[kizunaencounter]

For this experiment the goal was to find the % calcium in an unknown calcium carbonate sample by EDTA titration

First i standardized the EDTA solution with standard Ca2+ solution in triplicate

according to the reaction EDTA reacts with a 1:1 stiochiometry with Ca2+

For the first trial

29.01 ml of EDTA was used

inital CaCO3 solution
moles CaCo3 ?
CaCO3 MW=100.087g/mol

mol CaCO3= 1.0787g X 1molCaCO3/100.0869g/mol =.010778 mol CaCO3

.010778 mol /1L=.010778 M CaCo3

.010778 M CaCO3 X 25ml/1000ml=.000269 moles CaC03

.000269moles CaCO3 X 1mol EDTA/1molCaCO3=.000269 moles EDTA

.000269 moles EDTA/.02901 L=.009288 M EDTA

Trial # 2
 28.68 mL of EDTA was used

mol EDTA= .000269 mol EDTA

.000269/.02868L=.009379 M EDTA

Trial #3

28.72 mL of EDTA was used

mol EDTA=.000269

.000269/.02872=.009366 M EDTA

AVG moloarity =.009344 M EDTA

Then i had to prep and titrate an unknown calcium carbonate sample

0.251g of unknown calcium carbonate was transferred into a 250ml with deionized water and added some HCl
then 25.00ml of aliquot of the unknown was transferred into an 250ml Erlenmeyer flask
with metyel red indicator,6M NaH and ph 10 ammonia buffer and calmagite indicator
and then titrated with the standardization EDTA solution.

% CaO in Trial #1

1:1 ratio
0.251g of unknown calcium carbonate sample
20.51mL of EDTA

.02051 L X .009344 M EDTA =.000192 moles of EDTA

.000192 moles of EDTA =.000192 moles of CaO

.000192moles of CaO X 56.072 g/mol CaO =.010767 g CaO

.010767 g CaO/0.251g CaCO3 X 100 = 4.3%

Is this correct so far?

[Borek]

29.01 ml of EDTA was used
28.68 mL of EDTA was used
28.72 mL of EDTA was used

Why do you repeat calculations three times and average result, instead of averaging volumes and calculating concentration once? Result is the same, but the second approach is thrice faster.

.010767 g CaO/0.251g CaCO3 X 100 = 4.3%

Is this correct so far?

No, but you are close. You have titrated 25.00 mL, which contained only part of the original sample.

I just skimmed your calculations, not checking the math.

[kizunaencounter]

29.01 ml of EDTA was used
28.68 mL of EDTA was used
28.72 mL of EDTA was used

Why do you repeat calculations three times and average result, instead of averaging volumes and calculating concentration once? Result is the same, but the second approach is thrice faster.

.010767 g CaO/0.251g CaCO3 X 100 = 4.3%

Is this correct so far?

No, but you are close. You have titrated 25.00 mL, which contained only part of the original sample.

I just skimmed your calculations, not checking the math.

hmm so are you saying that i have to times the 25.00 mL answer by 10 since the original sample i made was 250ml and that would make the percent 43%
Which would make more sense only 4.3% didnt make much sense to me

[poca]

Yes you titrated 25ml of your unknown sample. Your actual sample is 250ml so times your answer by 10 for total