A hydrated form of copper(II) sulfate, CuSO4·nH2O(s), is heated to drive off all the waters of hydration. If we start with 9.40 grams of hydrated salt and have 5.25 grams of anhydrous CuSO4(s) after heating, find the number of water molecules, n, as- sociated with each CuSO4 formula unit.

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thomasdecabooter thomasdecabooter · Answer 1 · 2019-11-15T04:23:58+01:00

Answer:

The hydrated form of copper(II) sulfate = CuSO4*7H2O

Explanation:

Step 1: Data given

Mass of CuSO4·nH2O(s) = 9.40 grams

After heating, 5.25 grams of anhydrous CuSO4(s)

Step 2: Calculate mass of water

Mass of CuSO4·nH2O(s) - mass of anhydrous CuSO4(s) = mass of water

9.40 grams - 5.25 grams = 4.15 grams

Step 3: Calculate moles H2O

moles = mass/ Molar mass

moles = 4.15 grams / 18.02 g/mol

moles H2O = 0.23 moles

Step 4: Calculate moles CuSO4

Moles = 5.25 grams / 159.61 g/mol

Moles = 0.0329 moles

Step 5: Calculate number of water molecules

moles H2O / moles CuSO4 = 0.23/ 0.0329 = 7

1 mol of CuSO 4 combines with 7 mol of H2O

The hydrated form of copper(II) sulfate = CuSO4*7H2O