Question

Finely ground mineral (0.5864 g) was dissolved in 25 mL of boiling 4 M HCl and diluted with 175 mL H2O containing two drops of methyl red indicator. The solution was heated to 100C, and 50 mL of warm solution containing 2.0 g (NH4)2C2O4 were slowly added to precipitate CaC2O4. Then 6 M NH3 was added until the indicator changed from red to yellow, showing that the liquid was neutral or slightly basic. After slow cooling for 1 h, the liquid was decanted and the solid transferred to a filter crucible and washed with cold 0.1 wt% (NH4)2C2O4 solution five times until no Cl- was detected in the filtrate upon addition of AgNO3 solution. The crucible was dried at 105C for 1 h and then at 500 ± 25 C in a furnace for 2 h. Ca2+ + C2O42- --> CaC2O4.H2O(s) -->500C CaCO3(s) The mass of the empty crucible was 17.1278 g, and the mass of the crucible with CaCO3(s) was 17.467926 g. Find the wt% Ca in the mineral.

Answer 1

Given data,

Volume = 25 mL

Mass of empty crucible = 17.1278 g

Mass of crucible + CaCO3 = 17.467926 g

Mass of CaCO3 = [Mass of crucible + CaCO3 ] – [Mass of empty crucible]

                            = 17.467926 g – 17.1278 g

                            = 0.3401 g CaCO3

Mass of Ca present in the CaCO3

= (0.3401 g CaCO3 x  1 mol / 100.0869 g) x (1mol Ca/1 mol CaCO3)* x 40.078g / 1 mol Ca)

= 0.136 g

% Ca = [ mass of Ca / mass of mineral sample]*100%

          = 0.136 g / 0.5864 g] x 100%

= 0.231 x 100%

          = 23.19 % Ca