A 293.6-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 0.0833 g.

Question

A 293.6-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 0.0833 g.
Give the calcium content of the ground water in parts per million

DrBob222 · Answer

0.0833g Ca2P2O7/293.6 g sample = 0.0002837 g Ca2P2O7/1 g sample = 283.7 g/1,00,000 g sample = 283.7 ppm Ca2P2O7 = 2 x 283.7 ppm NH4CaPO4.7H2O = 567.4 ppm in the water

Anon · Answer

Can you explain why you changed 0.0002837g to 283.7g. The formula I have is (g solute/g solution) x 10^6 to find ppm. Also can you explain why you did 2 x 283.7? Is it to find the Calcium content because there are 2 of Ca?