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13 Dec 2019

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After weighing 25 microliters and 190 microliters (each in triplicate) with a P200, you found out that the standard deviation was smaller for the high volume measurements as compared to the lower ones. Therefore you would assume that your pipette is more accurate in its _________________ working range.

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Nelly StrackeLv2

17 Dec 2019

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I'm currently taking ochem part 2. I need to perform an acid-base extraction to separate out a mixture of solid organic compounds consisting of 50% 2-chlorobenzoic acid; 40% methyl 4-aminobenzoate; 10% 1,4-dibromobenzene. I need to create a procedure to first separate the compounds. I also need to recrystallize each compound after separating the compounds; therefore, I need to know which solvent to use for each compound. I also need to prove that I separated my compounds successfully using TLC (therefore I need to know which solvent to use for TLC). I need to do this as a micro-scale procedure (therefore, I will not be given large quantities of reagants). I'm given the reagants methanol, ethanol, methylene chloride, hexane, 1M NaOH, 6M NaOH, 1M HCl, 6M HCl, Anhydrous sodium sulfate, saturated sodium chloride, acetone, sodium bicarbonate, diethyl ether. I believe we will be given about 2 grams of the solid organic compound mixture.

Note: I want to run this acid-based extraction without using the separation funnel. Can I use a centrifuge tube instead?

I have a rough procedure of how to carry out the acid-base extraction that I've posted below, please look over my rough procedure of the acid-based extraction portion and tell me if the amounts of reagants look okay and if the process is okay. I also have accompanied this with a couple of follow up questions in regards to the acid-based extraction, recrystallization, and running the TLC all accompanied with subheadings.

OFFICIAL PROCEDURE:

Put 0.090g of solid organic mixture into a large test tube. Add 0.5 mL of diethyl ether at a time and stir until all dissolves. If needed, slowly add more diethyl ether solvent. Use minimal solvent.

Extracting methyl 4-aminobenzoate

1. Add this mixture in large test tube into a glass centrifuge tube (labeled âworking reactionâ).

2. Add 1mL 1.0 M HCl to the centrifuge tube. Shake gently and vent. Do this until layers separate. If needed, centrifuge.

Diethyl ether density: 0.713 g/cm3

Waterâs density: 1.00g/cm3 (aqueous layer)

Because diethyl ether has a lower density, it will be on the top layer and consist of the 1,4-dibromobenzene and the 2-chlorobenzoic acid. The bottom aqueous layer will consist of the conjugate acid of the organic base, methyl 4-aminobenzoate because H3O+ protonates the amine of the base to produce an ammonium salt, which is ionic and therefore aqueous in solution. (DRAW THIS OUT)

3. Extract out lower aqueous layer (consisting of the conjugate acid i.e. the ammonium salt, of methyl 4-aminobenzoate) into labeled test tube (ammonium salt).

Note: It is okay to extract a little more out from the bottom bleeding into the organic layer; throw that into the waste to ensure that all that is left are the other 2 organic compounds in diethyl ether. Parafilm.

4. Into the test tube with the ammonium salt, add 6.0 M NaOH DROP BY DROP until the solution returns to become BASIC, this has reproduced the organic base, methyl 4-aminobenzoate. Check with pH paper.

5. Collect the crude methyl 4-aminobenzoate precipitate by filtration using Buchner funnel.

6. Recrystallize from HOT WATER.

6. Filter the pure methyl 4-aminobenzoate precipitate by filtration using Buchner funnel.

Extracting 2-chlorobenzoic acid

1. The âworking reactionâ centrifuge tube now consists of the solvent (diethyl ether), 1,4-dibromobenzene, and 2-chlorobenzoic acid. Add 1 mL of 1.0 M NaOH. Shake and vent. Centrifuge if necessary. Again, as before, the lower layer is aqueous and the top layer is the organic layer (diethyl ether + 1,4-dibromobenzene)

NOTE: B/c NaOH was added, the 2-chlorobenzoic acid was deprotonated to yield a sodium salt, which produced its conjugate base. The salt is ionic and therefore soluble in water.

2. Extract out the bottom layer with Pasteur pipette (contains conjugate base of 2-chlorobenzoic acid) into test tube labeled sodium salt.

Note: It is okay to extract a little more out from the bottom bleeding into organic layer; throw that out into the waste to ensure that all that is left is 1,4-dibromobenzene in diethyl ether. Parafilm.

3. Into the test tube with the sodium salt, add 6.0 M HCl DROP BY DROP until the solution returns to become ACIDIC. This has now reproduced the organic acid, 2-chlorobenzoic acid.

4. Recrystallize with what solvent?

Extracting 1,4-dibromobenzene

1. The âworking reactionâ centrifuge tube should now only contain diethyl ether, 1,4-dibromobenzene with some potential water. Add sodium sulfate to dry out the remaining water. Extract the solution (leaving behind the sodium sulfate ppt) into a test tube labeled 1,4-dibromobenzene.

2. Place the test tube labeled â1,4-dibromobenzeneâ in a hot water bath and boil to that of the solvent (diethyl ether) at 40C.

3. Diethyl ether has a low boiling point of 34.6 C whereas 1,4-dibromobenzene has a high boiling point of 220.4C.

4. Therefore, the remaining organic layer in the âworking reactionâ centrifuge tube now only containing 1,4-dibromobenzene and diethyl ether can be separated by evaporating diethyl ether.

5. Recrystalize with what solvent?

Acid-Based Extraction Questions:

Q1: Is diethyl ether a good solvent for initially dissolving all three solvents for the acid-base extraction? And if so, how much solvent do I need to add? I thought minimal enough to dissolve the organic comopunds?

Q2: Are my rough amounts of reagants used correctly for the acid-base extraction?

Q3: Should I add anhydrous sodium sulfate to my last step of acid-base extraction which consists of the diethyl ether and 1,40-dibromobenzene because there might be some aqueous layer left?

Q4: Can I perform my acid-base extractions with a centrifuge tube instead of a separation funnel?

Q5: Should I start off with more grams of my organic mixture when I begin? Is 0.090 grams too small?

Recrystallization Question

Q6: Which solvents for each isolated crude compound should I use for recrystallization? Do I need to run a test for a suitable solvent? If so, which solvents are good to start with. I was thinking water and methanol.

a) I was thinking of using hot water as the hot solvent for methyl 4-aminobenzoate and for 2-chlorobenzoic acid. Is this correct to do so?

b) Which solvent should I also use to recrystallize 1,4-dibromobenzene?

TLC Question

Q7: Which solvent should I use to run the TLC for my organic compounds. After I know which solvent I will be using, where should I expect each compound to travel relative to the other organic compounds? (i.e. closest to the solvent line, middle of the solvent line, furthest from solvent line).

Miscellaneous Question

Q8: What's the point of telling me the percentages of the compounds in the mixture? Does that affect how much reagant I should be using during the acid-based extraction?

this is my chemistry lab discussion&conclusion
please help me revise it (mainly focus on conclusion ifpossible)
thank you!

Discussion
In the equilibrium experiment we were able to observe the directionof the equilibrium by looking at our chemical equation CH3COOH +C2H5OH <-> CH3COOC2H + H2O. The reaction in flasks b-eshifted left towards equilibrium, while flask f shifts to the righttowards equilibrium. Our Kc for flask b was 6.864, flask c was6.889, flask d was 5.695, and flask e was 98.73. Flask f was 6.833and our mean for our Kc was 25.003. The temperature for ourtitration was 22 degree Celsius.
For our titration for flask b we used 43.10mL of NaOH(equivalent to 0.043 moles of NaOH), flask c we used up 40.61 mL ofNaOH (equivalent to.04028 moles of NaOH), for flask d we used36.42mL of NaOH (equivalent to .03611 moles of NaOH), flask e weused 36.95mL of NaOH (equivalent to .03665 of NaOH), and flask f weused 22.10mL of NaOH (equivalent to .02192 moles of NaOH). We knowthat NaOH is one to one ration to HCl + acetic acid(CH3COOH). Since the moles HCl + moles acetic acid is equalto total moles of NaOH, then the moles HCl + acetic acid for flaskb-f is 0.043 moles, 0.04028 moles .03611 moles .03665 moles .02192moles. The moles of HCl in each flask were 0.01488. Therefore wewere able to subtract the total moles of NaOH with the moles HClleaving us with the final moles of acetic acid (CH3COOH). The molesacetic acid (CH3COOH) for flask b-f was 0.02787, 0.02540, 0.02124,0.02177, and 0.007042. The final moles of ethanol (C2H5OH) wereobtained by taking the initial moles of ethanol (C2H5OH) minus thechange of moles in acetic acid from initial to final. The finalmoles of ethanol (C2H5OH) in flasks b-f were 0.02787 moles, 0.02540moles, 0.03836 moles, 0.004309 moles, and 0.05808 moles. In orderto calculate the final moles of Ethyl acetate (CH3COOC2H) we had toadd the initial moles of ethyl acetate (CH3COOC2H) with the changeof moles of ethanol (C2H5OH) from initial to final. For flask b-fthe final moles of Ethyl acetate (CH3COOC2H) was 0.02319 moles,0.01544 moles, 0.01961 moles, 0.03654 moles, and 0.01042. In orderto obtain the final moles of H2O we had to add the initial moles ofwater (H2O) with the change of moles of ethanol (C2H5OH) frominitial to final. The final moles of water (H2O) are 0.2300 moles,0.2878 moles, 0.2366 moles, 0.2535 moles, and 0.2682 moles. Finallywe were able to obtain the equilibrium constant Kc. In thebeginning when we were calculating for the initial moles ofconcentrations we had to divide the mass percent by 100 to changeit back into its decimal form. So when we calculated Kc we had tomultiple each of the final moles by 100. We calculated Kc byeach of the final moles of products times 100 divided by each ofthe final moles of reactants times 100.
In conclusion our values of Kc were all between therange of 0.0001 to 1000. The data we recorded proves that they wereall at equilibrium. Our standard deviation for KcÂ¬ was 41.22percent. Our percent deviation was so high due to flask e which hada 98.73 for its Kc value. Knowing that our numbers are withinequilibrium range of 0.0001 to 1000, this proves that our Kc is valid. Some sources of errors were not accurately pipetting thecorrect of each solution into its proper flask. Other errorsinclude not accurately weighing out the KHP to its plus or minus0.0001 gram, contaminated beakers, flasks, pipet, burette, weightboat, and scoopula, also including not reading the exactmeasurement of NaOH titrated.

Everytime I make the graph based off my calculations in the chart it looks like this and I have no idea if it is even close? Please help ! For the questions I just need to understand how to start them and what info is nessiary to do so. very confused.

Data Analysis

Fill in the following table. Look below the table for instructions on how to calculate the values for each row of the table.

a	total volume (L)	0.065L
b	mol S₂O₃^2-_consumed	0.010 mol
c	mol I₃^-_produced	0.015M
d	[S₂O₃^2-] _consumed (M)	0.154M
e	[I₃-] _produced (M)	0.231M

Calculate the final volume of the reaction mixture after the contents of beaker B are added to beaker A. Report your answer in liters.

All S₂O₃^2- is consumed at the end of the reaction. Therefore, the moles of S₂O₃^2- consumed can be calculated using the equation below.
(moles S₂O₃^2-)_consumed = M_stock x (V_stock)

The I₃^- produced in reaction 1 reacts with S₂O₃^2- in reaction 2 as shown.
I₃^- (aq) 2S₂O₃^2- (aq) â 3I^- (aq) S₄O₆^2- (aq)

Therefore, for every mole of I₃^- produced, 2 moles of S₂O₃^2- have reacted.
(moles I₃^-)_produced = (moles S₂O₃^2-)_consumed / 2

[S₂O₃^2-]_consumed = (moles S₂O₃^2-)_consumed / (V_total)

Calculation is the same as d, but using I₃^- instead of S₂O₃^2-.

Fill in the following table. Look below the table for instructions on how to calculate the values for each column of the table.

	a	b	c	d	e	f	g	h	i
Trial	Ît	Î[I₃^-] / Ît (M/s)	log(Î[I₃^-] / Ît)	V 0.2 M KI added (mL)	[I^â]₀(M)	log[I^â]₀	V of 0.2 M (NH₄)S₂O₈ (mL)	[S₂O₈]₀ (M)	log[S₂O₈]₀
1	19	.011	-1.94	25	.077	-1.35	25	.077	-1.11
2	40	.0055	-2.24	25	.077	-1.11	12.5	.038	-1.42
3	85	.0030	-1.84	25	.077	-.1.11	6.25	.019	-1.72
4	42	.0055	-2.26	12.5	.038	-1.42	25	.077	-1.11

Conclusions

If a large amount of heat was released at the start of the reaction, what effect would this have on the rate measurements?

The rate of reaction would be much slower due to its exothermic reaction properties.

Click the box underneath the graph to show the trendline. It will automatically calculate your slope and intercept. Record them below.

Determine the value of q from the graph. Explain your answer.

What is the rate law for the reaction?

For each trial, calculate the rate constant. What is average value of the rate constant?

In this experiment, you assumed that [S₂O₈^2â] >> [S₂O₃^2â]. To find out if this assumption is correct, calculate the ratio of [S₂O₈^2â] / [S₂O₃^2â] for Trial 3. (In Trial 3, the [S₂O₈^2â] was lowest, and therefore the ratio [S₂O₈^2â] / [S₂O₃^2â] is the smallest).

In this experiment, you assumed that only a small amount of S₂O₈^2â was used during the time trial so that the concentration of this reactant did not change appreciably during the course of the reaction. Calculate how much S₂O₈^2â was used in Trial 3 and the percent remaining at the end of the reaction. Was this assumption valid?

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