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(1 u is equal to 1/12 the mass of one atom of carbon-12) Molar mass ( molar weight) is the mass of one mole of a substance and is expressed in g/mol.

That's a problem - i can't be having 210% composition by mass (the sample was 1g I calculated 2.1 g CaCO3 present). Molecular mass ( molecular weight) is the mass of one molecule of a substance and is expressed in the unified atomic mass units (u). Multiplying this by the molar mass yields 0.021mol * (40+12+48)g/mol = 2.1 grams CaCO3.

0.042 moles HCl reacted, therefore 0.021 mol CaCO3 were present in the sample. Hence the molar mass of C a C O 3: 40 + 12 + 48 100 g / m o l. The individual molar masses are as follows: C a 40 C 12 O 16 Since there are 3 moles of O, its molar mass is computed as 3 × 16 48. Therefore, molar ratio between CaCO3 and HCl is 1:2. In order to calculate the molar mass of C a C O 3 we have to summate the individual molar mass of each element. The reaction between CaCO3 and HCl is as follows:CaCO3 + 2HCl -> CaCl2 + CO2 +H2O To determine the molecular weight (molar mass) of carbon dioxide based on measurements of the pressure, temperature, volume and mass of a sample of the gas. 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. Molar mass 100.087 g/mol Appearance White powder. There were 50mL of HCl (1 mol/L) in the mixture originally 50mL * 1 mol/1000mL = 0.05 mol HCl total. Calcium carbonate Calcium carbonate Other names Limestone calcite aragonite chalk marble Identifiers CAS number 471-34-1 Properties Molecular formula CaCO3. 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. 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. The active ingredient in the antacid is CaCO3. 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.