Exp IMF Report

Results: Figure 1: Shows the temperature change during evaporation and also predicting temperature changes. Table 1: Temperature change during evaporation Substance Tmax (C) Tmin (C) Ethanol 23.87 14.53 14. 53 1-Propanol 23.75 19.52

Table 2: Comparing & predicting temperature changes during evaporation Substance Explanation Predicted Tmax (C) Tmin (C) T(C) < 4.23 1- Butanol It has a higher 23.31 22.66 molecular weight. than 1-propanol > 9.34 Pentane It has a higher 14.71 -0.84 molecular weight. than Ethanol > 9.34 Methanol It has a lower 24.03 8.63 molecular weight. than Ethanol < 15.55 Hexane It has a higher 23.80 8.60 molecular weight. than Pentane > 15.55  Acetone It has a lower lower 24.41 5.05 molecular weight. than Hexane < 9.34 Water It has a lower 23.75 20.18 molecular weight. than Ethanol

T(C)

9.34 4.23

T(C)

(Tmax - Tmin) 0.65

15.55

15.40

15.20

19.36

3.57

pg. 1

Figure 2: The graphs below shows the comparison between the temperature change during the experiment for Ethanol and Propanol. Ethanol has the greatest change in temperature. Ethanol Vs. Propanol 30

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ethanol

propanol

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Figure 3: The graphs below shows the comparison between the temperature change during the experiment for Butanol and Pentane. Pentane has the greatest change in temperature. Butanol Vs. Pentane 25

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-5 butanol

pentane

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Figure 4: The graphs below shows the comparison between the temperature change during the experiment for Methanol and Hexane. Methanol has the greatest change in temperature. Methanol Vs. Hexane 30 25 20 15 10 5 0

0

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hexane

Figure 5: The graphs below shows the comparison between the temperature change during the experiment for Acetone and Water. Acetone has the greatest change in temperature. Acetone Vs. Water 30

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water

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Alcohols versus their Molecular Weights 80

   )    l   o   m    /   g    (    t    h   g    i   e    W   r   a    l   u   c   e    l   o    M

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T(C)

Figure 6: The graph above shows that for alcohol, as the molecular weight decreases, the alcohols T(C) increases. Calculations: T(C)=

Tmax – Tmin =

Ethanol: 23.87 C  – 14.53C = 9.34C 1-Propannol: 23.75C – 19.52C = 4.23C 1-Butanol: 23.31C – 22.66C = 0.65C Pentane: 14.71C – (-0.84C) = 15.55C Methanol: 24.03C – 8.63C = 15.40C Hexane: 23.80C – 8.60C = 15.20C  Acetone: 24.41C – 5.05C = 19.36C Water: 23.75C – 20.18C = 3.57C

pg. 4

Conclusion: In conclusion the purpose of this experiment was to investigate the relationship of dispersion forces and hydrogen bonding forces in intermolecular attractions. Two types of organic compounds that were used in this experiment are Alkanes and Alcohols. As shown in the figures above each alkanes and alcohol will vary in temperature compared to each other. For example, Propanol has maintained a higher temperature compared to Ethanol during the experiment; however, Ethanol has shown the greatest change in temperature than Propanol. This information was gained when the experimenter calculated the

T(C)

by subtracting Tmax by Tmin. The

experimenter would like to add some improvements that would better help other experimenters improve their data / results for this experiment; such as clearly stating the size need for the filter paper to be properly placed on the probes. Another improvement for this experiment is that the experimenter should be timing the probes inside the liquid with a timer so that the experiment will not be affected by any incorrect timing. A third improvement that the experimenter would like to state is that the experimenter should be shown how to use the program LoggerPro before starting the experiment, so that no complications with the program will happen during the experiment.

Post Lab Questions: 1) Alkanes: Pentane < Hextane  Alcohols: 1-Butanol < 1-Propanol < Ethanol < Methanol 2) Large T values mean that the sample does evaporate readily. Large T values mean that the sample does not evaporate readily.

pg. 5

3) Looking at my formulas for the alkanes, the only difference between them would be the number of molecules of carbon and hydrogen. 4) The IMF that are present in the alkanes are London force. 5) Looking at my formulas for the alcohols, the only difference between them would be the number of molecules of carbon and hydrogen. 6) The IMF’s that are present in the alcohols are Hydrogen bonds and London force.

Work Cited: 1. Beran, J.A. (2014). Page 85-86 Laboratory Manual for Principles of General chemistry 2. (2010) Page 1-7 Intermolecular Forces: Evaporation & Intermolecular attractions Handout

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