PlatinumEssays.com - Free Essays, Term Papers, Research Papers and Book Reports
Search

Lab

By:   •  October 15, 2013  •  Essay  •  620 Words (3 Pages)  •  1,460 Views

Page 1 of 3

ss

HETP of column packing:

Reflux Ratio:

Figure 1; Standard Scale

Vials Temperture (Celsius) Mass of Empty Vial with Cap (g) Mass after Distillation (g) Mass of Distilled Solution (g)

HETP 81 9.009 9.116 0.107

Vial 1 81-84 18.851 20.562 1.711

Vial 2 85-96 17.554 28.320 10.766

Vial 3 97-106 17.556 22.149 4.593

Vial 4 107-111 16.980 22.765 5.785

Micro-scale

Height of Column Packing: __ mL

Mass of Cyclohexane from gas chromatography:

Mass of Toluene from gas chromatography:

Moles of Cyclohexane: 1.29 mol cyclohexane

Moles of Toluene: 0.0265 mol toluene

Total number of theoretical plates: 3.23

Number of plates by column: 2.23

HETP of column packing:

Reflux Ratio:

Figure 2; Micro Scale

Vials Temperture (Celsius) Mass of Empty Vials with Cap (g) Mass after Distillation (g) Mass of Distilled Solution (g)

HETP 81 8.803g 8.810 0.007

Vial 1 81-84 16.798 16.850 0.052

Vial 2 85-96 17.404 17.505 0.101

Vial 3 97-106 16.940 16.952 0.012

Vial 4 107-111 17.149g 17.156 0.007

Calculations

Standard Scale

HETP

A: (92.63384)(1.11) = (102.823562 g) / (84.16 g/mol cyclohexane) = nA = 1.22 mol cyclohexane

B: (7.36611)(1.05) = (7.7344155 g)/ (92.14 g/mol toluene) = nB = 0.0839 mol toluene

Total # theoretical plates = (log nA/log nB) / log ? = log (1.22/0.0839) / log(2.33) = 3.16

3.16-1 = 2.16 provided by the column.

Sample #1

A: (88.03853)(1.11) = (97.7227683 g) / (84.16 g/mol cyclohexane) = nA = 1.16 mol cyclohexane

B: (11.65914)(1.05) = (12.242097 g) / (92.14 g/mol toluene) = nB = 0.133 mol toulene

Molar % A: (100%)(nA / nA + nB) = (100%)(1.16 / 1.16 + 0.133) = 89.71%

Molar % B: (100%)(nB / nB + nA) = (100%)(0.133 / 0.133 + 1.16) = 10.29%

Avg. Temp: 81? +84? = 165/2 = 82.5?

Sample #2

A: (74.42797)(1.11) = (82.615046 g) / (84.16 g/mol cyclohexane) = nA = 0.982 mol cyclohexane

B: (25.57202)(1.05) = (26.850621 g) / (92.14 g/mol toluene) = nB = 0.291 mol toulene

Molar % A: (100%)(nA / nA + nB) = (100%)(0.982 / 0.982 + 0.291 ) = 77.14 %

Molar % B: (100%)(nB / nB + nA) = (100%)(0.291 / 0.291 + 0.982 ) = 22.86%

Avg. Temp: 85? + 96? = 181/2 = 90.5?

Sample #3

A: (40.16184)(1.11) = (44.5796424 g) / (84.16 g/mol cyclohexane) = nA = 0.53 mol cyclohexane

B: (59.83816)(1.05) = (62.830068 g) / (92.14 g/mol toluene) = nB = 0.68 mol toulene

Molar

...

Download:  txt (3.9 Kb)   pdf (66.7 Kb)   docx (10.3 Kb)  
Continue for 2 more pages »