Question

A 5.0 g sample of Cu(NO3)2-nH20 is heated, and 3.9 g of the anhydrous salt remains. What is the value of n?

I got as far as finding the molar mass of Cu(NO3)2, which is 187.5g/mol

Please help me further with this question.

Answer 1

Also please explain every step.

Answer 2

g Cu(NO3)2 = 3.9
g H2O = 5.0-3.9 = 1.1

mols Cu(NO3)2 = 3.9/187.5 = 0.0208
mols H2O = 1.1/18 = 0.0611

You want n mols H2O/1 mol Cu(NO3)2
So 1 mol Cu(NO3)2 = 0.0208/0.0208 = 1.000
n mols H2O = 0.0611/0.0208 = 2.94. Round that to a whole number of 3.0
Formula is Cu(NO3)2*3H2O

Answer 3

the answer is 2.9 converted into 3

Answer 4

I didn't understand....please explain in detail

Answer 5

I also did not understand.Please make it in a little bit easier way:)