# i know there's smart people out there

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##### My room is a mess
7+ Year Member
15+ Year Member

so won't ya please help me?? i have to do my quant lab, and for some reason, i'm trying to do the practice problem before i do the actual lab results, but i can't get the right answer. if anyone cares to take time out of their busy day to help me, it would be incredibly gratefulyl appreciated.

this ist he problem:
a 25.00 mL aliquot of unknown CaCO3 is titrated with EDTA to give a 22.50 mL equivalence point. Find the % CaO if the sample weigfhed .2200 g and was diluted to 250 mL in a vol. flask. The molarrity of EDTA is 0.0090 M, the FW of CaO is 56.072, FW of CaCO3 is 100.1

any help would be appreciated, even a general formula would be useful. thanks again.

##### My room is a mess
7+ Year Member
15+ Year Member
ok wait, can someone tell me if this is right?

.02250 x .0090 = 2..25x10^-4 = mol EDTA=mol CaCO3=mol CaO

10 x 2.025x10^-4= .002025 = mol Ca in aliquot

.002025(56.072) = .1135 g CaO

.1135/.2200 = 51.59%

does that look right to anyone?

#### DrPharaohX

##### Free...your...mind...
15+ Year Member
It's 3 AM right now (east coast) and I should be asleep, but I couldn't resist:

Here we go: (my final answer was 0.4430 % - notice it's NOT 44.30 %)

CaCO3: 0.22 g / 100.1 g x (25/250) = 2.198 x 10^-4

EDTA: 0.009 x 0.0225 = 2.025 x 10^-4

[This next step assumes that reaction is one to one ratio between CaCO3 and EDTA]:

mol CaO produced = (2.198 x 10^-4) - (2.025 x 10^-4) = 0.1738 x 10^-4 (which is 1.738 x 10^-5)

So....

1.738 x 10^-5 x 56.072 g = 9.74 x 10^-4 g of CaO produced

And...

9.74 x 10^-4 / 0.22 = 0.004430 which = 0.4430 %

From what I remember from that lab, the yield I got for this problem seems reasonable because the yields I remember getting for our problems were often < 1%