Chemistry 1?

Question

A student paid little attention to the size of his burner flame while heating a 0.979 g sample of CaC2O4*H2O. The final mass of the cooled solid was 0.670g.

A) was the reaction product anhydrous calcium oxalate? Briefly Explain

B) Assuming that the final product was pure, what was the final product? Explain.

Answer 1

A) no

the MW of CaC2O4•H2O is 145.12 g/mole
the MW of CaC2O4 is 128.12 g/mole

0.979 g of the hydrate is 0.979 g * mole/145.12 g = 0.00675 mole

this corresponds to 0.00675 mole * 128 g/mole = 0.865 g of the anhydrous material

B) Calcium Oxalate decomposes on heating

CaC2O4 --> CaCO3 + CO

MW CaCO3 = 100.087 g/mole

0.00675 mole * 100.087g/mole = 0.67 g

So the product is CaCO3

Answer 2

a) No, because had it been anhydrous calcium oxalate, the final weight would have been 0.858 g.

b) It is calcium carbonate. 0.979 g of calcium oxalate correspond to 0.00670 moles. Upon decomposition, you must obtain 0.00670 moles of an unknown compound. The molar weight of this compound would therefore be 100, which corresponds to that of calcium carbonate.

The reaction is, therefore:

CaC2O4-H2O -> CaCO3 + H2O + CO

Answer 3

A) No, the student charred away every thing.

B) The final product was calcium oxide.