Coordination Compounds

# Coordination Compounds - 3 Wet Samples The results of the...

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Roshni Sheth, Anne Tolene, Philip Pearson CH 118-M3 April 7, 2010 Coordination Compounds 1. The Formula By using the equation = C , the moles of Cu 2+ /grams of sample can be determined: . . × . = . - 0 12445 005 0 025 L 6 21E 4 mols . - . 6 21E 4 mols0 200g = . sample 0 0031 Then by using the equation (the amount of acid added) - (the amount of acid that did not react) = (the amount of NH 3 ), the moles of NH 3 /grams of sample can be determined. The average number of moles of NH 3 is 0.00708 mol. By dividing 0.00708/0.0031, x=2.28, which rounds to x=2. 2. The Class Data For group 2’s data, x=2.66, which rounds to x=3. For group 3’s data, x=14.4, which rounds to x=14. For group 4’s data, x=2.72, which rounds to x=3. Except for group 3, our x value was close to group 2 and 4’s value of 3. After assessing the class data, our group would change its value of x to 3, because it was a more common answer.

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Unformatted text preview: 3. Wet Samples The results of the experiment would be different if the copper ammine complex was not completely dry because the weight of the complex would be different. It would also include the weight of the methanol that did not dry away from the complex. Sample Calculation: M 1 V 1 = M 2 V 2 If the volume of the NaOH was not correct because the sample did not dry enough, then the volume of the HCl would also be incorrect. The substance must dry completely before analysis. 4. Exercise The absorbance of the new solution would be different because the color of the solution would be different. There would also be a different molarity, so using this complex instead of another would not affect the experiment’s observations....
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