Fda Vinegar Chemistry 1225 Lab

798 WordsJul 25, 20134 Pages
FDA Vinegar / Acetic Acid Analysis Objective- Determine the Acetic acid level contained within the vinegar sample Background- Per the FDA regulation which states that consumer vinegar may contain no less than four percent and no more than five percent Acetic Acid, we will determine the Acetic Acid content of a sample of the vinegar in question through titration. After standardization of our titrant, which in this case will be NaOH, we will use this along with the indicator Phenolphthalein, to titrate the vinegar to determine the concentration of Acetic Acid in the sample. Equipment and reagents- We will use a 50mL beaker, a graduated buret, spatula, droplet applicator, 125mL Erlenmeyer flask, hot plate, and a container of…show more content…
M1 NaOH= nNaOHVNaOH= 4.9×10-4mols 4.7mL=10.4×10-5 2. M2 NaOH= nNaOHVNaOH= 4.9×10-4mols 5.0mL=9.8×10-5 3. M3 NaOH= nNaOHVNaOH= 4.9×10-4mols 5.0mL=9.8×10-5 Mavg NaOH= M1×M2×M33=1.0×10-4M Titration of vinegar nNaOH= Mavg NaOH × VNaOH 1. n1 NaOH= 1.0×10-4M8.8 mL=8.8×10-4m 2. n2 NaOH= 1.0×10-4M8.5 mL=8.5×10-4m 3. n3 NaOH= 1.0×10-4M8.2 mL=8.2×10-4m nCH3COOH= nNaOH Mass of CH3COOH= nCH3COOH ×60gmol 1. Mass of CH3COOH= 8.8×10-4m ×60gmol = 5.28×10-2g 2. Mass of CH3COOH= 8.5×10-4m ×60gmol = 5.08×10-2g 3. Mass of CH3COOH= 8.2×10-4m ×60gmol = 4.9×10-2g Percent Weight CH3COOH in Vinegar = Mass of CH3COOHMass of Vinegar 1. Mass of CH3COOHMass of Vinegar = 5.28×10-2g1.004gmol ×100=5.26% 2. Mass of CH3COOHMass of Vinegar = 5.10×10-2g1.004gmol ×100=5.08% 3. Mass of CH3COOHMass of Vinegar = 4.9×10-2g1.004gmol ×100=4.91% Avg % Wght CH3COOH in Vinegar = Mass1+Mass2+Mass33= 5.26+5.08+4.913= 5.08% Results- The results of this examination is that the content percentage of Acetic Acid in the provided vinegar sample exceeds the FDA standards by .08 percent. While this clearly shows a result outside the provided guidelines, it is possible that the chance of error built into this method of testing could account for the excess. For instance, if there were an air bubble in the tip of the Buret during the titration, this would lead to the assumption of a higher level of NaOH needed to neutralize

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