In the case of an emergency the precautions which must be followed are listed below.
Eye contact: Immediately flush the eye with plenty of water. If irritation persists call for medical help.
Skin contact: Wash off with water. Remove any contaminated clothing. If the skin reddens or appears damaged, call for medical aid. Be aware that clothes soaked in butanol present a serious fire risk, so ensure that clothes (and anybody in them!) are kept well away from sources of ignition.
If swallowed: Wash out the mouth with water if the person is conscious. If a significant quantity has been swallowed call for immediate medical help.
To ensure the investigation is fair always weigh the spirit burner with the lid. This ensures all weights are accurate. Also be sure to keep the lid on the spirit burner when it is not in use to prevent vapour loss through evaporation. Make sure the same equipment is used each time. Use the same increase in temperature for each alcohol to ensure results are accurate. Keep the calorimeter the same height above the burner for each test. Try to keep the same conditions in each test. Remove the build up of soot in-between tests. Ideally there should be no build up of soot as this indicates incomplete combustion and therefore not all the energy used is being transferred to the water.
Results
Analysis
To work out the enthalpy change of combustion for the alcohols I can use the assumption that 1cm of water is equal to 1g. Using this assumption I can then use the equation
energy absorbed = MC▲T
Which means :
Amount of energy = mass of water x specific heat capacity x temperature rise
absorbed by water of water
Therefore for Methanol:
Energy absorbed = 200g x 4.2 gk x 15 = 12,600J
by water
This value is in Joules it would be more adequate in Kilo joules therefore
12,600 = 12.6 KJ
divide by a thousand 1000
Ethanol
Energy absorbed = 200 x 4.2 gk x 15 = 12,600J
by water
12,600 = 12.6KJ
divide by a thousand 1000
Propan-1-ol
Energy absorbed = 200 x 4.2 gk x 15 = 12,600J
by water
12,600 = 12.6KJ
divide by a thousand 1000
Butan-1-ol
Energy absorbed = 200 x 4.2 gk x 15 = 12,600J
by water
12,600 = 12.6KJ
divide by a thousand 1000
Pentan-1-ol
Energy absorbed = 200 x 4.2 gk x 16 = 13,440J
by water
13,440 = 13.44KJ
divide by a thousand 1000
I can then use this value to calculate the enthalpy change of combustion for the alcohols.
Enthalpy change of combustion is the enthalpy change when 1 mole of a fuel is burned completely in oxygen under standard conditions.
Therefore I will also need to calculate the number of moles burned for each fuel. To do this I need to know the relative molecular mass of each alcohol.
Methanol
Mr = 12+3+16+1
Mr = 32gmol
Number of moles = Mass of fuel burned
Molar mass
Number of moles = 1.3g _ = 0.04
32
Therefore.0.04 moles of methanol releases 12.6KJ of energy
1 mole of methanol will release 1 x 12.6
0.04
1 mole of methanol will release 315 KJ mol
Ethanol
Mr = 12+12+5+16+1
Mr = 46gmol
Number of moles = Mass of fuel burned
Molar mass
Number of moles = 1.05g = 0.023
46
Therefore 0.023 moles of ethanol releases 12.6KJ of energy
1 mole of ethanol will release 1 x 12.6
0.023
1 mole of ethanol will release 547.8 KJ mol
Propan-1-ol
Mr = 12+12+12+7+16+1
Mr = 60gmol
Number of moles =Mass of fuel burned
Molar mass
Number of moles = 0.92 = 0.015
60
Therefore 0.015 moles of propan-1-ol releases 12.6KJ of energy
1 mole of propan-1-ol will release 1 x 12.6
0.015
1 mole of propan-1-ol will release 840 KJ mol
Butan-1-ol
Mr = 12+12+12+12+9+16+1
Mr= 74gmol
Number of moles =Mass of fuel burned
Molar mass
Number of moles = 0.83 = 0.01
74
Therefore 0.01 moles of butan-1-ol releases 12.6 KJ of energy
1 mole of butan-1-ol will release 1 x 12.6
0.01
1 mole of butan-1-ol will release 1260 KJ mol
Pentan-1-ol
Mr = 12+12+12+12+12+11+16+1
Mr = 88gmol
Number of moles =Mass of fuel burned
Molar mass
Number of moles = 0.75 = 0.009
88
Therefore 0.009 moles of pentan-1-ol releases 12.6KJ of energy
1 mole of pentan-1-ol will release 1 x13.44
0.009
1 mole of pentan-1-ol will release 1493.3 KJ mol
My results and graph show a clear increase in the enthalpy of combustion as the alcohols get larger. The greater the number of carbon atoms in the chain the greater the enthalpy change of combustion. Methanol has the least number of Carbon atoms in the chain and also has the lowest enthalpy change of combustion. Pentan-1-ol has the greatest number of carbon atoms in the chain and also the highest enthalpy change of combustion. Therefore as the number of carbon atoms increases the enthalpy change also increases. This is because as the number of carbon atoms in a chain increases the number of bonds which can be broken also increases. Bond breaking is endothermic which means energy is given out to the surroundings. The more carbon atoms there are in a chain the more energy there will be given out to the surroundings through the breaking of bonds, therefore the greater the enthalpy change of combustion. Methanol has the least number of carbon atoms, therefore the least number of bonds broken which results in lower enthalpy change of combustion as less energy is given out to the surroundings. Pentan-1-ol has the greatest number of carbon atoms in the chain, therefore the most number of bonds broken which results in the highest enthalpy change of combustion as more energy is given out to the surroundings.
The alcohols increase in size by a CH group each time. This results in an increase of approximately 240KJ mol for the enthalpy change of combustion. This increase is attributed to the fact there is 1 more C-C bond and 2 more C-H bonds broken each time, which means the burning of the fuel is more exothermic so more heat will be given out to the surroundings and the enthalpy change will be greater. The bonds are broken and new bonds are formed to give the products water and carbon dioxide. As the alcohols increase in size by a CH group 1 more water and 1 more carbon dioxide are produced.
As you can see from the results table my results are much smaller than the data book values, possible explanations will be given for this in the Evaluation.
Evaluation
My results were considerably lower than the data book values for enthalpy change. This is because of heat loss due to errors in the investigation. It is inevitable that during this investigation numerous errors will occur. Some will be procedural errors others will be due to human error. These errors have reduced the accuracy of my results. Firstly human error may have occurred when taking readings and measurements, this is inevitable and little can be done to prevent it.
One source of procedural error is the electronic balance used for weighing the spirit burner. The electronic balance works to 2 decimal places therefore:
When weighing Methanol
Initial weight =185.75g to 2 decimal places therefore actual weight of the spirit burner is 185.75g + or – 0.005g
Weight after burning = 184.45g to 2 decimal places therefore actual weight of the spirit
burner is 184.45g + or – 0.005g
This gives a total precision error of + or - 0.01g , therefore initial weight of methanol burner may have been anywhere between 185.75g and 185.85g and weight after may have been anywhere between 184.45g and 184.55g. This inevitably leads to less reliable results.
Percentage error for methanol
Percentage error = total precision error x 100
of balance mass of fuel burned
Percentage error = 0,01 x 100 = 0.77%
1.3
Therefore there is a percentage error of 0.77% when weighing the methanol spirit burner.
When weighing Ethanol
Initial weight = 174.63g to 2 decimal places therefore actual weight of the spirit burner is
174.63g + or – 0.005g
Weight after burning = 173.58g to 2 decimal places therefore actual weight of the spirit
burner is 173.58g + or – 0.005g
This gives a total precision error of + or - 0.01g.
Percentage error for ethanol
Percentage error = 0,01 x 100 = 0.95 %
1.05
Therefore there is a percentage error of 0.95% when weighing the ethanol spirit burner.
When weighing Propan-1-ol
Initial weight = 187.73g to 2 decimal places therefore actual weight of the spirit burner is
187.73g + or – 0.005g
Weight after burning = 186.81g to 2 decimal places therefore actual weight of the burner is 186.81g + or – 0.005g
This gives a total precision error of + or - 0.01g.
Percentage error for Propan-1-ol
Percentage error = 0,01 x 100 = 1%
0.92
Therefore there is a percentage error of 1% when weighing the propan-1-ol spirit burner.
When weighing Butan-1-ol
Initial weight = 174.64g to 2 decimal places therefore actual weight of the spirit burner is
174.64g + or – 0.005g
Weight after burning = 173.81g to 2 decimal places therefore actual weight of the spirit burner is 173.81g + or – 0.005g
This gives a total precision error of + or - 0.01g.
Percentage error for butan-1-ol
Percentage error = 0,01 x 100 = 1.2%
0.83
Therefore there is a percentage error of 1.2% when weighing the butan-1-ol spirit burner
When weighing Pentan-1-ol
Initial weight = 170.15g to 2 decimal places therefore actual weight of the spirit burner is
170.15g + or – 0.005g
Weight after burning = 169.4gto 2 decimal places therefore actual weight of the spirit burner is 169.4g + or – 0.005g
This gives a total precision error of + or - 0.01g.
Percentage error for pentan-1-ol
Percentage error = 0,01 x 100 = 1.3%
0.75
Therefore there is a percentage error of 1.3% when weighing the pentan-1-ol spirit burner
There is also a precision error with measuring the volume of water. I used 200cm of water which I measured using a 200cm measuring cylinder. The actual volume of the water will have been 200cm + or – 1cm as the measuring cylinder only measures to the nearest 1cm.Therefore there is a precision error of 1cm when measuring the volume of H O
Percentage error for volume of H O
Percentage error = precision error x 100
Volume of H O
Percentage error = 1 x 100 = 0.5%
200
Therefore there is a percentage error of 0.5% when measuring the volume of water used.
There is also a precision error when recording the temperature of the water. The thermometer records the temperature to the nearest 1ºc therefore:
When measuring the temperature of the water for methanol
Initial temperature = 17ºc + or – 0.5ºc
End temperature = 32º+ or – 0.5º
This gives a total precision error of 1ºc
Percentage error = 1 x100= 6.7%
15
Therefore there is a percentage error of 6.7% when measuring the temperature.
When measuring the temperature of the water ethanol
Initial temperature =16ºc + or – 0.5ºc
End temperature = 31º+ or – 0.5ºc
This gives a total precision error of 1ºc
Percentage error = 1 x100= 6.7%
15
Therefore there is a percentage error of 6.7% when measuring the temperature.
When measuring the temperature of water for Propan1-ol
Initial temperature =20ºc + or – 0.5ºc
End temperature = 35º+ or – 0.5ºc
This gives a total precision error of 1ºc
Percentage error = 1 x100= 6.7%
15
Therefore there is a percentage error of 6.7% when measuring the temperature.
When measuring the temperature of water for Butna-1-ol
Initial temperature = 21ºc + or – 0.5ºc
End temperature = 36ºc + or – 0.5ºc
This gives a total precision error of 1ºc
Percentage error = 1 x100= 6.7%
15
Therefore there is a percentage error of 6.7% when measuring the temperature.
When measuring the temperature of water for pentan-1-ol
Initial temperature = 16ºc + or – 0.5ºc
End temperature = 32ºc + or – 0.5ºc
This gives a total precision error of 1ºc
Percentage error = 1 x100= 6.3%
16
Therefore there is a percentage error of 6.3% when measuring the temperature.
The above sources of error act to reduce the value of enthalpy change calculated.
All of the above sources of error are precision errors and therefore inevitable. Other sources of error include heat loss to the surroundings and the copper calorimeter. To reduce this an insulator could be used to insulate the copper calorimeter and prevent heat loss. More efficient draught shielding could be used to reduce heat loss to the surroundings. During the investigation soot appeared on the bottom of the calorimeter. This is carbon and is an indication that incomplete combustion has occurred. When incomplete combustion occurs carbon is produced where as with complete combustion oxygen is used to produce carbon dioxide. This is another source of heat loss. More accurate equipment could be used to reduce procedural errors and increase the reliability of my results.
