Lab 4_TLC Report
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Thin Layer Chromatography of Analgesics – Final Report
Report (70 points)
Commercial tablet used:
You: unknown #1
Partner: unknown #2
1.
Draw a representative TLC plate showing each standard compound (caffeine, acetaminophen, ibuprofen, and aspirin (acetylsalicylic acid)). Calculate the R
f value for each compound. (Show calculations). (10 pts)
All of the above Rf values are reported under 60 ethyl acetate: 40 hexane mixed solvent ratio conditions. Rf = distance traveled by the compound / distance traveled by the solvent.
2.
Also draw a representative plate of your
unknown tablet. Calculate the R
f value for each component. (Show calculations). (5 pts)
All of the above Rf values are reported under 60 ethyl acetate: 40 hexane mixed solvent ratio conditions.
3.
What compound(s) were in each of the two
tablets in your group? (5 pts)
unknown #1 has acetaminophen (after conducting an experimental plate that limits the drug selection down to acetaminophen/ibuprofen/and unknown 1; upon seeing the similarities between acetaminophen and unknown 1 in Rf, conducting another co-spotting test, to see whether acetaminophen would map on to the spot of unknown 1 exactly will give us a more definitive answer)
unknown #2 has ibuprofen (both share similar Rf values when compared across trials, similar Rf means both compounds have similar polarity in the given solvent, suggesting that these two may have been similar compounds)
4.
Describe what happened with the different solvent combinations: 60:40 ethyl acetate:hexane, hexane, and ethyl acetate.
Explain qualitatively why
you got these results with the different solvents. (What properties of the solvents led to this behavior?) (15 pts)
If 60:40 ethyl acetate:hexane mixed solvent served as the baseline (where ibuprofen had an Rf value of 0.80). Then in ethyl acetate solvent alone, compounds like aspirin, ibuprofen, and caffeine traveled the furthest (for instance, ibuprofen had an Rf of 0.87). On the other hand, these compounds traveled the smallest distance in pure hexane (i.e. ibuprofen only had an Rf value of 0.17). We got these results due to the intermolecular forces between the compounds and the solvents. The 60:40 mixture provides a balance between polar and nonpolar characteristics as it combines ethyl acetate (polar) with hexane (nonpolar). The mixture's moderate polarity allows compounds like ibuprofen, aspirin, and caffeine to travel
intermediate distances, creating the optimal separation between spots. Since ethyl acetate is a polar solvent, it has a strong affinity (dipole-dipole) for polar compounds like ibuprofen, aspirin, and caffeine, enabling them to travel easily along the plate. However, this ease of movement may lead to poor separation and thus it would be hard to differentiate between different compounds by their polarity if many spots overlap. Hexane, on the other spectrum,
is too nonpolar and leads to little to no movement of the compound samples because it fails to break the dipole-dipole interactions between the relatively polar compound of the silica gel. Thus, it would also be cumbersome to differentiate between the slightest separation. As such, the optimal solvent to use in this experiment is 60:40 ethyl acetate:hexane mixture to allow for optimal separation, visualization, and collection of the Rf values.
5.
A TLC was run in 70:30 ethyl acetate:hexane for compounds C and D. Compound C has an R
f
of 0.5 and compound D has an R
f
of 0.3 in this solvent system. (10 pts)
a.)
Draw the TLC plate. Which compound is more polar? Why?
A higher Rf value represents that the compound is less polar because of its increased interactions with the mobile phase (ethyl acetate and hexane) than the silica gel (stationary phase). Thus, the compound that is the more polar will interact with the stationary phase more and thus move a shorter distance, which, in this case, is compound D because (0.3 < 0.5). b.)
You decide to run another TLC on compounds C and D in 30:70 ethyl acetate:hexanes on a silica gel plate. Would the R
f
values for the compounds increase or decrease? Why?
The Rf values for the compounds would decrease. Since ethyl acetate is more polar than hexane, by increasing the ratio of hexane to ethyl acetate, you are decreasing the polarity of the solvent. With more nonpolar hexanes, it’s less likely to interfere with the polar silica gel (stationary phase); thus, making it unlikely to displace any of the compounds that are bonded to
the silica gel. As such, by increasing the level of hexanes in the solvent, the compounds will travel significantly less, lowering the Rf value, because they are interacting more with the stationary phase (silica gel) than the mobile phase (30:70 mix solvent), especially for the polar compounds that have a higher affinity for other polar molecules. 6.
In a TLC experiment, why should the spot not be immersed in the solvent in the developing chamber? (5 points)
The spot should not be immersed in the solvent in the beginning because there is a higher chance of
the compound being dissolved into the solution at the bottom of the beaker rather than traveling upwards in the plate (controlled by the IMFs between the compound and the mobile and stationary phase). This creates contamination and produces inaccurate results.
7.
Explain why the diameter of the initial spot should be as small as possible. (5 points)
The diameter of the initial spot should be as small as possible because, with a smaller dot, it’s more clear (with a more defined shape) when you see spot separation as the smaller dots
only have to travel a little to see the distinction. If the spots were larger, their movement upwards may interfere with one another and potentially cause mixing or streaking due to the
high concentration of compounds. With a smaller spot, you will also have higher accuracy and precision in determining the center of the spot and thus produce a more accurate Rf value than if the spots were large.
8.
Why is it important to not let the solvent run off the top of the plate? (5 points)
If you let the solvent run off from the top of the plate, then you can’t measure the total distance the solvent has traveled since there is no final measurable position (solvent line).
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Related Questions
show the complete solution and write answers in 3 significant figures.
DYE
CALCULATED Rf
Tartrazine (yellow #5)
?
Sunset yellow (yellow #6)
?
Allura red (red #40)
?
Brilliant blue (blue #1)
?
M&M's color
COMPOSITION
Orange
?
Green
?
Blue
?
Brown
?
CHROMATOGRAM BEFORE: (image 1)
CHROMATOGRAM AFTER: (image 2)
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a
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b
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c
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d
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Answer:
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1.26% w/w hydrocortisone
Solution of the given answer is needed. Thank you so much and God bless. Happy new year
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Column Chromatography
Alumina
Chromatography Mixture
9:1 Hexanes:Ether
8:2 Hexanes:Ether
1:1 Hexanes:Acetone
Amount Used
3.962 g
0.143 g
9.50 mL
9.50 mL
11.00 mL
Additional Observations (Color, etc.)
BIU X₂ X² →
BI IU X₂ X² →
BI IU X₂ X² →
BI IU X₂ X² →
BIU X₂ X² →
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Given the following chromatogra
What molecules in the mixture were separated successfully in Chromatogram A? In Chromatogram B?
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Please correct answer and don't use hand rating
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The following data are from the HPLC assay of paracetamol 500 mg tablets using the
calibration curve method:
- Weight of active ingredient in powdered tablets taken for analysis: 180 mg
• Initial extraction volume: 200 mL
- Dilution steps: 10 mL into 100 mL, then 5 mL into 100 mL
- Linear equation: y = 12000x - 50
%3!
- Area of chromatographic peak for sample is 4750.
What is the theoretical concentration in final solution in mg/100 mL? (2
decimal places). Please do not write the units, just numerical value.
Answer:
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Table #2 Worktable for BCA Assay
Standard
Water (μl)
Tube #
1
2
3
4
5
6
7
8
9
0
125
325
175
325
325
325
400
400
BSA source (μl)
300 stock
375 stock
325 stock
175 of tube #2
325 of tube #3
325 of tube #5
325 of tube #6
100 of tube #7
0
Protein (ug)
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appropriate standard. Pipette each of the protein samples directly into the water, do not
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My Answer: The HPLC analysis detects chromophores, compounds with multiple double bonds. Since water, ethyl acetate, and hexane all lack multiple pi bonds this will not be detected in the HPLC.
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Question 3 (b)
Given the chromatogram shown below, answer the following questions:
(b) Calculate the adjusted retention time for compound/peak 3 (in minutes).
3.
10
0.
5
8.
10
Time (min)
HO
OH
Benzoic acid
Phenol
Toluene
Benzene
Detector Signal (mAU)
20
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SHOW SOLUTION PROPRELY
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10:36 1
ull ?
Search
Expert Q&A
Done
Sample G.C. :
Simple Distillation
Fraction 3
1500-
1000-
500-
Time (min)
AM Simple- F3
Sample GC of products obtained from Simple
distillation of solution that is 40% ethyl acetate and
60% propyl acetate.
calculate the following for each sample gas
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a) retention time for each peak (label peaks)
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QUESTION 5
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R = 0.74. It is an acceptable resolution.
R = 1.36. It is an acceptable resolution.
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一
4
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mAU
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200
100
0
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1.04
1.24
2.4 2.6
2.8
LO
34
12
ww
3.0
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3.2 3.4 3.6
3.8 4.0
Minutes
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9
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