4.6 MOLECULAR
FORMULAS
2
3 Steps for determining Chemical
Formulas
1. Determine the percent composition of all elements.
2. Convert this information into an empirical formula
3. Find the true number of atoms/ elements in the
compound (Molecular Formula)
4.6 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
• Empirical Formula
- shows the ratio between atoms
Example: CH2O
• Molecular Formula
- shows the actual number of atoms
Example: C6H12O6
Steps to Determine Molecular Formula
1. List the given values (if the empirical formula is not
given, you must determine that before moving on).
2. Determine the molar mass for the empirical formula.
3. Divide the Molecular molar mass by the empirical
formula molar mass.
4. Calculate Molecular Formula by multiplying this
number by the empirical formula.
Example 1: 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 Molecular molar mass by
the empirical formula molar mass.
Molecular formula molar mass
Empirical formula molar mass
=
93.12 g/mol
31.04 g/mol
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
The empirical formula is not given so you must calculate it
now.
a. Calculate the mass of each element in a 100g sample
mC=40.03g mO=53.30g mH=6.67g
b. Convert Mass (m) into moles (n)
nC= 40.03g x 1mol/ 12.01g = 3.33 mol C
nH= 6.67g x 1 mol/ 1.01g = 6.60 mol H
nO= 53.30g x 1 mol/ 16.00g = 3.33 mol O
c. State the Amount Ratio
nC
: nH
: nO
3.33mol : 6.60mol : 3.33 mol
d. 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 2: Determine the molar mass for the
empirical formula
MEmpirical = 12.01g/mol + 2(1.01g/mol) + 16.00g/mol
= 30.03 g/mol
Step 3. Divide the molar mass by the empirical formula
molar mass.
Molar mass
= 180.18 g/mol
Empirical formula molar mass 30.03 g/mol
Step 4. 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 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?
List given values
C = 32.0%
H = 6.70%
Mformula = 75.08g/mol
1.
O = 42.6%
N = 18.7%
*** Empirical Formula is not given so you must now
calculate it!
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= 32.0g x 1mol/ 12.01g = 2.66 mol C
nH= 6.70g x 1mol/ 1.01g = 6.65 mol H
nO= 42.6g x 1mol/ 16.00g = 2.66 mol O
nN= 18.7g x 1mol/14.01g = 1.33 mol N
State the Amount Ratio
nC
: nH
: nO
2.66mol : 6.65mol : 2.6 mol
: nN
: 1.33mol
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
2. Determine the molar mass for the empirical
formula
MEmpirical = 75.08g
3. Divide the molar mass by the empirical formula
molar mass.
Molar mass
=
Empirical formula molar mass
75.08 g/mol
75.08 g/mol
4. 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
Assignment
• Proceed with the assignment that follows
Molecular Formula Check Point
1. The empirical formula of a compound is CH2. Its
molecular mass is 70g/mol. What is its molecular
formula?
2. A compound is found to be 40.0% carbon, 6.7%
hydrogen and 53.5% oxygen. Its molecular mass
is 60.0g/mol. What is its molecular formula?
Molecular Formula Check Point #2
1. A class of compounds called sodium
metaphosphates were used as additives to
detergents to improve cleaning ability. One of them
has a molecular mass of612g. Analysis shows the
composition to be 22.5% Na, 30.4% P, and 47.1 %
O. Determine the molecular formula of this
compound