How to Calculate the Mass of Anhydrous Cu(NO3)2 from Its Hydrate?

[Precursor]

Homework Statement
What is the mass of the anhydrous compound in $$Cu(NO_{3})_{2} * 2.5 H_{2}O$$
, in other words, the $$Cu(NO_{3})_{2}$$, if 4.875g of the hydrated compound is used?


The attempt at a solution
What I did was find the formula mass of $$Cu(NO_{3})_{2}$$ which turned out to be 187.57g
Cu= 1 * 63.55 = 63.55
N= 2 * 14.01 = 28.02
O = 6 * 16.00 = 96.00

The formula mass of the water is 45.05g
H= 2.5(1.01 * 2) = 5.05g
O= 2.5(16.00 * 1) = 40g

From here I set up a ratio:
4.875/232.62 = x/187.57
x = 3.931g of $$Cu(NO_{3})_{2}$$

Is this right?

 

[symbolipoint]

The method looks good.