Solubility

Of the alcohols tested 1-Butanol was found to contain the strongest intermolecular forces (IMF) of attraction, with Methanol containing the weakest. It was discovered through experimentation that Methanol induced the highest ?T of all alcohols tested, and that conversely 1-Butanol induced the lowest ?T. The atomic structure of all four alcohols is very similar, as starting with 1-Butanol a CH2 group is lost as you move from 1-Butanol to 1-Propanol to Ethanol and then again to Methanol. Each structure is fairly linear and contains an H-bond with Oxygen, so the real change is found in the loss of the CH2 group, this lowers the liquid’s Molecular Mass, thus lowering the London forces as you move from 1-Butanol through

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

Benzophenone was soluble in methyl alcohol whiles biphenyl was partially soluble. Benzophenone and biphenyl are both nonpolar molecules. Methanol is intermediately polar, allowing both nonpolar molecules to be dissolved as it is not too polar or nonpolar. However, biphenyl is less polar than benzophenone. Since methyl alcohol is somewhat polar, the more polar benzophenone is more soluble in methyl alcohol than biphenyl. Also, benzophenone can make hydrogen bonds. Thus, benzophenone is more soluble in methyl alcohol than

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

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.

B. Claim: As we go from methanol  ethanol  1-propanol  1-butanol the dispersion forces increase.

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.

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

By identifying the solubility of the unknown, it could lead to a closer interpretation as to what the functional group the unknown may be. Solubility is determined based on intermolecular attractive forces, such as hydrogen bonding, dipole-dipole, and London dispersion forces. Intermolecular attractive forces arise due to different electron environments in different molecules. For example, water molecules are good at dissolving

Methanol, ethanol, 1-propanol, and 1-butanol are alcohols that contain both London-dispersion forces and hydrogen bonds. Hydrogen

The purpose of using solubility analysis on an unknown is to narrow the possible unknowns given based on the solubility analysis. Acetone was a control for ketone and was soluble in water. Hexanal was a control for an aldehyde and was insoluble in water due to intermolecular forces, such as an increased amount of hydrophobic area rather than hydrophilic area. As a result, by identifying the solubility of the

In a test tube, 0.5mL of the sample will be added with 0.5 mL of water and shaken vigorously. Take note for its solubility by parts (0.5mL is one part). Keep adding parts of the solvent until the sample is soluble. If not, add until ten parts of the solvent and determine its solubility. To separate test tubes, water will be replaced with ethanol, chloroform, ether, and acetone as solvents. Same procedures were

Solubility – Very soluble (water), Freely soluble (methylene chloride, chloroform, alcohol), Slightly soluble (acetone) and Insoluble (ether).6 Melting point - 120°C or 248°F.5

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.