lab report 03
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Nalo Russell
Experiment 3 - Caffeine from Tea
Fill in the blanks in the given template. Use underlines, boldface, or parentheses to clearly
indicate the filled content within the context. Type up the report and save it as one single file in
the PDF format. Submit the PDF file by the deadline.
Key Observations
-
The appearance of
7.993 g (amount) of calcium carbonate in
300 mL
(volume) of
water in a 600-mL beaker:
brown, foggy, liquid (color, clarity and form).
-
15
(number) tea bags were added into the above mixture. The appearance of the
mixture:
dark brown, cloudy, and liquid
(color, clarity and form).
-
The resulting mixture was heated at a temperature of
~100 °C for
20 min. (time).
-
The approximate volume of the final mixture right before being transferred into a
separatory funnel:
~200 mL. Its appearance:
dark brown, clouding, and liquid (color,
clarity and form).
-
The appearance of the aqueous layer in the separatory funnel after the first round of
dichloromethane extraction:
translucent, slightly cloudy, and liquid form
(color, clarity
and form). It is the
bottom
layer (top or bottom).
-
The appearance of the organic layer in the separatory funnel after the first round of
dichloromethane extraction:
translucent, slightly cloudy, and in liquid form
(color,
clarity and form). It is the
top
layer (top or bottom).
-
The appearance of the interface between the organic layer and aqueous layer in the
separatory funnel after the first round of dichloromethane extraction:
distinct layers,
translucent, and sharp form (color, clarity and form).
-
The appearance of the combined organic layer before being dried with Na
2
SO
4
:
cloudy
liquid
(color, clarity and form). It was then dried with
~ 4 grams of Na
2
SO
4
for
20
min. (time).
-
Change of the appearance of Na
2
SO
4
(size and form) throughout the treatment:
from fine powder/ granular to larger, clumped/aggregated particles. .
-
The appearance of the resulting organic layer after being dried with Na
2
SO
4
:
colorless
(faint yellow), the clarity is clear and in liquid form
(color, clarity and form).
-
The Na
2
SO
4
was removed from the organic layer by
filtration .
-
The crude caffeine was obtained from the organic layer by using a
rotary evaporator
to
remove the organic solvent.
-
The crude caffeine was transferred to a tared vial with the aid of
~ 20 mL of
dichloromethane.
-
Crude caffeine (dried): amount
0.443 g , color
tan , form (and shape)
white
crystalline powder.
-
Power of the heating mantel used during the sublimation of the crude caffeine: between
40 - 65 %.
-
Sublimed caffeine: amount
0.8 g , color
yellow(ish) tan, form (and shape) powder
some parts were granulated , mp
222 - 227 °C.
Results (Show calculations. Assume 45 mg caffeine per tea bag. Apply the significant figure
rules)
-
Theoretical yield of caffeine (mg):
675 mg
-
Percent yield of the crude caffeine (percent crude yield = amount of the crude/ theoretical
yield x 100%):
6.562 %.
-
Percent yield of the purified caffeine (percent purified yield = amount of the purified /
theoretical yield x 100%):
1.1852 %.
-
Percent recovery yield of the sublimation (= amount of the sublimed / amount of the
crude x 100%):
18.0587 %.
Discussion
Suggest two major causes including one unavoidable cause for the low recovery yield of the
crude caffeine from tea.
(1) Incomplete extraction: caffeine extraction from tea leaves may not be thorough due to
limitations in the extraction method or conditions, such as insufficient extraction time or
improper temperature control. This may have caused a loss of material through
(2) Loss during purification: during the purification process, there could have been a loss of
caffeine due its solubility in the solvent, absorption into purification apparatus, or a loss
of material through transfer during the procedure.
Comment on the identity and purity of the purified caffeine based on the melting point of your
product in comparison to the melting point reported in the literature.
The identity and purity of the purified caffeine can be assessed by comparing its melting point to
the theoretical value. My recorded melting point was 222 - 227 °C, which is significantly lower
than the expected value of 236- 238 °C. Consistency in the melting point indicates the absence of
impurities. However, my melting point deviates significantly from the expectation, which
indicates the presence of impurities or a different crystalline compound within the caffeine.
Suggest two
practical
things you could do to improve the recovery yield and/or purity of the
caffeine without making any major change to the experimental procedure.
(1) Optimize extraction conditions: you can better optimize the extraction parameters with
the type of solvent, temperature precision, and extraction time. Experimenting with
different solvents, adjusting the temperature, and extending the extraction time can
possibly enhance the efficacy of caffeine extraction from tea leaves without altering the
fundamental extraction technique. For instance, using a solvent with a higher affinity for
caffeine or increasing the extraction temperature slightly within safe limits can improve
the overall yield without significant changes to the procedure.
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(2) Implement additional purification steps: incorporating additional purification steps, such
as recrystallization or possibly chromatographic methods, can help enhance the purity of
the recovered caffeine. Recrystallization involves dissolving the crude caffeine in a
minimum amount of hot solvent and then allowing it to slowly cool and crystallize. In
this way, promoting the formation of pure caffeine crystals while leaving the impurities
behind. Additionally, chromatographic methods like column chromatography or
thin-layer chromatography can further separate caffeine from impurities based on tier
different affinities for the stationary [hase, improving the purity of the final product
without altering the experimental protocol.
Post-Lab Questions for Experiment 3
1. n-Butanol is water soluble. But if you add salt to an aqueous solution of n-butanol, you can
extract the n-butanol into dichloromethane. Explain how this works.
When you add salt to an aqueous solution of n-butanol, the water becomes more polar due to the
presence of ions from the salt. This increased polarity of the water phase makes it less favorable
for nonpolar compounds, like n-butanol, to remain dissolved in it –– since like dissolves like. As
a result, the n-butanol molecules tend to aggregate and separate from the water phase.
Dichloromethane is an organic solvent that is immiscible with water and has a high affinity for
nonpolar compounds. When you introduce dichloromethane to the salted aqueous solution
containing n-butanol, the n-butanol molecules divide with a preference in the dichloromethane
phase due to their nonpolar nature. This partitioning occurs also because like dissolves like ––
where the nonpolar n-butanol molecules are more soluble in the nonpolar dichloromethane than
in the polar water phase.
2. In extraction of an aqueous mixture, an experienced organic chemist would prefer to use a
“heavier” organic solvent such as methylene chloride (density > 1) over a “lighter” solvent
such as ethyl acetate or diethyl ether (density < 1). This often has little to do with distribution
coefficient, volatility, miscibility, or reactivity. The chemist simply wants the organic layer as the
bottom layer in a separatory funnel. Provide two practical advantages of having the organic
layer as the bottom layer when conducting Experiment 3 - Caffeine from Tea.
Having the organic layer (methyl chloride) as the bottom layer in a separatory funnel during
experiment 3 can offer practical advantages:
(1) Ease of separation: with the organic layer at the bottom, it is easier to drain and separate
it from the aqueous layer. Gravity aids in the separation process, allowing for a cleaner
and more efficient transfer of the organic phase to another container. This ultimately
minimizes the risk of contamination or loss of the desired product during transfer.
(2) Safety considerations: methylene chloride, being more dense than water, tends to remain
at the bottom layer, resulting in the reduction of the likelihood of accidental spills or
splashes during handling. Overall, this improves safety in the laboratory by minimizing
the potential for exposure to hazardous chemicals. Also, if any water accidentally gets
into the organic layer, it will remain as a distinct layer at the top, making it easier to
detect and remove before continuing with the experiment.
3. The rotavap is not removing the methylene chloride (bp = 39.6°C) from your solution. This
often can be attributed to the problem of the vacuum and / or heating of the rotavap. Provide a
list of things to check for diagnosis of the cause of the problem.
If the rotavap is not effectively removing the methylene chloride from the solution, several
factors related to the vacuum and heating of the rotavap should be checked for diagnosis:
(1) Vacuum pump operation: ensure that the vacuum pump is functioning correctly. Check
for any leaks in the vacuum tubing, connections, or seals. Inspecting the vacuum gauge to
confirm that the necessary vacuuming level is being achieved in important for
functionality
(2) Vacuum control settings: verifying that the vacuum control settings are set appropriately
in order to optimize solvent evaporation.
(3) Rotavap bath temperature: check the temperature of the water bath in the rotavap. Ensure
that it is set to an appropriate temperature for the boiling point of methylene chloride.
Adjust the bath temperature as needed in order to facilitate solvent evaporation without
overheating.
(4) Bath fluid circulation: ensure that the bath fluid is circulating properly around the
condenser to efficiently cool the vaporized solvent. Checking for any blockages in the
circulation system that may impede heat transfer is important for proper function.
(5) Condenser coolant temperature: Verify that the coolant flow to the condenser is sufficient
and that the temperature of the coolant is low enough to condense the methylene chloride
vapor efficiently.
By systematically checking these factors, you can diagnose the cause of the problem with the
solvent removal in the rotavap and take appropriate corrective actions to improve its
performance.
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[References)
Use the References to access important values if needed for this question.
Taking logarithms and antilogarithms is necessary to solve many chemistry problems. For practice, complete the following table, where N is a number.
log N
7.91
1.778
-0.054
Submit Answer
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Save and Exit
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terface/acellus_engine.htm?ClassID=1467683063
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Et
onal Academy of Science. All Rights Reserved.
Mass Fe
(grams)
15.55
20.98
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Enter the answer that belongs in the green box.
Iron Ratio Cmpd 2
Mass O
(grams)
4.45
9.02
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Ksp = []^[]^
(b) Al2S3 (s)
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201591..docx
LAB EXP. #3 -..docx
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Substance
Physical Appearance
Odor
Density (g/mL)
Reacts with NaHCO3
1. Benzaldehyde
Clear liquid or almost pale yellow
It smelled like almonds
1.00
No, significant reaction
2. MTBE
Colorless
Fruity smell to it
1.236 -1.247
No chemical reaction occurred
3. Urea
White solid
A very strong order of ammonia
0.791
No reaction occurred
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Colorless
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1.335
No, did not react with sodium bicarbonate
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2NO(g) + 2H₂(g) →→→ N₂(g) + 2H₂O(l) + energy
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When the reaction takes place energy is
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Use the References to access important values if needed for this question.
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* Question Completion Status:
QUESTION 10
Name the following compounds:
1
H₂O
!
KF
Na2CO3
H₂SO4
PH3
SF6
SO3
Ca3(PO4)2
J
CO
Cu(NO3)2
N₂05
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
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