Formulas and Composition
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Transcript Formulas and Composition
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%
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%
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
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
% Composition of Hydrates
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
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
You practice – write the
molecular and empirical formulas
Copper (II) Phosphite
Sulfurous Acid
Copper (II) Chloride Dihydrate
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
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