Solubility Of Organic Chemistry Lab Essay

Take two test tubes, label each test tube according to solvent. Add 1 mL of distilled deionized water to the first test tube and 1 mL of ethanol to the second test tube. Next, add 1 mL of the unknown liquid to each test tube, shake for a small period of time, and observe patiently. During this observation, you will be able to determine whether the two liquids mix completely, slightly, or not at all. If the two liquids mix completely, then you should be able to see one liquid mixed together without a visible line indicating two different substances. This means the two liquids are considered soluble. If the two liquids

As indicated in Table 1, each of the given substances for this lab were soluble in ethyl acetate. However, the solubility results varied based on the structure of each tested compound. 4-bromoacetanilide and fluorene are both insoluble within solvents of HCl and NaOH. Fluorene will stay soluble within the organic layer since its is a purely made of carbon and hydrogen. 4-bromoacetanilide will act similarly given its nonpolarity from balanced branches and lack of acid/basic characteristics. 4-nitroaniline and 2,5-dichloroaniline travel solubly into the HCl/H2O solvent when added to the separatory funnel, yet form a precipitate after a highly concentrated NaOH is added. Both 4-nitroaniline and 2,5-dichloroaniline are slightly basic organic compounds, and when an acid (HCl) is added a proton will transferred to make them ionic. In their ionic state, both compounds become water-soluble. When NaOH is added to 4-nitroaniline and 2,5-dichloroaniline, it removes the proton causing the compounds to become insoluble in water once again. The same concept can be applied in an opposite fashion to the acid organic compounds of 3-chlorobenzoic acid and resorcinol; both become water-soluble when they lose a proton after mixed with NaOH. They go back to their water insolubility if you add an acid to donate a proton back to 3-chlorobenzoic acid and resorcinol. Therefore, after completing this lab,

Directions: Read/ Study all the lesson information in the 5.03 lesson then click the activity tab to perform two virtual labs. (There are recorded Teaching Videos for lesson 5.03. To view them click the “Help Sign” on the announcement page. Next scroll down to Lesson 5.03 stuff and you should see 5 part video links that will cover the lesson content.)

Samples of benzophenone, malonic acid, and biphenyl were each tested with water, methyl alcohol, and hexane. Benzophenone was insoluble in water as it is nonpolar while water is highly polar. Benzophenone was soluble in methyl alcohol, dissolving in 15 seconds, because methyl alcohol is intermediately polar as benzophenone is nonpolar. Methyl alcohol is polar but not as much as water. Thus, the nonpolar benzophenone was soluble in methyl alcohol. Benzophenone was partially soluble in hexane because hexane is nonpolar as is benzophenone. Thus, benzophenone was dissolved in hexane. Malonic acid was soluble in water because both malonic acid and water are polar. It took 25 seconds for malonic acid to dissolve in water. Malonic acid was soluble in methyl alcohol because malonic acid is polar and methyl alcohol is intermediately polar, allowing malonic acid to dissolve in the methanol in 15 seconds. Malonic acid was insoluble in hexane because hexane is nonpolar while malonic acid is polar. Biphenyl was insoluble in water as water is highly polar whilst biphenyl is nonpolar. Biphenyl was partially soluble in methanol which is intermediately polar whilst biphenyl is nonpolar, allowing it to dissolve a little. Biphenyl was soluble in hexane because both biphenyl and hexane are nonpolar molecules. Biphenyl dissolved in hexane in 10 seconds.

6. The solubility of the solids were tested using a micro tray, by placing them in water and oil to observe their polarity,

Substance A and B were weighed; Substance A weighed 0.502 g and substance B weighed 0.503 g. Both substances were put into two different test tube with approximately 8 ml of DI water into the test tub. Substance A and B were stirred and B dissolved while A did not. This shows that B is soluble in water compared to A. Thus, shows that B is soluble in water than A. The reason why B is soluble in water is because it has a higher dipole moment than A. With a higher dipole moment, it shows that it is soluble in water since it is polar and the bonds were easily broken.

Ionic compounds are soluble in water to a certain point depending on the compound. The level of solubility changes among different compounds. Some ionic compounds can completely dissolve in water and appear to be a homogeneous mixture. Although, some ionic compounds dissolve very little, and could be considered insoluble, since it does not dissolve fully. Depending on the compound, the level of solubility can be high or low. However, ionic compounds could dissolve to a certain degree. If the solution appears to be a heterogeneous mixture, many may assume through visual representation that it may be insoluble. As stated previously, the smallest amount of solubility should be considered. To confirm whether or not the substance is soluble, observe the efficiency when conducting electricity. Due to practical reasons, the slightest solubility could be considered insoluble by people.

Compare the solubility of various forms of panadol at body temperature in different pH conditions

The purpose of this experiment is to identify the periodic trends in the solubility of the alkaline earth metals and compare the results to that of lead

3) Explain the trend in the solubility of the three alcohols in hexane. (In your discussions, bring out the theoretical concepts on

A small beaker was placed under the arm of the distillation head to catch the distillate. Foil was wrapped around the neck of the round-bottomed flask and a wet paper towel was wrapped around the arm of the distillation head to create a condenser. The flask was heated gently so that the distillate dropped at a rate of two drops per minute. The temperature was recorded as every drop was collected. The distillation began at around 55.0 ℃. The distillation was stopped after 29 drops were collected to prevent the solution from being distilled to dryness. See attached data. The known boiling point of 1-butanol is 117.5 ℃ (Lemonds). The known boiling point of 1-propanol is 97 ℃ (Thiyagarajan). The known boiling point of acetone is 56 ℃ (Forss). The known boiling point of 2-butanone is 79.6 ℃ (Jiang). For unknown #3 the boiling point of the first substance seemed to be around 56 ℃ and the boiling point of the second substance seemed to be around 111 ℃. Therefore unknown #3 seemed to be a mixture of acetone and 1-butanol.

In this experiment, 0.31 g (2.87 mmol) of 2-methylphenol was suspended in a 10 mL Erlenmeyer flask along with 1 mL of water and a stir bar. The flask was clamped onto a hotplate/stirrer and turned on so that the stir bar would turn freely. Based on the amount of 2-methylphenol, 0.957 mL (0.00287 mmol) NaOH was calculated and collected in a syringe. The NaOH was then added to the 2-methylphenol solution and allowed to mix completely. In another 10 mL Erlenmeyer flask, 0.34 g (2.92 mmol) of sodium chloroacetate was calculated based on the amount of 2-methylphenol and placed into the flask along with 1 mL of water. The sodium chloroacetate solution was mixed until dissolved. The sodium chloroacetate solution was poured into the 2-methylphenol and NaOH solution after it was fully dissolved using a microscale funnel.

The purpose of the experiment is to oxidize a secondary alcohol (2-octanol) by using sodium hypochlorite (bleach) to produce 2-octanone. The starting material consisted of a sample of 2-octanol that was placed into a three-neck flask along with acetic acid and acetone creating an acidic solution. While monitoring temperature fluctuations to ensure a temperature of 400 Celsius was not reached, sodium hypochlorite slowly dripped from a separatory funnel into the acidic solution. Once this reaction reached its entirety, the solution was combined with sodium bisulfate to remove any of the remaining oxidizing agent. This solution was then tested and brought to a neutral pH using a sodium hydroxide solution. The reaction material was extracted using ether and was then washed with a saturated sodium chloride solution. The organic solution was then dried using magnesium sulfate and was then decanted and placed onto the rotovap. The produced weighed .599g and based on the infrared spectrum analysis (see Figure 1) preformed on the product it was determined to be 86.1% 2-octanol, which means .516g of 2-octanol was obtained in the final product.

The purpose of this lab was to study colligative properties. These properties are properties that are affected when a solute is added to a solvent. Thus, the amount is important, not the actual type of substance, for the colligative properties. A couple types of this property are the freezing point and boiling point of a substance. (1)

As mentioned in the discussion, olive oil, vegetable oil, crisco, and lard were soluble in nonpolar solvents and insoluble in polar solvents. This is due to the chemical composition of polar and nonpolar substances which results from the molecular shape as well as properties of dissolving solutes in solution. Polar substances are hydrophilic and contain polar Van Der Waals interactions (intermolecular forces) such as dipole-dipole forces, ion-dipole forces, and hydrogen bonding. Nonpolar substances are hydrophobic and contain non-polar Van Der Waals interactions. ‘Like dissolve like’ is the reason only polar solutes dissolve in polar solvents and why nonpolar solutes dissolve in nonpolar solvents. Molecules with similar polarity have similar intermolecular forces and therefore, can interact with each, or in this case dissolve9. Additionally, the solubility of a compound is determined by the length of the hydrocarbon chain. Long hydrocarbon chains such as the one found in oleic acid makes a compound more insoluble10. Therefore, since the lipids used in this experiment were hydrophobic substances and each lipid has long hydrocarbon chains, the results were consistent with the scientific literature and principles.