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Involves intermoleforces, dipole- dipole, ldf, and hydrogen bonds
Typology: Lab Reports
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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.
formula.
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
For each compound, calculate T, where T = Tfinal – Tinitial
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.
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.
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.
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.
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.
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