In an experiment, 34.8243g of copper (II) nitrate hydrate, Cu (NO3) 2•zH2O was heated to a constant mass of 27.0351g. Calculate the percent water in this hydrate?

What was the mass of water lost? What is the value of z (the number of waters in the formula) ?

1) The mass of water lost = 7.7892 grams

2) Z = 3: Cu (NO3) 2*3H2O

Explanation:

Step 1: Data given

Mass of copper (II) nitrate hydrate, Cu (NO3) 2•zH2O = 34.8243 grams

Mass of substance after heating = 27.0351 grams

Molar mass of Cu (NO3) 2 = 187.56 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate mass of water

The mass of water is the mass lost after heating.

Mass water = 34.8243 - 27.0351 = 7.7892 grams of water

Step 3: Calculate moles of Cu (NO3) 2

Moles Cu (NO3) 2 = Mass Cu (NO3) 2 / Molar mass Cu (NO3) 2

Moles Cu (NO3) 2 = 27.0351 grams / 187.56 g/mol

Moles Cu (NO3) 2 = 0.144 moles

Step 4: Calculate moles of H2O

Moles H2O = 7.7892 grams / 18.02 g/mol

Moles H2O = 0.432 moles

Step 5: Calculate Z

z = moles H2O / moles Cu (NO3) 2

Z = 0.432/0.144

Z = 3

This means we have 3 water molecules in the formula. This makes the formula ofthe hydrate: Cu (NO3) 2*3H2O