PART A: When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter

Question

PART A: When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 C . When a 0.275-g sample of caffeine, C8H10O2N4, is burned, the temperature rises 1.584 C . Using the value 26.38 kJ/g for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume.

PART B : Assuming that there is an uncertainty of 0.002 C in each temperature reading and that the masses of samples are measured to 0.001 g, what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

PART A: When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 C . When a 0.275-g sample of caffeine, C8H10O2N4, is burned, the temperature rises 1.584 C . Using the value 26.38 kJ/g for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume.

PART B : Assuming that there is an uncertainty of 0.002 C in each temperature reading and that the masses of samples are measured to 0.001 g, what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

Explanation / Answer

PART-A

When a 0.235 g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 Celcius. Using the value 26.38 kJ/g for the heat of combustion of benzoic acid...

calculate the heat released:
0.235 g benzoic acid @ 26.38 kJ/g = 6.1993 kJ

calculate the heat capacity of the calorimeter:
6.1993 kJ / temperature rises 1.643Celcius = 3.7732 kJ/ degree Celsius

find moles of a 0.275 g sample of caffeine, using molar mass:
a 0.275 g caffeine / 194.19 g/mol = 0.001416 moles

find heat released:
the temperature rises 1.584Celcius @ 3.7732 kJ/ degree Celsius = 5.9767 kJ

calculate the heat of combustion per mole of caffeine
5.9767 kJ / 0.001416 moles = 4220.86 kJ/mol
your answer, rounded to 3 sig figs would be:
4221 kJ/mol

PART-B

Assuming that there is an uncertainty of 0.002 Celcius in each temperature reading
0.002 C times 2 readings to get a dT = +/- 0.004 C

(0.004 C / dT of 1.643Celcius ) (times 100) = +/- 0.2435 % for benzoic
&
(0.004 C / dT of 1.584Celcius ) (times 100) = +/- 0.2525 % for caffeine

Assuming that there is an uncertainty of 0.001 g in each measured mass:
(0.001 g / 0.235 g sample of benzoic acid) times 100 = +/- 0.4255 % for benzene
(0.001 g / 0.275 g sample of caffeine) times 100 = +/- 0.3636 % for caffiene

total is +/- 1.29%

so 1.29% of 4221 kJ/mol = +/- 54.24 kJ

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