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Determining the water content of a hydrate crystal
When you deal with a hydrated crystal, one common task is to
determine the number of water molecules attached to each
crystal. For example, copper II sulfate pentahydrate is
CuSO4*5H2O. Therefore for every one molecule of copper II
sulfate there are five molecules of water attached.
In the lab, the normal way to find the water of hydration is to
heat the solid crystal until all the water evaporates (think back to
the reaction type lab when we heated the blue copper II sulfate
pentahydrate until it turned to a white solid). You weigh the compound before you heat it
and weigh it again after heating. All the mass that disappears should be water that boiled
off; conversely, all the solid that remains should be the anhydrous crystal (the chemical you
really want, like copper II sulfate).
Percent to mass
Mass to moles
Divide by smallest
Multiply ‘til whole
The problem then becomes an empirical formula problem
that jumps to the second line of the poem. The only
difference between this problem and all the other
empirical formula problems is that the pieces that you
are finding the moles of are NOT ELEMENTS. The
pieces are MOLECULES! One molecule is ALWAYS
H2O. The other molecule is the crystal that you want.
For example, you begin with 2.50 grams of a barium
hydroxide hydrate crystal. After heating, only 1.36 g of
the solid remains. This means that 1.14 g of water
evaporated.
Now that we have the mass of barium hydroxide that
remains (1.36 g) and the mass of the water that evaporated (1.14 g)
gsolid
gH 2O
molesBa(OH) 2 
molesH 2O 
FM
FM
1.36
1.14
 7.94e  3molBa(OH) 2
 .0633molH 2O
171.3
18
At this point, we have taken mass to moles. We divide by the smallest (which should be the
.0633molH
 2O  7.98H2O  8H2O
crystal).
So, the water of hydration number is
.00794molBa(OH)2 1Ba(OH)2 1Ba(OH)2
8. The formula is Ba(OH)2*8H2O. Multiplying ‘til whole is not usually necessary with water of
hydration.
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