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GROUP ASSIGNMENT (3)

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BY:

GROUP II

AMRI P. TAMBUNAN

HIDAYAT KESUMA

MASITHO PURNAMA SARI

SANSAH SINAGA

MATHEMATIC AND NATURAL SCIENCE OF FACULTY

STATE UNIVERSITY OF MEDAN

2011

QUESTION

1. Explain the extraction procedures by using separatory funnel is used to separate a
2. Explain the technique to be done if you want to separarate a mixture of benzoic acid and p-methoxy phenol when they are dissolved in dichloromethane
3. Ten grams of compound X is dissolved in 90 mL of water. The distribution coefficient for compound (X) between hexanes and water is 5 (K=5). How much of compound (X) will be in the hexanes if you extract it from the water with thre sequential extractions using 30 ml of hexanes extract ?
4. Consider the following solvent pairs. If mixed together, which pairs would form two layers? If they form two layers, which solvent would be on top?
1. hexanes and water
2. water and methylen chloride
3. hexanes and methylene chloride
4. methanol and hexanes
5. ethanol and water
6. acetone and toluene
5. You have 100 mL of water and add 4 mL of ether to it. You notice only one layer. Then, you add 6 mL more and see a small layer of a clear liquid on top of the water. Explain what is happening, give the reason.
6. You want to extract a compound from water, and have determined that the compound is equally soluble in both hexanes and diethyl ether. Which solvent would be the better choice for the extraction ? give the reason
7. The distribution coefficients for a compound are as follows K_cyclohexanes/water= 1.5; K_penatane/water= 11.2; and K_diethylether/water= 5.1. you have an aqueous mixture of the compound. Which solvent system, cyclohexane/water, pentane/water, or diethylether/water , would give the most efficient extraction into the organic solvent?
8. One gram of a compound requires the following quantities of solvent to dissolve : 47 mL of water 8.1 mL of chloroform, 370 mL of diethyl ether, or 86 mL of benzene. Calculate the solubility of the compound in these four solvents (as g/mL). estimate the partition coefficient of the compound between chloroform and water, ethyl ether and water, and benzene and water. Which slvent would you choose to extract the compound from an aqueous solution?
9. Five grams of compound A is dissolved in 90 mL of water. The distribution coefficient for compound A between hexanes and water is 5 Kcompound A_{(hexanes/water)} = 5
1. How much of compound A will be in the hexanes if you extract it from the water one time with 90 mL of hexanes?
2. How much of compound A will be in the hexanes if you extract it from the water with three sequential extractions using 30 mL of hexanes each time, and then combine the hexanes extracts?
3. Please make your conclution wheteher the extraction could be stop or has to be continue
10. You carefully purified phenanthrene from a very long, complicated procedures. Actually it took days. It is very important for you to hand the purified phenanthrene in to your Teacher in one hour to get a good grade. Without knowing, you accidentally put the phenanthrene into a vial which you thought was clean, but instead had a lot of NaCl clinging to the sides that contaminating the phenanthrene.
1. What can you do to quickly purify the phenanthrene again?
2. Explain the scientific reason to choose the technique to separate NaCl from phenanthrene?
11. A reaction work up for an aqueous reaction mixture calls for extraction first with diethyl ether and then a wash with saturated aqueous sodium chloride. What is the purpose of the saturated sodium chloride wash?
12. You have prepared an organic compound by reacting it with strong aqueous acid. The work up calls for extraction into methylene chloride, then a wash with 5 % NaHCO₃. What is the purpose of the NaHCO₃ wash?
13. What can you do if you do not know which layer is which in an extraction procedure?

SOLUTION

1. The extraction procedure :

1. Put the sep funnel in the ring on the ringstand.
2. Put the aqueous solution containing the compound you want into the sep funnel.
3. Add methylene chloride and shake.
4. Allow the layers to separate; the methylene chloride will be on botom and most of your compound will be in it, the aqueous layer is on the top.
5. Drain the lower methylene chloride layer and save it in a flask.
6. Leave the aqueous layer in the sep funnel and now add a fresh portion of methylene chloride to it.
7. Shake and then allow the layers to separate; the methylene chloride will be on bottom and most of your compound will be in it, the aqueous layer is on the top.
8. Drain the lower methylene chloride layer into the same flask as before (step 5) - this is where your compound is, in the methylene chloride.
9. You do not need the aqueous layer in the sep funnel anymore since most of the compound has been extracted into the methylene chloride.

2. Here is another mixture of benzoic acid and p-methoxyphenol, dissolved in dichloromethane:

NaOH was too strong a base, thus it does not differentiate the strong and weak organic acids. Use of weak inorganic base such as NaHCO₃ will differentiate between the compounds

Strong organic acids such as benzoic acid would be deprotonated and ionized, while weak organic acids such as phenols would NOT be deprotonated

Strong organic acids such as benzoic acid would be deprotonated and ionized, while weak organic acids such as phenols would NOT be deprotonated.

3. Ten grams of Compound A is dissolved in 90 mL of water. The partition coefficient for Compound A between hexanes and water is 5:

K_{Compound A(hexanes/water)} = 5

How much of Compound A will be in the hexanes if you extract it from the water with three sequential extractions using 30 mL of hexanes each time, and then combine the hexanes extracts?

K = 5 = x g in 30 mL hexanes/10-x g in 90 mL water

5 = 3x/10-x

x = 6.25 g

now, there is 10 – 6.25 g or 3.75 g of the compound in the water, therefore:

K = 5 = x g in 30 mL hexanes/3.75-x g in 90 mL of water

5 = 3x/3.75-x

x = 2.34375 g

now, there is 3.75 - 2.34375 g or 1.40625 g of the compound in the water, therefore:

K = 5 = x g in 30 mL hexanes/1.40625 -x g in 90 mL of water

5 = 3x/1.40625 -x

x = 0.879 g

Now, add the numbers in bold: 6.25 + 2.34375 + 0.879 g, and we get 9.47275 g.

4. It based on the immiscible

a. To hexanes and water: these compounds are immiscible so it will form two layers;

hexanes (d=0.68) will be on top

water (d=1.0)will be on the bottom.

b. To water and methylene chloride:these compounds are immiscible so it will form two layers;

water (d=1) will be on the top

methylene chloride (d=1.33) will be on the bottom.

c. To hexanes and methylene chloride: these compounds are miscible and it will not form two layers

d. To methanol and hexanes: these compounds are immiscible so it will form two layers;

hexanes (d=.68) will be on the top

methanol (d=.79) will be on the bottom.

e. ethanol and water: these solvents are miscible and will not form two layers

f. acetone and toluene: these solvents are miscible and will not form two layers

5. We know that Diethyl ether and water are called “immiscible”, immiscible is not an exact one. For example, Ether is soluble in water to a small extent (small amount of ether), as noted in the table in the middle of the miscibility section, 5.6 mL of ether is soluble in 100 mL of water. So, if only 4 mL of ether is added to 100 mL of water, we get that it will dissolve in the water. Once over 5.6 mL ether per 100 mL water is added, it will form a separate layer. Since ether is less dense than water, the ether will form a layer on top of the water.

6. The better choice is Hexanes because it is a lot less flammable than diethyl ether.

7. Look at the data here:

	given:	g/100 Ml	K*
water	1 g/47 mL	2.1
chloroform	1 g/8.1 mL	12.3	5.9
diethyl ether	1 g/370 mL	.27	.13
benzene	1 g/86 mL	1.2	.57

*K is calculated from K = C_{organic solvent/Cwater}

Chloroform would be the solvent of choice to extract the compound from an aqueous solution

Reason : because the partition coefficient for this pair of solvents is the highest.

8. The data can be summarized as follows:

	given:	g/100 mL	K*
water	1 g/47 mL	2.1
chloroform	1 g/8.1 mL	12.3	5.9
diethyl ether	1 g/370 mL	0.27	.13
benzene	1 g/86 mL	1.2	.57

1.water solubility = = 2.1

2.chloroform solubility = = 12.3

3.diethyl ether solubility = = 0.27

4. benzene solubility = = 1.2

*K is calculated from K = C_{organic solvent/Cwater}

Chloroform would be the solvent of choice to extract the compound from an aqueous solution because the partition coefficient for this pair of solvents is the highest.

9. Known : Five grams of Compound A is dissolved in 90 mL of water.

The partition coefficient for Compound A between hexanes and water is 5:

KCompound A(hexanes/water) = 5

Asked:

a) Compound A _(hexanes)=……..(if you extract it from the water one time with 90 mL of hexanes)?

Solution :

K = 5 = x g in 90 mL/5-x g in 90 mL

5 = x/5-x (90 mL cancels)

x = 4.2 g

b) Compound A _(hexanes) = ………(if you extract it from the water with three sequential extractions using 30 mL of hexanes each time, and then combine the hexanes extracts) ?

K = 5 = x g in 30 mL hexanes/5-x g in 90 mL water

5 = 3x/5-x

x = 3.1 g

now, there is 5 - 3.1 g or 1.9 g of the compound in the water, therefore:

K = 5 = x g in 30 mL hexanes/1.9-x g in 90 mL of water

5 = 3x/1.9-x

x = 1.2 g

there is 1.9-1.2 g or 0.7 g of the compound in the water, therefore:

K = 5 = x g in 30 mL hexanes/0.7-x g in 90 mL of water

5 = 3x/0.7-x

x = 0.44 g

Next, add the numbers in bold: 3.1 + 1.2 + 0.44 g, and you get 4.74 g.

c. Compare this value with the 4.2 g obtained with one extraction with 90 mL of hexanes (a, above), and you see why the efficiency of extraction is improved if you split the organic extraction solvent up into several portions and do multiple extractions.

10. You could purify the phenanthrene by adding water and hexanes because Phenanthrene is very soluble in organic solvents, like hexanes, while sodium chloride is soluble in water.Shaking, and allowing the layers to separate. Separate and save the organic layer containing the phenanthrene, throw away the aqueous layer containing the sodium chloride. Dry the organic layer to re-isolate the phenanthrene.

11. Saturated sodium chloride serves two functions in an extraction procedure. It pulls water from the organic layer to “dry” it, and it also helps force the organic compound into the organic layer. To some extent, the saturated sodium chloride will also pull inorganics from the organic layer into the aqueous layer.

12. The NaHCO₃ wash serves to neutralize the acid and to remove water-soluble polar compounds.

13. Drop a small amount of water into the neck of the separatory funnel. Watch it carefully: if it remains in the upper layer, that layer is the aqueous layer. If it sinks to the bottom of the upper layer to the interface between the two liquids, the bottom layer is the water layer.

If this does not work, remove a small amount of the top layer from the separatory funnel and place it in a test tube. Add a small amount of water to the test tube, mix it and allow it to settle: if you now see two layers, the top layer in the separatory funnel is the organic layer. If you see only one layer, the top layer in the separatory funnel is the aqueous layer.

1. Explain the extraction procedures by using separatory funnel is used to separate a substance

Answer:

a. Support the separatory funnel in a ring on a ringstand. Make sure stopcock is closed

b. Pour in liquid to be extracted

c. Add extraction solvent

d. Add ground glass stopper ( well grassed)

e. Pick up separatory funnel with the stopper in place and stopcock closed, and rock it once gently

f. Then, point the stem up and slowly open the stopcock to release excess pressure. Close the stopcock. Repeat the procedures until only a small amount of pressure is released when it is vented

g. Shake the separatory funnel vigorously

h. Vent frequently to prevent pressure buildup, which can cause the stopcock and perhaps hazardous chemicals from blowing out.

i. Let the funnel rest undisturbed until the layers are clearly separated

j. While waiting, Remove the stopper and place a beaker flask under the sep funnel

k. Carefully open the stopcock and allow the two layer to drain into the flask.

2. Explain the technique to be done if you want to separate a mixture of benzoic acid and p- methoxyphenol when they are dissolved in dichloromethane

Answer:

a. Don’t use NaOH because it will react with benzoic acid.

b. Use of weak acid like NaHCO₃ will differentative between the compounds

c. Strong organic acids would be deprotonated and ionized, while weak organic acids such as phenols would not be deprotonated

3. Ten grams of compounds (x) is dissolved in 90 mL of water. The distribution coefficient for compound (x) between hexanes and water is 5 (K=5). How much of compound (X) will be in the hexanes if you extract it from the water with three sequential extractions using 30 mL of hexanes each time, and then combine the hexanes extracts?

Answer:

Known: mass of compounds ( x ) = 10 g

Volume of water= 90 mL

K compound ( x ) = 5

Asked: amount of compound (X)….?

Solution: 90 mL = 3 times 30 mL

q =

q =

q = 0.047 %

amount of compound x that have extracted

m =

m= 0. 47 gram

4. Consider the following solvent pairs. If mixed together, which pairs would form two layers? If they from two layers, which solvent would be on top?

a. Hexanes and water

b. Water and methylene chloride

c. Hexanes and methylene chloride

d. Methanol and hexanes

e. Ethanol and water

f. Acetone and toluene

Answer

a. Hexanes and water will forms two layers because hexanes is organic solvent and soluble in water and water would be on the top layer.

b. Water and methylene chloride will forms layers because methylene chloride is organic solvent and soluble in water and water would be on the layer.

c. Hexanes and metheylene chloride will not forms two layers because both of them are organic solvents

d. Methanol and hexanes will not forms two layers because both of them are organic solvents

e. Ethanol and water will forms two layers because Ethanol is organic solvent and soluble in water and water would be on the top layer.

f. Acetone and toluene will not forms two layer.

1. The equation for the reaction of bromine and methanoic acid is

a. Suggest a way of keeping the concentration of methanoic acid virtually constant

b. The result of an experiment in which the concentration of bromine was monitored through out a reaction are shown in the table

Time ( s )	(/mol dm-3
0	0.01
25	0.009
50	0.008
75	0.007
100	0.0065
180	0.0050
240	0.0045
360	0.0030
420	0.0025
480	0.0020

Plot a graph of againts time show that this reaction is first order by working out three values for half life.

2. The half life of radioactive iodine 131 is 8.0 days what fraction of the initial amount of iodine 131 should be present in a patient after 24 days if none were to be eliminated through natural body process.

3. Explain why no external indicator is needed when titrating solution containing ethamidioated ions with potassium magnate (VII).

4. Calculated the volume of 0.02 mol dm^-3 potassium magnate (VII) solution need to completely oxidize 25.0cm³ if:

a. 0.01 moldm^-3 iron (II) sulphate solution

b. 0.2 moldm^-3 hydrogen peroxide solution

c. 0.1 moldm^-3 ethamidioate solution.

5. A 1,340 gr sample of iron are dissolved in dilute sulphuric acid to produce a solution of iron (II) sulphate this was titrated with 0.02 moldm^-3 potassium magnate (VII) solution and it was found that 28.75 cm³ was required. Calculated the percentage of iron in the iron ore.

Answer

1. Known : v olume bleach 2 cm³

0.05 moldm^-3

2.

Nt = 16.375 gr

3. Because the potassium manganate (VII) has ability of self indicating. So, there’s no needed to its solution.

4.

a. Known :

Concentration of KMnO₄ = 0.02 mol dm^-3

Volume = 25 cm³ = 25 x 10^-3 dm³

Concentration of FeSO₄ = 0.01 mol dm^-3

Solution :

Amount of KMnO₄ = 0.02 mol dm^-3 x 25 x 10^-3 = 5 x 10^-4 mol

From redox equation :

MnO₄^-_(aq) + 8H⁺_(aq) + 5Fe²⁺_(aq) à 5Fe³⁺_(aq) + 4H₂O_(l) + Mn²⁺_(aq)

1 mol of MnO₄^- will react with 5 mol of Fe³⁺

So, the amount of Potassium manganate (VII) = 5 . 10^-4 mol x 5 = 25 . 10^-4 mol = 2.5 x 10^-3 mol

4B. Known :

Concentration of KMnO₄ = 0.02 mol dm^-3

Volume = 25 cm³ = 25 x 10^-3 dm³

Concentration of H₂O₂ = 0.2 mol dm^-3

Solution :

Amount of KMnO₄ = 0.02 mol dm^-3 x 25 x 10^-3 dm³ = 5 . 10^-4 mol

From redox equations :

MnO₄^-_(aq) + 8H⁺_(aq) + 5O^-_(aq) à Mn²⁺_(aq) + 5/2O_2(aq) + 4H₂O_(aq)

1 mol of MnO₄^- will reacts with 5/2 mol of O₂

So, the amount of Potassium Manganate (VII) = 5 . 10^-4 mol x 5/2 = 1.25 x 10^-3 mol

4C

Known :

Concentration of KMnO₄ = 0.02 mol dm^-3

Volume = 25 cm³ = 25 x 10^-3 dm³

Concentration of FeSO₄ = 0.01 mol dm^-3

Solution :

Amount of KMnO₄ = 0.02 mol dm^-3 x 25 x 10^-3 = 5 x 10^-4 mol

From redox equation :

MnO₄^-_(aq) + 8H⁺_(aq) + 5C₂H₄O₂^-_(aq) à Mn²⁺_(aq) + 5/2 C₂H₄O_{2 (aq)} + 4H₂O_(aq)

1 mol of MnO₄^- will react with 5 mol of Mn²⁺

So, the amount of Potassium manganate (VII) = 5 . 10^-4 mol x 5 = 25 . 10^-4 mol = 2.5 x 10^-3 mol