Question

A 267.8-g sample of ground water is analyzed for calcium. The Ca²⁺ in the sample is first precipitated and filtered-off as NH₄CaPO₄·7H₂O. This precipitate is dried and heated, releasing water and ammonia to yield anhydrous calcium pyrophosphate (Ca₂P₂O₇). The mass of Ca₂P₂O₇ obtained is 0.0437 g.
Give the calcium content of the ground water in parts per million (to three significant figures).

Answer 1

Ans :

Mol of Ca2P2O7 = mass / molar mass

= 0.0437 g / 254.053 g/mol

= 1.72 x 10^-4 mol

so mol of NH₄CaPO₄·7H₂O will be : 2 x 1.72 x 10^-4 mol = 3.44 x 10^-4 mol

then mass of Ca in water = mol x molar mass

= 3.44 x 10^-4 mol x 40.078 g/mol

= 0.01379 g

ppm concentration of calcium in ground water = ( mass calcium / mass water) x 10⁶

= (0.01379 g / 267.8 g ) x 10⁶

= 51.5 ppm