1. Calcium containing sample

Suppose the concentration of EDTA is 0.05180 M

A sample weighing 0.2537 g is titrated with EDTA and requires 28.42 mL

The number of mmoles of EDTA used is     28.42 mL x 0.05180 mmol/mL = 1.472 mmol

The complex between Ca²⁺ and EDTA is 1 : 1, i.e., 1 mmol of EDTA can complex 1 mmol of Ca²⁺

    Therefore, the number of mmol of Ca²⁺ is equal to the number of mmol of EDTA, i.e., 1.472 mmol

    The number of mmol of CaCO₃ is equal to the number of mmol of Ca²⁺, i.e., 1.472 mmol

The molar mass of CaCO₃ is   40.078 + 12.011 + 3 x 15.9994 = 100.087 g/mol = 100.087 mg/mmol

    Therefore, 1.472 mmol of CaCO₃ weighs:    1.472 mmol x 100.087 mg/mmol = 147.3 mg = 0.1473 g

The mass percent of the original sample (which weighed 0.2537 g) which was CaCO₃ was

      mass % CaCO₃ =  100 x 0.1473 / 0.2537 = 58.06 %

The number of mmoles of EDTA used is     35.91 mL x 0.05180 mmol/mL = 1.860 mmol

The number of mmol of Ca²⁺ is equal to the number of mmol of EDTA, i.e., 1.860 mmol

The number of mmol of CaCO₃ is equal to the number of mmol of Ca²⁺, i.e., 1.860 mmol

1.860 mmol of CaCO₃ weighs:    1.860 mmol x 100.087 mg/mmol = 186.2 mg = 0.1862 g

The mass percent of the original sample (which weighed 0.3226 g) which was CaCO₃ was

      mass % CaCO₃ =  100 x 0.1862 / 0.3226 = 57.72 %

From the two results above, we can calculate an average mass % of CaCO₃ in our sample.

    Avg Mass % = ( 58.06 + 57.72 ) / 2 = 57.89 %

The average deviation is:

        ( | 58.06 - 57.89 | + | 57.72 - 57.89 | ) / 2 = 0.17 %

Our result could be reported as:    Mass % CaCO₃ = 57.89 ± 0.17 %

The percent deviation of our result is 100 x  0.17 / 57.89 = 0.29 %