[kizunaencounter]

Procedure:
1.Unknown calcium solution obtained in 125 ml flask.
2.Pipetted 25.00 ml of unknown into 400 ml beaker,diluted with 75 ml of 0.1M HCl and 5 drops methyl red indicator
3.25 ml of ammonium oxalate solution added to the beaker
4.filtered,put in oven cooled weighed, etc

Ca2+(aq) + C2O42-(aq) -----> CaC2O4(s)

For Trial # 1
Crucible clean dried weight=15.5572g
after analysis crucible weight=16.1992g
mass of calcium oxalate mono-hydrate precipitate=.6420g

Determine the molarity of Ca2+ in the original unknown solution
moles of calcium oxalate precipitated=.6420g/128.10g/mol=0.050 mol CaC2O4
moles of Ca2+=ratio is 1:1 Ca2+:CaC2O4=.0050 mol CaC2O4
molarity=0.050 mol Ca2+/0.125ml=0.040=0.040 X 5= .20M Ca2+ in the original solution

did i miss anything in my calculations? 

[Dan]

Your answer is correct but you have typed out some of the numbers incorrectly - you've put 0.050 when you mean 0.0050 for moles of calcium oxalate in the titre, but judging from the calculations, you've put the correct number into the calculator.

In the last part, you have gone about it in an unusual (but not incorrect) way. I'd have said...

concentration = moles/volume

moles in titre = 0.0050
titre volume = 25 mL

concentration = 0.0050/0.025 = 0.20 M

...rather than using the volume of the original solution (0.125 L) and applying a scaling factor afterwards (multiplying by 5).