Empirical & Molecular Formulas
• Law of Constant Composition a
compound contains elements in a certain
fixed proportions (ratios), regardless of how
the compound is prepared or where it is
found in nature.
• If you have one molecule of methane gas,
you will always have 1 carbon atoms and 4
hydrogen atoms.
Empirical Formula
• Empirical Formula is the formula that gives the
lowest ratio of atoms in a compound. It does not
necessarily tell you the exact number of each type
of atom.
• Example 1: The percent composition of a
compound is 69.9 % iron and 30.1% oxygen.
What is the empirical formula of a compound?
Example 1: The percent composition of a
compound is 69.9 % iron and 30.1% oxygen.
What is the empirical formula of a compound?
Step 1: List the given values
Fe=69.9% and O = 30.1%
Step 2: Calculate the mass (m) of each element in a
100g sample.
mFe= 69.9 x 100g = 69.9g
100
mO= 30.1 x 100g = 30.1g
100
Step 3: Convert Mass (m) into moles (n)
nFe= m/M = 69.9g/55.86g/mol = 1.25 mol Fe
nO= m/M = 30.1g/16.00g/mol = 1.88 mol O
Step 4: State the Amount Ratio
nFe
:
nO
1.25mol : 1.88 mol
Step 5: Calculate lowest whole number ratio
1.25mol : 1.88 mol When you don’t get a whole number,
1.25mol 1.25 mol multiply entire ratio by 2, 3, 4 etc.
until you get a whole number
1
: 1.5
2
: 3
Empirical Formula
is Fe2O3
Example 2: The percent
composition of a compound is
21.6% sodium, 33.3% chlorine,
and 45.1% oxygen. What is the
empirical formula of the
compound?
Step 1: List the given values
Cl=33.3%, Na = 21.6% and O = 45.1%
Step 2: Calculate the mass (m) of each element in a
100g sample.
mCl= 33.3 x 100g = 33.3g Cl
100
mNa= 21.6 x 100g = 21.6g Na
100
mO= 45.1 x 100g = 45.1g O
100
Step 3: Convert Mass (m) into moles (n)
nCl= m/M = 33.3g/35.5g/mol = 0.94 mol Cl
nNa= m/M = 21.6g/23.0g/mol = 0.94 mol Na
nO= m/M = 45.1g/16.00g/mol = 2.82 mol O
Step 4: State the Amount Ratio
nFe
: nNa
: nO
0.94mol : 0.94mol : 2.82 mol
Step 5: Calculate lowest whole number ratio
0.94mol : 0.94mol : 2.82 mol
0.94mol : 0.94mol : 0.94 mol
1 : 1: 3
Empirical Formula
is NaClO3
Molecular Formula
• Molecular Formula of a compound tells you
exact number of atoms in one molecule of a
compound. This formula may be equal to
the empirical formula or may be a multiple
of this formula.
• To determine, you need:
– The empirical formula
– The molar mass of the compound
Molecular Formula
- shows the actual number of atoms
Example: C6H12O6
Empirical Formula
- shows the ratio between atoms
Example: CH2O
The empirical formula of a compound is
CH3O and its molar mass is 93.12g/mol.
What is the molecular formula?
Step 1: List given values
Empirical Formula=CH3O
Mcompound = 93.12 g/mol
Step 2: Determine the molar mass for
the empirical formula, CH3O.
MEmpirical = 12.01g/mol + 3(1.01g/mol) + 16.00g/mol
= 31.04 g/mol
Step 3. Divide the molar mass by the
empirical formula molar mass.
Molecular formula molar mass = 93.12 g/mol
Empirical formula molar mass 31.04 g/mol
=3
Step 4. Calculate Molecular Formula by
multiplying this number by the empirical
formula.
Molecular formula = x (empirical formula)
3 x CH3O
Therefore, the molecular formula is C3H9O3
Example 2: The percent composition of a
compound is determined by a combustion and
analyzer is a 40.03% carbon, 6.67% hydrogen, &
53.30% oxygen. The molar mass is 180.18g/mol.
What is the molecular formula
Step 1: List given values
C= 40.03%, O=53.30%, H=6.67%
Mcompound = 180.18 g/mol
Step 2: Calculate the mass of each element in
a 100g sample
mC=40.03g
mO=53.30g
mH=6.67g
Step 3: Convert Mass (m) into moles (n)
nC= m/M = 40.03g/12.01g/mol = 3.33 mol C
nH= m/M = 6.67g/1.01g/mol = 6.60 mol H
nO= m/M = 53.30g/16.00g/mol = 3.33 mol O
Step 4: State the Amount Ratio
nC : nH
: nO
3.33mol : 6.60mol : 3.33 mol
Step 5: Calculate lowest whole number ratio
3.33mol : 6.60mol : 3.33 mol
3.33mol : 3.33mol : 3.33 mol
1 : 2: 1
Empirical Formula
is CH2O
Step 6: Determine the molar mass for the
empirical formula
MEmpirical = 12.01g/mol + 2(1.01g/mol) + 16.00g/mol
= 30.03 g/mol
Step 7. Divide the molar mass by the empirical
formula molar mass.
180.18 g/mol
Molar mass
Empirical formula molar mass
= 30.03 g/mol
=6
Step 8. Calculate Molecular Formula by multiplying
this number by the empirical formula.
Molecular formula = x (empirical formula)
6 x (CH2O)
Therefore, the molecular formula is C6H12O6
Example 3: The percent composition of a
compound is determined by a combustion and
analyzer is a 32.0% carbon, 6.70% hydrogen,
42.6% oxygen & 18.7% nitrogen. The molar mass is
75.08g/mol. What is the molecular formula?
Calculate the mass of each element in a 100g sample
mC=32.0g mO=42.6g mH=6.70g mN=18.7g
Convert Mass (m) into moles (n)
nC= m/M = 32.0g/12.01g/mol = 2.66 mol C
nH= m/M = 6.70g/1.01g/mol = 6.65 mol H
nO= m/M = 42.6g/16.00g/mol = 2.66 mol O
nN= m/M = 18.7g/14.01g/mol = 1.33 mol N
State the Amount Ratio
nC
: nH
: nO
: nN
2.66mol : 6.65mol : 2.6 mol:
1.33mol
Step 5: Calculate lowest whole number ratio
2.66mol : 6.65mol : 2.6 mol:
1.33mol
1.33mol : 1.33mol : 1.33 mol: 1.33mol
2 : 5: 2: 1
Empirical Formula
is C2H5O2N
Determine the molar mass for the empirical
formula
MEmpirical = 75.08g
Divide the molar mass by the empirical formula
molar mass.
Molar mass
Empirical formula molar mass
75.08 g/mol
75.08 g/mol
=
=1
Calculate Molecular Formula by multiplying this
number by the empirical formula.
Molecular formula = x (empirical formula)
1 x (C2H5O2N)
Therefore, the molecular formula is C2H5O2N