Thursday, 23 May 2013

22:21 - No comments


EXPERIMENT 5: EXPERIMENT USING LIPID




INTRODUCTION
 Triglyceride is an ester composed glycerol and three fatty acids. Saponification number primarily serves to determine the proportion of fatty acid esters in the sample. In this experiment we prepare soap from vegetable oil. Vegetable oil are esters of carboxylic acids which have a high molecular weight and contain the alcohol, glycerol. Chemically, these fats and oils are called triglycerides. The principal acids in vegetable oils can be prepared from the natural triglycerides by alkaline hydrolysis (saponification). The saponification reaction is a base usually NaoH or KOH hydrolysis of triglycerids to make three salts (soap) and glycerol. The molecules crystallize differently depending on the base used. Soap is made by the saponification reaction. It is an exothermic chemical reaction which happens when fatty acids react with base. This process involves boiling the fats together with the base. The crude soap obtained from the saponification reaction contains sodium chloride, sodium hydroxide, and glycerol. These impurities are removed by boiling the crude soap curds in water and re-precipitating the soap with salt



MATERIALS:

Triglyceride sample: e.g. Coconut oil, corn oil, palm oil, margarine, butter
Solvent (1:1 ethanol/ether)
0.5M KOH
Phenolphthalein
0.5M HCl


 PROCEDURE



1.Saponification of tryglyceride
\



2. Application : making Soap



RESULT:


1.Saponification of triglyceride

Sample
Blank
(V HCl used in liter)
Sample (V HCl used in liter)
Mol for blank
Molarity x V HCl (blank)
Mol for sample
Molarity x V HCl (sample)
Mol of reacted KOH or MolKOH
Saponification number = MolKOH x 56.1 x 103 gram of fat
Palm
0.025
0.025
0.013
0.012
0.0003
14.0 g
Corn
0.026
0.024
0.013
0.012
0.0010
56.1 g
Sunflower  
0.027
0.022
0.014
0.011
0.0030
168.3 g

Table 1: Result of Saponification Number of Different Types of Oils

2.Application : Making Soap





DISCUSSION:

1.      Saponification of triglycerides

Triglycerides or triacylglycerols are composed of three fatty acids each in ester linkage with a single  glycerol. Since the polar hydroxyls of glycerol and the polar carboxylates of the fatty acids are bound in ester linkages, triacyl glycerols are non polar hydrophobic molecules, which are insoluble in water. Saponification is the hydrolysis of fats or oils under basic conditions to afford glycerol and the salt of the corresponding fatty acid.
The saponification number is the number of milligrams of potassium hydroxide required to neutralize the fatty acids resulting from the complete hydrolysis of 1 g of fat. It gives information concerning the character of the fatty acids of the fat. The longer the carbon chain, the less acid is liberated per gram of fat hydrolyzed. It is also considered as a measure of the average molecular weight or chain length of all the fatty acid present. The long chain fatty acids found in fats have low saponification value because they have a relatively fewer of carboxylic functional groups per unit mass of the fat and therefore high molecular weight. The principle in this experiment is fats (triglycerides) upon alkaline hydrolysis (either with KOH or NaOH) yield glycerol and potassium or sodium salts of fatty acids (soap).
In this experiment, there are 3 types of oil or fat are used such as palm oil, corn oil and sunflower oil. The result shows that sunflower oil has the highest saponification number. So that, it means that sunflower oil has the shortest chain of fatty acid and lower molecular weight. Corn oil is the shorter fatty acid chains and  the long fatty acid chains is the palm oil with high molecular weight. However, all the results obtained are differ from the theoretical value. The actual saponification number of these oils is shown in the Table 2:

Sample (Oil)
Saponification Number
Palm oil
196-205
Sunflower oil
188-194
Corn oil
188-193

Table 2: Actual Saponification Number

Actually, the highest saponification number should be the palm oil and followed by the sunflower oil and the lowest is the corn oil. From the saponification number, we will know that palm oil has the shortest fatty acid chains while the corn oil has the long fatty acid chains. In this experiment, the result obtained is differ from the theory due to error occurred while carrying out the experiment. The titration process may have contributed to an error for this result as the volume of HCl used in the titration might be excess after the indicator pink color turn to colorless.

Application : making soap

Making soap was a long process.  Since water and oil do not mix, this mixture had to be continuously stirred and heated sufficiently to keep the fat melted. Slowly a chemical reaction called saponification would take place between the fat and the hydroxide which resulted in a liquid soap.  When the fat and water no longer separated, the mixture was allowed to cool.  At this point salt, such as sodium chloride, was added to separate the soap from the excess water.  The soap came to the top, was skimmed off, and placed in wooden molds to cure.  It was often aged many months to allow the reaction between the fat and hydroxide to run to completion.  Poorly make soap could contain excess alkali and could dry and chap people's skin
Soaps have hydrocarbon and ionic ends. The hydrocarbon ends are hydrophobic and non-polar. The ionic ends are hydrophilic and polar. When both oil and water present and mix with the soap. The hydrophobic part of the soap mixes with the oil, while the hydrophilic part of the soap mixes with the water. The attraction of the polar end of the soap is strong enough to pull the grease molecules into the water. Therefore, the grease molecules associated with the non-polar ends are pulled into the water along with the soap molecules. The soap and grease combine and arrange in the form of micelles. Finally the micelles will disperse in the water.   

QUESTION:
1.      What is the relationship between saponification and phase (liquid / solid) of a triglyceride?

For any chemical reaction to occur, including saponification, the reactant should be in the same phase. Hydroxide will almost entirely exist in the aqueous phase, and the triglyceride will be almost entirely in the organic phase. Increasing stirring will create more interface for the reaction to proceed. As the reaction goes, glycerin and fatty acid salts will be produced, which can have appreciable solubility in both phases. So, the regents will increase in concentration in the phase most favored by the other as the reaction progresses. The higher the concentration of the reactants, the more opportunity they have to react and the faster the reaction will go.
2.      Why do triglycerides with longer fatty acids have a lower saponification number than those with shorter fatty acids?

Triglycerides are composed of three fatty acids linked to glycerol. The fatty acids may be saturated or unsaturated. Triglycerides with longer fatty acid have a lower saponification number because the number of carboxylic functional groups per unit mass is low. It results in a higher molecular weight. So, less volume of HCl is needed per gram of fat hydrolyzed. KOH volume to hydrolyzed the ester bond also less as the saponification number is low.

3.     Why is the difference in the molar amount of HCl used to neutralize the control and the amount of HCl used to neutralize the sample equivalent to the molar amount of KOH used to saponify the test sample?

Because one mole of HCl reacts with 1 mole of KOH. Thus the test sample will require more acid to neutralize it because it contains more alkali than the control.

4.   Why do soaps disperse grease?
Soap contain hydrophilic (‘water-loving’) and hydrophobic (‘water-hating’). The dispersing of grease of soap is determined by its polar and non-polar structures in conjunction with an application of solubility properties. The long hydrocarbon chain non-polar hydrophobic (repelled by water). The “salt’ end of soap molecule is ionic and hydrophilic (water soluble). When grease or oil (non-polar hydrocarbons) are mixed with a soap-water solution, the soap molecules work as a bridge between polar water molecules and non-polar oil molecules. Since soap molecules have both properties of non-polar and polar molecules, the soap can act as an emulsifier. An emulsifier is capable of dispersing one liquid into another immiscible liquid. This means that oil doesn’t naturally mix with water, soap can suspend oil/grease in such a way that it can be removed. Therefore, the soap will form micelles and trap the fats within the micelle. Since the micelle is soluble in water, it can easily be washed away.


CONCLUSION

In a conclusion, the highest saponification number indicates its fatty acid chain length in triglyceride is lower. Soap is produced by undergo the saponification reaction

REFERENCE











Tuesday, 14 May 2013

09:05 - 1 comment

Experiment 4: Experiment on Vitamin C


GUIDE 4: EXPERIMENT ON VITAMIN C

Introduction
We will measure the amount of vitamin C in many different types of foods. The chemical reaction we will use to measure the amount of vitamin C uses one of its functions in the body. Vitamin C involves in our cells oxidation-reduction reactions. Vitamin C can react with iodine. Therefore we will measure the amount of vitamin C by adding iodine to our food extracts until the vitamin C can bind no more iodine. Iodine in excess of the vitamin C will react with a starch solution you will add to the extract to produce a bluish-black color. The addition of a chemical to measure another chemical is called a titration.

1.             Measuring Vitamin C using starch-iodine test.
Materials
1.      Food sources of vitamin C: for example juices, extraction of plants, flowers, fruits, grains, and vegetables, vitamin C tablet or cooked/treated food sample (boiled/refrigerated/grilled)
2.      Starch solution (1%): Mix 1 g starch in 100 ml boiling H2O. Boil for one minute while stirring. Stir until completely dissolved (this solution will be cloudy).
3.      Iodine solution: Mix 0.6 g potassium iodide in 500 ml H2O. Mix 0.6 g iodine in 50 ml of ethyl alcohol. These two iodine solutions should be mixed well before combining. Combine the two iodine solutions and add an additional 450 ml of H2O.
4.      Hydrochloric Acid (HCl) 1 M, (5 ml)
5.      Blender
6.       Filter/ cheesecloth

Procedure

1.        Preparing the vitamin C extracts:
i. Chop food material into small pieces and place into blender.
ii. Add 100 ml of distilled water to the blender.
iii. Blend using the highest speed until the material is thoroughly ground.
iv. Strain the ground extract
v. Measure 30 ml of the strained extract into a 250 ml Erlenmeyer flask or beaker.
2.    Measuring vitamin C in the food sample:
i. Place 30 mL of the food extracts solution in a 250 ml flaskor beaker.
ii. Add 2 drops of the 0.1 M HCl to the flask.
iii. Add 5 ml of the starch solution to the flask.
iv. Fill a burette with the iodine solution.
v. Record the initial volume reading.
vi. Add the iodine solution in 1 ml increments to the flask while swirling the flask.
vii. Add iodine until the solution stays blue-black for 15 seconds.
viii. Record the volume reading on the burette.

3.    Comparing cooked food and raw food’s vitamin C
Does the way you prepare your food affect the vitamin C available to be ingested? Vitamin C is a water-soluble vitamin. Would cooking food by boiling in water affect the vitamin C content? If vitamin C is lost during the cooking process, where does it go? What types of experiments could you design to test your hypothesis? You will be testing your hypothesis to determine if vitamin C content is changed during cooking or if different ways of food preparation yield different amounts of vitamin C.

i. Food can be prepared according to your creativity. For example, you can boil or steam or place in a freezer. You can also prepare the food by different exposing time to heat etc.
ii. Chop food material into small pieces and place into blender.
iii. Obtain your data using the same method in previous section.
iv. Record the volume reading on the burette.
v. Compare the relative amounts of ascorbic acid present in the samples you are testing.
vi. Compare your results with those of other members of the class. What do the results show?

2.      Application: Magic Writing

Materials
Beaker
Iodine
Lemon/Lime juice
Notebook paper
Cup
Art brush

Procedure:
STEP A: IODINE SOLUTION
1. Pour 100 ml water into a 500ml-beaker.
2. Add 10 ml of Iodine to the water and stir.

STEP B:
1. Cut a section from the notebook paper.
2. The paper must fit inside a 500ml-beaker

STEP C: VITAMIN C SOLUTION
1. Squeeze the juice of the lemon/lime into another beaker

STEP D:
1. Dip the art brush into the lemon/lime juice
2. Write a message on the piece of paper.
3. Allow the juice to dry on the paper.
4. Submerse the paper in the iodine solution in the bowl.


 Result

PART: 1. Measuring Vitamin C using starch-iodine test.


Table 1

Sample: Broccoli
Ascorbic acid:
[1 mg/ml = 75.3 ml Iodine]


Condition
(Broccoli)

Reading I
(ml)

Reading II
(ml)
Average Vitamin C concentration in    30 g (mg/ml)
Vitamin C concentration in       100 g (mg/ml)
Fresh

48.9
41.2
0.60
2.00
Water Bath
(60 °C)
46.5
40.4
0.58
1.93
Boil
(150 °C)
38.7
33.4
0.48
1.60


Calculation:

Fresh Broccoli

Reading I = (1 mg/ml ÷ 75.3 ml) x 48.9 ml = 0.65 mg/ml
Reading II = (1 mg/ml ÷ 75.3 ml) x 41.2 ml = 0.55 mg/ml
Average vitamin C  in 30g = 0.60 mg/ml
Vitamin C in 100g = (0.60 mg/ml ÷ 30g) x 100 = 2 mg/ml
                       
Water Bath(60 oC)

Reading I = (1 mg/ml ÷ 75.3 ml) x 46.5 ml = 0.62 mg/ml
Reading II = (1 mg/ml ÷ 75.3 ml) x 40.4 ml = 0.54 mg/ml
Average vitamin C  in 30g = 0.58 mg/ml
Vitamin C in 100g = (0.58 mg/ml ÷ 30g) x 100 = 1.93 mg/ml

Boil 150oC(hot plate)

Reading I = (1 mg/ml ÷ 75.3 ml) x 38.7 ml = 0.51 mg/ml
Reading II = (1 mg/ml ÷ 75.3 ml) x 33.4 ml = 0.44 mg/ml
Average vitamin C  in 30g = 0.48 mg/ml
Vitamin C in 100g = (0.48 mg/ml ÷ 30g) x 100 = 1.60 mg/ml


PART: 2. Application: Magic Writing

The paper turns a blue-puple except where the message was written. The words are outlined by the dark background. This is due to the starch combines with the iodine forming iodine-starch molecules. These molecules are blue-purple in color. Vitamin C combines with iodine to form a colorless molecule. The area covered with lemon juice remains unchanged because the paper is coated with vitamin C from the lemon.

Discussion

Broccoli is a plant in the cabbage family, whose large flower head is used as a vegetable. Broccoli is high in vitamin C and dietary fiber. It also contains multiple nutrients with potent anti-cancer properties. 

Vitamin C is an organic molecule known as ascorbic acid. Vitamin C is a water-soluble vitamin that occurs naturally in many fruits and vegetables. It can be found in oranges, limes, lemons,  cabbage, strawberries, tomatoes, spinach, other leafy vegetables, or other citrus fruits. The indicator  was added to the sample, and the indicator will produce a dark blue color when the endpoint has been reached. The amount of iodine added from the burret will be an indication of the amount of vitamin C. Vitamin C is a sensitive molecule that is altered in the presence of heat.  It is not stable to heat, so cooking fruits and vegetables destroys much of their vitamin C content.

From the Table 1, fresh broccoli contains a high concentration of vitamin C which is 17.9 mg/ml. The lowest concentration of vitamin C is boil Broccoli at 150°C with vitamin C concentration is 14.37mg/ml. Heat decreases the vitamin C level. Vitamin C is damaged by heat by increasing its rate of oxidation.

The heat actually preserves vitamin C if the temperature is below 70 degrees Celsius, because the high heat used kills the enzyme ascorbic acid oxidase, found in fruits and vegetables, before much Vitamin C is oxidized. Since this enzyme catalyzes the oxidation process, this is actually serve to protect the vitamin. However, if the temperature rises above 70 degrees Celsius, then the vitamin C will be damaged.

Conclusion

Fresh Broccoli has the highest concentration of Vitamin C which is 17.9mg/ml. The lowest concentration of vitamin C is boil Broccoli at 150oC which is 14.37mg/ml.