paraphrase

Abstract

In this lab the experimental 
molar mass of biphenyl was determined by calculating the molality 
of cyclohexane solvent. The freezing point depression method is used to find the freezing temperature of the pure solvent. The final molar mass of biphenyl is 148.6 and the percent error was calculated to be 3.6%.

Introduction

The purpose of this lab was to use the freezing point depression method to determine the molar mass of biphenyl. Dissolving a solute in a solvent can cause the freezing temperature to decrease according to the number of moles in the solute. Therefore, this property depends on the ratio of solute to solvent in a mixture. It is called the freezing-point depression, which is a colligative property. The following equation describes the property.

is the change of temperature in the freezing point depression of the solution. The temperature constant is for t- cyclohexane which is the solvent that used in this experiment. The molality is the ratio of the number of moles in the solute over the solvent in kg. The freezing temperature of pure cyclohexane was determined by placing 3mL of the substance in a test-tube that Then it was placed in 250 mL beaker that filled with ice and water. A temperature probe was in the test-tube to monitor the temperature change, which is then plotted. In order to calculate the molar mass of biphenyl, approximately 0.1g were measured and added to the cyclohexane. Likewise, the solution was then placed in the ice and water and data was collected for about 10 minutes. This process gave the freezing temperature of the solution. This was indicated in the graph due the gradual linear decrease of the temperature. Using the freezing temperature of pure cyclohexane and the solution the change of temperature can be calculated. Therefore, the molality of cyclohexane can be calculated with the equation below:

The weight of the solvent can be calculated by subtracting the mass of the solution, the mass of the test-tube and the mass of the biphenyl. By manipulating the equation above the number of moles of biphenyl can be determined. Finally to find the molar mass of biphenyl the following equation was used.

The relationship above gave the experimental molar mass of biphenyl. The value calculated was then compared to the actual value, which is 154.2

Data

Table 1 Mass of test tube and solvent before experiment

Table 2 Data collected from the experiment

Figure 1 Graph of Trial 1 for the pure cyclohexane to find the freezing point

Figure 2 Graph of Trial 2 for the pure cyclohexane to find the freezing point

Figure 3 Graph of Trial 1 of Solution of biphenyl and cyclohexane to find the freezing point.

Figure 4 Graph of Trial 2 of Solution of biphenyl and cyclohexane to find the freezing point.

Results

Table 3 Calculated results from the experiment

Discussion

The molar mass of biphenyl was determined to be 148.6(table 3). The average freezing point of pure cyclohexane was found 7.5 C shown on figure 1and 2. The graph illustrates the freezing point since the temperature becomes steady at approximately 7.5 . Moreover, table 1 shows the mass measurements used to conduct the experiment. The entire mass of the solution was important to measure in order to find the mass of the cyclohexane. Since, the solution consists of biphenyl and cyclohexane, the mass of cyclohexane was determined to be 2.25 g. The number of moles biphenyl was calculated to be moles that was then divided by the actual mass used. The accepted molar mass of biphenyl is 154.2 and the experimental value is 148.6 which result in 3.6% error.

The percent of error is reasonable and acceptable since several sources of error may have occurred. When transferring the biphenyl to the test-tube some particles of biphenyl stayed on the weighing boat. This will slightly alter the value of the molar mass. Moreover, stirring the solution with the copper stirrer can change the freezing point depression, since the temperature will constantly increase and decrease.

When more biphenyl was added to the solution, the freezing point depression decrease. This shows as the solute increases the freezing temperature decrease. Therefore, illustrates the colligative property, which depends on the number of molecules present not the type of molecules used in the solution.

Losing some of the solid biphenyl during transfer to the test tube will result in effect the measured molecular weight because it is calculated by dividing mass by number of moles. The mass of biphenyl used to calculate the molecular weight will be less than the mass that actually used in the equation because it was lost during the transfer, which mean the molecular weight will be less.

Appendix

Figure 5 A sample calculation to find the molar mass of biphenyl and the percent error.

Latest

Time

(min)

Temperature

(°C)

173

174

175

176

177

178

179

180

181

182

183

184

185

186

187

188

189

190

191

192

193

194

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207

1.72

1.73

1.74

1.75

1.76

1.77

1.78

1.79

1.80

1.81

1.82

1.83

1.84

1.85

1.86

1.87

1.88

1.89

1.90

1.91

1.92

1.93

1.94

1.95

1.96

1.97

1.98

1.99

2.00

2.01

2.02

2.03

2.04

2.05

2.06

8.6

8.5

8.4

8.4

8.3

8.2

8.1

8.1

8.0

7.9

7.8

7.8

7.7

7.7

7.6

7.5

7.5

7.4

7.4

7.3

7.3

7.3

7.3

7.3

7.4

7.4

7.4

7.4

7.4

7.4

7.4

7.4

7.5

7.5

7.4

Using Freezing-Point Depression to Find Molecular Weight

0 1234

10

20

30

Time (min)

T

e

m

p

e

r

a

t

u

r

e

(

°

C

)

Linear Fit for: Latest | Temperature

Temp = mt+b

m (Slope): -37.27 °C/min

b (Y-Intercept): 65.21 °C

Correlation: -0.9978

RMSE: 0.1348 °C

Linear Fit for: Latest | Temperature

Temp = mt+b

m (Slope): -0.6710 °C/min

b (Y-Intercept): 8.586 °C

Correlation: -0.9328

RMSE: 0.1243 °C

Time: 1.548 min

Temperature: 7.5085

Temperature: 7.5468

Latest

Time

(min)

Temperature

(°C)

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

185

186

187

188

189

190

1.55

1.56

1.57

1.58

1.59

1.60

1.61

1.62

1.63

1.64

1.65

1.66

1.67

1.68

1.69

1.70

1.71

1.72

1.73

1.74

1.75

1.76

1.77

1.78

1.79

1.80

1.81

1.82

1.83

1.84

1.85

1.86

1.87

1.88

1.89

8.9

8.8

8.7

8.7

8.6

8.5

8.4

8.3

8.3

8.2

8.1

8.1

8.1

8.0

8.0

7.9

7.9

7.8

7.8

7.7

7.7

7.7

7.6

7.6

7.6

7.5

7.5

7.4

7.4

7.4

7.4

7.3

7.3

7.3

7.2

Using Freezing-Point Depression to Find Molecular Weight

02468 10

0

5

10

15

20

Time (min)

T

e

m

p

e

r

a

t

u

r

e

(

°

C

)

(1.385, 7.403)

Linear Fit for: Latest | Temperature

Temp = mt+b

m (Slope): -20.14 °C/min

b (Y-Intercept): 35.41 °C

Correlation: 0

RMSE: 0.3099 °C

Linear Fit for: Latest | Temperature

Temp = mt+b

m (Slope): -0.9852 °C/min

b (Y-Intercept): 8.886 °C

Correlation: 0

RMSE: 0.05113 °C

Time: 1.388 min

Temperature: 7.4624

Temperature: 7.5184

Latest

Time

(min)

Temperature

(°C)

104

105

106

107

108

109

110

111

112

113

114

115

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119

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122

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138

1.03

1.04

1.05

1.06

1.07

1.08

1.09

1.10

1.11

1.12

1.13

1.14

1.15

1.16

1.17

1.18

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1.20

1.21

1.22

1.23

1.24

1.25

1.26

1.27

1.28

1.29

1.30

1.31

1.32

1.33

1.34

1.35

1.36

1.37

3.5

3.4

3.4

3.3

3.3

3.3

3.2

3.2

3.2

3.2

3.2

3.1

3.1

3.1

3.0

3.0

3.0

3.0

3.0

3.0

3.0

3.0

3.0

3.0

3.0

3.0

2.9

2.9

2.9

2.9

2.9

2.9

2.8

2.8

2.8

Using Freezing-Point Depression to Find Molecular Weight

02468 10

0

5

10

Time (min)

T

e

m

p

e

r

a

t

u

r

e

(

°

C

)

(0.382, 4.443)

Linear Fit for: Latest | Temperature

Temp = mt+b

m (Slope): -12.78 °C/min

b (Y-Intercept): 15.96 °C

Correlation: -0.9967

RMSE: 0.1583 °C

Linear Fit for: Latest | Temperature

Temp = mt+b

m (Slope): -0.6872 °C/min

b (Y-Intercept): 3.526 °C

Correlation: -0.9590

RMSE: 0.1367 °C

Time: 1.031 min

Temperature: 2.7800

Temperature: 2.8171

Latest

Time

(min)

Temperature

(°C)

140

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

1.39

1.40

1.41

1.42

1.43

1.44

1.45

1.46

1.47

1.48

1.49

1.50

1.51

1.52

1.53

1.54

1.55

1.56

1.57

1.58

1.59

1.60

1.61

1.62

1.63

1.64

1.65

1.66

1.67

1.68

1.69

1.70

1.71

1.72

1.73

3.5

3.5

3.5

3.4

3.4

3.4

3.4

3.4

3.3

3.3

3.3

3.3

3.2

3.2

3.2

3.2

3.2

3.2

3.2

3.2

3.1

3.1

3.1

3.1

3.1

3.1

3.1

3.0

3.0

3.0

3.0

3.0

3.0

3.0

3.0

Using Freezing-Point Depression to Find Molecular Weight

02468 10

0

5

10

15

Time (min)

T

e

m

p

e

r

a

t

u

r

e

(

°

C

)

(1.034, 2.095) (Δt:1.73Δy:0.1)

Linear Fit for: Latest | Temperature

Temp = mt+b

m (Slope): -18.10 °C/min

b (Y-Intercept): 22.10 °C

Correlation: -0.9958

RMSE: 0.1830 °C

Linear Fit for: Latest | Temperature

Temp = mt+b

m (Slope): -0.6830 °C/min

b (Y-Intercept): 4.017 °C

Correlation: -0.9727

RMSE: 0.08233 °C

Time: 1.036 min

Temperature: 3.3405

Temperature: 3.3087