Synthesis of zinc Iodide

Synthesis of zinc Iodide - (Moles zinc/moles...

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Synthesis of Zinc Iodide Author: Austin Kelly Lab Partner: Daniel Swanson Lab Instructor: Sveta Enman Chem 104b Section 70 Date work performed: September 5, 2007 Date work submitted: September 12, 2007 Calculations
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In this lab we used a series of calculations to determine the molecular formula of the Zinc Iodide. A couple of basic equations used are: Atomic weight= (mass of atoms in grams/moles of atoms) Molecular weight= (mass of molecules in grams/moles of molecules). In order to calculate the moles of Iodine atoms reacted from the weight of Iodine used, we converted the 0.498g of Iodine we used into moles which equals 0.0078 mols used. To calculate the weight of Zinc that reacted, we simply subtracted the ending amount (0.321g) from the beginning amount(0.502g) to show that 0.181g of Zinc reacted Using the results from the previous calculation, we converted the remaining weight into moles of Zinc to show that 0.0028 moles of zinc reacted. We then expressed the moles of Zinc and Iodine as a ratio:
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Unformatted text preview: (Moles zinc/moles iodine)= (.0028mol/.0078mol)=(1/3) Here is the equation from 7-1 rewritten with the appropriate numbers in place of the variables: ___________________________________________________ Results Ÿ Weight of zinc used:0.502g Ÿ Weight of iodine used:0.498g Ÿ Dry weight of unreacted zinc remaining at the end of experiment:0.321g Ÿ Dry weight of anhydrous zinc iodide:0.021g Ÿ Moles of iodine reacted:0.0078moles Ÿ Moles of zinc reacted: 0.0028moles Ÿ Calculated ratio of zinc to iodine: 1:3 Ÿ Proposed equation for the formation of zinc iodide based on my experimental results: ZnI +3I à Zn + 3I à ZnI3 Ÿ Experimental moles of ZnI2 obtained: 6.58X10^-5 moles Ÿ Theoretical Moles of ZnI2 based on moles of Iodine reacted: 6.58X10^-5 moles Ÿ Theoretical moles of ZnI2 based on the moles of zinc reacted: 4.71X10^-5 moles Ÿ Percent yield: 71.58% yield...
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Synthesis of zinc Iodide - (Moles zinc/moles...

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