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Laboratory: Evaporation and Intermolecular Attractions

January 25, 2010

Quim 3134
Section: 131

I. Introduction

For this laboratory called Evaporation and Intermolecular Attractions it is important to know the meanings of various concepts. Intermolecular forces are the attractive and repulsive forces among the molecules, atoms or ions in a sample of matter. There aredifferent types of intermolecular forces. Among these we have ion-dipole which occurs when an ion and a nearby polar molecule attracts each other. There is also the dipole-dipole force which forms when the positive pole of one molecule attracts the negative pole of another. In another situation the partially positive H of one molecule is attracted to the partially negative lone pair on the N, O orF of another molecule and the hydrogen bond forms. The London or dispersion force is responsible for the condensed states of nonpolar substances. We can distinguish all these types of forces in nature, like the production of chemicals, solutions, food... For example hydrogen bonds are present in the structure of biopolymers such as protein and nucleic acids. Also the actions of many enzymes(protein that speed all metabolic reactions) that are adjudicated to the hydrogen bonds. In this experiment we will determine the volatility or evaporation of several organic compounds (alkanes and alcohols). These are: methanol, ethanol, n-propane, n-hexane, 1-propanol and 1-butanol. The main purpose of this experiment is to determine the temperature changes caused by the evaporation of the liquidsalready mentioned and relate the temperature changes to the strength of intermolecular forces. An example of these forces that can be seen in our daily life is the insects that can walk on water. This is possible because the network of hydrogen bonds are strong enough to support the insect. A technique that will be used consists of using temperature probes( that will be soaked in the liquid andexposed to the air) to detect the temperature change.

II. Objectives

Determine the temperature change for a liquid
Test the molecular structure of alcohols and alkanes for the relative strength and presence of two intermolecular forces
Observe which liquid is more volatile
Determine which alcohol and alkane has a higher and lower evaporation rate

III. Equations

4pIV. Data and results

A. Table of temperature change
Substance t1(˚C) t2(˚C) (˚C)
ethanol 20.3˚C 14.850˚C 5.45˚C
1-propanol 21.16˚C 16.810˚C 4.35˚C
1-butanol 20.1˚C 18.500˚C 1.6˚C
n-pentane 22.53˚C
20.186˚C 2.344˚C
methanol 21.34˚C 10.9˚C 10.44˚C
n-hexane 22.4˚C 20.372˚C 2.028˚C

B. Graph for temperature change for several liquids

Light blue- ethanol
Grey- methanol
Dark blue- n-hexane

C. Questions
1. The liquids, n-pentane and 1-butanol had nearly the same molecular weights but different values because 1-butanol has more dispersion force. It is also polar in comparison to n-pentane that is nonpolar. Another reason for this difference is that 1-butanol can form hydrogen bonds between itsmolecules which results in a stronger attraction and a slower evaporation rate.

2. The alcohol that had the strongest intermolecular force of attraction between its molecules was the 1-butanol because it has the largest molecules and strongest dispersion force. This results in a slower evaporation rate. In contrary, methanol had the weakest intermolecular force of attraction given that it has aweak dispersion force and a faster evaporation rate.

3. The n-hexane had a stronger force of attraction between its molecules given that it has larger molecules and a stronger dispersion force which results in a lower evaporation rate. N-pentane had a weaker attraction given that it has smaller molecules and a weaker dispersion force which results in a higher evaporation rate.

4. Graph...
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