A 7.0g sample of cu (no3) 2•nh20 is heated, and 4.3g of anhydrous salt remains. what is the value of n

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.