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Chemistry
Date
Dec 6, 2023
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Leuna Sarah CHEM 223
Crystallization Experiment
Lab Partner: Benjamin
Introduction:
Crystallization is a cost-effective and efficient method for purifying solids, but it requires patience
and keen observation. It's both an art and a science, with finding the right solvent often being a
trial-and-error process. A good solvent dissolves the substance when hot and sparsely when
cold. For mixed solids, determining the right amount of solvent is crucial, as too much can
prevent precipitation. Crystallization with a solvent pair involves dissolving in a highly soluble
solvent, slowly adding a less soluble one, and cooling for crystallization. While a single solvent
is ideal for reproducibility, mixtures are often necessary.
Chemical and Equipment:
hot plate, 125 and 250 ml erlenmeyer flasks, Büchner funnel, filter paper, spatula, Norit, 250 ml
beaker, boiling chips, ice bath, glass rod, unknown compound 19, melting point apparatus,
capillary tubes, thermometer, glass pestle, watch glass and salicylic acid.
Procedure:
We started by obtaining a sample of the unknown impure mixture assigned to us and then
measuring its melting point. To prepare for recrystallization, we set up the required equipment in
the fume hood. We boiled approximately 150 mL of water, as it may take some time. We took
5-10 mL of this hot solvent and placed it in a flask on a hot plate for heating. We weighed out
around 2 g of the impure unknown and put it in a 125 mL Erlenmeyer flask, recording the weight
as precisely as possible. We gradually added the hot solvent to the flask, stirring continuously
and using the minimum amount required by paying close attention to any changes in the solid
content. We removed it from the hotplate, added a small amount of decolorizing carbon
(Norite®), and gently swirled the solution. We boiled it slowly for about 5 minutes, then filtered
the hot solution using a preheated funnel and filter paper. W kept the solution and apparatus hot
throughout the filtration. We added slightly more water than necessary to prevent premature
crystallization. We covered the flask with a watch glass, allowing it to cool gradually, first at room
temperature and then in an ice bath. Once crystallized, we collected the crystals carefully in a
Büchner funnel on filter paper, rinsed the flask, and washed the crystals with cold water. Suction
was applied, followed by pressing the crystals with a clean glass stopper. Finally, we heated the
crystals in the oven for a thorough drying process. We measured the weight, yield, and melting
point of the recrystallized material after it has been dried. We used the melting point range and
the anticipated melting points to determine the identity of the unknown compound 19 to be
salicylic acid. For absolute confirmation of the compound's identity, we performed a 1:1 mixed
melting point test by comparing it with a known reference sample of salicylic acid.
Results:
Mass of impure mixture: 1.92 g
Melting point of impure mixture: 150-155
℃
Mass of recrystallized pure compound: 0.92 (total mass) - 0.34 (filter paper mass) = 0.58 g
Melting point of pure compound: 161-162
℃
(Trial 1) and 160-161
℃
(Trial 2)
1:1 mix with salicylic acid melting point: 162-163
℃
Unknown compound 19 is therefore salicylic acid.
% recovery = (0.58/1.92) x 100 = 30.21%
Discussion:
The laboratory procedure was straightforward and hands-on, yet we meticulously observed
each stage. Crystallization is a purification technique used in chemistry to obtain pure solid
compounds from a solution. It involves dissolving an impure solid in a hot solvent, removing
insoluble impurities, cooling the solution to allow crystals to form, collecting and washing the
crystals, and finally drying and analyzing the purified compound. The choice of solvent and
careful temperature control are critical for the success of the process, which is commonly used
for isolating and purifying chemicals in laboratory settings. One intriguing insight gained from
this lab was the significance of the interaction between the solvent and the solid during the
crystallization process. Grasping this aspect proved pivotal in achieving favorable outcomes and
successfully completing the crystallization procedure.
However, like many experiments in science, ours was not completely perfect either. We
accomplished a low percent recovery of our unknown compound after the recrystallization to
remove impurities. Thus a few possible reasons for this occurrence could be: the solid did not
fully dissolve in the hot solvent, not all of the impurities were removed during recrystallization,
some of the solid was lost during the filtration step if the filter paper was not properly fitted or if
the vacuum filtration was not done properly, the solvent may have evaporated, the solution was
not cooled slowly or allowed to sit for a sufficient time, mistakes in measurements, calculations,
or handling of the equipment, solid lost during the transfer of materials between containers or
during washing steps and our smaller crystals may have be more challenging to collect and
handle, leading to lower yields.
To improve the recovery in recrystallization, it's essential to carefully consider each step of the
process, including the choice of solvent, temperature control, filtration techniques, and handling
of the crystals, to minimize the potential sources of loss or inefficiency. Consequently this was a
great learning process and experiment.
Post-lab questions:
Pavia:
1a. Graph the solubility of A versus temperature. Use the data given in the table. Connect the
data points with a smooth curve.
1b. Suppose 0.1 g of A and 1.0 mL of water were mixed and heated to 80°C. Would all of
substance A dissolve?
Yes. Because at 80 °C 0.17 g of A is dissolved in 1.0 ml of water. So, 0.1g of A dissolved in 1.0
ml of water is completely dissolved when heated at 80 °C.
1c. The solution prepared in (b) is cooled. At what temperature will crystals of A appear?
Solubility of A of 0.1g is near 56 °C on the curve. Therefore, below 56 °C, crystals should start
appearing.
1d. Suppose the cooling described in (c) were continued to 0°C. How many grams of A would
come out of solution? Explain how you obtained your answer.
According to the graph, 0.1 g is soluble in 1.0 ml of water at 80 °C and at 0 °C, 0.015 g is
soluble in 1.0 ml of water. If the solution is cooled from 80°C to 0 °C, then we will get: 0.1g -
0.015g = 0.086g of A from the solution.
2. What would likely happen if a hot saturated solution were filtered by vacuum filtration using a
Büchner funnel? (Hint: The mixture will cool as it comes in contact with the Büchner funnel)
The solution becomes oversaturated and forms solid precipitates, causing it to become trapped
at the upper part of the Büchner funnel.
6. A solid substance A is soluble in water to the extent of 10 mg/mL of water at 25°C and 100
mg/mL of water at 100°C. You have a sample that contains 100 mg of A and an impurity B.
a. Assuming that 2 mg of B are present along with 100 mg of A, describe how you can purify A if
B is completely insoluble in water. Your description should include the volume of solvent
required.
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Since B cannot be dissolved, we can try dissolving substance A and filter out substance B using
a filter paper. At 100°C, substance B has a solubility of 100mg/mL and requires only 1 mL of
water as a solvent to dissolve substance A. Then substance A can be cooled to form crystals.
b. Assuming that 2 mg of the impurity B are present along with 100 mg of A, describe how you
can purify A if B has the same solubility behavior as A. Will one crystallization produce pure A?
(Assume that the solubilities of both A and B are unaffected by the presence of the other
substance.)
Since there is less of B than A, and both have the same solubility, we can try dissolving both
substances in 1mL of water at 25°C. This should dissolve all of substance B and only 10% of
substance A since the solubility is 10g/mL at 25°C. So one crystallization should be enough to
produce crystal growth of A since all of B has been removed through the use of this method.
c. Assume that 25 mg of the impurity B are present along with 100 mg of A. Describe how you
can purify A if B has the same solubility behavior as A. Each time, use the minimum amount of
water to just dissolve the solid. Will one crystallization produce absolutely pure A? How many
crystallizations would be needed to produce pure A? How much A will have been recovered
when the crystallizations have been completed?
The method from question B can be used here. 1 mL can be used to dissolve 10 g of A and 10
g of B which would result in remaining 90 g for A and 15 g for B. Through repeating the same
method from question B, 80 g of A and 5 g of B can be retrieved. Finally, repeating the method
again will give 70 g of A and 0 g of B resulting in pure A. It will require more than one
crystallization process to produce pure A as the process has to be done more than once to
remove impurities and to make sure that all of substance B has dissolved before producing
crystals.
Lab manual:
The solubility of benzoic acid in water is 6.8 g/100 mL at 95 oC and 0.3 g/100 mL at 25 oC. You
are given 10 g of benzoic acid. Calculate the amount of water that you would use to recrystallize
the sample. If the crystals are collected at 25 oC what is the maximum possible recovery?
Max % recovery = (10g – 0.441g) / 10g x 100 = 95.6%
Maximum possible recovery = 9.569 g
Pre-lab:
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