Finding the percent of calcium carbonate in commercial antacid by using back titration I'm an IB Chemistry student and I had to do a back titration lab where 1g of a powdered antacid tablet were mixed with 50mL 1M HCl, and the mixture was titrated using 0.2 M NaOH; the NaOH was to neutralize the leftover HCl that hadn't reacted with the antacid. The active ingredient in the antacid is CaCO3.

It took me about 40mL of NaOH to titrate the leftover HCl to endpoint; 40 mL at 0.2 mol/L translate to 0.008 moles of NaOH. In the equation for reaction between NaOH and HCl the ratio of the two reactants is 1:1; therefore there were 0.008 mol of unreacted HCl in the mixture. 

There were 50mL of HCl (1 mol/L) in the mixture originally; 50mL * 1 mol/1000mL = 0.05 mol HCl total. Subtracting from this the molar quantity of unreacted (leftover) HCl yields: 0.05 - 0.008 = 0.042 mol HCl that did react with the CaCO3 in the antacid.

The reaction between CaCO3 and HCl is as follows:CaCO3 + 2HCl --> CaCl2 + CO2 +H2O
Therefore, molar ratio between CaCO3 and HCl is 1:2. 0.042 moles HCl reacted, therefore 0.021 mol CaCO3 were present in the sample. Multiplying this by the molar mass yields 0.021mol * (40+12+48)g/mol = 2.1 grams CaCO3.

That's a problem - i can't be having 210% composition by mass (the sample was 1g; I calculated 2.1 g CaCO3 present). So somewhere along the line i made a mistake, but I've checked my math so I don't know if it was a calculation error or if my method is dead wrong? Can anyone help me with this?
In chemical analysis usually do three times the same experimet, so you can minimize experimental errors.

After that, in order to find where is the error you can check:

Has everyone in you class use the same HCl and NaOH solutions (taken from the same bottle i.e.?

The antiacid tablets are all the same (given by your teacher)? If don't, could you post the composition?

Have you measured correctly the volumes used? To obtain that 210%, it seems tha you have used around 25 mL of HCl solution instead 50 mL.