Burning fuels.

New Bonds:

C = O x 4 = 740 x 4=   2960 kj           5570

H – O x 8 = 464 x 8= +3712 kj         -6672

                                    -6672             -1102   2 = - 551 kj/mol 

 ETHANOL

C2 H5 OH + 3O2                      2CO2 + 3 H20

        H    H

 H – C – C – O – H + 3(O=O)              2 (C=O) + 3 (H – O – H)

   

        H    H  

Bonds Broken: 

C – H x 5= 413 x 5=  2,065

C – C x 1= 347 x 1=     347

C – O x 1= 335 x 1=     335

O – H x 1= 464 x 1=     464

O = O x 3 = 498 x 3=1,494                                                                                                                                                          

                                  4,705

New Bonds     

C = O x 4= 740 x 4=   2960 kj

O – H x 6= 464 x 6= +2784 kj    

                                   -5744

          4705

      -  5744

        -1039 kj/mol

 

PROPANOL

2C3 H7 OH + 9O2                 6CO2 + 8H2O

        H    H    H

2(H  C    C     C  O   H) + 9 (O=O)           6(C=O) + 8 (H – O – H)

        H    H   H

Bonds Broken

C – H x 14 = 413 x 14 = 5782

C – C x 4   = 347 x  4  = 1388

C – O x 2   = 335 x 2   =   670

O – H x 2   = 464 x 2   =   928

O = O x 9   = 498 x 9   = 4482

                                      13250

New Bonds:

C = O x 12 = 740 x 12 =   8880 kj

O – H x 16 = 464 x 16 = +7424 kj

                                       -16304

             13250

            -16304

   -3054     2 = -1527 kj/mol

BUTANOL

C4 H9 OH + 6O2                      4CO2 + 5H2O

         H       H      H       H

   

H       C        C      C       C    O     H + 6 (O=O)            4(C=O) +5 (H – O – H)

             

          H       H      H       H

Bonds Broken

C – H x 9 = 413 x 9 = 3717

C – C x 3 = 347 x 3 = 1041

C – O x 1 = 335 x 1 =   335

O – H x 1 = 464 x 1 =   464

O = O x 6 = 498 x 6 = 2988        

                                     8545

New Bonds

C = O x 8   = 740 x 8   =   5920 kj

O – H x 10 = 464 x 10 = +4640 kj

                                      -10,560

           8545

        -10560

          -2015 kj/mol

DELTA  H  VALUE

Graph Of Results

Diagram Of Equipment

Method

1. I am going to set up the equipment as shown above to proceed with the experiment.
2. I am going to weigh 1gm of fuel starting with methanol, followed by ethanol, Propanol and Butanol.
3. I am going to measure 50 cm3 of water and transfer it into the colorimeter.
4. I will take the temperature of the water using a thermometer, and then I shall light the fuel and use it to heat the water in the colorimeter.
5. When the fuel goes out, I will stir the water softly and then take the temperature using the thermometer.
6. Then I shall write down the results into my table after I have taken the temperature.
7. I will repeat steps 1-6 with all 4 fuels
8. For each of the fuels I shall repeat the experiment at least twice.

   Safety

For my own safety I will need to do the following

• Wear goggles at all times to protect my eyes,
• Keep the fuel away from the copper colorimeter
• Don’t spill fuel along workbenches.
• Keep all lose items of clothing tucked in.
• The lids on the alcohols must be kept on at all times to prevent evaporation or any spillages.
• Bunson burner will also need to be kept on safety flame at all times.

  Fair Test

   In order to make the experiment a fair test I will need to have:-

• The same glass of fuel each time
• Same distance between the fuel and the colorimeter
• Same temperature
• Same mass of fuel
• Always 50cm3 of water
• Same thickness of metal
• Same colorimeter
• Always 1gm of fuel
• Start with coldwater each time

   Obtaining results

 Table Of Results

 

Prediction

     

For my prediction I have used lots of background knowledge and done some preliminary experiments. As you can see I have worked out the theoretical energy release for each alcohol and from these I would predict that as the number of carbon atoms go up so will the heat of combustion. I believe that the results will be proportional to the amount of carbon atoms. I believe this because every time you add an extra carbon bond you also add two more hydrogen bonds, which means that the relative molecular mass will increase as well. I also predict that this will be a constant amount. When I achieve my results I will need to figure out the actual amount of energy given out and the molecular mass plays a big part in this.

 

Analysing

I think my results tables and graphs clearly show the pattern that I have found in this experiment. That is that heat combustion does increase when the amount of carbon atoms increases. A table that answers the question more specifically and clearly is shown below:

 

How many grams in one mole of each fuel?

I believe that my results do show a positive correlation and do show that the more carbon atoms there are the heat of combustion goes up. In this case my prediction is right but not totally right. I said that it would be a constant amount. Another reason for these results is that the molecular length becomes longer in the bigger molecules increasing the surface area hence allowing more energy to be released. These results do support my initial prediction.

Energy calculations

Methanol 1: 4.2 x 50 x 10 =  2100

                2: 4.2 x 50 x 9 =  +1890

                                              39902= 1995

Ethanol 1: 4.2 x 50 x 11 =   2310

             2: 4.2 x 50 x 10 = +2100

                                            44102= 2205

Propanol 1: 4.2 x 50 x 14 =   2940

               2: 4.2 x 50 x 16 = +3360

                                              63002= 3150

Butanol 1: 4.2 x 50 x 15 = 3150

             2: 4.2 x 50 x      =

Evaluation

My results could never have been as accurate as the theoretical calculations because there were many experimental errors involved.

• Sound and light energy could have been lost into the room.
• The beaker that the water was being held in would have used up some of the heat energy to heat itself up along with the rubber clamp.
• Also the size of the tin can was not exactly 500 ml so much of this space would have been wasted along with the fact that the alcohol containers had varying amounts of alcohol in them to start with along with varying sizes of wicks.
• This all contributed to the fact that the flame coming from the alcohol was varying in size so was sometimes not even touching the tin can.
• The room temperature would also have acted as a cooling agent.

One of the less important factors could have been if there was a lacking of oxygen leading to incomplete combustion. Then the oxygen molecules would only form with one carbon molecule producing carbon monoxide but I doubt this actually happened.                                                      

   This was a very difficult experiment top conduct in a classroom because there are lots of potential ways of losing heat because everything likes to gain heat energy. I think the thing that hindered our results the most was the fact that gusts of air and convection currents were taking the heat away from the experimental area and there was no way to stop this. As you can see from my results and my graph I did have a few anomalous results and I believe this to be in the combustion of Butanol. I think one of the things that is helping to this is that my Propanol experiment could also have been a bit anomalous but with the two together my best fit line shows Butanol to be the main culprit. If you loom at my results carefully for Butanol you can tell that I did change my spirit holder after the first experiment. The next two results do drop drastically and this could have been because of the fact that the wick sizes were different on each spirit burner.

   If I could do the experiment again I would try to find a better container for holding the water because tin might gain heat energy a lot more easily than other metals so this would cut down on my experimental errors.

   Ensure that there was no windows open because during our experiment there may have been and sudden gushes of wind could have taken some of the heat energy away.

   

Conclusion

According to the aim we planned to find out which one out of the following four fuels releases the most energy in to the atmosphere when burnt.

The equipment we used to perform the experiment were:

• Clamp
• Thermometer
• Stand
• Crucible
• Measuring cylinder
• And a Copper colorimeter

   We used Ethanol, Methanol, Propanol and Butanol assemble the experiment.

To make sure we had a fair test, we had to make sure that we had the:

Same mass of fuel

Same distance between the stand and copper colorimeter

Same temperature

Always 50cm3 of water

Same thickness of metal

Same colorimeter

And start with cold water all the time

To continue the same fair test, we performed the experiment twice with all the four fuels.

    After conducting the experiment we found out the Butanol was the best fuel according to the table of results.

Butanol released the most energy into the when it is burnt because, it had more carbon atoms, and according to the results Butanol released 240.87 kj/ mol.

    Later on we realized that there was a similar pattern between the results we had and the hypothesis, this was that the more carbon atoms there are in a compound, the more energy is released into the atmosphere.

  As we proceeded with the experiment we came across two types of reactions:

ENDOTHERMIC: A reaction that we must give it energy. This is when bonds break, and heat energy is put in.

EXOTHERMIC: When new bonds are made energy is given out.

    We can prove this by using scientific knowledge because our scientific knowledge tells us that a chemical reaction takes place in two steps.

The chemical bonds in the reactants are broken to release the atoms, which make the reactants.

    But for this to happen we need to provide energy. And the energies are as I said ENDOTHERMIC and EXOTHERMIC.

We also found out that the prediction agrees with the conclusion because we predicted that the more carbon atoms a compound has the more energy it will release. Later on we investigated the experiment further and found out that

Energy Joules/gm = 4.2 x 50 x temperature increase

4.2 is needed to increase the temperature of 1gm of water by 1C.

 Using this we found out that Butanol released 3255 Joules/gm.

 After this I can conclude that my initial prediction was actually right but I didn’t allow for all of the experimental errors. I conclude that carbon atoms in alcohols do have an effect on the heat of combustion. As the amount of carbon atoms go up the heat of combustion does. This is because every time you add another carbon atom you are also adding 15 onto the relative atomic mass that plays a big part in calculating the end results.