Tuesday, March 29, 2011

Vapor Pressure and Enthalpy of Vaporization of Water


Introduction
The purpose of this lab was to determine the enthalpy of vaporization of water. The vapor pressor was determined at increments of five degree ranging from 80º C - 50º C. The vapor pressure of a pure substance is dependent on the enthalpy of vaporization, the gas constant, the temperature, and a constant according to the Clausius-Clapeyron equation (eq 1).
P=(-∆Hvap/RT) + C (1)
In Equation 1, ∆Hvap  is the enthalpy of vaporization, R is the gas constant, T is the temperature and C is a constant.
The Clausius-Clapeyron equation was put into linear form (eq 2).
lnP=lnA -∆Hvap/RT (2)
The Ideal Gas Law (eq 3) was used to calculate the partial pressures of air for each temperature.
Pair = Nair RT/V (3)
Dalton’s Law of Partial Pressures (eq 4) was used to calculate the vapor pressor of water at different temperatures.
Pwater = Patm - Pair (4)
Procedure
A 10 mL graduated cylinder was filled with 9.4 mL water and inverted into a beaker full of water. The beaker was heated to 80º C and as it was cooled the volume of the air trapped in the graduated cylinder was measured every 5º C. It was then rapidly cooled to 5 ºC and the air volume was measured. For each temperature reading the partial pressure of air was calculated using the Ideal Gas Law. The vapor pressure was calculated using Daltons Law of Partial Pressures. Plots were created and the enthalpy of vaporization and the pressure of water at 65º C were calculated using the line of best fit from the partial pressure of water versus 1/temperature plot (fig. 2) and the Clausius-Clapeyron equation (eq 1).
Detailed procedures can be found in reference 1.
Results 
Table 1 gives the raw data collected with 0.2 mL subtracted from the volumes to take into account the space taken up by the graduated cylinder.
Table 1. Volume and corresponding temperature data

Volume (L)
Temperature (K)
0.0076
353.15
0.0068
348.15
0.0062
343.15
0.0057
338.15
0.0051
333.15
0.0052
328.15
0.0048
323.15
0.0036
278.15
The Ideal Gas Law was used to calculate the partial pressure of air at each temperature and volume shown in Table 1. 
Figure 1 shows the vapor pressure of water and the temperature of water.
pastedGraphic.pdf
Figure 1. The vapor pressure versus the temperature of the water.
The natural log of the vapor pressure versus the inverse temperature of water was used to linearize and plot the data according to the Clausius-Clapeyron Equation (eq 2), this can be found in Figure 2.
pastedGraphic_1.pdf
Figure 2. Plot of linearized data according to Equation 2
The best-fit line and Equation 2 were used to calculate the heat of vaporization to be 37.45 kJ/mol, and the pressure of water at 65ºC to be 183.2 torr. 
Discussion
The heat of vaporization of water was calculated to be 37.45 kJ/mol. When compared to the value of 44.02 kJ/mol (ref 1) it has 15 % relative error. The pressure of water at 65ºC was calculated to be 183.2 torr, which has a 2% relative error when compared to the value of 187.5 torr (ref 2). The error of the heat of vaporization is likely caused by inaccurate volume reading due to foggy glass and rotation of the graduated cylinder.
References
1. General Chemistry Experiments: A Manual for Chemistry 204, 205, and 206, Department of Chemistry, Southern Oregon University: Ashland, OR, 2010
2. Brown, LeMay, Burnsten, Murphy. Chemistry: The Central Science. 11th Edition. Pearson Prentice Hall. Upper Saddle River, NJ. 2009.

Thursday, March 3, 2011

Acid Analysis by Titration Determination of Molecular Weight

Table 1. Data from standardization of NaOH titrant with KHP.

Trial
Mass KHP (g)
Initial Burette Volume (mL)
Finial Burette Volume (mL)
NaOH Concentration (M)
1
0.1133
0.23
20.15
0.02785
2
0.1148
20.15
40.21
0.02802
3
0.1118
20.97
40.38
0.02820
4
0.1104
1.48
20.97
0.02774
The average NaOH concentration was calculated to be 0.02795 M with a standard deviation of 0.00020.
Table 2. Data from molar mass determination of unknown number 519A.

Trial
Mass KHP (g)
Initial Burette Volume (mL)
Finial Burette Volume (mL)
Molar Mass (g/L)
1
0.2175
0.48
18.70
426.7
2
0.2037
18.70
35.31
438.7
3
0.1982
0.28
16.63
433.7
The average molar mass the unknown was determined to be 433.0 g/L with a standard deviation of 6.1.

Tuesday, March 1, 2011

The Ideal Gas Law


Introduction
The purpose of this lab was to investigate relationships between the properties of gas, including pressure, volume, moles, and temperature, and then use the Ideal Gas Law (eq 1) to explain observations made. The lab consisted of a pressure-volume experiment and a pressure-temperature experiment.
Equation 1 is the The Ideal Gas Law, where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature. 
PV=nRT (1)
In equation 2, Boyle’s Law, pressure multiplied by volume equal a constant.
PV=Constant (2)
In equation 3, Charles Law, pressure divided by temperature equal a constant.
P/T=Constant (3)
Procedure
Detailed procedures may be found in reference 1.
Results 
Figure 1 shows the pressure compared with the volume from part one. 
pastedGraphic.pdf
Figure 1. The pressure of air measured by controlling the volume.
The average volume was 0.014 L ± 0.004 and the average pressure was 1.4 atm ± 0.4. The curve of Figure 1 means the pressure was inversely related to volume. 
Figure 2 shows the pressure compared with 1/volume of the air from the same data. 
pastedGraphic_1.pdf
Figure 2. The pressure of air versus the inverse volume.
The data in Figure 1 was used to solve Equation 1 for the number of moles of air in the syringe, which was calculated to be 7.71 x 10-4 moles. The pressure was multiplied by the volume for each data point according to Boyle’s Law (eq 2). The results from each data pair were constant, confirming Boyle’s Law.  The value of R, calculated using equation 1 and the slope of the trendline in Figure 2, was 0.0757 L atm/mol K.
Figures 3 and 4 show the relationship of temperature versus pressure of the data obtained from the pressure-temperature experiment. Figure 3 is atmospheres versus degrees Celsius, Figure 4 is atmospheres versus degrees Kevin. pastedGraphic_2.pdf
Figure 3. The relationship of pressure (Atmospheres) versus temperature (Celsius) of air.
pastedGraphic_3.pdf
Figure 4. The relationship of pressure (Atmospheres) versus temperature (Kelvin) of air.
It is known that pressure and temperature are normally directly related 1 (eq 3), which was tested using the data from Figure 4. Pressure was divided by temperature for each data point to determine if they equal one constant value. The calculated values remained relatively constant, they got smaller as the temperature and pressure increased, which agrees with equation 3.
By setting the pressure (y value) equal to zero and solving for the volume (x value) in the trendline of figure 4, the pressure was calculated to be 12.1 atm. Then trendline was solved for the pressure when the temperature equaled 200 and 400 degrees K. It was found that when the temperature was doubled that the pressure also doubled. 
Discussion
The Ideal Gas Law was tested by observing a pressure-volume relationship and a temperature-pressure relationship. The R value was calculated with the data in Figure 1 to be 0.0757 L atm/mol K. This value is smaller than the accepted R value of 0.08206 L atm/mol K.1 Error in this value may be due to composition of the air in the room that the lab was done deviated from where the accepted value was determined. 
Temperature would double if pressure was doubled. When temperature increases the amount of kinetic energy in molecules, in the form of heat, increases with with pressure. With the increase of kinetic energy the molecules increase in velocity and collide more frequently.
References
1. General Chemistry Experiments: A Manual for Chemistry 204, 205, and 206,Department of Chemistry, Southern Oregon University: Ashland, OR, 2010

Tuesday, February 15, 2011

Molar Mass Determination by Freezing Point Depression


Introduction
The purpose of this lab was to use freezing point depression techniques to determine the molar mass of camphor. The pure solvent cyclohexane was cooled to determine its freezing point (Tf (solvent)). Then a solution of cyclohexane and camphor was cooled until frozen and its freezing point was determined (Tf (solution)).
Equation 1 was used to calculate the freezing point depression (∆Tf).
∆Tf = Tf(solvent) – Tf(solution) (1)
The freezing point depression is equal to the cryoscopic constant (Kf) multiplied by the 
molality (m) of the solute particles (eq 2). 
∆T= Kf m = Kf (mol solute/kg solvent) (2)
Equation 3 was then used to determine the molar mass (mm) of the solvent.
MM= (Kf)(g solute)/(kg solvent)(∆Tf) (3)
Procedure
A test tube was sealed with a rubber stopper and placed in a beaker. It was weighed on a top loading balance and then 0.26 g cyclohexane was added. A salt-ice bath was created and then a temperature probe was placed into the test tube. The data collection was started on Logger Pro and the cyclohexane was stirred with the probe while the test tube was stirred in the salt-ice bath simultaneously. Once the cyclohexane was frozen the data collection was stopped. In Logger Pro lines of linear fit were put onto the cooling and frozen sections of the data and they were interpolated to find the freezing point. Two camphor/cyclohexane solutions were created to do trials one and two, and the same procedure was done with each as the cyclohexane.
With equation 1, the results of the graphical analysis were used to calculate the freezing point depression. The molar mass of camphor was calculated for the two trials using equation 2  and the average molar mass was determined.
Detailed procedures may be found in reference 1.
Results 
The data collected to determine the freezing points of the solvent and solution were put into graphs which can be found in Figures 1, 2, and 3.
pastedGraphic.pdfFigure 1. Cooling curve of cyclohexane solvent.
pastedGraphic_1.pdfFigure 2. Cooling curve of the solution in trial one.
pastedGraphic_2.pdfFigure 3. Cooling curve of the solution in trial two.
Equations 1, 3 and the data collected from trials (Table 1) were used to calculate the molar mass for the two trials.
Table 1. Data from cyclohexane/camphor solution trials.

Trial
Mass Cyclohexane (g)
Freezing Point of Solvent (°C)
Mass Camphor (g)
Freezing Point of Solution (°C)
1
5.07
7.335
0.26
1.101
2
5.03
7.335
0.26
1.855
The average molar mass of trials one and two was calculated to be 172 g/mol.
Discussion
The mean molar mass of camphor was calculated to be 172 g/mol, which is higher than the known value of 152.17 g/mol.1 The most likely cause for this lack of accuracy would be from the freezing point of the solvent, which was used to determine ∆Tin the calculations of both trials. A ∆Tf value that is too large would cause the calculated molar mass to also be too large.
The calculated freezing point of cyclohexane 7.0585 °C is larger than the known value of 6.5 °C.1 This error is likely due to a calibration difference of the temperature probes.
References
1. General Chemistry Experiments: A Manual for Chemistry 204, 205, and 206, Department of Chemistry, Southern Oregon University: Ashland, OR, 2010

Tuesday, February 8, 2011

Titration Analysis of Weak Acid Solutions: Potassium Hydrogen Phthalate and Citric Acid in Fruit Juice


Introduction
The purpose of this lab was to determine the citric acid concentration in commercial fruit juice. This was achieved by titration of a sample of fruit juice with standardized sodium hydroxide. Sodium hydroxide is hygroscopic and had to be standardized, which was achieved using KHP (hydrogen phthalate, KHC8H4O4)(eq 1).
KHC8H4O4(aq) + NaOH(aq) pastedGraphic.pdf KNaC8H4O4(aq) + H2O(l) (1)
In the second part of this lab the standardized base was used to determine the concentration of citric acid in lemon juice (eq 2). 
H3C6H5O7(aq) + 3NaOH(aq) pastedGraphic_1.pdf Na3C6H5O7(aq) + 3H2O(l) (2)
The neutralization point of the base and acid were found with titration methods.
Procedure
A standard solution of sodium hydroxide was created by dissolving 1.05 g of sodium hydroxide pellets into approximately 500 mL water. Four samples of approximately 0.4 g KHP were weighed and each was put into separate Erlenmeyer flasks. Approximately 50 mL of distilled water and three drops phenolphthalein were added to each sample. The buret filled with the sodium hydroxide solution and it was titrated into each sample until the indicator turned pink. The molarity of each trial was calculated. The average and standard deviation of the values that agreed within 
0.0005 M were calculated.
Distilled water, indicator, and lemon juice were put into Erlenmeyer flasks to create three samples. Each sample was titrated and the amount of sodium hydroxide required to turn the indicator pink was recorded. The molarity of citric acid in the juice sample, the weight/weight percent and the weight/volume percent of citric acid were calculated for each of the trials, along with the averages and standard deviations. A pipet was used to measure 2.00 mL of lemon juice, which was weighed and it’s density calculated.
Detailed procedures can be found in reference 1.
Results 
Table 1 shows the mass of KHP, the initial buret volume, the final buret volumes, and the concentration of sodium hydroxide for each trial done during the standardization of sodium hydroxide.
Table 1. Data collected during standardization of NaOH.

Sample
Mass KHP (g)
Initial Buret Volume (mL)
Final Buret Volume (mL)
[NaOH]
1
0.4178
0.51
44.11
0.04692
2
0.4051
0.41
42.01
0.04768
3
0.4072
0.39
42.38
0.04749
4
0.4097
0.62
42.99
0.04735
The average [NaOH] of samples 2, 3, and 4 was 0.04751 ± 0.00017 M.
Table 2 shows the volumes of juice, the initial buret volumes, and final buret volumes  from the titration of lemon juice with standardized sodium hydroxide.
Table 2.  Data collected during the titration of lemon juice with standardized NaOH.

Sample
Juice Volume (mL)
Initial Buret Volume (mL)
Final Buret Volume (mL)
1
2.00
0.65
32.69
2
2.00
0.97
32.21
3
2.00
0.32
32.28
Table 3 shows the citric acid concentration, the weight/weight percent, and the weight/volume percent of citric acid in the final solution. 
Table 3. Data calculated from titration of lemon juice with standardized NaOH.

Sample
Citric Acid Concentration (mol/L)
Wt/Wt % Citric Acid
Wt/Vol % Citric Acid
1
0.2537
3.891
4.874
2
0.2474
3.794
4.752
3
0.2531
3.881
4.862
Average
0.2514
3.855
4.830
Standard Deviation
0.0034
0.053
0.067
References
1. General Chemistry Experiments: A Manual for Chemistry 204, 205, and 206, Department of Chemistry, Southern Oregon University: Ashland, OR, 20010