Calculate the number of moles of H2O removed and the moles of anhydrous CuCl2 remaining in the crucible?

Question

Having trouble with a chem problem, thank you for the help!

Answer 1

Presumably your lab experiment involved heating copper (II) chloride dihydrate in a crucible to remove the water.  At some point you would have weighed an empty crucible (tare wt).  Then you would have added the copper hydrate salt, and reweighed (gross wt). The mass of the added hydrate salt is obtained by subtracting the tare wt from the gross wt (= net wt).  Then you would have heated the crucible at some temperature above 100 C until the mass of the crucible plus dried salt remained constant (the crucible would be cooled in a dessicator before weighing).   The mass of the dried salt is obtained by subtracting the tare wt from the mass of the crucible plus dried salt.  The mass of water is obtained by subtracting the mass of the dried salt from the mass of the hydrate salt.  The mass of the water can be converted to moles by dividing by the gram formula mass of water (18.01528 g/mol).  The mass of the dried copper salt can be converted to moles by dividing by the gram formula mass of CuCl₂ (134.4620 g/mol).

Hopefully if you divide your value for moles of water by moles of anhydrous salt, you will get a ratio of 2 to 1.