Ethanol ( C2H5OH )
Qualitative data
No soot found on the boiling tube.The color of the flame was light orange. Incomplete transmission of heat is assumed due to the movements caused in the laboratory, which affected the flame and therefore the transmission.
Qualitative data
Propanol ( C3H7OH )
Qualitative data
Just a little bit of soot was formed on the boiling tube.The color of flame is light orange. Incomplete transmission of heat is assumed due to the movements caused in the laboratory, which affected the flame and therefore the transmission.
Qualitative data
Buthanol ( C4H9OH )
Qualitative data
A lot of soot was seen on the boiling tube. The flame was light orange. Incomplete transmission of heat is assumed due to the movements caused in the laboratory, which affected the flame and therefore the transmission.
Qualitative data
Pentanol ( C5H12OH)
Lot of soot was formed on the boiling tube.The color of the flame was light orange .Incomplete transmission of heat is assumed due to the movements caused in the laboratory, which affected the flame and therefore the transmission.
Qualitative dataQualitative data
Data processing
Combustion of Methanol
2CH3OH + 3O2 ------> 2CO2 + 4H2O
Because in this reaction heat energy is transformet to water, in order to do the calculations,we must know the heat energy that has been given out. This can be calculated in a following wasy :
Q = m (H2O) * c (H2O) * ΔT (H2O)
Q – heat energy
M – mas of water
C- heat capacity of water , the value is 4.20 J g-1 K-1
ΔT – change in temperature
In the method it has been said that the volume of water is 40 cm ³. Therefore mass of water can be calcukated from here, if we assume that the density of water is ρ (H2O) = 1.0 g cm-3 . From here we can find the mass of water :
m (H2O) = ρ (H2O) * V (H2O) = 1.0 g cm-3 * 40.0 cm3 = 40.0 g
The change in temperature is found in the following way :
ΔT (H2O) = Tmin – Tmax = 24.0 °C – 50.0 °C = 26.0°C
ΔT (H2O) = 26.0 K
Knowing this , we can calculate the total amount of energy that has been transformed to water :
Q = 40.0 g * 4.20 J g-1 K-1 * 26 K
Q = 4368 J = 4.36 kJ
Mass of Methanol used in experiment :
m (CH3OH) = 1.59 g ( as decided )
Molar mass of methanol is needed for further calculations and it can be calculated in a following way :
M (CH3OH) = Ar (C) + Ar (O) + 4*Ar (H) = 12.01 + 16.00 + 4*1.01 =
= 32.05 g mol-1
From here the amount of burned methanol can be calculated :
n (CH3OH) = m (CH3OH)M (CH3OH) = 1.59 g32.05 g mol-1 = 0.0496 mol
The enthalpy of combustion of methanol from here is :
ΔHc (CH3OH) = Heat energy transformed to the waterAmount of methanol burnt = 4.36 kJ0.0496mol
ΔHc (CH3OH) = - 87.9kJ mol-1
Combustion of ethanol
C2H5OH(l) + 3 O2(g) =Heat=> 2 CO2(g) + 3 H2O(g)
The same formulas have to be used for these calculations.
Because in this reaction heat energy is transformet to water, in order to do the calculations,we must know the heat energy that has been given out. This can be calculated in a following wasy :
Q = m (H2O) * c (H2O) * ΔT (H2O)
Q – heat energy
M – mas of water
C- heat capacity of water , the value is 4.20 J g-1 K-1
ΔT – change in temperature
In the method it has been said that the volume of water is 40 cm ³. Therefore mass of water can be calcukated from here, if we assume that the density of water is ρ (H2O) = 1.0 g cm-3 . From here we can find the mass of water :
m (H2O) = ρ (H2O) * V (H2O) = 1.0 g cm-3 * 40.0 cm3 = 40.0 g
The change in temperature is found in the following way :
ΔT (H2O) = Tmin – Tmax =24.0 °C – 56.5 °C = 32.5°C
Therefore , ΔT (H2O) = 32.5K
Knowing this , we can calculate the total amount of energy that has been transformed to water :
Q = 40.0 g * 4.20 J g-1 K-1 * 32.5 K
Q = 5460 J = 5.40 kJ
Mass of Methanol used in experiment :
m (CH3OH) = 1.96 g ( from Figure 1)
Molar mass of ethanol is needed for further calculations and it can be calculated in a following way :
M (C2H5OH) = 2*Ar (C) + Ar (O) + 6*Ar (H) = 2*12.01 + 16.00 + + 6*1.01= 46.08 g mol-1
From here the amount of burned methanol can be calculated :
n (CH3OH) = m (CH3OH)M (CH3OH) = 1.96 g46.08 g mol-1 = 0.0425mol
The enthalpy of combustion of methanol from here is :
ΔHc (CH3OH) = Heat energy transformed to the waterAmount of methanol burnt = 5.40kJ0.0425mol
ΔHc (CH3OH) = - 127.05kJ mol-1
Combustion of propanol
C3H7OH (l) + 412 O2 (g) 3 CO2 (g) + 4 H2O (g)
The same formulas have to be used for these calculations.
Because in this reaction heat energy is transformet to water, in order to do the calculations,we must know the heat energy that has been given out. This can be calculated in a following wasy :
Q = m (H2O) * c (H2O) * ΔT (H2O)
Q – heat energy
M – mas of water
C- heat capacity of water , the value is 4.20 J g-1 K-1
ΔT – change in temperature
In the method it has been said that the volume of water is 40 cm ³. Therefore mass of water can be calcukated from here, if we assume that the density of water is ρ (H2O) = 1.0 g cm-3 . From here we can find the mass of water :
m (H2O) = ρ (H2O) * V (H2O) = 1.0 g cm-3 * 40.0 cm3 = 40.0 g
The change in temperature is found in the following way :
ΔT (H2O) = Tmin – Tmax = 24.0 °C – 74 °C = 50°C
ΔT (H2O) = 50K
Knowing this , we can calculate the total amount of energy that has been transformed to water :
Q = 40.0 g * 4.20 J g-1 K-1 * 50 K
Q = 8400J = 8.40 kJ
Mass of Methanol used in experiment :
m (CH3OH) = 1.81g ( from Figure 1)
Molar mass of propanol is needed for further calculations and it can be calculated in a following way :
M (C3H7OH) = 3*Ar (C) + Ar (O) + 8*Ar (H) = 3*12.01 + 16.00 + + 8*1.01 = 60.11 g mol-1
From here the amount of burned methanol can be calculated :
n (CH3OH) = m (CH3OH)M (CH3OH) = 1.81 g60.11 g mol-1 = 0.0301mol
The enthalpy of combustion of methanol from here is :
ΔHc (CH3OH) = Heat energy transformed to the waterAmount of methanol burnt = 8.40kJ0.301mol
ΔHc (CH3OH) = - 279.06kJ mol-1
Combustion of butanol
C4H9OH + 6O2 → 4CO2 + 5H2O + heat
The same formulas have to be used for these calculations.
Because in this reaction heat energy is transformet to water, in order to do the calculations,we must know the heat energy that has been given out. This can be calculated in a following wasy :
Q = m (H2O) * c (H2O) * ΔT (H2O)
Q – heat energy
M – mas of water
C- heat capacity of water , the value is 4.20 J g-1 K-1
ΔT – change in temperature
In the method it has been said that the volume of water is 40 cm ³. Therefore mass of water can be calcukated from here, if we assume that the density of water is ρ (H2O) = 1.0 g cm-3 . From here we can find the mass of water :
m (H2O) = ρ (H2O) * V (H2O) = 1.0 g cm-3 * 40.0 cm3 = 40.0 g
The change in temperature is found in the following way :
ΔT (H2O) = Tmin – Tmax = 24.0 °C – 76 °C = 52°C
Therefore , ΔT (H2O) = 52K
Knowing this , we can calculate the total amount of energy that has been transformed to water :
Q = 40.0 g * 4.20 J g-1 K-1 * 52 K
Q = 8736J = 8.73 kJ
Mass of Methanol used in experiment :
m (CH3OH) = 1.29g ( from Figure 1)
Molar mass of butanol is needed for further calculations and it can be calculated in a following way :
M (C4H9OH) = 4*Ar (C) + Ar (O) + 10*Ar (H) = 4*12.01 + 16.00 + + 10*1.01 = 74.14 g mol-1
From here the amount of burned methanol can be calculated :
n (CH3OH) = m (CH3OH)M (CH3OH) = 1.29 g74.14 g mol-1 = 0.017mol
The enthalpy of combustion of methanol from here is :
ΔHc (CH3OH) = Heat energy transformed to the waterAmount of methanol burnt = 8.73kJ0.017mol
ΔHc (CH3OH) = - 513.52kJ mol-1
Combustion of pentanol
C5H11OH (l) + 712 O2 (g) 5 CO2 (g) + 6 H2O (g)
The same formulas have to be used for these calculations.
Because in this reaction heat energy is transformet to water, in order to do the calculations,we must know the heat energy that has been given out. This can be calculated in a following wasy :
Q = m (H2O) * c (H2O) * ΔT (H2O)
Q – heat energy
M – mas of water
C- heat capacity of water , the value is 4.20 J g-1 K-1
ΔT – change in temperature
In the method it has been said that the volume of water is 40 cm ³. Therefore mass of water can be calcukated from here, if we assume that the density of water is ρ (H2O) = 1.0 g cm-3 . From here we can find the mass of water :
m (H2O) = ρ (H2O) * V (H2O) = 1.0 g cm-3 * 40.0 cm3 = 40.0 g
The change in temperature is found in the following way :
ΔT (H2O) = Tmin – Tmax = 24.0 °C – 85.0°C = 61°C
ΔT (H2O) = 61K
Knowing this , we can calculate the total amount of energy that has been transformed to water :
Q = 40.0 g * 4.20 J g-1 K-1 * 61K
Q = 10248J = 10.24 kJ
Mass of Methanol used in experiment :
m (CH3OH) = 1.50g ( from Figure 1)
Molar mass of pentanol is needed for further calculations and it can be calculated in a following way :
M (C5H11OH) = 5*Ar (C) + Ar (O) + 12*Ar (H) = 5*12.01 + 16.00 + + 12*1.01 = 88.17 g mol-1
From here the amount of burned methanol can be calculated :
n (CH3OH) = m (CH3OH)M (CH3OH) = 1.50 g88.17 g mol-1 = 0.0170mol
The enthalpy of combustion of methanol from here is :
ΔHc (CH3OH) = Heat energy transformed to the waterAmount of methanol burnt = 10.24kJ0.0170mol
ΔHc (CH3OH) = - 602.35kJ mol-1
Conclusion
The hypothesis was the greater the number of carbon atoms in the chain the greater the enthalpy change of combustion.The explanation of this according to my previous knowledge was that the more bonds there are holding the carbon, oxygen and hydrogen atoms together, more energy will be required to break them apart. For example Ethanol has the formula C2H5 OH. The walues of each bonds are presented bellow, and by just looking at them and thinking about the chain lenght of ethanol,methanol,propanol,butanol and pentanol respectively, it can be concluded that pentanol will have the longest chain, therefore the strongest one and so will need more energy to break down ( see the values ).
TYPE OF BOND ENERGY REQUIRED TO BREAK THE BOND (kj)
C-H 410
C-O 360
O-H 510
O=O 496
C=O 740
C-C 350
To separate C-H bond 410 joules of energy need to be applied . There are five such bonds in ethanol so 410 is multiplied by five to get 2050 joules. The same calculations are done for other alcohols.
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 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 the length of the carbon chain increases the enthalpy of combustion ( the sequence of chemical reactions between a and an accompanied by the production of and conversion of chemical specie ) of a primary alcohol increases . This means that as the number of carbon- carbon bonds increase, with the increase in the carbon chain length, more energy will be needed to break those bonds.Once they are broken more energy will be released during the combustion , because more carbon- oxygen bonds ( within carbon- oxides ) are being formed at that time.
Therefore ,the larger the alcohol molecule, the more bonds will be broken and formed, and therefore the more heat will be produced.
Errors and uncertanities
Experimental values of the combustion :
Methanol = - 87.9kJ mol-1
Ethanol = - 127.05kJ mol-1
Propanol = - 279.06kJ mol-1
Butanol = - 513.52kJ mol-1
Pentanol = - 602.35kJ mol-1
According to the data taken from the booklet ,accepted values of combustion for those five alcohols are the following :
Methanol: - 726 kJ/mol
Ethanol: - 1367 kJ/mol
Propan-1-ol: - 2021 kJ/mol
Butan-1-ol: - 2676 kJ/mol
Pentan-1-ol: -3330.63 kJ/mol
According to the results it is clear that there were some mistakes / errors in the experiment. Experimental and theoretical values are different and this can be evaluated in the following way :
In this experiment of combustion, the alcohols were burnt and the dana was collected. As they were being burnt, the bonds were being broken and the final outcome is the dana colected in the experimentAs the length of the carbon chain increases the enthalpy of combustion of a primary alcohol increases in its value,as s as the number of carbon- carbon bonds increase, parallel with the increase in the carbon chain length, more energy will be needed to break those bonds, but once the bonds are broken more energy will be released during the combustion process since more carbon- oxygen bonds ( within carbon-oxides ) are being made. The molecular length becomes longer in the bigger molecules increasing the surface area hence allowing more energy to be released.
Therefore,the conclusion is that the larger the alcohol molecule the more heat is produced during combustion.
Evaluation
There were not any major errors with the experiment, although there
were many factors that suggest some errors. The major problem was the heat loss to the surrounding.This has largely affected the results There was also soot produced on the tubes with a bigger alcohol molecule alcohols, which suggests an incomplete combustion . Burning of soot means that more energy is gained.
One reason why the results obtained were inaccurate is that some of the heat liberated was used to vaporise the liquid prior to its combustion. A suitably designed burner could get around this - using several fine holes rather than one big
hole to get better mixing of the vapour with the oxygen. Another modification might be to use oxygen enriched air or even pure oxygen to ensure the complete combustion of the alcohol.
Each time the cylinder was used, some of the content has still remained into it. To prevent this the pipet should be used.
Flame was not concentrated at the same point throughout because of the movements. To prevent this, the setup should be in the glass box ( the isolated space ) , so that less heat is lost.
At the end of each experiment soot (carbon) could be seen on the base of the copper can. This means that not all the energy from the alcohol heated the water. Also if there were a gradual build up of soot on the copper can, it would mean that it the thickness between the water and the flame was increasing. This could again have affected the results.
Temperature measured might not have been accurate every time, as well as the mass weighted. Stopwatch was also a problem ,because it was hard to look at the time and take the measurements at the same time. To prevent errors, it would be good to work in pairs .
Suggestions
Taking more data would be advisable .
Burette should be used as a more accurate measuring device.
To help keep the flame where it should be , some sort of a foil or an isolated space should be costructed to keep the heat .
Soot could have been wiped ,to reduce the possibility of this effecting the results.
The burners ,of course, need to be of the same surface area.
The workplace should not be bussy .
A pipette could be used to make sure that all the liquid is collected.
To provide a better evidence about the link between concentration and
the enthalpy of alcohols the following could be done :
Instead of investigating the affect of increasing the length of the chainthe position of the
O-H group could be investigated. For the purpose of this alcohols such as Butan-2-ol and Pentan-2-ol and Hexan-2-ol should be used and compared to, Butan-1ol and Pentan-1-ol and Hexan-1-ol.
The constants and variables could be changed , for example varying the distance between the spirit burner and copper can or volume of water.
by
zenica (student)
