+  CuSO4 .5H2O(s) copper sulfate hydrate (before heating) EXAMPLE HYDRATE-CUPRIC SULFATE PENTAHYDRATE The coefficient of 5 indicates five moles of water are attracted to one mole of CuSO4 CuSO4 .5H2O(s)  CuSO4(s) + 5H2O(g) NOTE – your task is to find the coefficient in the formula of your unknown! The dot indicates the water is tightly attracted to the salt part of the hydrate H2O H2O H2O CuSO4 Solid-WHITE anhydrous H2O CuSO4 BLUE H2O  + H2O H2O H2O H2O H2O CuSO4 is the Salt part of the hydrate These waters are independent gas particles and will effuse into the atmosphere CuSO4 .5H2O(s) copper sulfate hydrate (before heating) CuSO4(s) copper sulfate anhydrous (after heating)

OVERVIEW OF HYDRATE CALCULATIONS CALCULATION OF THE WATER OF HYDRATION COEFFICIENT FROM YOUR DATA, FIND MASS OF WATER LOST DURING HEATING (SUBTRACT THE MASS OF THE BEAKER CONTAINNING HYDRATE BEFORE HEATING FORM BEAKER AFTER HEATING). CALCULATE THE MOLAR MASS OF WATER AND CONVERT THE WATER TO MOLES FROM YOUR DATA, FIND MASS OF ANHYDROUS SALT ( SUBTRACT EMPTY BEAKER FROM BEAKER CONTAINNING THE ANHYDRATE AT END OF LAB). 4. CALCULATE THE MOLAR MASS (GFM) OF YOUR ANHYDROUS SALT AND CONVERT YOUR MASS TO MOLES. CALCULATE THE MOLAR MASS OF WATER AND CONVERT THE WATER TO MOLES. COMPARE THE MOLES OF ANHYDROUS SALT TO MOLES OF WATER BY DIVIDING BY THE SMALLEST, THIS SHOULD GIVE THE WATER OF HYDRATION COEFFICIENT.

71.3g - 70.0 g = 1.3 g OF ANHYDROUS SALT (CuSO4 ) SAMPLE DATA for dehydration of CuSO4 . X H2O EMPTY BEAKER = 70.0 g BEAKER + HYDRATE = 72.0 g BEAKER AFTER FINAL = 71.3 g HEATING (CONTAINS ANHYDROUS SALT) 0.70 GRAMS WATER WAS LOST DURING HEATING. To calculate the water mass, subtract the beaker before heating (contains hydrated salt, from the beaker after heating (contains the anhydrous salt). The mass lost is water gas that escaped your beaker 71.3g - 70.0 g = 1.3 g OF ANHYDROUS SALT (CuSO4 ) To calculate the mass of the anhydrous (“without water”) salt simply subtract the empty beaker from the beaker after heating. After the heating cycle the hydrate has been dehydrated and is termed the anhydrous salt.

CALCULATE THE MOLAR MASS OF THE H2O Formula subscripts element Atomic mass Subtotal for the element H 2 X 1.00794 = 2.01588 O 1 15.99 = 15.99 + 18.00588 g/mol

CALCULATE THE MOLAR MASS OF THE ANHYDROUS SALT CuSO4 element Atomic mass Subtotal for the element Cu 1 X 63.546 = 63.546 S 32.06 = 32.06 O 4 15.99 = 63.96 + 159.566g/mol Formula subscripts

CALCULATE THE MOLES OF THE MOLES (H2O) = 0.0388761 mol H2O in hydrate element Atomic mass Subtotal for the element H 2 X 1.00794 = 2.01588 O 1 15.99 = + 15.99 18.00588 g/mol MOLES = MASS/GFM MOLES (H2O) = 0.70g/18.00588 g/mol MOLES (H2O) = 0.0388761 mol H2O in hydrate

CuSO4 MOLES = MASS/GFM MOLES (CuSO4) = 1.3g/156.566 g/mol CALCULATE THE MOLES THE ANHYDROUS SALT CuSO4 element Atomic mass Subtotal for the element Cu 1 X 63.546 = 63.546 S 32.06 = 32.06 O 4 15.99 = 63.96 + 159.566g/mol MOLES = MASS/GFM MOLES (CuSO4) = 1.3g/156.566 g/mol MOLES (CuSO4) = 0.0083037 mol CuSO4 in hydrate

MOLES (H2O) = 0.0388761 mol H2O in hydrate MOLES (CuSO4) = 0.0083037 mol CuSO4 in hydrate 0.0388761 mol H2O = 4.68 which rounds to 5 0.0083037 mol CuSO4 Now that we have the moles of the water and anhydrous salt, we divide by the smallest number of moles to get an integer ratio. NOTE, only round your FINAL answer, never round during the calculation. The water of hydration is always rounded to the nearest integer. Therefore the mole ratio of water to anhydrous salt is 5:1, X, the water of hydration is 5.