Question

A 0.4939 g sample containing calcium carbonate was homogeneously precipitated as CaC₂O₄×H₂O (MW = 146.12). The precipitate was filtered and dried. The glass filtering crucible containing the precipitate weighed 30.8352 g and the empty crucible weighed 30.4190 g. Calculate the percent by weight of calcium carbonate and of calcium in the sample.

Answer 1

Ans. #1. Calculate moles of CaC₂O₄ obtained:

Mass of CaC₂O₄ obtained = Mass of crucible plus precipitate – Mass of Empty crucible

                                                = 30.8352 g – 30.4190 g

                                                = 0.4162 g

Moles of CaC₂O₄ precipitated = Mass / Molar mass

                                                = 0.4162 g / (128.0976 g/ mol)

                                                = 0.003249 mol

#2. Calculate Mass of CaCO₃ and Ca²⁺ in sample:

Balanced reaction:    CaCO₃(s) + H₂C₂O₄ (aq) -----> CaC₂O₄(s) + H₂O(l) + CO₂(g)

According to the stoichiometry of balanced reaction, 1 mol CaC₂O₄ is produced by 1 mol CaCO₃.

So,

            Moles of CaCO₃ in sample = 0.003249 mol = Moles of CaC₂O₄ precipitate

Since, 1 mol CaCO₃ consists of 1 mol Ca²⁺, so the moles of Ca²⁺ must be equal to that of CaCO₃.

Thus,

            Moles of Ca²⁺ = 0.003249 mol = Moles of CaCO₃

Now,

            Mass of CaCO₃ in sample = Moles x Molar mass

                                                            = 0.003249 mol x (100.0872 g/ mol)

                                                            = 0.3252 g

Mass of Ca²⁺ in sample = 0.003249 mol x (40.078 g/ mol) = 0.1302 g

#3. Calculate % mass

# % CaCO₃ in sample = (Mass of CaCO3/ Mass of sample) x 100

                                    = (0.3252 g / 0.4939 g) x 100

                                    = 65.84 %

# % Ca²⁺ in sample = (Mass of Ca²⁺/ Mass of sample) x 100

                                    = (0.1302 g / 0.4939 g) x 100

                                    = 26.37 %