Zinc Iodide Synthesis Lab Report

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Name: Daniel Voskresenskiy Post-Lab Report Lab Name: 68th & Zinc - Lab #6 Course: CHEM106 LB SEC: 052 Date: March 18, 2013 Lab Partner: Justin Nus Observations of Reactants (Table 1) Chemical 0.17 M Acetic Acid Buffer 10 Unknown Zinc Compound Physical Appearance Aqueous; clear; colorless Aqueous; clear; colorless; not viscous Fine; Powdery; White; Solid Chemical Zinc Iodide (ZnI2) Calmagite NaEDTA Physical Appearance Solid; powder; fine; white Aqueous; brown/red; not viscous; opaque Fine; Powdery; White; Solid Procedure - Preparing a Solution of EDTA Before titration, a titrant must successfully be made - the substance that will react with zinc ion. Purpose: To find out how much zinc is a sample of zinc iodide we will react zinc…show more content…
For zinc ion to react, the NaEDTA must also be an ion in the solution which means that the large salt must be dissolved in water. Procedure 1) Weighed out 3.64 of NaEDTA on an electronic beam balance 2) Added this mass of NaEDTA using wide mouth funnel to a 250 mL volumetric flask. 3) Rinsed the funnel with a squirt bottle containing deionized water making sure none of the solid remained in the funnel. 4) Added about 100-200 mL of deionized water to the volumetric flask containing the solid NaEDTA. 5) Swirled the solution until the NaEDTA (s) dissolved entirely. 6) After dissolving the solid, deionized water was added to the volumetric flask to make 250mL of EDTA solution. 7) This solution (from Step 6) was then transferred to a clean 250 mL Erlenmeyer flask and placed inside the desk for safe keeping; using a cork wrapped tightly in Parafilm. The disassociation of NaEDTA in water is expressed by the following equation: Na2H2EDTA (s) + H2O (l) yields EDTA4- (aq) + 2Na+(aq) + 2H+(aq) Na2H2EDTA (s) has a molar mass of 372.24 g/mol. Procedure - Reacting EDTA with your Zinc ion in Zinc Iodine Purpose; The…show more content…
Percent error is most commonly caused by human error at various parts of the experiment. Question 5; Part 2; Empirical formula for our experiment (commercial): Let's suppose we have 19.6g Zn and 80.4g I. 19.6g Zn x 1mol Zn/65.39g Zn= 0.2997mol Zn 80.4g I x 1mol I/126.9g I= 0.6336mol I (0.2997mol)Zn + (0.6336mol)I --> (?)ZnI2 (0.2997mol)/(0.2997)Zn + (0.6336mol)/(0.2997)I (1mol)Zn + (2.11mol)I ~ 1:2 ratio Empirical formula for this problem: Let's suppose we have 78.8g I and 21.2g of Zn. 78.8g I x 1mol I/126.9g I = 0.621mol I 21.2g Zn x 1mol Zn/65.39g Zn = 0.324mol Zn (0.324mol)Zn + (0.621mol)I --> (?)ZnI2 (0.324mol)/(0.324)Zn + (0.621mol)/(0.324)I (1mol)Zn + (1.92mol)I ~ 1:2

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