion of fructose using the data from this experiment. Since this experiment is carried out under conditions of constant volume, we are measuring ∆E. Your answer should be in kJ/mol and entered to 3 sig. fig.

∆E=?

Answer 1

Given that mass of fructose = 0.54 g

molar mass of fructose = 180 g/mol

moles of fructose n = mass / molar mass = 0.003 mol

heat capacity of the calorimeter = 2.69 kJ/^oC

dT = 3.16 ^oC

heat capacity of the calorimeter c=  q/dT

   q = c dT = 2.69 kJ/^oC x 3.16 ^oC = 8.5004 kJ

∆E = q/n

= 8.5004 kJ / 0.003 mol

= 2833.5 kJ/mol

  ∆E = 2833.5 kJ/mol

Therefore, ∆E = 2833.5 kJ/mol