1) Calculate the formula weight for the compounds of the formula [(Cu(NH3)x)]SO4

Question

1) Calculate the formula weight for the compounds of the formula [(Cu(NH3)x)]SO4*H20, where x=1^-6

2) Assuming x=1, how much CuSO4*5H20 would you need to prepare 6g of copper ammine complex?

3) How much of the above solution is required to prepare 25 mL of a .050 M aqueous stock solution?

4) Determind the moles of ammonia in the following cobalt ammine complex: A sample of .1500 g of the cobalt ammine commplex us placed in a flask and 25.00 mL of .200 M HCl is added. A few rops of bromocresol green is added and the ititration with .100 M NaOH required 16.42 mL.

Thank you!

Explanation / Answer

The molar mass of CuSO4*H2O is 177.62 g/mol. The molar mass of NH3 is 17.03 g/mol; therefore, the amount of ammine in the complex is:

(249.72 g/mol) = (177.64 g/mol) + x(17.03 g/mol)
(76.71 g/mol) = x(17.03 g/mol)
x = 4

The compound with formula Cu(NH3)4SO4*H2O has a molar mass of 245.75 g/mol, so let's consider the titration to see if this calculation is validated.

Using the titration (although the experimental set-up is completely wrong) you'd expect that the following reaction takes place:

(NH4)+(aq) + OH-(aq) ---> NH3(aq) + H2O

Since the number of moles of ammonium used are the same as that of hydroxide, then the number of moles of NaOH used is equivalent to the number of moles of ammonium present:

(0.01642 L)x(0.100 M) = 1.64x10^-3 moles NH3

The number of moles of complex is:

(0.1500 g) x [(1 mole)/(249.72 g)] = 6.007x10^-4 moles

Dividing the number of moles of NH3 with the number of moles of complex gives the number of equivalent NH3 molecules per molecule of Cu(NH3)xSO4*H2O:

(1.64x10^-3)/(6.007x10^-4) ~ 3

However, the compound Cu(NH3)3SO4*H2O only weights 228.72 g/mol.

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