Answer:
6.5
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 187.56 18.02
Cu(NO₃)₂·nH₂O ⟶ Cu(NO₃)₂ + nH₂O
m/g: 7.0 4.3
1. Moles of Cu(NO₃)₂
Moles of Cu(NO₃)₂ = 4.3 g × (1 mol/187.56 g)
Moles of Cu(NO₃)₂ = 0.0229 mol
2. Mass of H₂O
Mass of Cu(NO₃)₂·nH₂O = mass of Cu(NO₃)₂ + mass of H₂O
7.0 = 4.3 + x
7.0 - 4.3 = x
2.7 = x
3. Moles of H₂O
Moles of H₂O = 2.7 g × (1 mol/18.02 g)
Moles of H₂O = 0.150 mol
4. Value of n
The molar ratio is 1 mol (NO₃)₂ = n mol H₂O
n = moles H₂O/moles Cu(NO₃)2
n = 0.150/0.0229
n = 6.5
This answer does not make sense, because the maximum value of n in hydrated copper(II) nitrate is 6.
