A 7.5 g sample of Cu(NO3)2.nH2O is heated. After the water has been driven off, 5.8 g of Cu(NO3)2 remains. What is the value of n in the empirical formula of this hydrate?

Answer 1

Loss in mass = mass H2O = 7.5-5.8 = ?
Change that to mols H2O. mols = grams/molar mass.
mols Cu(NO3)2 = 5.8g/molar mass.

mols from above for H2O/mols for Cu(NO3)2 = n in the formula.

Answer 2

Round that answer for n to a whole number.