Chemistry Lab Report

1106 words - 4 pages

The goal of the project was to characterize an "unknown" organic acid in order to make a proper identification of the acid, while learning proper techniques for scientific measurement and analysis of error.

In order to ensure the most accurate data, a purification was performed by the process of recrystallization. To perform the recrystallization the powder was dissolved in a minimal amount of hot ethanol/H2O solvent that allowed the unknown powder to crystallize properly when cooled. This process allowed for the removal of soluble impurities when suction filtered. A sample of the unknown acid was weighed at 8.24 g, and it was found that 164ml of a 40% ethanol, 60% H20 solvent dissolved the 8.24 g of unknown acid when heated. The beaker containing the dissolved acid was then placed in a beaker containing ice, allowing the unknown acid to recrystallized. After vacuum filtration, the recovered unknown was dried and weighed at 6.92 g. The percent recovery was determined by the following calculation: (8.24--6.92)8.24 x 100% = 16% loss = 84% recovery of unknown.

The melting point of the unknown was determined for both pure (recrystallized) and impure (original) powders. To perform thee melting point procedure the powders were placed in small tubule. The tubule was placed in a Mel-temp device and the melting range was recorded. The data is present in the form of a range in the following chart:

"Melting Point"

Substance Starting point (°c) Ending Point (°c)

Pure Acid (recrystallized)

Pure Acid 2nd Range 125.3

7 129.2
6

Impure Acid (original) 124.1 128.2

In order in determine the equivalent weight of the unknown acid, a base was needed at a known or "standardized" molarity. To achieve this, a basic solution of NaOH was prepared and titrated to an endpoint equilibrium with an acidic solution of a known molarity, the acid being KHP (204.24 g/mole). Phenolphthalein was used as an endpoint indicator to determine the molarity of the NaOH solution, which was found to be approximately .0979 M. The data collected from the titration follows in the table below.

"NaOH Titration"

Trial I II III

Mass KHP (g) .403 .402 .411

Final NaOH 21.02mL 41.04mL 30.01mL

Initial NaOH .89mL 21.02mL 9.50mL

Volume NaOH Total 20.13mL 20.02mL 20.60mL

Molarity NaOH .0980 M .0980 M .0977 M

The molarity of each trial was found using the formula below; following the formula is a sample calculation:

Molarity of NaOH Solution = Mass KHP (g) / [(204.34 g/mole) x (Delivered NaOH (ml)]

.403g /[(204.24g/mole)x(0.02013L)] = .0980 moles/liter

To ensure the best possible data, the error of the collected data was determined. This error was found to be: (9.80-9.77)/9.77 = .003 x 100% = .31% error.

The equivalent weight of the unknown was determined by titration with NaOH. By discovering the amount of NaOH required to react with the unknown acid, it was possible to determine the equivalent weight of 98.4 g/mole. Three...

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