Candle Rings
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combustion of a candle lab HELP PLEASE!?
my datas:
* initial mass of the candle = 10.17g
* initial temperature of the water = 18.5 ‘C
* highest temp. reached by the water = 38.5 ‘C
* final mass of the candle = 9.45g
the purpose of my lag was to determine the amount of heat released by the burning of a paraffin wax candle.
we had to lit a candle just below the ring stand. and place an can of coke filled with water in the ring. we then had to heat up the water using the heat of the candle.
QUESTION: based on your experimental data, calculate the amount of heat that would be released by one mole of tricontane
i’m just so lost right now 
please help!
the balanced chemical equation:
2 C30H62 + 61 O2 -> 60 CO2 + 62 H2O
thank youu!
sorry! mass of water = 200g!
sorry but one more QUESTION:
what would be the thermochemical equation for the combustion of tricontane?
would it be…:
2 C30H62 + 61 O2 -> 60 CO2 + 62 H2O + the amount of heat released by one one (or is it two moles?:S) mole of tricontane?
do you have the mass of water???? or was the coke can full- that is contained 12 ounces? if yes, than can convert 12 ounces to mass.
The energy to heat the water uses the equation:
Q=mCpdT
Q= heat input- in calories of kcalories
m=mass of water
Cp= heat capacity of water 1 cal/gm Deg C
dT= the change in temperature
Once you calculate Q, this approximates the heat released by burning the candle.
heat release= Q/dm units are cal/gm or kcal/gm
dm= change in mass of the candle
or use Q/change in moles= Q/0.74
Tricontane is C30H62- found the formula by doing a google search
MW tricontane= 30 * 12 + 62 * 1= 422 gms/mole
change in candle mass= 10.17-9.45= 0.72 gms
0.72 gms * 1mole/422 gms=0.0017
Another way to do it is to use the chemical formula and do a theoretical calculation for the heat of combustion.
C30H62 + O2 –> 30CO2 + 31H2O
1 mole C30H62 gives 30 moles CO2 which is 30 moles of Carbon .. so 0.0017 moles of C30H62 * 30 moles C/mole C30H62= .051 moles Carbon
1mole C30H62 gives 31 moles H2O which is 31 moles of H2
so 0.0017 moles C30H62 * 31 moles H2/mole C30H62
= 0.0527 moles H2
Heat of combustion for Carbon= 32.8 MJ/kg
Hc= 32.8 MJ/kg * 1kg/1000gms * 12 gms/mole= 0.393 MJ/mole = 393 kJoules/mole C * 0.051 moles C= 20.1 kJoules
Heat of combustion for H2=121 MJ/kg
Hh= 121 MJ/kg * 1kg/1000gms * 2gms/mole =0.242MJ/mole H2
Hh= 242 kJoules/mole
Heat of combuston= 20.1 kJ/mole + 242 kJ/mole= 261 kJ/mole C30H62
Answer= 261 kJ/mole C30H62
261 kJ/mole * .0017 moles C30H62= 0.44kJoule heat release during experiment
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