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Evaporation and Intermolecular Attractions Remote Lab, Lab Reports of Chemistry

Butler County Community CollegeChemistry

Involves intermoleforces, dipole- dipole, ldf, and hydrogen bonds

Typology: Lab Reports

2019/2020

Uploaded on 11/16/2020

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Evaporation and Intermolecular Attractions – Remote Lab
Purpose: To study temperature changes caused by the evaporation of several liquids and
relate the temperature changes to the strength of intermolecular forces of attraction.
In this experiment, temperature probes are placed in various liquids. Evaporation occurs when
the probe is removed from the liquid’s container. (See the figure, below.) This evaporation is an
endothermic process that results in a temperature decrease. The magnitude of a temperature
decrease is related to the strength of intermolecular forces of attraction. Those molecules which
experience stronger attraction to their neighbors will require the most energy to disrupt those
attractions and escape into the gas phase. As you might imagine, it is more difficult for those
molecules to escape; therefore, the rate of evaporation (number of molecules/second) is the
lower, thus resulting in a smaller temperature change.
You will encounter two types of organic compounds in this experiment—alkanes and alcohols.
The two alkanes are pentane, C5H12, and hexane, C6H14. In addition to carbon and hydrogen
atoms, alcohols also contain the -OH functional group. Methanol, CH3OH, and ethanol,
C2H5OH, are two of the alcohols that we will use in this experiment. You will examine the
molecular structure of alkanes and alcohols for the presence and relative strength of two
intermolecular forces—hydrogen bonding and dispersion forces.
Directions: Do the following, as directed, on this document.
1. 1. Complete Table II:
a. a. Molar Mass? You can calculate this on your own from the molecular
formula.
b. b. To help you with this column, read Section 7.3.
Table II: Compound Information
Compound Molecular
Formula
Molar Mass
(g/mol)
Hydrogen Bond
(Yes or No)
Ethanol C2H5OH 46.0 Yes
1-propanol C3H7OH 60.0 Yes
1-butanol C4H9OH 74.1 Yes
n-pentane C5H12 72.1 No
Methanol CH3OH 32.0 Yes
n-hexane C6H14 86.1 No
a. 2. Complete Table III:
b. a. I conducted this experiment and my temperature data is listed in Table III.
For each compound, calculate T, where T = Tfinal – Tinitial
c. b. Note: All T values should be negative as evaporation is an endothermic
process. In this experiment, the filter paper is soaked with each compound and
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Evaporation and Intermolecular Attractions – Remote Lab Purpose: To study temperature changes caused by the evaporation of several liquids and relate the temperature changes to the strength of intermolecular forces of attraction. In this experiment, temperature probes are placed in various liquids. Evaporation occurs when the probe is removed from the liquid’s container. (See the figure, below.) This evaporation is an endothermic process that results in a temperature decrease. The magnitude of a temperature decrease is related to the strength of intermolecular forces of attraction. Those molecules which experience stronger attraction to their neighbors will require the most energy to disrupt those attractions and escape into the gas phase. As you might imagine, it is more difficult for those molecules to escape; therefore, the rate of evaporation (number of molecules/second) is the lower, thus resulting in a smaller temperature change. You will encounter two types of organic compounds in this experiment—alkanes and alcohols. The two alkanes are pentane, C 5 H 12 , and hexane, C 6 H 14. In addition to carbon and hydrogen atoms, alcohols also contain the -OH functional group. Methanol, CH 3 OH, and ethanol, C 2 H 5 OH, are two of the alcohols that we will use in this experiment. You will examine the molecular structure of alkanes and alcohols for the presence and relative strength of two intermolecular forces—hydrogen bonding and dispersion forces. Directions: Do the following, as directed, on this document.

1. 1. Complete Table II:

a. a. Molar Mass? You can calculate this on your own from the molecular

formula.

b. b. To help you with this column, read Section 7.3.

Table II: Compound Information Compound Molecular Formula Molar Mass (g/mol) Hydrogen Bond (Yes or No) Ethanol C 2 H 5 OH 46.0 Yes 1-propanol C 3 H 7 OH 60.0 Yes 1-butanol C 4 H 9 OH 74.1 Yes n-pentane C 5 H 12 72.1 No Methanol CH 3 OH 32.0 Yes n-hexane C 6 H 14 86.1 No

a. 2. Complete Table III:

b. a. I conducted this experiment and my temperature data is listed in Table III.

For each compound, calculate T, where T = Tfinal – Tinitial

c. b. Note: All T values should be negative as evaporation is an endothermic

process. In this experiment, the filter paper is soaked with each compound and

then wrapped around the temperature probe. Most of the time, when I removed the filter paper from the probe, it was cold to the touch. The pentane filter paper actually had ice crystals on it: the water vapor in the air condensed on the -7C paper and crystallized. Table III: Evaporation Data Compound Tf (C) Ti (C) T (Tf – Ti) (C) Ethanol 13.0 24.7 -11. 1-propanol 17.8 24.5 -6. 1-butanol (^) 22.6 23.6 - n-pentane -7.3 23.2 -30. methanol 6.5 24.2 -17. n-hexane 5.6 24.1 -18.

a. 3. Answer the following questions on this document with

complete sentences. Do not be vague. Be specific and detailed! (If a question asks for data, be sure to give the actual value, don’t just say high or low.) Please highlight or boldface your answers.

b. a. What is the relationship between surface area and LDF (London

Dispersion Forces)? Explain why. The surface area of non-polar particles has a big effect on the strength of the LDF between them. The larger the surface area, the stronger the LDF between them will be.

c. b. Two of the liquids, n-pentane and 1-butanol, had nearly the same molar

masses, but significantly different T values. Explain the difference in T values of these substances, based on their intermolecular forces. Be sure to compare and contrast: what is the same? Different? The difference is that butanol has a smaller T value because of the presence of hydrogen bonding between the molecules. Butanol has a stronger intermolecular force than pentane because there is no hydrogen bonding in pentane, only London dispersion forces. Out of the four alcohols, 1-butanol has the weakest intermolecular forces. Also, out of the 2 alkanes n-pentane has the weakest intermolecular forces.

d. c. Using the T data, list the alcohols in order of increasing

intermolecular forces of attraction. (Be sure to give the T values with each compound.) Explain the order of the list by comparing and contrasting the intermolecular forces for the alcohols: what is the same? Different? -1-butanol (-1), propanol (-6.7), ethanol (-11.7), methanol (-17.7) Methanol, ethanol, 1- propanol, and 1-butanol, all have hydrogen bonding. The smaller the molecule the greater the hydrogen bonding. Methanol is the smallest, it will have the strongest intermolecular forces. 1-butanol will have the smallest intermolecular forces.

e. d. Which of the alkanes studied has the stronger intermolecular forces of

attraction? The weaker? Explain. Compare/contrast intramolecular forces and use the T data. n-hexane is the largest and n-pentane is the weakest. N-hexane has a T value of -18.5, while n-pentane has a T value of -30.5. In non-polar molecules, the larger molecules have the greater intermolecular forces because