Volumetric Analysis Chemistry Lab Report Essays - 980 Words

Introduction: The purpose behind (the first step in) this experiment is to show that similarly to week 1, the molarity of an acid or base in solution can be determined (so long as one value’s is known) using titration. In this case though, finding the molarity of the acid used in the reaction is then used to determine the percent of that acid in a vinegar solution and compared to the standard value for % acid present in vinegar. The second part of the experiment was to see if by titrating a solution of NaOH and an unnamed mystery acid, you could find the molar mass of the unknown acid (solving the mystery). It must be understood that the number of moles of the reacting NaOH and the number of moles of the product NaX acid, must both†¦show more content†¦Each sample was used in separate titrations. Each titration was carried out using the same methods as before. Once the molar mass of each sample was found the values were then averaged and the deviation was found. To prove that the m olar mass was accurate the deviation had to be within 1%. Calculations: VINEGAR 1) Molarity of acid: a = vinegar b = NaOH Run #1: Ma * (5.00mL)Va = (0.02116)Mb * (14.1)(Vf - Vi) Ma = 0.5967 Run #2: Ma * (5.00mL)Va = (0.02116)Mb * (14.1)(Vf - Vi) Ma = 0.5967 2) Average Molarity of Acid: (Ma1 + Ma2)/2 = (0.5967 M) 3) Density (g/L) (Avg. M) * MM(CH3COOH) (0.5967M) * (65.05 g/mol) = 35.85 g/L 4) Percent Acid ((g/L)/(given density)) * 100 ((35.85 g/L)/(1005 g/L)) * 100 = 3.57% UNKNOWN ACID 1) Moles of NaOH ((Vf - Vi)/1000) * M Run #1: ((31.11 – 0.51)/1000) * 0.2116m = .006475 Run #2: ((31.35 – 0.38)/1000) * 0.2116m = .006553 2) Calc Moles of Acid because 1:1 ratio moles acid equals moles NaOH moles acid = Run 1: .006475 and Run #2: .006553 3) Molar Mass (grams used/moles acid) Run #1: (1.3160g/.006475) = 203.24 g/mol Run #2: (1.3276g/.006553) = 202.59 g/mol 4) Average Molar Mass (203.24 + 202.59)/2 = 202.92 g/mol 5) Average Deviation ((202.92 – 203.24) – (202.92 – 202.59))/2 = .005 6) % Deviation (.005/202.92) * 100 = .0025 which is lt;1% Conclusion/Discussion: Using titration as a method to calculate theShow MoreRelatedChem 103 Project Lab Essay1359 Words   |  6 PagesThe data gathered and calculated in the experiment accurately portrayed the way the reactions would have taken place. 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