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Name: ID: Date: Lab 14: Melting Points of Lipids Objectives Observe and predict melting points of fats based on their chemical structure Understand the relationship between intermolecular forces and phase changes . Background Lipids are biomolecules that are mostly nonpolar, therefore they are insoluble in water. This is a fairly general definition, and there are many different types of lipids including: triglycerides - involved in fatty acid metabolism membrane lipids - the main component of cell membranes in mammals steroids - involved in cell signaling and also a component of cell membranes vitamins - vital nutrients for cellular function . . . . Fatty acids are a component of both triglycerides and membrane lipids. They have a basic structure of a carboxylic acid attached to a long carbon chain (usually 10-24 carbons in length). They can be saturated (all C-C single bonds) or unsaturated (some C-C double bonds). O HO НО saturated unsaturated The main intermolecular forces (IMFs) that are present in lipids are dispersion forces. You may have heard that dispersion forces are weak, however this is not entirely true. Dispersion forces depend on a molecules surface area. If a molecule is large and has a large surface area, the dispersion forces it exhibits are rather strong. When IMFs are strong, molecules are highly attracted to one another. IMFs have an affect many aspects of a compound: chemical reactivity, solubility, and phase changes. We will explore the effect the strength of IMFs has on phase changes. Melting point is the normal temperature at which a substance changes from solid to liquid. The phase change from solid to liquid is endothermic; it requires an input of energy. How much energy depends on the attractive forces between molecules, the IMFs. If IMFs are strong, molecules are held together tightly and more energy will be required to undergo this phase change. More energy required leads to a higher temperature melting point. If IMFs are weak, molecules are not held as tightly together; less energy will be required in the solid to liquid phase change. Less energy required leads to a lower temperature melting point.
Date: Name: ID: Let's look at an example of two lipids, fatty acids. Fatty acid length of carbon chain melting point 0 12 43.2°C HO lauric acid 18 69.3°C HO stearic acid Lauric acid has a smaller surface area than stearic acid. We can use the length of the carbon chain to help in this determination. Stearic acid's larger surface area means it exhibits stronger dispersion forces, its molecules are highly attracted to one another and more energy will be required to move them apart in the solid to liquid phase change. Therefore, we expect stearic acid will have a higher melting point, and we see in the 3rd column of the table that it, in fact, does. Length of the carbon chain is not the only factor in determining the strength of dispersion forces in fatty acids. We also need to take into account if the fatty acid is saturated or unsaturated and the degree to which it is unsaturated. In today's lab you will determine the effect that double bonds have on the melting point of fatty acids and therefore on the dispersion forces. Materials hot plate 250 ml beaker test tubes test tube tongs thermometer butter vegetable oil (corn or soybean) olive oil coconut oil margarine ice Procedure 1. Prepare a hot water bath by filling a 250 ml beaker half full and setting it on a hot plate. Do not boil the water, you only need to bring it to about 75°C. 2. Clean, dry, and label five test tubes 1-5. 3. Obtain the fat and oil samples from the freezer or ice chest. All samples should be solid. 4. Place a small amount of the appropriate sample from the table below into the test tube. Be careful when adding the samples - you need to get the sample to the bottom of the test tube. Test tube Fat/Oil 1 butter 2 Crisco (hydrogenated soybean and palm oils) 3 coconut oil 4 margarine 5. Insert the thermometer into test tube 1. Using the tongs, place test tube 1 into the water bath. Observe the sample as it melts and record the melting point in your lab notebook. 6. Thoroughly clean the thermometer. 7. Repeat steps 5-6 on the remaining test tubes. 8. Discard of the fats and oils in the trash or as directed by your instructor.
9. When cleaning the test tubes, put a little soap in each and a small volume of water. Shake the test tube and pour out the small volume. Add another small volume of water and repeat until the test tube is clean and there is no greasy residue. This is a more efficient way to clean rather than filling the tubes over and over and rinsing with water. Remember, we are in a severe drought.
Fats/Oils Primary FA Structure Butter (Palmitic acid) ОН Soybean oil (linoleic acid) OH FO OH Olive oil (Oleic acid) Oleic Acid Coconut oil (lauric acid) OH Margarine (hydrogenated Oleic acid) OH
ID: Date: Name: Report Sheet Fat/Oil Melting point (°C) Butter 36 Soybean oil -16 Olive Oil -6 Coconut oil 24 43 Margarine
Post-Lab Questions 1. Rank the fatty acids tested in order of increasing melting point. 2. What effect does the presence of double bonds have on the melting point of a fatty acid? 3. What effect does the presence of double bonds have on the strength of dispersion forces exhibited by a fatty acid?
Pre-Lab Questions 1. Look up the primary fatty acids (FA) in each of the following fats or oils. Identify the length of their carbon chain. Identify them as saturated, monounsaturated, or polyunsaturated. a. butter i. primary FA: ii. length of carbon chain: iii. degree of saturation: b. olive oil i. primary FA: ii. length of carbon chain: iii. degree of saturation: C. soybean oil i. primary FA: ii. length of carbon chain: iii. degree of saturation: d. coconut oil i. primary FA: ii. length of carbon chain: iii. degree of saturation: e. margarine i. primary FA: ii. length of carbon chain: iii. degree of saturation:
Fats/Oils Primary FA Structure Butter (Palmitic acid) OH Soybean oil (linoleic acid) ОН OH Olive oil (Oleic acid) Oleic Acid Coconut oil (lauric acid) mo OH Margarine (hydrogenated Oleic acid) OH
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