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Topic: Calcium Carbonate + 2 Hydrochloric acid lab uncertainty  (Read 5160 times)

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MuskieMan33

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Calcium Carbonate + 2 Hydrochloric acid lab uncertainty
« on: July 08, 2011, 08:55:52 PM »
So in my Gen Chem II class we did a lab for CaCO3 + 2 HCl  H2CO3 + CaCl2
We are to find the % of "junk" in the Tums/Equate antacids and determine which is the better buy. It doesn't seem hard, although I'm not sure if my calculations are correct. If not, I'd like to fix my mistake. I previously did a similar titration with vitamin C and found that "pure vit. C" tabs were anything but.
The three things we tested were pure calcium carbonate powder, Tums, and Equate brand antacids. For example, when doing the Tums portion of the lab I ground a tablet and added 0.1513 g of the tums tablet to 10 ml of 0.484 M HCl. Then titrated with 0.260 M NaOH. The initial tums tablet was completely neutralized by the excess of HCl, then doing a back titration I am supposed to be able to find the % of calcium carbonate and junk in the tablet sample by neutralizing the remaining HCl with NaOH.
Here's the part I'm not completely sure if I did right. The calculations. I believe I should go as followed:

moles HCl - moles NaOH = moles HCl reacted with CaCO3 / 2 (because HCl to CaCO3 ratio is 2:1) = moles CaCO3 in sample (then convert to grams)

initial sample size (in grams) - CaCO3 reacted (grams) = weight of "filler/junk" in sample (grams) / initial sample = % of junk in sample

Using the numbers gathered from titration.
0.00484 mol HCl - 0.003692 mol NaOH = 0.001148 mol HCl that reacted with CaCO3 / 2 (2:1 ratio) = 0.000574 mol CaCO3

0.000574 mol CaCO3 x (100.06 g / 1 mol) = 0.057434 g CaCO3

0.1513 g initial sample - 0.0574 g CaCO3 reacted = 0.0939 g filler / 0.1513 g initial sample x 100% = 62.06% junk

Is my math right or am I mixing up something small like dividing by the 2:1 ratio if I'm supposed to multiply? I appreciate everyone who takes the time to read this, it would mean a lot to me to get some other opinions on it before I hand it in.

MuskieMan33

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Re: Calcium Carbonate + 2 Hydrochloric acid lab uncertainty
« Reply #1 on: July 08, 2011, 10:52:44 PM »
Upon further review of online material and noticing Borek's signature, I have to thank him, and maybe others will too as it could possibly help answer my dilemma. After reading an article I feel more confident in my calculations, perhaps other chem students, professors, and enthusiasts can give more feedback about my concern after looking over some other sources.

The website is http://www.titrations.info/back-titration

specifically the following example, thanks to "ChemBuddy"

"1.435 g sample of dry CaCO3 and CaCl2 mixture was dissolved in 25.00 mL of 0.9892 M HCl solution. What was CaCl2 percentage in original sample, if 21.48 mL of 0.09312 M NaOH was used to titrate excess HCl?"

Calculations:
"During titration 21.48×0.09312=2.000 mmole HCl was neutralized. Initially there was 25.00×0.9892=24.73 mmole of HCl used, so during CaCO3 dissolution 24.73-2.000=22.73 mmole of acid reacted. As calcium carbonate reacts with hydrochloric acid 1:2 (2 moles of acid per 1 mole of carbonate), original sample contained 22.73/2=11.37 mmole of CaCO3, or 1.137 g (assuming molar mass of CaCO3 is 100.0 g). So original sample contained 1.137/1.435×100%=79.27% CaCO3 and 100.0-72.27%=20.73% CaCl2."

However, I am not calculating my data in mmol simply because I'm used to using moles per liter, just another thing for me to trip over. once again, I'd appreciate any feedback and input, even if I may be wrong. I want to learn it, not just get by. Thanks everyone!