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# Gravimetric Analysis Problem

7 replies to this topic

### #1 Spliknot13

Spliknot13

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Posted 30 April 2013 - 12:48 AM

A 1.5176 g sample of CaCO3 was dissolved in an acidic solution. The calcium was precipitated as CaC2O4H20(146.11 g/mol) and the ignited precipitate at 230 degrees Celcius was found to weigh 0.8249. What is the percentage of CaO(56.08 g/mol) in the sample?

I was reading the method on solving this using the gravimetric factor, but did not quite understand the concept behind it thus unable to solve the problem,  also thus the ignited precipitate mean that its free of H20? (ignited precipitate=CaC2O4?). Im already exhausted on thinking about this and hope that people here would give me other insights towards this problem.

### #2 breizh

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Posted 30 April 2013 - 12:55 AM

Hi,

* CaCo3 is reacting with oxalic acid (H2C2O4) to produce  calcium oxalate then the cacium oxalate is "cooked" to produce CAO +CO2 +H2O .

** CaCO3 ---> CaO+CO2

hope this helps

Breizh

Edited by breizh, 30 April 2013 - 01:22 AM.

### #3 Spliknot13

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Posted 30 April 2013 - 01:31 AM

Thanks for the help. the answer in the book says that its 23.8 %. could you explain how you arrived at 97.16%? I still have trouble understanding this.

### #4 breizh

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Posted 30 April 2013 - 01:43 AM

theoritically : 1 mole CaCo3 is giving 1 mole of CAO >>>  CaCO3 ( 100 g/mol) >>>> 1.5176/100 = 0.015176 mol of CACO3 or CAO >>>>

0.015176 *56 = 0.849856 g CaO

to compare with 0.8249

0.8249/0.849856= 97.06 %

### #5 Spliknot13

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Posted 30 April 2013 - 02:04 AM

but I think that the %CaO in the sample should be calculated as  (weight of CaO in the sample)/(weight of sample)? Im really confused.

### #6 breizh

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Posted 30 April 2013 - 02:24 AM

My answer is related to the purity of the sample . others may answer your query !

Breizh

### #7 MrShorty

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Posted 30 April 2013 - 10:10 AM

also thus the ignited precipitate mean that its free of H20? (ignited precipitate=CaC2O4?).

Yes. the idea behind the baking/burning of the precipitate is to drive off any water and CO2 from the precipitate to convert CaC2O4*H2O to CaO.

The calculation Breizh did, as he noted, is related to the purity of the sample. Sometimes I've seen this called a % recovery calculation. You compute the quantity of CaO that the reaction should produce assuming the original CaCO3 is 100% pure and none of the Ca is lost over the course of the analysis (Breizh showed 0.85 g). Compare to the actual amount of CaO recovered, and you get 97% recovered. This number means that, either the original CaCO3 was not 100% pure (actually about 97%), or some of the CaO was lost over the course of the analysis.

but I think that the %CaO in the sample should be calculated as  (weight of CaO in the sample)/(weight of sample)?

I think you are correct, that this is the kind of percentage they are looking for, but I'm not sure how they got 24%, because mass CaO/mass CaCO3 is bout 54%. Theoretically, CaCO3 (CaO*CO2) should be 56% CaO (weight of CaO/weight of CaCO3). CaO would be 38-39% of the hydrated precipitate.

I'm missing something, I'm sorry. Maybe these ramblings will help in some way, though.

### #8 Spliknot13

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Posted 01 May 2013 - 01:14 AM

I think I figured it out. The ignited precipitate is indeed CaC2O4(128.11g/mol) then gram CaO = (0.8249/128.11)x56.08 = 0.3611 g CaO

% CaO = (0.3611g CaO/1.5176 g sample)x100 = 23.79 %

Anyways, thanks for the help guys.