Question

In a lab, we mixed 0.50 g CuSO4*5H2O with 0.250g glycine. Using the molar mass of CuSO4*5H2O we found that we had 2.002x10^-3 moles. Using molar mass of glycine we found that we had 3.33x10^-3 mols of glycine. What is the limiting reagent? What is the molar mass of the new compound we created, Cu(glycine)2 *H2O (note for molar mass, each glycine loses the acidic proton in order to be a bidentate ligand? What is the mass of Cu(glycine)2*H2O based on our limiting reagent? (we recovered 0.753g of Cu(glycine)2*H2O

Answer 1

Molar mass of CuSO4*5H2O is 249.68 g/mol and that of glycine (NH₂‐CH₂‐COOH) is 75.07 g/mol.

So moles of CuSO4*5H2O =0.5/249.68 = 2.002x10^-3 mol and that of glycine = 0.25/75.07 = 3.33x10^-3 mol.

The balanced equation is CuSO4*5H2O + 2NH₂‐CH₂‐COOH -----> Cu(glycine)2 *H2O +H2SO4+4H2O

Since CuSO4*5H2O and glycine react in 1:2 molar ratio, so for reacting with 3.33x10^-3 mol of glycine moles of CuSO4*5H2O required = 3.33x10^-3 /2 mol= 1.666x10^-3 and available amount is more than required, so CuSO4*5H2O is in excess and glycine is limiting reagent.

Molar mass of Cu(glycine)2*H2O is 229.68 g/mol, so mass of Cu(glycine)2*H2O created =1.666x10^-3 mol x 229.68g/mol =0.382 g. So mass of product Cu(glycine)2*H2O based on the limiting reagent is 0.382g.

Comment if any doubt.