Question

An impure sample of calcium sulphate (CaSO₄.2H₂O) was analyzed by gravimetric analysis of calcium. A (4.9000x10^-1) gram sample was dissolved in water and the calcium was precipitated, filtered  and   weighed   as   Ca₃(PO₄)₂.   A (2.31x10^-1) gram precipitate was obtained. Calculate the percent by mass of calcium in the sample. [Answer in 3 significant figure]

Answer 1

Given,

Mass of Ca3(PO4)2 filtered = 0.231 g

Molar Mass of Ca3(PO4)2 = 310.2 g / mol

=> Moles of Ca3(PO4)2 = 0.231 / 310.2 = 7.45 x 10^-4 moles

1 mole of Ca3(PO4)2 contains 3 moles of Ca

=> Moles of Ca that was precipitated = 7.45 x 10^-4 x 3 = 2.234 x 10^-3 moles

Molar Mass of calcium = 40.08 g / mol

=> Mass of Ca = 2.234 x 10^-3 x 40.08 = 0.0895 g

Mass of Ca precipitated = Mass of Ca present in the impure sample

=> Mass of Ca present in the impure sample = 0.08955 g

Mass of sample = 0.49 g

% by mass of Ca in the sample = 0.08955 x 100 / 0.49 = 18.27 %