viernes, 23 de enero de 2015

Intermolecular forces (Evaporation rate)




Objective


To relate the intensity of the intermolecular forces to a measurable property. 
Associate the intensity of the intermolecular forces to structural characteristics of the molecules.

Hypothesis


We believe that, when doing the experiment, the substances with stronger intermoleculas forces (hydrogen bonds) will evaporate slower, or, in other words, their change in temperature is lower and therefore, it will take more time to reach the temperature it needs to evaporate. This happens due to the fact that intermolecular forces will hold the molecules together and prevent them from evaporating.

Consequently, those molecules with weaker intermolecular forces will have a higher temperature change which means they will evaporate quicker because there are less forces holding the molecules together and preventing them from evaporating.

Materials


  • Group of substances 1 (similar molecular weight): diethyl ether, pentane, methyl acetate, butanone, butanol, propanoic acid.
  • Group of substances 2 (homologous series): methyl acetate, ethyl acetate, propyl acetate, butyl acetate.
  • Temperature probe
  • LoggerPro or PC interface
  • Pieces of filter paper (2.5x2.5 cm2)
  • Small rubber bands which were too big, so a wire was used.
  • Test tubes in a rack (12)
Security

  • Use goggles.
  • Diethyl ether is extremely volatile and flammable. Manipulate it always in a ventilated place and away of flames and sparks.
  • All these substances are volatile. Avoid inhalation.

Method
  1. Take one piece of paper and envelope with it the tip of the probe. Secure it with a rubber band.
  2. Take 5 mL of one of the substances in a test tube.
  3. Dip the probe in the tube and let the paper soak.
  4. Take the probe out, put it horizontally and begin to acquire data (about 4 minutes, a data each 3 seconds).


  1. Return the remaining liquid to its container.
  2. Repeat the whole procedure for each substance.


Results

With the data we obtained, we have set up a table to organise the drop of temperature (difference between the highest temperature value and the lowest) and the substance to which the drop of temperature belongs:

Maximum drop of temperature depending on the substance
Chemical name
Maximum drop of temperature (ºC)
Butyl Acetate
4,2
Propyl Acetate
6, 43
Ethyl Acetate
11,37
Methyl Acetate
15,35
Butane
1,69
Ethanol
21,57

For example, in Butyl Acetate, the highest value in its temperature was 22,87ºC and the lowest was 18,67ºC so, we substract the second value to the first one and we obtain the maximum drop of temperature, 4,2ºC.

Later, we laid out a table with the substances with a similar molecular weight (which we found looking on the Internet) and their maximum drop of temperature, which resulted in the following table:
Maximum drop of temperature depending on the molecular weight

Chemical name
Molecular mass (amu)
Maximum drop of temperature (ºC)
Methyl acetate
74, 078
4,2
Ethyl acetate
88,105
6, 43
Propyl acetate
102,132
11,37
Butyl acetate
116,158
15,35
Afterwards, with these data, we elaborated a graph in which the dependent variable was the maximum drom of temperature (y axis) whereas the independent variable was the molecular mass (x axis):


Questions 
  • Find out what relationship exists between the drop of temperature and the intermolecular forces.

The relationship between Van der Waal's forces and temperature is that (especially in the lipid-water ones), the intermolecular force is directly proportional to the temperature. So, if we decrease the temperature, the intermolecular force will decrease too.

In dipole-dipole forces, as the temperature decreases, the force increases because, as the rotational energy of the molecule decreases, the attraction is stronger. In other words, the poles of the molecules move at a slower rate and, as they are during more time in the same place, the two opposite poles of the molecules are in contact during more time.

Finally, in hydrogen bonding, the bond gets stronger when the temperature decreases. For example, when water freezes; as the temperature becomes lower and due to the kinetic theory (particles move slower), it forms a more rigid and stable network of hydrogen bonds.

This means that when there is a molecule with only Van der Wal forces, there will be a small drop in temperature; if there is a chemical substance with dipole-dipole and Van der Waal forces there will be a higher drop in temperature and lastly, a substance with all three types of intermolecular forces will have the highest drop in temperature (because it needs more energy to separate the molecules apart)


  • Find out what intermolecular forces have each substance
Chemical name
Type of intermolecular force
Butyl acetate
Dipole-dipole and Van der Waals
Ethyl acetate
Dipole-dipole and Van der Waals
Methyl acetate
Dipole-dipole and Van der Waals
Propyl acetate
Dipole-dipole and Van der Waals
Ethanol
Hydrogen bonding (therefore, it also has dipole-dipole and Van der Waal’s)
Butane
Van der Waals


Conclusions

During this experiment, we have learnt the relationship between the evaporation rate, the drop of temperature, intermolecular forces and molecular weight; First of all, the relationship between the evaporation rate and the intermolecular force is that, the faster a substance evaporates, the less intermolecular forces it has (van der waal's or dipole-dipole) because the substance needs less energy to overcome the weaker IMF. Consequently, if a substance evaporates slower, it means that it has a higher intermolecular force (hydrogen bonding for example) as the chemical needs more energy to overcome the stronger IMF.

Secondly, we have also learnt the relationship between the drop of temperature and intermolecular forces which is stated above: however, to summarise the answer, we can say that the smaller the drop of temperature is, the lower IMF a substance because it means that it needs less energy to overcome the intermolecular force and therefore, the IMF must be weaker.

Finally, the pattern that you can observe in the graph or, in other words, the relationship between evaporation rate (drop of temperature) and molecular mass is inversely proportional; the higher the drop of temperature is, the smaller the molecule is (lower molar mass). This is due to the fact that if you have two molecules very similar, for example heptane and octane (where octane is bigger and heavier), as heptane weighs less, its molecules can more easily escape and become a gas because they require less energy to overcome the forces between molecules. Smaller molecules need less energy because, for example, is it easier to lift a paper or to move a 10 kilogram weigh? It is easy to lift the paper because it weighs less, just like it is easier that heptane becomes a gas rather than octane.


Evaluation



Regarding the method, there were several problems that we are going to list below:
  1. One of the problems of this experiment is related to the use of the rubber band. We needed one in order to put the filter paper in the substance to be able to measure the evaporation rate. The problem was that the rubber band was way too big so we needed to use another material. We used a piece of wire but the wire was hard to wrap around and we had to put it various times until it stayed and didn’t fall. Because of this, we lost time and, within a couple of times, we had to repeat the process of wrapping the wire because the paper had fallen. The solution we have thought for this problem is to buy smaller (and thinner) elastic bands.
  2. Another problem the method of the experiment has is related with with the time the paper had to soak, because the method is a bit vague since it just says “Dip the probe in the tube and let the paper soak.” But if one of the pieces of paper was wetter than other the evaporation rate would have been slower. To solve this problem, we think the method should specify the time we have to dip the probe, for example, 1 minute.
  3. Furthermore, the size of the paper was not specified either so we have the same problem as above, the paper is bigger in some cases so there is more liquid to evaporate and therefore, the data can be incorrect. The solution is to put in the method the size of the paper, for example, a length and width of 3 cm.  
  4. There was a problem regarding the method in relation with time. We did not have enough time to finish the experiment due to the fact that there were too many substances and too little time (where it is mathematically impossible to read the experiment, understand it, and finish it within the laboratory class).  The solution we have thought is to remove some substances from the experiment, or have longer laboratory classes.
  5. The last problem we have seen in the method of the experiment is also related with time. Substances with weak intermolecular forces will evaporate quick and therefore the drop of temperature may be quite big (due to the fact that it evaporates and then, as it is wholly evaporated, the temperature starts to rise again) and, substances with strong intermolecular forces will evaporate slower which will have as a consequence, a substance that has not evaporated completely in the short period of time we leave the probe out of the water and this will cause that there may be a smaller drop of temperature than the one of the substance with weak intermolecular forces (which, in reality, should be the opposite, the drop of temperature of a substance with strong intermolecular forces is bigger than the one of a substance with weak intermolecular forces). To solve this problem, we can only think of leaving the probe out of the water for a longer period of time, for example, 10 minutes, which will lead to the problem stated above (not enough time to finish the experiment) so we will need also the solutions of the problem above to solve this one: we need more time so we will need either less substances or longer classes.

However, even though all these are problems with the method, we think that:

  • If problem 1 was solved, we would save time so maybe we wouldn't have been short of time and would have done all of the substances in the method.
  • Problems 2 and 3 cause that the results are not homogeneous so, if they were solved, the method would be more reliable.
  • Even though problem 5 is a problem, after looking up the intermolecular forces present in each substance, we discovered that this problem did not affect our results because the substance with the highest drop of temperature is the one with hydrogen bonding present whereas the substance with the lowest drop of temperature has only Van der Wall forces.

Bibliography


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