Evaporation and Intermolecular Attractions Lab

September 27, 2017 | Author: Chaylen Jade | Category: Evaporation, Solution, Chemical Bond, Intermolecular Force, Gases
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Chaylen Andolino Partner: Bethany Rudolph Make-up Lab Date: 5/25/2012 Evaporation and Intermolecular Attractions Lab Purpose: The purpose of this lab was to observe temperature changes caused by the evaporation of ethanol, 1- propanol, 1- butanol, n- pentane, methanol, and n- hexane, and then relate the temperature changes to the strength of intermolecular forces of attraction. Background: Evaporation is the process where a liquid solution changes into the gaseous state. Solutions can evaporate at any given temperature and evaporation occurs at the surface of the solution instead of through the solution like boiling. Evaporation relates with kinetic energy because the average velocity of the particles in a solution are raised because of temperature increase then the particle’s individual velocity will rise and more collisions will occur. As more collisions occur it will reach a point where the particles of the solutions surface will begin to be collided and hit with more force and the particles will be removed from the liquid and the surface which causes those particles to become gaseous. Physical properties like melting point, boiling point, viscosity, solubility, and evaporation are related to the strength of attractive forces between molecules. These attractive forces are called intermolecular forces. Hydrogen bonds are the strongest bond between molecules out of the three. When a hydrogen bond is present it becomes more difficult to break that bond between the molecules since the strength of the bond between the molecules is strong it then becomes more difficult for a solution in a liquid state to become gaseous which would entail bonds to be broken in turn would result in a slower evaporation rate. The weakest intermolecular force would be the London dispersion bond since the bond is the weakest and it is less difficult to break apart the bonds in substances that have dispersion bonds present. Since the bonds are easier to break apart, solutions that consist of dispersion forces have a faster evaporation rate; the easier it is to break the bonds the less energy is required to do so therefore the rate of evaporation is quickened. Dipole-dipole forces are stronger than dispersion but weaker than hydrogen bonding which means that its rate of evaporation is between hydrogen bonding and dispersion. In this experiment two types of intermolecular forces, hydrogen bonding and dispersion forces, can help to understand the evaporation rates of alcohols and alkanes.

All of the substances are flammable, so the person conducting the experiment should be careful. Once the probe has been in the vial for the time recommended and is taken out, the vial should immediately be closed. Not only will this prevent dangerous spills, but it will also prevent others from getting sick from the strong smell of the alcohols. Glasses should also be worn as the substances could be harmful to eyes.

Prediction: My partner and I predicted that as we went down the table, the changes in temperature would increase because that was the pattern we saw so far with the first two substances. We also thought that butanol would have around 12.5 because it has hydrogen bonding like 1- propanol, but has a higher molecular weight. N- pentane was predicted to have higher than ethanol and propanol because its molecular weight is higher than each. We thought that methanol would be higher because of the hydrogen bonds and since the molecular weight is lower compared to ethanol. N- hexane was predicted to have highest because although there is no hydrogen bonding, the molecular weight is higher than n-pentane. Variables: Independent Variable: - ethanol, 1- propanol, 1- butanol, n- pentane, methanol, and n- hexane Dependent Variable: -rate of evaporation of the substance Controls: -length of paper -how long the probe stays in the vial Procedure Alterations: We made a few changes to the procedure, such as doing the last four trials all at the same time. This may have made it more difficult to make sure all samples were taken out of the liquids at the same time, and also made it more difficult to keep track of which sample was which. Another change was that the vials of the substances were placed in a small beaker before the probes were put it in to make sure that they did not fall over. We also taped the probes above the sink instead of over the edge of the table to ensure that they did not get hit by people walking by the table.

RESULTS: Substance

t₁

t₂

ethanol

21.4

12.2

1- propanol 21.2

10.3

ΔT t₁-t₂ .

. .

. .

.

.

.

.

1- butanol

22

18.5

n- pentane

22.3

9

.

.

.

methanol

22

2.3

.

.

.

n- hexane

21.8

7.8

.

.

or the ernier temperature pro e the uncertainty is

.

.

Qualitative Data: -The paper became completely saturated in the liquid that it in immersed in. - Some of the samples evaporated more quickly, I could see the paper dry. -The substances smelled very strong.

Conclusion: The purpose of this lab was to study the temperature changes of four different alcohols and two alkanes when they were evaporated and compare the results to the strength of intermolecular forces of attraction. Not all of our predictions were correct. We had guessed that the changes in temperature would continue rising at a steady rate. However, this was not the case. We had predicted that 1butanol would have a higher temperature change compared to methanol, but not to the same extent that we collected. From the data in the table, it can be said that although both alcohols have hydrogen bonding, the one with the lower molecular weight has a much higher change in temperature. Our predictions for the alkanes were almost exactly correct. Overall, the lab went pretty well. The data that my partner and I collected was mostly consistent with the data that other groups in the class collected. Our data had a few differences compared to the rest of the class. Our change in temperature for methanol (19.7 C) was high in comparison to other data that the rest of the class got (13.2 C) for an unknown reason. Our results for pentane was very low (13.3 C), while the majority of the class got about (18 C). The data collected showed that as intermolecular forces become stronger, evaporation becomes a slower process. There are numerous sources and possibilities of error that may have occurred during the data collection. One random error that may have altered data could be a breeze or wind that sped up the rate of evaporation. Also, if the sizes of the paper were not consistent, some substances may have had longer evaporation rates if the paper was longer. We caused a few systematic errors. The first was that the paper wrapped around pentane unraveled during the evaporation timing, which could have sped up the process. Also, we did not measure exactly 5 centimeters off of the edge of the sink, so that may have caused our results to be slightly different than other groups. Other groups in the class did not follow the procedure exactly and either cooled substances on the side of the table or help the probes in the air instead of taping them down. These errors make it so that not all of the groups had consistent processes, so it is harder to compare the data. Data Questions (from lab): 1) The difference in change of temperature for n- pentane and 1- butanol is due to their intermolecular attractions. They have nearly the same molecular weights but 1- butanol has hydrogen bonding between the molecules, which means strong attractions and low evaporation rates. 2) The alcohol with the strongest intermolecular forces would be 1- butanol because it has the strongest attractions between the particles due to its hydrogen bonding. Its molecules are also the largest which means that it would also have stronger forces than the other alcohols. The alcohol with the weakest intermolecular forces would be methanol it is lighter and has weaker hydrogen bonding when compared to 1- butanol.

3) N- hexane is the alkane with the strongest intermolecular forces. This is because like 1butanol, its particles are largest so it has stronger forces between the molecules. Because of this, there is less of a temperature change and lower evaporation rate. N- pentane has weaker attractions because it has a lower molecular weight compared to n- hexane. 4) 25 20 15

ΔT

32.05 10

46.08 60.11

5

74.14

0 0

20

40

60

80

Molecular Weight

Sample Calculations: Substance

t₁

t₂

ethanol

21.4

12.2

ΔT t₁-t₂ .

.

.

The ΔT was determined y su tracting the final temperature from the initial temperature. The uncertainty for the temperature probes was ± .2 C, so that was added on at the end of the answer. 21.4 – 12.2 = 9.2 ± .2 C

References: IB HL Chemistry Textbook http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/H/HydrogenBonds.html http://itl.chem.ufl.edu/2045/lectures/lec_g.html

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