{"id":117535,"date":"2023-08-28T23:23:32","date_gmt":"2023-08-28T23:23:32","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=117535"},"modified":"2023-08-28T23:23:33","modified_gmt":"2023-08-28T23:23:33","slug":"chem-212-lab-exams-and-reports-organic-chemistry-verified","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2023\/08\/28\/chem-212-lab-exams-and-reports-organic-chemistry-verified\/","title":{"rendered":"CHEM 212 Lab Exams and Reports Organic Chemistry (Verified)"},"content":{"rendered":"\n

Lab 4: Separation of Organic Compounds \u2013 Column
Chromatography CHEM 212 Organic Chemistry Laboratory<\/p>\n\n\n\n

Background:
Chromatographic separation has an important role in organic synthesis laboratory.
Column Chromatography is used to separate larger quantities of material than thin layer
chromatography. The purpose of this experiment is to establish the identity of a unknown
compound (mixed with benzophenone) by separating of mixtures by column chromatography
and by analysis of individual fraction by thin layer chromatography. The unknown compound is
one of the following: biphenyl, naphthalene, or acenaphthene.
Key Experimental Details\/Observations:
The first step of this experiment is the column packing procedure. First, the column
apparatus was set up securely and completely vertical, a small cotton ball was added into the
bottom to plug the tip of the column. Then, using a small funnel, 1.803 grams of silica gel was
added to about 5.00 cm in height. To ensure proper packing of the column, 14.9 mL of hexanes
was allowed to elute through the silica gel. Applying pressure allowed for the removal of air
bubbles and further elution of the gel. After the silica gel became clear, the column was packed
with 1.00 cm of sand. Then, 1.0 mL of hexanes was added to the column to moisten the bed.
The second part of this experiment began after our T.A. loaded 0.4 mL of the sample
solution \u201cA8\u201d into the column. The second step of this experiment is the mobile phase
preparation procedure. A 25.3 mL solution of pure 100% hexanes was prepared. In addition, a
25.0 mL solution of 25% dichloromethane\/75% hexanes was prepared, and a 25.0 mL solution
of 50% dichloromethane\/50% hexanes was prepared. Each solution was put in a separate
Erlenmeyer flask and labeled accordingly.<\/p>\n\n\n\n

CHEM Lab 8 Organic Chemistry
Provide the expected melting point of caffeine and the boiling points of dichloromethane, ethanol, acetic
acid and ethyl acetate. (5 pts.)
MP Caffeine: 234.2\u00b0C-236.3\u00b0C
BP DCM: 103.6\u00b0C-107.5\u00b0C
BP Ethanol: 72.3\u00b0C-75.3\u00b0C
BP Acetic Acid: 117.8\u00b0C-118.9\u00b0C
BP Ethyl Acetate: 77.1\u00b0C
Provide an outline on how to perform a recrystallization. (10 pts.)
Recrystallization is a procedure for purifying an impure compound in a solvent. The
method of purification is based on the principle that the solubility of most solids increases with
increased temperature. This means that as temperature increases, the amount of solute that can be
dissolved in a solvent increases. As the solvent cools, the solution becomes saturated with the solute
and the solute crystallizes out (a solid). As the crystal develops, impurities are excluded from the crystal
lattice, thereby completing the purification process. The crystals can then be collected, washed, and
dried.<\/p>\n\n\n\n

Complete the following based on the video you watched: (11x 2 = 22 pts.)
Isolation of caffeine from tea leaves in the lab involved a
Solid-liquid extraction followed by a Liquid-liquid extraction
Mass of tea leaves used 10.0 g
Mass of calcium carbonate used 4.00 g
Moles of calcium carbonate used 0.04 mol
Crude weight of extracted caffeine 161mg
Melting point of extracted, crude caffeine 212.1\u00b0C-225.6\u00b0C
Melting point of extracted, recrystallized caffeine 233.9\u00b0C-235.2\u00b0C
Expected melting point of pure caffeine 234.2\u00b0C-236.3\u00b0C
During the extraction you noticed two distinct (top and bottom) layers: which one of them is the
dichloromethane layer? Which property of dichloromethane was responsible for this observation? (4
pts.)<\/p>\n\n\n\n

CHEM 212 Lab 2 Report Separation of Organic
Compounds \u2013Column Chromatography
Background:
Chromatographic separation is an important technique used to separate a compound from
a mixture. Column chromatography uses solid-liquid absorption and elution to separate the
compounds. The objective of this lab was to preform column chromatography to separate an
unknown compound from benzophenone through a series of addition of elution solvents that
increased in polarity, from there analysis of the individual fractions, collected during the column
chromatography, was preformed via thin layer chromatography, and the unknown compound\u2019s
identity was established via a mixed melting point and TLC. The unknown compound could have
been either biphenyl, naphthalene, or acenaphthene.
Key Experimental Details and Observations:
This experiment consisted of two parts, the first being column chromatography and thin
layer chromatography analysis, and the second being mixed melting point and TLC comparison
to identify an unknown compound. In the first part with column chromatography, packing the
column was the first step. The first thing to be done was to add a small amount of cotton snugly
to the bottom of the glass column to act as a plug. Then 5.40 cm of silicagel was added to the
glass column and washed with 49.1 mL of hexanes until the powder turned into a gel (around 10
complete washes), and 1.21 cm of sand was added on top of the silicagel and hexanes were run
through the column again to ensure the materials inside the column were packed down. The<\/p>\n\n\n\n

instructor proceeded to load 0.4 mL of unknown mixture C directly into the column. After the
unknown mixture was loaded into the column a series of 3 solvents were added in order from
lower polarity to higher polarity, and fractions of around 5 mL were collected continuously
through the additions of the solvents in labelled test tubes. The first solvent added was 25.5 mL
of 100% hexanes, then 25.9 mL of 25% DCM\/hexanes, and finally 50.5 mL of 50%
DCM\/hexanes. In total 15 fractions were collected, and a TLC plate labeled with a matrix was
used to perform a spot analysis of the fractions to determine which fractions were appearing
under the UV light. Then TLC analyses with different elution of 100% hexanes and 50%
DCM\/hexanes were performed on the fractions to figure out which ones were polar and which
were non-polar. Once that was discovered the non-polar fractions were added into a roundbottomed flask and the polar ones were added into a different round-bottomed flask. In the
second part of the lab the solvents were removed from the two groups of liquids (non-polar and
polar) using a rotovap and then mixed melting point analysis (via Mel-Temp machine) and thin
layer chromatography were used to determine the identity of the unknown compound C.
Results:
Table 1: List and Amounts of Fractions Combined
Fractions Polarity Amount (mL)
1 and 2 Non-polar 14.9 mL
3-13 Polar 68.5 mL
When performing the TLC spot test for the 15 fractions collected, fractions 1-5 produced
spots visible under the UV light along with fractions 6-8, 10-11, and 13 which also produced
spots under the UV light but less intense and not as filled in spots as fractions 1-5 did. It was
decided to run TLC analyses for spots 1, 2, 3, 5, 6, 11, and 13 in 100% hexanes and 50%
DCM\/hexanes respectively. For the 100% hexanes TLC all of the fraction samples moved
slightly, and had a retention .06. In addition to this slight movement fractions 1 and 2 had two<\/p>\n\n\n\n

CHEM 212: Organic Chemistry Lab 6 – Separation of Organic Compounds \u2013 Extraction<\/p>\n\n\n\n

    \n
  1. What is an extraction typically used for? (5 pts.)
    The function of an extraction is to separate and purify organic compounds.<\/li>\n\n\n\n
  2. What type of extraction was being performed in the lab? (5 pts.)
    The type of extraction being performed was an acid\/base extraction to separate an unknown neutral
    organic compound from an organic acid or an organic base.<\/li>\n\n\n\n
  3. Why should you look up the density of an organic solvent before using it in an extraction? (5 pts.)
    As density increases, denser solvents will accumulate at the bottom of the funnel, and the lighter
    density solvents will stay at the top of the separation funnel. An increase in the density of organic
    solvent, which is more dense than water, i will accumulate and form a layer below the water.<\/li>\n\n\n\n
  4. What is meant by the term \u2018immiscible\u2019 and why is this factor important when performing
    an extraction? (5 pts.)
    Immiscible liquids do not mix together so they form visible layers. For extraction to take place, both
    liquids have to be immiscible so they do not mix and form a homogenous mixture.<\/li>\n\n\n\n
  5. What were the two types of compounds that were being separated from each other in
    this experiment? (5 pts.)
    In this experiment the two types of compounds being separated from each other were the ionizable
    and non-ionizable compounds.<\/li>\n\n\n\n
  6. Name the technique that will be used to identify the unknown once it is isolated. (5 pts.)
    The technique used to identify the unknown is melting point analysis.<\/li>\n<\/ol>\n\n\n\n

    CHEM Lab 8 Report Organic Chemistry
    Dayna Cox
    01070394
    Background:
    Lab 8
    The goal of Lab 8 is to extract Limonene using steam distillation (separating
    compound with different boiling points). Two techniques will be used: steam
    distillation and liquid-liquid extraction. Steam distillation is purifying a liquid in
    steam conditions due to the liquid being difficult to purify in a normal temperature
    state. Liquid-liquid extraction will then be performed to separate the organic and
    aqueous layer.
    Key Experimental Details:
    Dried orange zest is provided. 5.01g of orange zest is obtained and put into a 100mL
    round bottom flask along with 50mL of DI water to soak for 10 minutes. After
    soaking is complete add 2 drops of antifoam agent and a couple of boiling chips into
    flask (3 boiling chips were added). The distillation flask should be wrapped in
    aluminm foil to conserve heat (the top and bottom should be covered). A hot plate
    should be on heat to boil the water with orange zest. Fill receiving flask with 25mL
    of water and mark the level with sharpie, this is only so we can measure 25mL in
    the flask (this is in separate flask). After marking 25mL of water on the flask, pour
    out the water. Assembling of the steam distillation is to be set up and distillate
    should start collecting in receiver flask in about 30 minutes after heat is turned on.
    25mL of distillate should collect into the receiver, this can be observed by the
    sharpie line of 25mL. This will determine the distillation of limonene being
    complete. Transfer distillate to separatory funnel. Gently mix the two layers and
    allow them to separate. Isolate the organic layer in Erlenmeyer flask and dry using
    anhydrous sodium sulfate. Decant and store into organic layer in test tube; seal with
    cork and parafilm, then store in drawer.
    Next week:
    The limonene sample is to be analyzed in Ethyl Acetate. This is to be compared to
    standard sample of Limonene in Ethyl Acetate (its provided). Comparing the results
    of the two GC\u2019s runs you should be able to calculate the amount of Limonene
    present in your distillate in mg\/mL.
    Results:
    Compound Volume Boiling Point
    Distillate collected 25mL<\/p>\n\n\n\n

    CHEM 212: Organic Chemistry
    Lab 6 \u2013 Report Separation of Organic Compounds \u2013 Extraction
    INTRODUCTION:
    Extraction is another method that is used in lab to separate and also purify organic
    compounds. Organic compounds can be separated from each other via the partitioning of them
    between two solvents, one being the organic compound and one being the aqueous, of different
    densities in a separatory funnel. The purpose of this lab was to separate an unknown neutral
    organic compound from either an organic acid or organic base through the performance of
    acid\/base extractions. Then the unknown compound\u2019s identity was confirmed by carrying out
    melting point and mixed melting point analyses.
    KEY EXPERIMENTAL DETAILS AND OBSERVATIONS:
    For this experiment an unknown mixture (unknown C2) was provided and was massed at
    0.595 grams. The unknown compound was a white powder with some slight clumping. The
    sample was dissolved in 20 mL of DCM in a beaker and this mixture was poured into the
    separatory funnel and the beaker was rinsed with approximately 5.0 mL of DCM and added to
    the separatory funnel. 9.98 mL of a strong base, sodium hydroxide (NaOH), was added to the
    mixture in the separatory funnel, and mixed. The addition of the NaOH caused separation into an
    aqueous layer and an organic layer. The organic layer was first drained from the separatory
    flask, being denser than the aqueous layer it was on the bottom, and the aqueous layer was also
    drained into a separate flask labeled \u201cbasic.\u201d Then the organic layer was added back to the
    separatory<\/p>\n\n\n\n

    Chem 212 Lab 4 Separation of Spinach Pigments
    Organic Chemistry
    Lab
    Separation of Spinach Pigments \u2013 Thin Layer Chromatography (TLC) Worksheet<\/p>\n\n\n\n

      \n
    1. What is the ideal Rf value in a TLC? (3 pts.)
      The ideal Rf (ratio to front) value in a TLC is 0.3-0.5; Rf= compound distance\/solvent front<\/li>\n\n\n\n
    2. In a mixture of polar and non-polar compounds, which one should have a higher Rf value?
      (3 pts.)
      Non-polar compounds generally have a higher Rf value than polar compounds because nonpolar molecules travel further as they do not hold onto the polar silica at the origin spot as
      well as polar molecules; the polar molecules will stick to it and not travel as far, thus having a
      lower Rf value than non-polar compounds.<\/li>\n\n\n\n
    3. If compound A has an Rf value of 0.3 and compound B has an Rf value of 0.5 in a
      particular solvent system, which one interacts strongly with the stationary phase? (3 pts.)
      Because the stationary phase is usually polar, it will interact strongly with the polar
      compound, i.e., compound A.<\/li>\n\n\n\n
    4. If a solvent system X shows 3 spots for an unknown mixture and another system Y shows 5
      spots which solvent system will you choose? (3 pts.)
      I would choose solvent system Y that shows 5 spots because it shows better separation than
      solvent system X.<\/li>\n\n\n\n
    5. If compound C has an Rf value of 0.1 and compound D has an Rf value of 0.12 in 10% EtOAc
      (ethyl acetate) \u2013 90 % hexanes mixture, you will have to run the TLC in a new solvent system
      to achieve a better separation. Among the following two solvents which one will you choose
      and why? (6 pts.)
      a. 100 % hexanes
      b. 20% EtOAc-hexanes mixture
      I would choose B (20% EtOAc-hexanes mixture) because A (100% hexanes) would cause
      the compounds to not travel very far and have no difference in distance.<\/li>\n\n\n\n
    6. Fill in the blanks with the appropriate information: (10
      pts.) Method A<\/li>\n<\/ol>\n\n\n\n
        \n
      • Mass of spinach leaves used 0.529 g<\/li>\n\n\n\n
      • Volume of DCM used 4 mL<\/li>\n\n\n\n
      • Volume of water used initially 3.9 mL<\/li>\n\n\n\n
      • Volume of water used to wash the organic layer 4.5 mL<\/li>\n\n\n\n
      • Volume and color of extract A 3.9 mL; color is a light green
        Method B<\/li>\n\n\n\n
      • Mass of spinach leaves used 0.521 g<\/li>\n\n\n\n
      • Weight of anhydrous MgSO4 0.506 g<\/li>\n\n\n\n
      • Weight of sand used 1.053 g<\/li>\n\n\n\n
      • Volume of acetone used 2 mL<\/li>\n\n\n\n
      • Volume and color of extract B 1.1 mL; slightly darker green than Method A<\/li>\n<\/ul>\n\n\n\n
          \n
        1. Report the composition of the solvent mixtures you prepared for TLC analysis below:
          (12 pts.)<\/li>\n<\/ol>\n\n\n\n
            \n
          • Mixture 1
            Solvent Volume % Composition
            Hexane 3 mL 50%
            Ethyl Acetate 3 mL 50%<\/li>\n\n\n\n
          • Mixture 2
            Solvent Volume % Composition
            Hexane 4 mL 67%
            Ethyl Acetate 2 mL 33%<\/li>\n\n\n\n
          • Mixture 3
            Solvent Volume % Composition
            Cyclohexane 3 mL 60%
            DCM 2 mL 40%<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

            Lab 4: Separation of Organic Compounds \u2013 ColumnChromatography CHEM 212 Organic Chemistry Laboratory Background:Chromatographic separation has an important role in organic synthesis laboratory.Column Chromatography is used to separate larger quantities of material than thin layerchromatography. The purpose of this experiment is to establish the identity of a unknowncompound (mixed with benzophenone) by separating of mixtures […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-117535","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/117535","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=117535"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/117535\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=117535"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=117535"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=117535"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}