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OVERVIEW OF HYDRATE CALCULATIONS CALCULATION OF THE WATER OF HYDRATION COEFFICIENT 1.FROM YOUR DATA, FIND MASS OF ANHYDROUS SALT ( SUBTRACT EMPTY BEAKER.

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Presentation on theme: "OVERVIEW OF HYDRATE CALCULATIONS CALCULATION OF THE WATER OF HYDRATION COEFFICIENT 1.FROM YOUR DATA, FIND MASS OF ANHYDROUS SALT ( SUBTRACT EMPTY BEAKER."— Presentation transcript:

1 OVERVIEW OF HYDRATE CALCULATIONS CALCULATION OF THE WATER OF HYDRATION COEFFICIENT 1.FROM YOUR DATA, FIND MASS OF ANHYDROUS SALT ( SUBTRACT EMPTY BEAKER FROM BEAKER CONTAINNING THE ANHYDRATE AT END OF LAB). 1.CALCULATE THE MOLAR MASS (GFM) OF YOUR ANHYDROUS SALT AND CONVERT YOUR MASS TO MOLES. 2.FROM YOUR DATA, FIND MASS OF WATER LOST DURING HEATING (SUBTRACT THE MASS OF THE BEAKER CONTAINNING HYDRATE BEFORE HEATING FORM BEAKER AFTER HEATING). 3.CALCULATE THE MOLAR MASS OF WATER AND CONVERT THE WATER TO MOLES. 4.COMPARE THE MOLES OF ANHYDROUS SALT TO MOLES OF WATER BY DIVIDING BY THE SMALLEST, THIS SHOULD GIVE THE WATER OF HYDRATION COEFFICIENT.

2 SAMPLE DATA for dehydration of CuSO 4. X H 2 O 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 (CuSO 4 ) 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.

3 Cu1X 63.546 = 63.546 S1X 32.06 = 32.06 O4X 15.99 = 63.96 + 159.566 g/mol CALACULATE THE MOLAR MASS OF THE ANHYDROUS SALT CuSO 4 Formula subscripts element Atomic massSubtotal for the element

4 element Atomic massSubtotal for the element H2X 1.00794 = 2.01588 O1X 15.99 = 15.99 + 18.00588 g/mol CALACULATE THE MOLAR MASS OF THE H 2 O Formula subscripts

5 element Atomic massSubtotal for the element H2X 1.00794 = 2.01588 O1X 15.99 = + 15.99 18.00588 g/mol CALACULATE THE MOLES OF THE H 2 O MOLES (H 2 O) = 0.70g/18.00588 g/mol MOLES = MASS/GFM MOLES (H 2 O) = 0.0388761 mol H 2 O in hydrate

6 Cu1X 63.546 = 63.546 S1X 32.06 = 32.06 O4X 15.99 = 63.96 + 159.566g/mol CALACULATE THE MOLES THE ANHYDROUS SALT CuSO 4 element Atomic mass Subtotal for the element MOLES (CuSO 4 ) = 1.3g/156.566 g/mol MOLES = MASS/GFM MOLES (CuSO 4 ) = 0.0083037 mol CuSO 4 in hydrate

7 MOLES (H 2 O) = 0.0388761 mol H 2 O in hydrate MOLES (CuSO 4 ) = 0.0083037 mol CuSO 4 in hydrate 0.0388761 mol H 2 O = 4.68 which rounds to 5 0.0083037 mol CuSO 4 Now that we have the moles of the water and anhydrous salt, we divide by the smallest number of moles to get in integer ratio. Therefore the mole ratio of water to anhydrous salt is 5:1, X, the water of hydration is 5.


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