Analysis of Tetraammine Copper(lI) Sulfate Monohydrate: Calculations Part I: Mol

Question

Analysis of Tetraammine Copper(lI) Sulfate Monohydrate: Calculations Part I: Moles of Cu"2 per Gram of Complex 1. Determine the concentration (molarity) of Cu' from the slope of the Beer's Law Plot. Determine the concentaon (mlarity oC from the slope of the Bers Law Plot. 2. Determine the number of moles of Cu"2 in the spectroscopy sample from the volume of the solution (data) and the molarity (calculated above). 3. Determine the number of moles of Cu2 per gram of the complex from the number of moles (calculated above) and the mass (data). 1. For each 'good' titration, determine the number of moles of HCl used in the titration from volume of the solution (data) and the molarity (recorded on HCI container). For each, determine the moles of NH, from the moles of HCl (calculated above) and the mole ratio for the neutralization (one mole acid to one mole base). 2. 3. For cach, determine the number of moles of ammonia per gram of complex from the number of moles (calculated above) and the mass (data). 4. Determine the average number of moles of ammonia per gram of complex from the values above +2 Divide the number of moles of ammonia per gram of complex (Step 11-4) by the number of moles of copper(I) per gram of complex (Step 1-3). 1. 2. Write the chemical formula for the complex.

Explanation / Answer

Ans:

Part-I: b) No moles of cu(II) ions = molarity * volume =0165*0.020=0.0033moles.

           c) i) No of moles of Cu(II) ions per gram of complex

                              245 g (Copper Complex mol wt) contains 58g

                               1 g of complex conatains ?   = 1*58/245 = 0.236moles

                 ii) No of moles of Cu(II) ions (as per mass)

                              245 g (Copper Complex mol wt) contains 58g

                                0.267 g of complex conatains ?   = 0.267*58/245 = 0.063moles

Part II) a) no of moles of HCl = 0.508 * 0.025 = 0.0127 moles

             b) As per reaction between Hcl and NH3 is 1:1, so no moles of NH3 is 0.0127moles and mole ratio is 1

            c) i) No of moles of ammonia ions per gram of complex

                              245 g (Copper Complex mol wt) contains 17g

                               1 g of complex conatains ?   = 1*17/245 = 0.06moles

              ii) No of moles of ammonia ions as per given mass

                              245 g (Copper Complex mol wt) contains 17g of ammonia

                               0.819 g of complex conatains ?   = 0.819*17/245 = 0.056moles

Determination of mole ratio of ammonia to cu(II) =0.06/0.236=0.25

Chemical formula is [Cu(NH3)4(H2O)n]SO4   n=1

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