last updated on Tue, Feb 27, 2001

30L, Oxalate Complex Experiment

Part 1

1) mw of (NH4)2Fe(SO4)2•6H2O = 392.14 g/mol

moles (NH4)2Fe(SO4)2•6H2O = 1.6 g / 392.14 g/mol = 4.1x10-3 moles

moles of oxalic acid = (1 M) x (6 mL) = 6 mmoles or 6x10-3 moles

moles of ammonium iron(II) sulfate hexahydrate is less than the moles of oxalic acid. Thus, the ammonium iron(II) sulfate hexahydrate is the limiting reagent.

2) Answer depends on group size.

Part 2

1. See page 402 of Oxtoby for balancing Oxidation reduction reactions.

Half reaction 1: (reduction)

MnO4– (aq) 8 H+ (aq) + 5 e– ---> Mn2+ (aq) + 4 H2O (l)

Half reaction 2: (oxidation)

C2O42- (aq) ---> 2 CO2 (g) + 2 e–

Net Reaction:

2 MnO4– (aq) + 5 C2O42- (aq) + 16 H+ (aq) ---> 2 Mn2+ (aq) 10 CO2 (g) + 10 H2O (l)

2) 23.56 mL x 0.01052 M = 2.4785x10-4 moles EDTA

Assumming that calcium has six coordination sites, then one mole of EDTA complexes to one mole of calcium present. Thus, 2.4785x10-4 moles EDTA = 2.4785x10-4 moles calcium.

Since one tenth of the original sample was used, multiple by 10
thus,

2.4785x10-3 moles calcium in the original sample

moles calcium = moles of marble

thus,

mw of marble = (0.2481 g) / (2.4785x10-3 moles) = 100.1 g/mole

Part 3

3) note: neutralization equivalent is identical to equivalent weight

Ni2+(NH3)mBn•pH2O

30.00 mL x 0.1013 N = 3.039 mmole HCl added to complex

6.30 mL x 0.1262 N = 0.79506 mmole NaOH used to back titrate

thus, amount of HCl reacted = 3.039 mmole - 0.79506 mmole = 2.244 mmoles = 2.244x10-3 moles

thus the equivalent weight of the complex = 0.185g / (2.244x10-3 mol) = 82.4 g/mole.

4) Assumptin: the metal has six ligands

also, n = 2 (since the complex must be neutral and Br– is the counter ion)

thus, if (m + p) = 6

Thus the equivalent weight = (atomic wt of Ni + 2 x Atomic wt of Br + m x mw of NH3 + mw of p•H2O) / m

= (58.7 + 2 x 79.7 + m (17.0) + p (18.0) ) / m

= (218.5 + m(17.0) + p(18.0) ) / m

Possible Combinations:

m = 6, p = 0, equivalent weight = 53.4 g/mole
m = 5, p = 1, equivalent weight = 64.3 g/mole
m= 4, p =2, equivalent weight = 80.6 g/mole (this value is closest)
m = 3, p = 3, equivalent weight = 107.8 g/mole
etc.

thus,

the best formula = Ni2+(NH3)4Br2•2H2O