Honors Chemistry Polar vs. Non‐polar Solvents Name ____________________
Period _____ Introduction: You have learned that covalent bonds are formed by the sharing of electrons. The sharing can be equal or unequal. This depends on the electronegativities of the atoms involved. The bonds may be polar or non‐polar. Molecules may be polar or non‐polar depending on the types of bonds and the arrangement of the bonds, that is, the shape of the molecule. If all the bonds in a molecule are non‐polar, the molecule is definitely non‐polar. If all of the bonds in a molecule are polar, then the arrangement of the bonds determines the polarity of the molecule. If the bonds are arranged symmetrically, the electron cloud in the molecule will be evenly distributed and the molecule will be non‐polar. If the bonds are asymmetrically arranged, the electron cloud in the molecule will not be evenly distributed. One end of the molecule will be slightly negative, the other end will be slightly positive, and the molecule will be polar. In this activity, you will carry out a simple test for a polar and non‐polar liquid. Then in Part B you will use these substances to study the solubility of polar and non‐polar substances. Part A Materials burettes (2) animal fur water beakers (2) petroleum ether ring stand plastic wrap glass rod ebonite rod buret clamp Procedure 1. Prepare a table like the one shown below to record your observations. Test Water Pet Ether glass ebonite 2. Set up the apparatus as shown in the figure below. 3. a. Rub the glass rod with plastic wrap to produce a positive charge on the rod. b. Open the burette containing water to allow a fine stream of water to run into the beaker. c. Put the glass rod near the stream of water. Record your observations. 4. Repeat step 3 using an ebonite rod rubbed with animal fur to produce a negative charge on the rod. 5. Repeat steps 3 and 4 but this time using the burette containing petroleum ether instead of water in 3b. Part A. Questions and Conclusions 1. Which substance, water or petroleum ether, is more affected by the charged rods? 2. Which is the polar substance? 3. Explain why the polar substance is affected in a similar way by differently charged rods. 4. What two factors determine the polarity of a molecule? 5. Explain how a compound with all polar bonds could be a non‐polar molecule. Part B The polar or non‐polar nature of a molecule has an important influence on its behavior. For example, a common ionic solute like salt will dissolve readily in some solvents but not in others. What factors determine whether a solute will or will not dissolve in a particular solvent? You may have noticed that some paints are oil‐
based and some are water‐based. Water is used to clean brushes and rollers used with water‐based paints. Paint thinner is used for oil‐based paints. Both water and paint thinner are solvents. One is polar; the other is nonpolar. In this next activity, you will try to dissolve a polar compound, glycerol, and a nonpolar molecule, iodine, in different solvents. The solvents you will use for most of the activity are water, which is polar, and petroleum ether, which is non‐polar. A solid solute that dissolves in a solvent is said to be soluble. A different term is used when the solute is liquid. A liquid solute that dissolves in a solvent is said to be miscible. Miscible means that the solute is soluble, or completely mixed, in the solvent. A liquid solute that is not soluble in the solvent is said to be immiscible. At the end of this activity, you will use an emulsifying agent to make an immiscible mixture miscible. The resulting mixture is called an emulsion. Materials test tubes water test tube rack glycerol stoppers unknown liquids medicine dropper vegetable oil petroleum ether vinegar solid iodine egg yolk Procedure Part 1 (1st data table) 1. Prepare a data table for Part B. 2. Fill four small test tubes one‐fourth full of petroleum ether. Label the test tubes. 3. Fill four small test tubes one‐fourth full of water. Label the test tubes. 4. Add one or two small iodine crystals to one test tube of petroleum ether and to one test tube of water. Stopper the test tubes and shake well. 5. Observe carefully to see whether the solute, the iodine, dissolves in either or both of the solvents. 6. Repeat steps 4 and 5 with glycerol instead of iodine. 7. Repeat steps 4 and 5 with the unknown liquids, #1 and #2, instead of the iodine. Use about a half dropperful of unknown. Part 2 (2nd data table) 8. Add one‐fourth of a test tube of vinegar to one‐
fourth of a test tube of vegetable oil. Stopper the test tube and shake well. Observe whether the two are miscible or immiscible. 9. Add a half dropperful of egg yolk to the mixture from step 8. Stopper the test tubes and shake well. Observe whether the two are miscible or immiscible. Part B. Questions and Conclusions 1. Iodine molecules are non‐polar. In which solvent was the iodine soluble? 2. Glycerol is polar. In which solvent did glycerol dissolve? 3. Are unknowns #1 and #2 polar or non‐polar? What evidence do you have? 4. In what type of solvents are polar solutes most soluble? 5. In what type of solvents are non‐polar solutes most soluble? 6. What can you conclude about the solubility of polar or non‐polar solutes in polar or non‐polar solvents? 7. Were the oil and vinegar miscible or immiscible when they were first mixed? 8. Were the oil and vinegar miscible or immiscible after the egg yolk was added? What food is made this way – egg yolk holding vinegar and oil together? 9. What is the emulsifying agent in this activity?