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?

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.

Round that answer for n to a whole number.

To determine the value of n in the empirical formula of this hydrate, we need to find the mass of water that was initially present and then subtract it from the total mass.

1. Calculate the mass of water lost:
Mass of initial sample - Mass of remaining Cu(NO3)2
7.5 g - 5.8 g = 1.7 g

2. Convert the mass of water lost to moles:
Moles = mass / molar mass

The molar mass of water (H2O) is approximately 18.015 g/mol.
Moles of water = 1.7 g / 18.015 g/mol ≈ 0.094 mol

3. Determine the molar ratio between Cu(NO3)2 and water:
Looking at the empirical formula of Cu(NO3)2.nH2O, we can see that the ratio of Cu(NO3)2 to water is 1: n.

Since 1.7 g of water corresponds to 0.094 mol, we can determine the value of n by dividing the moles of water by the moles of Cu(NO3)2:
n = Moles of water / Moles of Cu(NO3)2
= 0.094 mol / 0.094 mol
= 1

Therefore, the value of n in the empirical formula of this hydrate is 1. The empirical formula of this compound would be Cu(NO3)2.H2O.