Suppose 0.3000g of an unknown blue coordination compound was dissolve in water.

Question

Suppose 0.3000g of an unknown blue coordination compound was dissolve in water. The exact formula was not known, but Cu2+ and NH3 were present in the compound. Exactly 39.12 mL of 0.123 M HCL was needed to titrate the NH3 that was present. Then 24.40 mL of 0.0500 M Na2S2O3 was used to titrate the I3- produced when KI was added after the HCl titration. Calculate the moles of NH3 and moles of Cu present in 0.3000 g of this salt. Calculate the mole ratio of Nh3 to Cu. What is the % by mass of Nh3 and % by mass of Cu in the unknown compound?

Explanation / Answer

NH3 + HCl ---> NH4Cl

moles of NH3 present = moles of HCl used

                                    = 0.123 M x 39.12 ml

                                    = 4.812 mmol

mass of NH3 present = 4.812 mmol x 17 g/mol/1000 = 0.082 g

mass% NH3 in sample = 0.082 x 100/0.3 = 27.27%

moles of Cu2+ present = moles of Na2S2O3 used

                                     = 0.05 M x 24.40 ml

                                     = 1.22 mmol

moles ratio (NH3/Cu) = 4.812/1.22 = 4

mass of Cu2+ present = 1.22 mmol x 63.546 g/mol/1000

                                    = 0.077 g

mass% Cu in sample = 0.077 x 100/0.3 = 25.67%

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