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Exp. 3 Lab Report

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Experiment #3 Intermolecular Forces: Solutions 1

Experiment #3 Lab Report

Daniela Garcia

Palm Beach State College

CHM1046L
Experiment #3 Intermolecular Forces: Solutions 2

Pre-Laboratory Questions
1. Using your knowledge of molecular structure, identify the main intermolecular force

in the following compounds. You may find it useful to draw Lewis structures

to help with the answers.

Compound IMF Compound IMF


PF3 Dipole-Dipole HF Hydrogen Bonding
CH2O Dipole-Dipole, KBr Dipole -dipole
(formaldehyde) Dispersion dispersion forces

2. Explain how dipole forces within molecules would cause them to be attracted to one

another.

Dipole forces are attracted to one another because they have opposite charges which

means partially positive and negative side. They are weaker than ionic bonds so, the have a

significant effect when they are close together.

3. Rank the following compounds from lowest (1) to highest (4) boiling point. Explain

why!

Calcium carbonate, methane (CH4), methyl alcohol (CH3OH), dimethyl ether (CH3OCH3)

1 methane (CH4),
2 dimethyl ether (CH3OCH3)
3 methyl alcohol (CH3OH),
4 calcium carbonate
The compounds with lower boiling points have weaker intermolecular forces, while the

compounds with higher boiling points have stronger intermolecular forces.

4. Explain why the nonpolar molecules usually have much lower surface tension than

polar molecules.
Experiment #3 Intermolecular Forces: Solutions 3

The surface tension is needed to stretch the surface of a liquid against the intramolecular forces.

Non-polar molecules have much lower surface tension than polar molecules because they have

lower intramolecular attraction.


Experiment #3 Intermolecular Forces: Solutions 4

Report Sheet

Test Tube # Chemical 1 Chemical 2 Results (S, P, I)


1 Acetone 1 Butanol S
2 Acetone CCl4 S
3 Acetone Ethanol S
4 Acetone Hexane S
5 Acetone Water S
6 Butanol CCl4 S
7 Butanol Ethanol S
8 Butanol Hexane S
9 Butanol Water P
10 CCl4 Ethanol S
11 CCl4 Hexane S
12 CCl4 Water I
13 Ethanol Hexane P
14 Ethanol Water S
15 Hexane Water I

Mass of KNO3 was weighed at 10.001 g

Trial # Total Mass of H2O Initial Concentration of


Crystallization KNO3 (g KNO3/ 100
Temperature (C) g H2O)
1 5g 85 (
10.001 𝑔 𝐾𝑁𝑂3 100
)( )=
5 𝑔 𝐻2 𝑂 5
40.004
2 10 g 49 10.001 𝑔 𝐾𝑁𝑂3 100
)( 10 ) =
10 𝑔 𝐻2 𝑂
10.001
3 15 g 35 10.001 𝑔 𝐾𝑁𝑂3 100
)( 15 ) =
15 𝑔 𝐻2 𝑂
4.445
4 20 g 25 10.001 𝑔 𝐾𝑁𝑂3 100
)( 20 ) =
20 𝑔 𝐻2 𝑂
2.500
Experiment #3 Intermolecular Forces: Solutions 5

Graph of Data
Experiment #3 Intermolecular Forces: Solutions 6

Post-Laboratory Questions

1. Explain your results in terms of IMFs for the insoluble liquids.

Water and hexane are insoluble liquids because the dispersion forces that affect solubility is

when non polar molecules are soluble with a non polar solvent. In this case, water is a polar

liquid, but hexane is not so that is why they cannot dessolved .

2. Explain your results in terms of IMFs for the soluble liquids.

Soluble liquids are miscible, so they can dissolve. Acetone is attracted by dipole- dipole

interactions. Acetone can accept hydrogen bonds so, water has hydrogen bonds. Tha is the

reason why acetone and water are soluble.

3. Explain your results in terms of IMFs for the slightly soluble liquids.

The slightly soluble are materials that have low solubility. Butanol and water have intermolecular

forces because they can partially dissolve but not entirely.

4. Using the graph of your data, extrapolate the mass amount of KNO3 that would be soluble

in 100 g H2O at 24 C.

Because the mass amount of KNO3 at 25 C is 2.500 g, the mass amount of KNO3 at 24 C

25 24
is approximately 2.400 g. ( 2.5 = , x = 2.400)
𝑥

5. Using the graph of your data, extrapolate the mass amount of KNO3 that would be soluble

in 100 g H2O at 100 C.

Because the mass amount of KNO3 at 85 C is 40.004 g, the mass amount of KNO3 at

85 100
100C is approximately 47.064 g. ( 40.004 = , x = 47.064)
𝑥

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