Water of Crystallization

Water of crystallization (WC) is the number of molecules of water present in one mole of a hydrated salt. Example, in iron (II) tetraoxosulphate (VI)-heptahydrate [FeSO4.7H2O], there are seven molecules of water attached to a molecule of FeSO4, that is its water of crystallization.
Some other examples of salts with water of crystallization include:
MgSO4.7H2O - Magnesium tetraoxosulphate(VI)-heptahydrate, (Epsom Salt)
CaSO4.2H2O - Calcium tetraoxosulphate(VI)-dihydrate (Gypsum)
Na2CO3.10H2O - Sodium trioxocarbonate(IV)-decahydrate (Washing Soda)
CuSO4.5H2O - Copper (II) tetraoxosulphate(VI)-pentahydrate (Blue Vitriol)
ZnSO4.7H2O - Zinc tetraoxosulphate(VI)-heptahydrate (White Vitriol)
Cu(NO3)2.3H2O - Copper (II) trioxonitrate(V)-trihydrate
From the above examples, it can be observed that each salt has a definite number of molecules of water of crystallization attached to it.
Therefore, it can also be defined as the definite amount of water some substances chemically combine with to form salt from their aqueous solution, as shown in the formation of CuSO4.5H2O:
CuSO4(s) + 5H2O(l) ---> CuSO4.5H2O(s)
In the equation, five molecules of water of crystallization combine with a molecule of CuSO4 to form the pentahydrate.
However, the molecules of water of crystallization are loosely attached to the hydrated salt and can easily dissociate on heating, leaving behind the anhydrous form of the salt. The water molecules can also be lost through efflorescence, a process whereby hydrated salts lose their water of crystallization when exposed to the atmosphere over a period of time.
The definite number of molecules of water of crystallization present in the hydrated salt gives it a definite crystalline structure and heating it above a particular temperature leads to the loss of the water of crystallization leaving behind a residue of the anhydrous salt, which is amorphous (formless or noncrystalline) in nature. Most times, this is usually characterized by a change in colour.
Generally,
Hydrated Salt ----> Anhydrous Salt + Water of Crystallization
crystalline                   amorphous
Determination of Amount of Water of Crystallization in a Hydrated Salt
CuSO4.5H2O(s) -----> CuSO4(s)   +   5H2O(l)
blue hydrated            white anhydrous
(crystalline)                (amorphous)
Now, let's apply a little mole concept by calculating their molar masses:
Molar mass of CuSO4.5H2O = (1*Cu) + (1*S) + (4*O) + 5[(2*H) + (1*O)]
= (1*63.5) + (1*32) + (4*16) + 5[(2*1) + (1*16)]
= 63.5 + 32 + 64 + 5(18)
= 159.5 + 90
= 249.5 g/mol
Similarly, molar mass of CuSO4 = 159.5 g/mol
and, molar mass of 5H2O = 90 g/mol
           CuSO4.5H2O(s) -----> CuSO4(s)   +   5H2O(l)
Mole ratio        1                :             1             :        5
Molar mass    249.5                     159.5                90
From the above stoichiometry, it implies that 1 mole of the pentahydrate on decomposition will produce 1 mole of the anhydrous salt and 5 moles of water, and vice-versa.
Also, 249.5g of CuSO4.5H2O will decompose to give 159.5g of CuSO4 and 90g of H2O. This implies that given the masses of the hydrated salt and its anhydrous component or water of crystallization, we can easily calculate the percentage by mass of the water of crystallization present in it as shown below:
Mass of hydrated salt = Mass of anhydrous salt + Mass of WC ………(i)
Therefore,
Mass of WC = Mass of hydrated salt - Mass of anhydrous salt ………(ii)
Then,
Percentage by mass of water of crystallization
= (Mass of WC / Mass of hydrated salt) x 100 ………(iii)
Or
[(Mass of hydrated - Mass of anhydrous) / Mass of hydrated] x 100 ………(iv)
So, for CuSO4.5H2O, the % by mass of the water of crystallization content will be:
= (90/249.5) x 100
= 0.36 x 100
= 36.1 %
While, the % by mass of the anhydrous content will be:
= (159.5/249.5) x 100
= 0.64 x 100
= 63.9 %
The above can also be obtained by subtracting the % by mass of the water of crystallization from 100%, i.e,
100% - 36.1% = 63.9%
From the above analysis, we can deduce that in a given hydrated salt,
(Mass of WC / Mass of Anhydrous Salt) = (% by Mass of WC / % by mass of Anhydrous Salt) ………(v)
Hence, for CuSO4.5H2O, equation (v) will give:
90g/159.5g = 36.1%/63.9% = 9/16
Also,
(Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt) = No. of molecules of water of crystallization ………(vi)
Equation (vi) above can be reduced to:
No. of moles of WC / No. of moles of anhydrous salt = No. of molecules of water of crystallization ………(vii)
Similarly,
(Reacting mass of WC/Reacting mass of anhydrous salt) = (Mass of WC in 1 mole of hydrated salt/Molecular mass of anhydrous salt) ………(viii)
Equations (vi), (vii) or (viii) is used when the complete formula of the hydrated salt is not given. Usually, the molecular mass of the anhydrous salt will be given.
Examples
Question 1
5.00g of hydrated salt of barium when heated to a constant weight gave 4.26g of anhydrous salt with a molecular weight of 208. Calculate:
a) the percentage by mass of water of crystallization
b) the number of molecules of water of crystallization in the hydrated salt.
Answer
Mass of hydrated salt = 5.00 g
Mass of anhydrous salt = 4.26 g
Mass of water of crystallization  = 5.00 - 4.26
= 0.74 g
Molecular mass of anhydrous salt = 208
a) % by mass of water of crystallization
= (Mass of WC/Mass of hydrated salt) x 100
= (0.74/5.00) x 100
= 0.148 x 100
= 14.8 %
b) Molecular mass of H2O = (2*H) + (1*O)
                                                = (2 x 1) + (1 x 16)
                                                = 2 + 16
                                                = 18
No. of molecules of water of crystallization = (Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt)
No. of WC = (0.74/18)/(4.26/208)
                   = 0.0411/0.0205
                   = 2.005, approximately 2.00
Therefore, the hydrated barium salt has 2 molecules of water of crystallization.
Note that the formula of the hydrated salt is not provided in this question, hence, there is no need for the equation of reaction. Equation (vi), therefore, provides the best solution to the problem.
Question 2
Some copper (II) tetraoxosulphate (VI) pentahydrate was heated at 120°C with the following results:
Mass of crucible = 10.00g
Mass of crucible + CuSO4.5H2O = 14.98 g
Mass of crucible + residue = 13.54 g
How many molecules of water of crystallization were lost? [H = 1, Cu = 63.5, O = 16, S = 32]
Answer
Mass of crucible = 10.00 g ………(a)
Mass of crucible + hydrated salt = 14.98 g ………(b)
Mass of crucible + anhydrous salt = 13.54 g ………(c)
Mass of hydrated salt = (b) - (a)
                                        = 14.98 - 10.00
                                        = 4.98 g …………(d)
Mass of anhydrous salt = (c) - (a)
                                           = 13.54 - 10.00
                                           = 3.54 g …………(e)
Mass of water of crystallization lost = (d) - (e)
                                                                = 4.98 - 3.54
                                                                = 1.44 g
No. of molecules of WC lost = (Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt)
but,
Molecular mass of H2O = 18
Molecular mass of anhydrous CuSO4 = 159.5 (see calculations above)
Substitute the values into the equation above:
No. of molecules of WC lost = (1.44/18) / (3.54/159.5)
                                                  = 0.08/0.022
                                                  = 3.64
Therefore, approximately 4 molecules of water of crystallization were lost.
Question 3
0.499g of CuSO4.xH2O when heated to constant weight gave a residue of 0.346g. Find the value of x. [Cu = 63.5, S = 32, O = 16, H = 1]
Answer
Hydrated salt = CuSO4.xH2O
                CuSO4.xH2O(s) -----> CuSO4(s)   +   xH2O(l)
Mass of hydrated salt = 0.499g
Mass of anhydrous residue = 0.346g
Mass of water of crystallization lost (x) = 0.499 - 0.346 = 0.153 g
Molar mass of anhydrous CuSO4 = 159.5 g/mol
Molar mass of H2O = 18 g/mol
(Always endeavour to show the full workings)
No. of molecules of WC lost (x) = (Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt)
No. of molecules of WC lost (x) = (0.153/18) / (0.346/159.5)
                                                        = 0.0085/0.0022
                                                        = 3.86
Therefore, x is approximately 4
Question 4
1.34g of hydrated sodium tetraoxosulphate (VI) was heated to give an anhydrous salt weighing 0.71g. What is the formula of the hydrated salt? [Na = 23, S = 32, O = 16, H = 1]
Answer
Proposed formula of hydrated salt = Na2SO4.xH2O
On heating, it decomposes according to the equation
                 Na2SO4.xH2O(s) -----> Na2SO4(s)   +   xH2O(l)
Mass of hydrated salt = 1.34 g
Mass of anhydrous salt = 0.71 g
Mass of water of crystallization = 1.34 - 0.71
                                                         = 0.63 g
No. of molecules of WC lost (x) = (Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt)
but,
Molar mass of Na2SO4 = (2*Na) + (1*S) + (4*O)
                                          = (2*23) + (1*32) + (4*16)
                                          = 46 + 32 + 64
                                          = 142 g/mol
No. of molecules of WC lost (x) = (0.63/18) / (0.71/142)
                                                       = 0.035/0.005
                                                       = 7
Since x = 7, therefore, the formula of the hydrated salt is Na2SO4.7H2O
Question 5
A hydrated salt of formula MSO4.xH2O contains 45.3% by mass of water of crystallization. Calculate the value of x. [M = 56, S = 32, O = 16, H = 1]
Answer
Hydrated salt = MSO4.xH2O
Percentage by mass of WC = 45.3 %
Percentage by mass of MSO4 = 100 - 45.3
                                                     = 54.7 %
Mass of anhydrous salt, MSO4 in the hydrated salt
= (1*M) + (1*S) + (4*O)
= (1 x 56) + (1 x 32) + (4 x 16)
= 56 + 32 + 64
= 152 g
Mass of WC, xH2O, in the hydrated salt = x[(2*H) + (1*O)]
                                                                      = x[(2 × 1) + (1 × 16)]
                                                                      = x(2 + 16)
                                                                      = 18x g
Using equation (v),
(Mass of WC / Mass of anhydrous salt) = (% by mass of WC / % by mass of anhydrous salt)
Hence,
18x/152 = 45.3/54.7
18x/152 = 0.828
        18x = 0.828 x 152
               = 125.86
            x = 125.86/18
               = 6.99, approximately 7
Question 6
Calculate the percentage of water of crystallization in ZnSO4.7H2O [Zn = 65, S = 32, O = 16, H = 1]
Answer
Hydrated salt = ZnSO4.7H2O
We will make use of equation (iii), i.e,
Percentage by mass of water of crystallization
= (Mass of water of crystallization / Mass of hydrated salt)
but,
Mass of ZnSO4.7H2O = (1*Zn) + (1*S) + (4*O) + 7[(2*H) + (1*O)]
                                       = (1 x 65) + (1 x 32) + (4 x 16) + 7[(2 x 1) + (1 x 16)]
                                       = 65 + 32 + 64 + 7(18)
                                       = 161 + 126
                                       = 287 g
Mass of WC in hydrated salt = 7 x 18
                                                   = 126 g
Therefore,
Percentage by mass of water of crystallization
= (126/287) x 100
= 0.439 x 100
= 43.9 %
Do These
Question 1
24.6g of iron (II) tetraoxosulphate (VI) crystals when heated gave 12.0g of the anhydrous salt. If it lost its water of crystallization due to the heating, find the:
a) percentage mass of the water of crystallization
b) molecules of water of crystallization in the hydrated salt. [Fe = 56, S = 32, O = 16, H = 1]
Question 2
If Na2CO3.10H2O contains 62% of water of crystallization, what mass of the hydrated salt are expected to be obtained from 7.38g of anhydrous sodium tetraoxosulphate (VI). [Na = 23, S = 32, O = 16, H = 1]
Question 3
18.50g of white vitriol,  ZnSO4.7H2O, were heated in a crucible to a constant mass, first at a temperature of 120°C, and then at 140°C. The residue that was obtained at 120°C weighed 12.70g and that obtained at 140°C weighed 10.38g. Calculate the stages of dehydration reached (molecules of water of crystallization lost) at 120°C and 140°C respectively. [Zn = 65, S = 32, O = 16, H = 1]
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