The chemical composition can be
expressed as the mass percent of
each element in the compound.
Carry out % mass to the
hundredths place.
Example: Determine the percent
composition of C3H8.

Assume you have one mole of the substance.
3 C = 3(12.01) = 36.03 g
%C 
36.03g
100  81.68%
44.11g
8 H = 8(1.01) = 8.08 g
%H 
8.08g
100  18.32%
44.11g
Total MW = 44.11 g
Example: Determine the percent
composition of iron(III) sulfate.

Iron(III) sulfate = Fe2(SO4)3
2 Fe = 2(55.85) = 111.70 g
3 S = 3(32.07) = 96.21 g
12 O = 12(16.00) = 192.00 g
Total MW = 399.91 g
%Fe 
111.70g
100  27.93%
399.91g
%S 
96.21g
100  24.06%
399.91g
%O 
192.00g
100  48.01%
399.91g
Hydrated Compounds
Some compounds exist in a “hydrated” state.
Some specific number of water molecules are
present for each molecule of the compound.
Example: oxalic acid (COOH)2 can be obtained
in the laboratory as (COOH)2•2H2O.
Note: the dot in (COOH)2•2H2O
shows that the crystals of oxalic
acid contain 2 water molecules
per (COOH)2 molecule.
Example: oxalic acid (COOH)2 can be obtained
in the laboratory as (COOH)2•2H2O.

The molar mass of (COOH)2 = 90.04 g/mol
90.04
% anhydrous molecule 
100  71.41%
126 .08

The molar mass of (COOH)2•2H2O = 126.08 g/mol
36.04
% water 
100  28.59%
126 .08
Water can be driven out of a hydrated compound
by heating it to leave an “anhydrous” (without
water) compound.

Example: A 7.0 g sample of calcium nitrate,
Ca(NO3) 2•4H2O, is heated to constant mass.
How much anhydrous salt remains?

% mass anhydrous salt =
164.10g
100  69.48%
236.18g

Mass of anhydrous salt remaining =
.6948 7.0g  4.9 g
Hydration Number
Some molecules attach themselves to water
molecules. This is done in set numbers, depending on
the molecule. For example, Magnesium sulfate
attaches to 7 water molecules. We say it’s hydration
number is 7
MgSO47H2O
(Magnesium Sulfate Heptahydrate)
Anhydrides
A compound that is normally a hydrate and
has lost its hydration water is said to be
anhydrous and is called an anhydride.
BaCl22H2O
Barium Chloride Dihydrate
BaCl2
Barium Chloride Anhydride
-or-
Anhydrous Barium Chloride
Finding the Hydration Number
The hydration number can be conveniently
found by heating the compound and measuring its
mass loss. This mass loss is usually due to the
hydration water molecules being driven off.
For example…
A 15.35 g sample of Strontium nitrate,
Sr(NO3)2nH2O, is heated to a constant mass of
11.45 g. Calculate the hydration number.
Sample Data:
Mass Hydrate
15.35g
Mass Anhydride
11.45g
Mass of Water (mass loss)
3.90g
Calculations
Moles Anhydride
11.45g Sr(NO3 ) 2

1 molSr(NO3 ) 2
 0.0541molSr(NO3 ) 2
211.64g Sr(NO3 ) 2
Moles Water
3.90g H 2O
1 mol H 2O

 0.216 mol H 2O
18.02g H 2O
Molar ratio… mol H O
0.216 H 2O
2

 3.99  4
molSr(NO3 ) 2
0.0541molSr(NO3 ) 2
Hydration Number is 4, Sr(NO3)24H2O
(Strontium Nitrate Tetrahydrate)