Question

A)A bomb calorimeter, or constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and the energy content of foods.
Since the "bomb" itself can absorb energy, a separate experiment is needed to determine the heat capacity of the calorimeter. This is known as calibrating the calorimeter.
In the laboratory a student burns a 0.392-g sample of bisphenol A (C₁₅H₁₆O₂) in a bomb calorimeter containing 1140. g of water. The temperature increases from 25.00 °C to 27.40 °C. The heat capacity of water is 4.184 J g^-1°C^-1.
The molar heat of combustion is −7821 kJ per mole of bisphenol A.

C₁₅H₁₆O₂(s) + 18 O₂(g) ------>15 CO₂(g) + 8 H₂O(l) + Energy

Calculate the heat capacity of the calorimeter.
heat capacity of calorimeter =_________ J/°C

B)

A bomb calorimeter, or a constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and the energy content of foods. In an experiment, a 0.5574 g sample of benzil (C14H10O2) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.379×103 g of water. During the combustion the temperature increases from 26.36 to 28.99 °C. The heat capacity of water is 4.184 J g-1°C-1. The heat capacity of the calorimeter was determined in a previous experiment to be 847.1 J/°C. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of benzil based on these data. C14H10O2(s) + (31/2) O2(g) ------->5 H2O(l) + 14 CO2(g) + Energy Molar Heat of Combustion = _________ kJ/mol C) A bomb calorimeter, or a constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and the energy content of foods. In an experiment, a 0.9900 g sample of β-D-fructose (C6H12O6) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.292×103 g of water. During the combustion the temperature increases from 21.82 to 24.12 °C. The heat capacity of water is 4.184 J g-1°C-1. The heat capacity of the calorimeter was determined in a previous experiment to be 991.1 J/°C. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of β-D-fructose based on these data. C6H12O6(s) + 6O2(g) -------> 6 H2O(l) + 6 CO2(g) + Energy Molar Heat of Combustion = ____________ kJ/mol All three problems are a set for one question on my homework, so please answer all three. Thank you so much!

Answer 1

a) Moles of Bisphenol= mass/molar mass

Molar mass of bisphenol (C15H16O2)= 15*12+16*1+2*16= 180+16+32= 228 g/mole

Moles of Bisphenol= 0.392/228 mole=0.001719 moles

Molar heat of combustion = -7821 Kj/mole

1 mole of Bisphenol liberates 7821 KJ

0.001719 moles of Bisphenol liberates 7821*0.001719 Kj of heat=13.45 KJ of heat=13.45*1000 Joules

This heat is taken by both Calorimeter and water

Let C= heat capacity of Calorimeter

Hence heat liberated= heat taken by calorimeter+ heat taken by water

= heat capacity of calorimeter* temperature difference +mass of water* specific heat of water* temperature difference

13.45*1000= C*(27.4-25.0)+1140*4.184*(27.4-25)

C=834.41 J/deg.c

b)

Molar mass of Benzil= 14*12+10*1+2*16= 210 g/mole

Moles of Benzil= mass/molar mass= 0.5574/210 =0.002654 moles

The heat liberated is taken by water and calorimeter

Let C= heat capacity of Calorimeter

Hence heat liberated= heat taken by calorimeter+ heat taken by water

= heat capacity of calorimeter* temperature difference +mass of water* specific heat of water* temperature difference

Heat liberated= 847.1*(28.99-26.36)+1.379*10³*4.184*(28.99-26.36)=17402.28 J

This much heat is liberated by burning of 0.002654 moles of Benzil

Heat of combustion = 17402.28 J/0.002654 moles = 6557001 J/mole =-6557.001 Kj/mole

Since heat is liberated, heat of combustion becomes negative.

c)

Molar mass of beta-D- fructose= 6*12+12*1+6*16=180 g/mole

Moles of beta-D-fructose= mass/molar mass =0.99/180=0.0055 moles

The heat liberated is taken by water and calorimeter

Let C= heat capacity of Calorimeter

Hence heat liberated= heat taken by calorimeter+ heat taken by water

= heat capacity of calorimeter* temperature difference +mass of water* specific heat of water* temperature difference

991.1*(24.12-21.82)+1.292*1000*4.184*(24.12-21.82)=14712.7 J

Heat liberated per mole = 14712.7/0.0055=2675037 J/mole=2675.037 Kj/mole

Heat of combustion = -2675 KJ/mole