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Wednesday, December 1

  1. page Extraction and Separation of a Mixture edited Introduction- Liquid-liquid extractions are essentially the only kind of extraction performed in t…
    Introduction-
    Liquid-liquid extractions are essentially the only kind of extraction performed in the organic labs. The "liquid-liquid" phrase means that two liquids are mixed in the extraction procedure. The liquids must be immiscible: this means that they will form two layers when mixed together, like oil and vinegar do in dressing. Some compounds are more soluble in the organic layer and some compounds are more soluble in the aqueous layer. The ionic substance will be in the aqueous layer,usually lower layer, the nonionic ones in the organic layer, upper layer.
    This is a pretty good intro. I'd be pleased to also see you discuss the acid/base chemistry of the process we are doing today, in addition to what you have here.
    Procedure-
    ...
    p. . pages??
    For this lab one would want to use the micro-scale glassware
    Dissolve roughly 0.18 g of the mixture in 2 mL of MTBE in a reaction tube.
    ...
    Weight of Unknown Mixture- 0.181 g
    Contents of Tubes:
    ...
    dimethoxybenzene: 32.0°C-40.7°C These are your experimental values, right?
    Tube 2, Benzoic Acid: 117.5°C-119.6°C
    Tube 3, 4-tert-butyl-phenol 84.6°C-88.4°C
    ...
    0.00g
    0.026g
    The table above would be nice, but what's with all the zeros?
    I would like to know what your product looks like.

    Percent Recovery:
    Tube 1, dimethoxybenzene:
    ...
    Tube 3,4-tert-butyl-phenol
    0.026g / 0.181g * 100 = 14% Recovered
    calculations OK. sig figs OK.
    Analysis-
    I guess we've got an incomplete report here.
    This report earns the following scores for: format (1/2) style (0.5/2) data (2/3) quality of result (1/1) quality of reported data (1/1) conclusion (0/2) error (0/1) post-lab Q (2 freebie points) for a total of 7.5/14. Note that some scores may vary from this score, based on participation levels.

    (view changes)
    5:11 pm

Sunday, November 28

  1. page Extraction and Separation of a Mixture edited ... Liquid-liquid extractions are essentially the only kind of extraction performed in the organic…
    ...
    Liquid-liquid extractions are essentially the only kind of extraction performed in the organic labs. The "liquid-liquid" phrase means that two liquids are mixed in the extraction procedure. The liquids must be immiscible: this means that they will form two layers when mixed together, like oil and vinegar do in dressing. Some compounds are more soluble in the organic layer and some compounds are more soluble in the aqueous layer. The ionic substance will be in the aqueous layer,usually lower layer, the nonionic ones in the organic layer, upper layer.
    Procedure-
    Williamson K.L., 2003. Macroscale and Microscale Organic Experiments 4th Edition. Boston (MA): Houghton Mifflin Company, p. .
    For this lab one would want to use the micro-scale glassware
    Dissolve roughly 0.18 g of the mixture in 2 mL of MTBE in a reaction tube.
    (view changes)
    9:22 am
  2. page Extraction and Separation of a Mixture edited ... 0.026g / 0.181g * 100 = 14% Recovered Analysis- {jpg1.jpg} jpg1.jpg
    ...
    0.026g / 0.181g * 100 = 14% Recovered
    Analysis-
    {jpg1.jpg} jpg1.jpg
    (view changes)
    9:21 am

Tuesday, November 23

  1. page Extraction and Separation of a Mixture edited ... In the exactly same way, neutralize the contents of tube 3 with concentrated HCL (no carbon di…
    ...
    In the exactly same way, neutralize the contents of tube 3 with concentrated HCL (no carbon dioxide is released in this neutralization), heat the tube to bring most of the material into a solution, allow to cool slowly, then remove the solvent, and re crystallize the phenol from boiling water.
    Data-
    General Observations-
    When doing this lab one would want to pay special attention to the amount
    Weight of solution addedxUnknown Mixture- 0.181 g
    Contents of Tubes:
    Tube 1, 1,4 dimethoxybenzene: 32.0°C-40.7°C
    Tube 2, Benzoic Acid: 117.5°C-119.6°C
    Tube 3, 4-tert-butyl-phenol 84.6°C-88.4°C
    Table 1: Mass of Product Crystals
    Object/Crystal Substance
    Tube 1, dimethoxybenzene
    Tube 2, benzoic acid
    Tube 3, 4-tert-butylphenol
    Filter Paper / Container
    g
    0.00g
    0.g
    Filter Paper / Container + Product
    g
    0.00g
    0.g
    Product Mass
    0.019g
    0.00g
    0.026g
    Percent Recovery:
    Tube 1, dimethoxybenzene:
    0.019g / 0.181g * 100 = 10% Recovered
    Tube 2,benzoic acid:
    0.00g / 0.181g * 100 = 0% Recovered
    Tube 3,4-tert-butyl-phenol
    0.026g / 0.181g * 100 = 14% Recovered

    Analysis-
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    (view changes)
    7:16 pm

Friday, November 19

  1. page Extraction and Separation of a Mixture edited ... Liquid-liquid extractions are essentially the only kind of extraction performed in the organic…
    ...
    Liquid-liquid extractions are essentially the only kind of extraction performed in the organic labs. The "liquid-liquid" phrase means that two liquids are mixed in the extraction procedure. The liquids must be immiscible: this means that they will form two layers when mixed together, like oil and vinegar do in dressing. Some compounds are more soluble in the organic layer and some compounds are more soluble in the aqueous layer. The ionic substance will be in the aqueous layer,usually lower layer, the nonionic ones in the organic layer, upper layer.
    Procedure-
    For this lab one would want to use the micro-scale glassware
    Dissolve roughly 0.18 g of the mixture in 2 mL of MTBE in a reaction tube.
    The add 1 mL of a saturated aqueous solution of sodium to the reaction tube.
    ...
    Collect the crystals on a tared piece of paper, allow them to dry thoroughly, and determine the percent recovery of the acid.
    In the exactly same way, neutralize the contents of tube 3 with concentrated HCL (no carbon dioxide is released in this neutralization), heat the tube to bring most of the material into a solution, allow to cool slowly, then remove the solvent, and re crystallize the phenol from boiling water.
    Data-
    General Observations-
    When doing this lab one would want to pay special attention to the amount of solution addedx
    Analysis-

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    (view changes)
    3:28 pm

Thursday, November 18

  1. page Extraction and Separation of a Mixture edited ... Collect the crystals on a tared piece of paper, allow them to dry thoroughly, and determine th…
    ...
    Collect the crystals on a tared piece of paper, allow them to dry thoroughly, and determine the percent recovery of the acid.
    In the exactly same way, neutralize the contents of tube 3 with concentrated HCL (no carbon dioxide is released in this neutralization), heat the tube to bring most of the material into a solution, allow to cool slowly, then remove the solvent, and re crystallize the phenol from boiling water.
    {jpg1.jpg} jpg1.jpg
    (view changes)
    6:34 pm
  2. page Sublimation edited ... The lab experiment followed the procedures as described in Williamson (2003). The unknown subs…
    ...
    The lab experiment followed the procedures as described in Williamson (2003). The unknown substance was kept warm and the collection site was kept cold. Aluminum foil was also used, which kept the crystals from forming on the inside of the filter flask. The percent recovery of 74.5% suggests that some product was lost in the sublimation process; however this percentage is large enough to adequately determine the unknown substance from melting point data. The experimental melting point data was determined used a MelTemp apparatus, which was 123.9 degrees Celcius. In Table 7.3 in Williamson (2003) on page 124, the sublimation unknowns given in which are within the range to the experimental results are both Acetanilide (mp 114 degrees Celcius) and Benzoic Acid (mp 122 degrees Celcius). The experimental unknown value was 123.9 degrees Celcius, which is 1.9 degrees Celcius warmer than the theoretical melting point of Benzoic Acid. Based on the procedures followed, the percent recovery and close proximity of the melting point values; the unknown product is concluded with fairly high confidence to be Benzoic Acid.
    this is all good.
    error? Please address in a very deliberate way.
    Post Lab Questions:
    What is lyophilization and why is it used?
    (view changes)
    11:58 am
  3. page Sublimation edited ... 0.038 123.9 The always report experimental melting points as temperature ranges. The pe…
    ...
    0.038
    123.9
    Thealways report experimental melting points as temperature ranges.
    The
    percent recovery
    ...
    is 74.5%. sig figs here: you have two to work with, not 3.
    Comparing the
    ...
    (see Analysis). this is correct!
    General Observations:
    The unknown substance before sublimation had the appearance of small white crystals. The process of sublimation required heat, time and a cold point for the gaseous product to collect and turn directly into solid crystals. When heated in the hot sand bath from below, the solid turned to gas, and the cold tip of the test tube worked as a point for the gas to collect and solidify. Not long after placing the container in the hot sand bath the crystals started to vaporize, within ten minutes of the start of sublimation, vapors started to form crystals on the cold tip of the test tube. After heating for approximately 27 minutes, sublimation had completed, leaving a substance with the appearance of small, clear/white, somewhat prismatic symmetry, with a low density, appearing like a spiderweb.
    Analysis:
    The lab experiment followed the procedures as described in Williamson (2003). The unknown substance was kept warm and the collection site was kept cold. Aluminum foil was also used, which kept the crystals from forming on the inside of the filter flask. The percent recovery of 74.5% suggests that some product was lost in the sublimation process; however this percentage is large enough to adequately determine the unknown substance from melting point data. The experimental melting point data was determined used a MelTemp apparatus, which was 123.9 degrees Celcius. In Table 7.3 in Williamson (2003) on page 124, the sublimation unknowns given in which are within the range to the experimental results are both Acetanilide (mp 114 degrees Celcius) and Benzoic Acid (mp 122 degrees Celcius). The experimental unknown value was 123.9 degrees Celcius, which is 1.9 degrees Celcius warmer than the theoretical melting point of Benzoic Acid. Based on the procedures followed, the percent recovery and close proximity of the melting point values; the unknown product is concluded with fairly high confidence to be Benzoic Acid.
    this is all good.
    Post Lab Questions:
    What is lyophilization and why is it used?
    ...
    prepared meals. good answer.
    (view changes)
    11:57 am

Wednesday, November 17

  1. page Sublimation edited Organic Chemistry Lab #4: Sublimation Introduction: ... phase diagram (see below). (Figure 1)…
    Organic Chemistry Lab #4: Sublimation
    Introduction:
    ...
    phase diagram (see below).(Figure 1). Sublimation is
    ...
    of sublimation include,include dehydration of
    {http://www.chemistry-blog.com/wp-content/uploads/2009/02/phasediagramco2.png}
    Figure 1: Phase diagram.
    
    Procedure:
    Below are the steps of the Sublimation lab experiment. Detailed lab instructions are described in the following citation: Williamson K.L., 2003. Macroscale and Microscale Organic Experiments 4th Edition. Boston (MA): Houghton Mifflin Company, p. 120-125.
    ...
    of an impure unknown.unknown into a flask.
    Assemble small scale vacuum sublimation apparatus (Figure 2).

    Close the flask with a rubber pipette bulb, and then place ice water in the centrifuge tube.
    Cautiously warm the flask until sublimation starts, and then maintain that temperature throughout the sublimation.
    ...
    to collect toon the tube.
    Once sublimation is complete, remove the ice water from the centrifuge tube and replace it with water at room temperature.
    ...
    and the precent recovery, and frompercent recovery.
    Record
    the melting point identify(degrees Celcius) using a MelTemp apparatus, comparing with values in Williamson (2003) to determine the name of the unknown.
    {http://www.chemistryland.com/CHM130W/LabHelp/Experiment5/sublimationApparatus.jpg}
    Figure 1:2: Example of
    Data:
    The identification number of the unknown substance is 241-7-8. The tables below summarized the weights and melting points of the unknown before and after sublimation.
    ...
    0.038
    123.9
    The percent recovery of the unknown substance is equal to the weight of the substance recovered divided by the weight of the substance before sublimation multiplied by 100. The percent recovery
    ...
    in Williamson (2003) on page 124, the closedclosest melting point
    ...
    be Benzoic Acid.Acid (see Analysis).
    General Observations: Sublimation
    The unknown substance before sublimation had the appearance
    of a solid takessmall white crystals. The process of sublimation required heat, time
    ...
    turn directly to ainto solid crystal.crystals. When heated
    ...
    bath from belowbelow, the solids turnsolid turned to gas, and the cold
    ...
    test tube worksworked as a
    ...
    ten minutes after that it started to become apparent thatof the start of sublimation, vapors were startingstarted to form
    ...
    test tube. After heating for approximately 27 minutes, sublimation had completed, leaving a substance with the appearance of small, clear/white, somewhat prismatic symmetry, with a low density, appearing like a spiderweb.
    Analysis:
    The lab experiment followed the procedures as described in Williamson (2003). The unknown substance before sublimation was small white crystals. After heating for about 27 minutes sublimation had completed leaving a substance withkept warm and the collection site was kept cold. Aluminum foil was also used, which kept the crystals from forming on the appearanceinside of small, clear/white, somewhat prismatic symmetry, with a low density (like a spiderweb).the filter flask. The percent recovery of 74.5% suggests that some product was lost in the sublimation process; however this percentage is large enough to adequately determine the unknown substance from melting point data. The experimental melting point data was determined
    ...
    a MelTemp apparatusapparatus, which was 123.9 degrees Celcius. In Table 7.3 in Williamson (2003) on page 124, the sublimation unknowns given in which are within the range to be the same as benzoic acid.experimental results are both Acetanilide (mp 114 degrees Celcius) and Benzoic Acid (mp 122 degrees Celcius). The experimental unknown value was 123.9 degrees Celcius, which is 1.9 degrees Celcius warmer than the theoretical melting point of Benzoic Acid. Based on the procedures followed, the percent recovery and close proximity of the melting point values; the unknown product is concluded with fairly high confidence to be Benzoic Acid.
    Post Lab Questions:
    What is lyophilization and why is it used?
    (view changes)
    8:39 pm

Tuesday, November 16

  1. page Extraction and Separation of a Mixture edited Introduction- Liquid-liquid extractions are essentially the only kind of extraction performed in …
    Introduction-
    Liquid-liquid extractions are essentially the only kind of extraction performed in the organic labs. The "liquid-liquid" phrase means that two liquids are mixed in the extraction procedure. The liquids must be immiscible: this means that they will form two layers when mixed together, like oil and vinegar do in dressing. Some compounds are more soluble in the organic layer and some compounds are more soluble in the aqueous layer. The ionic substance will be in the aqueous layer,usually lower layer, the nonionic ones in the organic layer, upper layer.
    Procedure-
    Dissolve roughly 0.18 g of the mixture in 2 mL of MTBE in a reaction tube.
    The add 1 mL of a saturated aqueous solution of sodium to the reaction tube.
    Mix the contents of the tube thoroughly by pulling the two layers into a Pasteur pipette and expelling them forcefully into the reaction tube. Repeat process for roughly three minutes, then allow the layers to separate completely and then draw off the lower layer. (Usually the ionic substance will be in the aqueous layer)
    Add another 15 mL of sodium bicarbonate solution to the tube, mix the contents as before, and add the lower layer to reaction tube two.
    Add 2 mL of ether to tube 2, mix it thoroughly, remove the ether layer, and discard it. (Backwashing-removes any organic material the might contaminate the contents)
    Add 1 mL of 3 M aqueous sodium hydroxide to tube 1, shake the mixture thoroughly, allow the layers to separate, draw off the lower layer using a clean Pasteur pipette, and place it in reaction tube 3.
    Then to tube 1, add saturated sodium chloride solution, mic, remove the aqueous layer, and then add to the ether anhydrous calcium chloride pellets until the during agent no longer clumps together.
    Wash it off with ether after the drying process is finished.
    Then with hydrochloric acid, carry out neutralization, by adding drops of HCL and testing the solution with litmus paper.(Be extremely careful because much carbon dioxide is released in the neutralization)
    Add a boiling stick to the tube and very cautiously heat the tube to bring most of the solid carboxylic acid into solution.
    Allow the tube to cool slowly to room temperature and then cool it in ice.
    Once crystallization occurred, use the Hirsch funnel to remove the crystals from the solution.
    Collect the crystals on a tared piece of paper, allow them to dry thoroughly, and determine the percent recovery of the acid.
    In the exactly same way, neutralize the contents of tube 3 with concentrated HCL (no carbon dioxide is released in this neutralization), heat the tube to bring most of the material into a solution, allow to cool slowly, then remove the solvent, and re crystallize the phenol from boiling water.

    (view changes)
    8:47 pm

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