Suppose a chemistry problem requires a determination of the percentage of calcium carbonate, CaCO3, in a sample. Assume that 0.500 grams (g) of impure CaCO3 is dissolved in exactly 50 milliliters (ml) of 0.0985 molar (M) hydrochloric acid (HCL) solution.
Note that "M" is molar concentration, or molarity, defined as the number of moles of a substance per liter of solution or millimoles per ml of solution.
Assume the excess HCL is titrated by 0.1050 M sodium hydroxide (NaOH) solution to an end-point indicated at 6.00 mls. Further assume that the sample contains no other substances that react with HCL or NaOH. Write the equations of the chemical reactions as follows: CaCO3 + 2HCL = CaCL2 + H2O + CO2; HCL + NaOH = NaCL + H2O.
Note the molar ratio of the dissolution reaction is 1 mmol CaCO3 / 2 mmol HCL.
Calculate total millimoles (mmol) of HCL by multiplying the volume of HCL by the molarity, as follows: 50.00 ml x 0.0985 mmol HCL / ml = 4.925 mmol HCL. Calculate the number of millimoles of HCL that reacted with NaOH, as follows: 6.00 ml x 0.1050 mmol / ml = 0.630 mmol.
Determine the number of millimoles of HCL that reacted with CaCO3 by subtracting the number of millimoles of HCL that reacted with NaOH from total millimoles of HCL: 4.925 -- 0.630 = 4.295 mmol. Determine the number of millimoles of CaCO3 in the sample by multiplying millimoles of HCL that reacted with CaCO3 by the molar ratio of the reaction: 4.295 mmol HCL x 1 mmol CaCO3 / 2 moles HCL = 2.148 mmol.
Calculate the weight of CaCO3 in the sample by multiplying millimoles of CaCO3 by its formula weight, or molar mass, as follows: 2.148 mmol x 100.1 mg / mmol CaCO3 = 215.0 mg CaCO3.
Calculate the percent CaCO3 in the sample as follows: [215 mg CaCO3 / 500 mg sample] x 100 = 43.0 %.