Title: Intermolecular Forces and the Evaporation of Liquids
Experiment Number: 9
Date: 25th October, 2018
Name- Aryan Shetty*
Section Number- 580
Lab Partners- Thomas Rod, Andrew
Summary:
The objective of the experiment was to identify the relationship between the size and structure of a chemical molecule and the corresponding strength of its molecules. This was done by observing the evaporation and the vapor pressure of numerous organic liquids including alcohols and alkanes. The evaporation time and the vapor pressures were identified at multiple different temperatures to ensure that the relationships inferred withstand different temperatures and pressures.
The scope of the experiment was restricted to dispersion/London forces, dipole-dipole forces, and hydrogen bonding. The different intermolecular forces have different applications. For example, dipole-dipole forces dictate the nature of bonding between metal atoms and carbonyl groups and are responsible for the enhanced boiling point of water. Hydrogen bonding accounts for the cleaning action of soaps and detergents as well as the structural integrity of carbohydrates and DNA molecules amongst numerous others. Lastly dispersion forces are used in dispersive adhesion. Hence the significance of the experiment is that by identifying the relationship between intermolecular forces and their constituent factors (shape,size etc.) can their usage in daily life be fully maximized with regard to aspects such as solubility, evaporation etc.
The procedural part of the experiment was carried out in two parts. In the first part of the experiment the temperature change of pairs of organic solvents – ethanol/propanol, butanol/pentane, methanol/hexane- and the corresponding time period , was recorded on logger pro and the graph was obtained. This was done by connecting two vernier probes (with their ends wrapped in filter paper) into the organic solvents. The evaporating liquids were placed under the exhaust hoods in order to prevent the vapors from accumulating in the lab. Hence in the first part the evaporation times of the organic solvents was noted down. In the second part of the experiment, the vapor pressures of ethanol and methanol were noted down at four different temperature ranges. This was done using temperature probes and pressure sensors. The alcohols were heated by immersing them into water baths (heated beforehand to reach the appropriate temperature). As methanol and ethanol are flammable, care was taken to ensure that they were not directly heated. This was done by using placing the alcohols in an Erlenmeyer flask with a stopper in the neck of the flask, along with a syringe.
The general result of the experiment is that in part A it demonstrates the linear direct proportionality between molecular mass and boiling point, verified by the heavier solvents taking longer to evaporate. This is because as the molecular size increases the intermolecular force increases as well- resulting in a higher boiling point. In part B of the experiment it showcases that pressure and volume are directly proportional as ratified by Gay Lussac’s Law.
Results:
Temperature change is inversely proportional to the evaporation time
Figure 1- Evaporative Change in Temperature vs Time: Ethanol(RED) vs Propanol(GREEN):
This graph displays the temperature change and its time period of ethanol and propanol. Temperature Change (Propanol)=5.6°C, Temperature Change(Ethanol)=9.7°C. From this graph it can be inferred that ethanol evaporates faster than propanol.
Figure 2- Evaporative Change in Temperature vs Time: Butanol(RED) vs Pentane(GREEN):
This graph displays the temperature change and its time period of butanol and pentane. Temperature Change (Butanol)=4.6°C, Temperature Change(Pentane)=16.8°C. It can be inferred that pentane evaporates faster than butanol.
Figure 3: Evaporative Change in Temperature vs Time: Methanol(RED) vs Hexane(GREEN):
This graph displays the temperature change and its time period of hexane and methanol. Temperature Change (Hexane)=17.2°C, Temperature Change(Methanol)=17.9°C. It can be inferred that hexane and methanol have similar evaporation times.
Table 1: Methanol- Temperature vs Vapor Pressure:
Latest: Event No.
Temperature (K)
Vapor Pressure (kPa)
1
296.1049824
13.15
2
302.105493
17.94
3
282.5440058
6.261
4
288.6214939
8.97
From this graph the linear relationship between temperature and vapor pressure in the case of methanol is observed. Hence it is inferred that they are directly proportional. The trendline equations and R^2 value for both organic solvents are displayed as well.
Table 2: Ethanol- Temperature vs Vapor Pressure:
Latest: Event No.
Temperature (K)
Vapor Pressure (kPa)
1
294.9304823
4.211
2
297.7472562
5.689
3
290.3782021
3.296
4
286.5738384
2.38
From this graph the linear relationship between temperature and vapor pressure in the case of ethanol is observed. Hence it is inferred that they are directly proportional. The trendline equations and R^2 value for both organic solvents are displayed as well.
Table 3: Change in Temperature vs Molecular Weight – Alcohols:
Serial Number (S.No.)
Molecular Weight (MW)(g/mol)
Temperature Change (°C)
1
32.04
17.9
2
46.08
9.7
3
60.11
5.6
4
74.14
4.6
Calculations:
General Equations:
Celsius to Kelvin:
K=273.15+°C
Eg: Tc=5.67809 °C
Tk=Tc+273.15= 278.82809
2) Calculating air pressure:
P1/T1=P2/T2, where P1=atmospheric pressure, T1= initial room temperature, T2=temperature at which pressure is to be found, P2=partial pressure at T2
Vapour Pressure = Measured Pressure-Air Pressure
Room Temperature= 22.7 °C
Room Pressure= 100.973 kPa
Methanol Temperature= 22.9549824 °C
Ethanol Temperature= 21.78048226 °C
Table 4: Temperature Change of Organic Liquids:
Liquid
MW (g/mol)
Tmin (°C)
Tmax (°C)
Delta T (°C)
Methanol
32.04
5.443
22.63
17.2
Ethanol
46.08
12.29
21.99
9.7
1-Propanol
60.11
15.97
21.60
5.6
1-Butanol
74.14
18.23
22.79
4.6
n-Pentane
71.15
4.07
20.86
16.8
n-Hexane
86.20
5.75
23.67
17.9
Table 4: Methanol – Measured/Calculated Pressures & Temperature:
Temperature
(°C)
Temperature
(K)
Measured Pressure (kPa)
Air Pressure (kPa)
Vapor Pressure (kPa)
22.954982
296.11
114.12515
100.98
13.15
28.955492
305.11
120.9540
103.01
17.94
9.394
282.54
102.6086
96.34
6.26
15.47149
288.62
107.38954
98.42
8.97
Table 5: Ethanol – Measured/Calculated Pressures & Temperature:
Temperature(°C)
Temperature(K)
Measured Pressure (kPa)
Air Pressure (kPa)
Vapor Pressure (kPa)
21.78048
294.93
104.7820
100.57
4.21
24.5972
297.75
107.2209
101.53
5.69
17.22820
290.38
102.3145
99.01
3.30
13.42383
286.57
100.1016
97.72
2.38
Discussion:
It is observed that n-pentane and 1-butanol have similar molecular weights but different change in temperatures – 1-butanol(4.6 °C) , n-pentane(16.8 °C). Butanol has a much lower change in temperature due to its stronger intermolecular forces relative to pentane. Because butanol posses an hydrogen/OH group it has greater hydrogen bonding between its molecules. Butanol also possesses dispersion forces of attraction between its molecules. Pentane has ion induced dispersion/Van der Waals forces between its molecules which results in minimal distance between molecules. Although it had been observed that intermolecular force of attraction is directly proportional to the mass of the molecule, the strength of the bonds has a greater impact and causes the anomaly in the temperature change. Hydrogen bonding is stronger than dispersion forces – hence, 1-butanol has a lower change in temperature because it has much stronger intermolecular forces of attraction as compared to pentane.
It has been inferred over the course of the experiment that the strength of the intermolecular forces is inversely proportional to the temperature change as measured in part A. Hence by this hypothesis it is identified that the alcohol with the strongest intermolecular forces is butanol whereas methanol is the alcohol with the weakest intermolecular forces.
Both alkanes possess weak London dispersion forces. As previously mentioned it can be identified that n-pentane has weaker intermolecular forces due to its increased temperature change. Hence n-hexane has stronger intermolecular forces due to its reduced temperature change. Furthermore, intermolecular dispersion forces increase with molecular weight- this is another contributing factor as to how n-hexane has greater intermolecular forces.
The graph of temperature change vs molecular weight is presented above in the results section of the assignment. From the graph it is inferred that an increase in molecular weight corresponds to a decreasing temperature change and thereby an increasing evaporation time/increasing intermolecular forces. This is because a greater molecular mass results in an increase in surface area of the molecule. An increase in surface area is accompanied by an increase in evaporation temp and a decrease in the temperature change. The inverse relationship between temperature change and molecular weight can be applied to both alcohols and alkanes. Alcohols (as previously mentioned) possess decreased temperature change, increased evaporation times, greater intermolecular forces of attraction(dispersion forces and hydrogen bonding). Alkanes possess increased temperature changes, decreased evaporation time, weaker intermolecular forces of attraction (only weak dispersion forces).
In part B the linear relationship between vapor pressure and temperature is observed. This is demonstrated by the vapor pressure vs temperature graphs for ethanol and methanol presented in the results section. This can be explained by the kinetic theory of gases as follows: with an increase in temperature the molecules of the vapor possess greater thermal energy and thereby greater kinetic energy and thus move with a greater velocity. Furthermore the number of intermolecular collisions increases. Vapour pressure is the result of molecular collisions with the container, and since the collisions increase with temperature- the vapour pressure increases with temperature.
From the vapour pressure vs temperature graphs of ethanol and methanol it is observed that methanol possesses greater vapour pressure than ethanol at room temperature. It has been previously established that vapour pressure is inversely proportional to intermolecular forces. As ethanol possesses greater molecular weight it has increased intermolecular forces of attraction. Therefore it has a lower vapour pressure as fewer molecules go into the vapor state (due to increased intermolecular forces of attraction) and the vapor pressure exerted is relatively less.
Errors that arise may be systematic, random, or blunders. Systematic errors such as improper calibration of pressure sensors and temperature probes may lead to incorrect readings. Observational errors such as parallax while noting down readings may occur. An example of a random error would be a leak in the stopper used in the flask while measuring vapour pressure. Furthermore the flask should be completely dry and should have no water droplets inside. An increase in the liquid inside the flask would mean that there would be a greater number of molecules that can transcend into the vapour state, thereby resulting in greater vapor pressure. Hence a dry flask is necessary to prevent mismeasurement of ethanol/methanol vapour pressures.
In conclusion by measuring the vapor pressures and the temperature changes of organic molecules, multiple relationships have been inferred as mentioned. Greater molecule mass results in greater intermolecular forces, which subsequently results in decreased vapour pressure. Hence it was established that molecular weight and temperature change are inversely proportional.