Formulas and Composition
Percent Composition
Percent composition lists a percent each
element is of the total mass of the
compound
 H2O
 Total mass is 18.02
 Mass of O is 16, so O is 16/18.02 or
88.79%
 Mass of H is 2.02, so H is 2.02/18.02 or
11.21%
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You Practice - Iron (III) Sulfate
Fe2(SO4)3
 2*Fe = 111.70 g/mol
 3*S = 96.21 g/mol
 12*O = 192.00 g/mol
 Total mass = 399.91 g/mol
 Fe = 111.70/399.91 => 27.93%
 S = 96.21/399.91 => 24.06%
 O = 192.00/399.91 => 48.01%
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Hydrates
Some compounds come with water attached
 These are known as hydrates
 The water is not part of the molecule, just
attached to the molecule
 A hydrate has a specific amount of water
attached to each molecule
 Iron (II) Sulfate is FeSO4* Iron (II) Sulfate is
FeSO4*7H2O
 There are 7 water molecules attached to one
Iron (II) Sulfate
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Molar Mass of Hydrates
Finding the molar mass will include the
mass of the attached water molecules
 Iron (II) Sulfate is FeSO4*7H2O
 Fe + S + 4*O + 14*H + 7*O
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% Composition of Hydrates
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Iron (II) Sulfate is FeSO4*7H2O
2*Fe = 111.70 g/mol
3*S = 96.21 g/mol
12*O = 192.00 g/mol
7*H2O = 126.14 g/mol
Total mass = 526.05 g/mol
Fe = 111.70/526.05 => 21.23%
S = 96.21/526.05 => 18.29%
O = 192.00/526.05 => 36.50%
H2O =126.14/526.05 => 23.98%
Molecular Formula
So far, we have been dealing with molecular
formulas
 These are the complete formula for a
molecule of the ionic compound
 Ionic compounds are always written in the
reduced form – we use NaCl, not Na2Cl2
 This is not true for covalent molecules
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Glucose is C6H12O6
 This is the molecular formula as is shows
the complete number of each element in the
molecule
 This formula can be reduced to CH2O
 The reduced formula, if it is not the
molecular formula, is called the empirical
formula
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You practice – write the
molecular and empirical formulas
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Copper (II) Phosphite
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Sulfurous Acid
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Copper (II) Chloride Dihydrate
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Sucrose (C12H22O12)
Using % Composition to find the
empirical formula
Hydrogen Peroxide has a molar mass of
34.02 g/mol
 H is 5.94%, O is 94.06%
 Using this, we can assign masses to each
element based on the %
 94.06 g O and 5.94 g H
 Now we convert these to moles by dividing
by molar mass
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94.06 g O (1 mol O/16 g O) = 5.88 mol O
5.94 g H (1 mol H/1.01 g H) = 5.88 mol H
Therefore the empirical formula is HO
Using this, and the molar mass, we can find the
molecular formula
x * (HO) = 34.02
x* (17.01) = 34.02
x=2
Formula = H2O2
You Try It
This compound is 12.63% Li, 29.15% S,
and 58.22% O. Its molar mass is 109.92
g/mol.
 Find the empirical and molecular formulas
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