Intermolecular Forces

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  • Topic: Oxygen, Macromolecule, Intermolecular forces
  • Pages : 4 (971 words )
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  • Published : April 11, 2013
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Summary:

Experiment 9 is conducted to identify the importance of intermolecular forces and how they affect molecules. Intermolecular forces are forces between molecules that determine whether the molecule is a solid, liquid, or gas under standard conditions. In our lab, we measured the maximum and minimum temperature reached and time it took to reach it of alcohols and alkanes. In addition, we measured the vapor pressure of the liquid at different temperatures. Through Experiment 9, we concluded that molecules with hydrogen bonds or long chains in the structural formula have stronger intermolecular forces, and that as temperature increases the vapor pressure of a substance increases exponentially. Results:

Figure [ 1 ] (Temp1: Ethanol; Temp2: n-Propanol)

Figure [ 2 ] (Temp1: 1-Butanol; Temp2: n-Pentane)

Figure [ 3 ] (Temp1: Methanol; Temp2: n-Hexane)

Chart [ 1 ]

Chart [ 2 ]

Chart [ 3 ]

Table [ 1 ]
Alcohols| Molecular Weight (g)| ∆T (°C)|
Ethanol| 46.068| 9.17|
1-Propanol| 60.09| 6.72|
1-Butanol| 74.12| 3.35|
Methanol| 32.04| 15.615|
| | | |
Calculations:
1. Calculating Molecular Weight
a. Substance: Ethanol
b. Formula: C2H5OH
c. C- 12.0107g; H- 1.0079g; O- 15.9994g
d. 2(12.0107g)+6(1.0079g)+1(15.9994g)= 46.0682g
2. Finding Change in Temperature
e. ∆T(°C)=Maximum T(°C)-Minimum T(°C)
f. Ethanol
i. Max. T: 21.82°C
ii. Min. T: 12.65°C
g. 21.82°C-12.65°C=9.17°C
3. Changing Celsius to Kelvin
h. °C+273.15=K
i. 23.7°C+273.15= 296.85K
4. Finding Corrected Air Pressure (P2)
j. P2T2=P1T1
k. P1=Atmospheric Pressure 99.76kPa; T1=Room Temperature (22.9℃/296.05K) l. P2296.85K=99.76 kPa296.05 K
m. P2=(99.76 kPa*296.85K)/296.05K
n. P2=100.03kPa
5. Finding Vapor Pressure
o. Abs(Measured Pressure-Corrected Air Pressure)=Vapor Pressure p. 101.70kPa-100.03kPa=1.67kPa...
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