calculate the __ for the combustion reaction?

Question

A 0.608-g sample of quinizarin (C14H8O4) is burned in a bomb calorimeter and the temperature increases from 25.70 °C to 28.19 °C. The calorimeter contains 1.18×103 g of water and the bomb has a heat capacity of 943 J/°C. The heat capacity of water is 4.184 J g-1°C-1. Based on this experiment, calculate ΔE for the combustion reaction per mole of quinizarin burned.

kJ/mol

Answer 1

I have no problem with the previous answer, except that it uses Q in two ways: 1 to abbreviate the compound quinizarin, and the other to signify heat. At least that's my take on it. I thought I'd provide another answer which may or may not be easier for you to follow.

Molar mass of quinizarin (C14H8O4) = 240 g/mol (add up all the atomic masses)

heat = q = mC∆T + C_cal∆T

m = mass of water = 1180 g

C = specific heat of water = 4.184 J/g/degree

∆T = change in temperature = 28.19 - 25.70 = 2.49 degrees

C_cal = heat capacity of the calorimeter = 943 J/degree

Substituting these values, we get...

q = (1180 g)(4.184 J/g/deg)(2.49 degree) + (943 J/degree)(2.49 degree)

q = 12,293 J + 2348 J

q = 14,641 J = heat from 0.608 g of quinizarin

moles quinizarin = 0.608 g x 1 mol/240 g = 0.0025 moles

14,641 joules/0.0025 moles = 585,640 J/mole = 586 kJ/mole (to 3 significant figures) = ∆E combustion

Answer 2

To figure out the ΔH of combustion/ mole of Q , you need to find the heat released and divide by the moles of Q. The formula is ΔH_C = -Q/n where Q is heat into the calorimeter and water and n is moles of Q, but you can just reason it out as illustrated in the first sentence.

1) .608 g Q x 1mole/ M g of Q (divide by molar mass of Q to get moles). You are given the formula for Q, so you know the molar mass, M, in g/mole).

2) Q into the water/calorimeter = m_wC_w(T_f-T_i) + C_Cal(T_f- T_i)

where m_w is mass of water and C_w is the specific heat of water, T_f is the final temperature of the water, and T_i is the initial temperature.

It is assumed that the calorimeter has the same temperature as the water at both ends of the process. Not that the specific heat means units of J/gK which requires a mass and a ΔT to multiply to get Joules. The C_cal is a property of the specific calorimeter (it's size, geometry, materials, etc.) and has units of J/K which does not require a mass to multiply it.

Good luck!