EXAMPLE HYDRATE-CUPRIC SULFATE PENTAHYDRATE
.
CuSO4 5H2O(s)
 CuSO4(s) + 5H2O(g)
The coefficient of 5
indicates five moles of
water are attracted to
one mole of CuSO4
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
Cu
H2O
CuSO4
H2O
BLUE
H2O
H2O
H2O
H2O
H2O
CuSO4 is the
Salt part of
the hydrate

CuSO4
SolidWHITE
anhydrous
+
H2O
H2O
H2O
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
1.
2.
3.
4.
5.
6.
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).
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.
.
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)
71.3g
-
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
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
Atomic mass
element
H
2 X
Subtotal for the
element
1.00794
2.01588
=
O
1 X
15.99
15.99
=
+
18.00588 g/mol
CALCULATE THE MOLAR MASS OF THE ANHYDROUS SALT
CuSO4
Atomic mass
element
Subtotal for the
element
Cu
1 X
63.546
=
63.546
S
1 X
32.06
=
32.06
O
4 X
15.99
=
63.96
Formula subscripts
+
159.566g/mol
CALCULATE THE MOLES OF THE
H 2O
element
Atomic mass
H
2
X
1.00794 =
O
1
X
15.99
=
Subtotal for the element
2.01588
+ 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
CALCULATE THE MOLES THE ANHYDROUS SALT
CuSO4
Atomic
mass
element
Subtotal for
the element
Cu
1
X
63.546 =
63.546
S
1
X
32.06
32.06
O
4
X
=
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
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.
0.0388761 mol H2O = 4.68 which rounds to 5
0.0083037 mol CuSO4
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.