Chapter 2
Empirical &
Molecular
Formulas
1
Types of Formulas
The formulas for compounds can be expressed as an
empirical formula and as a molecular (true) formula.
Empirical
CH
CH
CO2
CH2O
Molecular (true)
C2H2
C6H6
CO2
C5H10O5
Name
acetylene
benzene
carbon dioxide
ribose
2

An empirical formula represents the
simplest whole number ratio of the
atoms in a compound. They are
determined experimentally by finding
the mass of each element in a
compound.

The molecular formula is the true or
actual ratio of the atoms in a
compound.
3
Learning Check EF-1
A. What is the empirical formula for C4H8?
1) C2H4
2) CH2
3) CH
B. What is the empirical formula for C8H14?
1) C4H7
2) C6H12
3) C8H14
C. What is a molecular formula for CH2O?
1) CH2O 2) C2H4O2
3) C3H6O3
4
Solution EF-1
A. What is the empirical formula for C4H8?
2) CH2
B. What is the empirical formula for C8H14?
1) C4H7
C. What is a molecular formula for CH2O?
1) CH2O 2) C2H4O2
3) C3H6O3
5
Learning Check EF-2
If the molecular formula has 4 atoms of N,
what is the molecular formula if SN is the
empirical formula? Explain.
1) SN
2) SN4
3) S4N4
6
Solution EF-2
If the molecular formula has 4 atoms of N,
what is the molecular formula if SN is the
empirical formula? Explain.
3) S4N4
If the actual formula has 4 atoms of N, and
S is related 1:1, then there must also be 4
atoms of S.
7
Learning Check EF-3
A compound has a formula mass of 176.0 and an
empirical formula of C3H4O3. What is the
molecular formula?
*First we find the molar mass of the compound C3H4O3. Does it equal 176g/mol?
We then divide the given molecular mass of the compound by the molar mass
of the empirical formula.
The Molecular Formula is always a whole number multiplier of the
empirical formula.
1) C3H4O3
2) C6H8O6
3) C9H12O9
8
Solution EF-3
A compound has a formula mass of 176.0
and an empirical formula of C3H4O3. What is
the molecular formula?
2) C3H4O3 = 88.0 g for the EF
176.0 g
=2
88.0
= C6H8O6
9
Learning Check EF-4
If there are 192.0 g of O in the
molecular formula, what is the true
formula if the EF is C7H6O4?
1) C7H6O4
2) C14H12O8
3) C21H18O12
10
Learning Check- EF4
If there are 192.0 g of O in the
molecular formula, what is the true
formula if the EF is C7H6O4?
3) C21H18O12
192 g O
= 3 x O4 or 3 x C7H6O4
64.0 g O in EF
11
Solution EF-4
If there are 192.0 g of O in the
molecular formula, what is the true
formula if the EF is C7H6O4?
3) C21H18O12
192 g O
= 3 x O4 or 3 x C7H6O4
64.0 g O in EF
12
Finding the Molecular Formula
A compound is Cl 71.65%, C 24.27%, and
H 4.07%. What are the empirical and
molecular formulas? The molar mass is
known to be 99.0 g/mol.
1. State mass percents as grams in a
100.00 g sample of the compound.
Cl 71.65 g C 24.27 g
H 4.07 g
13
2. Calculate the number of moles of each
element.
71.65 g Cl = 2.02 mol Cl
35.5 g Cl
24.27 g C
12.0 g C
=
2.02 mol C
4.07 g H
1.01 g H
=
4.04 mol H
14
Why moles?
Why do you need the number of moles of
each element in the compound?
15
3. Find the smallest whole number ratio by
dividing each mole value by the smallest
mole values:
Cl: 2.02
=
1 Cl
2.02
C:
2.02
2.02
=
1C
H:
4.04
=
2H
2.02
4. Write the simplest or empirical formula
CH2Cl
16
5. EM (empirical mass)
= 1(C) + 2(H) + 1(Cl) =
49.5
6. n = molar mass/empirical mass
Molar mass
EM
=
99.0 g/mol = n = 2
49.5 g/EM
7.Molecular formula
(CH2Cl)2
= C2H4Cl2
17
Learning Check EF-5
Aspirin is 60.0% C, 4.5 % H and 35.5 O.
Calculate its simplest formula. In 100 g
of aspirin, there are 60.0 g C, 4.5 g H,
and 35.5 g O.
18
Solution EF-5
60.0 g C = ______ mol C
4.5 g H = _______mol H
35.5 g O = _______mol O
19
Solution EF-5
60.0 g C = 5.00 mol C
12.0 g C
4.5 g H
= 4.5 mol H
1.01 g H
35.5 g O = 2.22 mol O
16.0 g O
20
Divide by the smallest # of moles.
5.00 mol C
=
________________
______ mol O
4.5 mol H
______ mol O
=
________________
2.22 mol O =
________________
______ mol O
Are are the results whole numbers?_____
21
Divide by the smallest # of moles.
5.00 mol C
=
___2.25__
2.22 mol O
4.5 mol H
2.22 mol O
=
___2.00__
2.22 mol O =
___1.00__
2.22 mol O
Are are the results whole numbers?_____
22
Finding Subscripts
A fraction between 0.1 and 0.9 must not be
rounded. Multiply all results by an integer
to give whole numbers for subscripts.
0.5 x 2
0.333 x 3
0.25 x 4
0.75 x 4
=
=
=
=
1
1
1
3
23
Multiply everything x 4
C: 2.25 mol C
x 4 = 9 mol C
H: 2.0 mol Hx 4 = 8 mol H
O: 1.00 mol O
x 4 = 4 mol O
Use the whole numbers of mols as the
subscripts in the simplest formula
C9H8O4
24
Learning Check EF-6
A compound is 27.4% S, 12.0% N and 60.6
% Cl. If the compound has a molar mass of
351 g/mol, what is the molecular formula?
25
Solution EF 6
0.853 mol S /0.853 = 1 S
0.857 mol N /0.853 = 1 N
1.71 mol Cl /0.853
= 2 Cl
Empirical formula = SNCl2 = 117.1 g/EF
Mol. Mass/ Empirical mass
351/117.1 = 3
Molecular formula = S3N3Cl6
26