Hydrates
Compounds that have water
molecules attached
A hydrate is generally an ionic compound which
has a certain number of water molecules
attached.
Their formulas are written:
CuSO4 *5H2O
The * means associated with the compound but
they are not bonded with intramolecular forces.
Names for Hydrates
The “hydrate” adds to the name of the ionic
compound by the following:
Copper (II) sulfate pentahydrate
Penta is for the five water molecules associated
with the ionic compound and hydrate is for the
water molecules.
Words to use for the number of water
molecules
Mono
Di
Tri
Tetra
Penta
Hexa
Hepta
Octa
Nona
Deca
Practice Decoding
Copper (I) sulfite
monohydrate
Cu2SO3 *H2O
Cobalt (II) fluoride
tetrahydrate
CoF2*4H2O
Lithium nitrate trihydrate
LiNO3 * 3H2O
Hydrates
Because the water molecules are associated
with the formula units of the ionic compounds,
they increase the mass.
You must include the mass of the associated
water molecules when figuring the gram
formula mass (not molecular because it is ionic,
but it is the same process).
Hydrates
The water molecules can easily be removed
from the hydrates by heating (because they are
intermolecular forces not intramolecular forces).
Lab this week is about % water of a hydrate (or
one of the two labs).
Percentage
A 15.67 g sample of a hydrate of magnesium
carbonate was heated, without decomposing
the carbonate, to drive off the water. The mass
was reduced to 7.58 g. What is the formula of
the hydrate?
1) Determine mass of water driven off:
15.67 minus 7.58 = 8.09 g of water
2) Determine moles of MgCO3 and water:
MgCO3 ⇒ 7.58 g / 84.313 g/mol = 0.0899 mol
H2O ⇒ 8.09 g / 18.015 g/mol = 0.449 mol
3) Find a whole number molar ratio:
MgCO3 ⇒ 0.0899 mol / 0.0899 mol = 1
H2O ⇒ 0.449 mol / 0.0899 mol = 5
MgCO3 · 5H2O
Problem
A hydrate of Na2CO3 has a mass of 4.31 g before
heating. After heating, the mass of the
anhydrous compound is found to be 3.22 g.
Determine the formula of the hydrate and then
write out the name of the hydrate.
1) Determine mass of water driven off:
4.31 minus 3.22 = 1.09 g of water
2) Determine moles of Na2CO3 and water:
Na2CO3 ⇒ 3.22 g / 105.988 g/mol = 0.0304 mol
H2O ⇒ 1.09 g / 18.015 g/mol = 0.0605 mol
3) Find a whole number molar ratio:
Na2CO3 ⇒ 0.0304 mol / 0.0304 mol = 1
H2O ⇒ 0.0605 mol / 0.0304 mol = 2
Na2CO3 · 2H2O
sodium carbonate dihydrate
Problem
A 5.00 g sample of hydrated barium chloride,
BaCl2 · nH2O, is heated to drive off the water.
After heating, 4.26 g of anhydrous barium
chloride, BaCl2, remains. What is the value of n
in the hydrate's formula?
1) Calculate moles of anhydrous barium
chloride:
4.26 g / 208.236 g/mol = 0.020458 mol
2) Calculate moles of water:
5.00 minus 4.26 = 0.74 g 0.74 g / 18.015 g/mol =
0.041077 mol
3) Determine whole number ratio:
0.041077 / 0.020458 = 2
4) Formula is:
BaCl2 · 2H2O
Percent Error
Percent error is the determination of how far off
your values are from the currently accepted value.
|experimental value – theoretical value| *100% = %error
Theoretical value
Theoretical values are looked up from standards.
Or it is calculated from conversion factors not
measured in lab.
Your value is the experimental value.