Empirical and Molecular
Formulas
Determining Empirical Formulas


Definition: the lowest whole number ratio of elements in
a compound.
Steps to determine empirical formulas:
1) find moles of each element [find the moles of H20 if
compound is hydrated].
2) divide by smallest number of moles to get a whole
number.
3) multiply by a common whole number, if necessary.
Examples:
Calculate EF of a compound with 25.9% N and 74.1% O.
1)
%=assume mass
mass mols
25.9g N x 1 mol N = 1.85 mols N
14.007g N
74.1g O x 1 mol O = 4.63 mols O
16 g O
2)
N: 1.85 = 1
1.85
O: 4.63 = 2.5
1.85
3) Multiply by common whole number
N: 1 x 2 =2
O: 2.5 x 2 = 5
= N2O5
Just remember this!!
Percents to grams
Grams to moles
Divide by smallest
Multiply ‘til whole
… examples continued
•
Given: 36.5g Na, 25.4g S, and 38.1g O
36.5g Na x 1 mol Na = 1.59 mols Na
22.989g Na
25.4g S x 1 mol S = .79 mols S
32g S
38.1g O x 1 mol O = 2.4 mols O
16g O
Na: 1.59 = 2
.79
S: .79 = 1
.79
O: 2.4 = 3
.79
= Na2SO3
An oxide of aluminum is formed by the reaction of
4.151g of aluminum with 3.692g of oxygen. Find EF.
4.151g Al x 1 mol Al = 0.1539 mol Al atoms
26.98 g Al
3.692g O x 1 mol O = 0.2308 mol O atoms
16 g O

0.1539 mol Al = 1.000 mol Al atoms
0.1539
0.2308 mol O = 1.500 mol O atoms
0.1539
1.500 O x 2 = 3.000 = 3 O atoms
1.000 Al x 2 = 2.000= 2 Al atoms
= Al2O3
Determining Molecular Formulas


Molecular formulas show the number of atoms in a compound. In
order to determine the molecular, you must have the empirical
formula first.
The molecular mass of molecule will always be given.
Multiple = molecular mass x
empirical formula (whole number)
•
The molecular formula is always an integer
multiple of the empirical formula.
Examples:

Determine the molecular formula of a compound whose
EF is CH2O and molecular mass is 120 g/mol
120 g/mol = 4
30 g/mol
-Distribute that 4 throughout the empirical formula =
C4H8O4
…continued

A compound is 64.9% carbon, 13.5% hydrogen, and
21.6% oxygen. Its molc mass is 74 g/mol. What is its MF?
64.9g C x 1 mol C = 5.40 mols C = 4 C
12.01 g C
13.5g H x 1 mol H = 13.37 mols H = 10 H
1.01 g H
1.35
21.6g O x 1 mol O = 1.35 mols O = 1 O
16.0 g O
1.35
74 g/mol = 1
74 g/mol
= C4H10O

Ex) a compound is 54.5% carbon, 9.1% hydrogen, and
36.4% oxygen. It’s molc mass is 88 g/mol. What is its
molecular formula?
54.5 C x 1 mol C = 4.54 mols C
12.01 g C
9.1 g H x 1 mol H = 9 mols H
1.01 g H
36.4 g O x 1 mol O = 2.28 mols O
16 g O
4.54 mols C = 2 C
2.28 mols
9 mols H = 4H
2.28 mols
2.28 mols O = 1O
2.28 mols
= C4H8O2

Ex) A compound has an empirical formula of
ClCH2 and a molecular weight of 98.96 g/mol.
What is it’s molecular formula?
Mass Cl + mass C + 2(mass H)
Mass of empirical unit= 35.45 +12.00 + 2(2.008) =
49.47 g/mol
98.96 g/mol = 2.000
49.47 g/mol
= Cl2C2H4