Download Empirical Formulas

Empirical Formulas &
Molecular Formulas
Empirical Formulas
Molecular Formulas
Types of Formulas
The formulas for compounds can be
expressed as an empirical formula and as a
molecular(true) formula.
Empirical
Molecular (true)
Name
CH
C 2H 2
acetylene
CH
C 6H 6
benzene
CO2
CO2
CH2O
C5H10O5
carbon dioxide
ribose
Empirical Formulas
Write your own one-sentence definition for
each of the following:
Empirical formula
Molecular formula
• An empirical formula represents the
simplest whole number ratio of the
atoms in a compound.
• The molecular formula is the true or
actual ratio of the atoms in a
compound.
Empirical Formulas
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
Solution
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
Molecular Formula
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
Solution
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.
Empirical and Molecular Formulas
To obtain the molecular formula, you
must first have an empirical formula.
You must also have the molar mass
or be given enough information to
determine the molar mass. With
these two values, you can determine
the molecular formula.
Empirical and Molecular Formulas
molar mass
EF mass
=
a whole number = n
n = 1 molar mass = empirical mass
molecular formula = empirical formula
n = 2 molar mass = 2 x empirical mass
molecular formula =
2 x empirical formula
molecular formula = or > empirical formula
Empirical
Formula
Empirical
Mass
Molecular
Formula
Molecular
Mass
Molecular Formula
A compound has a formula mass of 176.0
and an empirical formula of C3H4O3. What
is the molecular formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9
Solution
A compound has a formula mass of 176.0
and an empirical formula of C3H4O3. What is
the molecular formula?
2) C6H8O6
C3H4O3 = 88.0 g/EF
176.0 g
88.0
=
2.00
Molecular Formula
If there are 192.0 g of O in the molecular
formula, what is the true formula if the
Empirical Formula is C7H6O4?
1) C7H6O4
2) C14H12O8
3) C21H18O12
Solution
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
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
2. Calculate the number of moles of each
element.
71.65 g Cl x 1 mol Cl = 2.02 mol Cl
35.5 g Cl
24.27 g C x
1 mol C
12.0 g C
= 2.02 mol C
4.07 g H x
1 mol H
1.01 g H
=
4.04 mol H
Why moles?
Why do you need the number of moles
of each element in the compound?
Why moles?
Why do you need the number of moles
of each element in the compound?
Recall Avogadro’s number! A mole
contains equal numbers of particles. That
means we are comparing atoms on an
equal basis.
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:
H:
2.02
2.02
=
1C
4.04
=
2H
2.02
4. Write the simplest or empirical formula
CH2Cl
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
49.5 g/EM
7.Molecular formula
(CH2Cl)2
= C2H4Cl2
= n=2
Empirical Formula
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.
Solution
60.0 g C x
4.5 g H
___________= ______ mol C
x ___________ = _______mol H
35.5 g O x ___________ = _______mol O
Solution
60.0 g C x
1 mol C
=
5.00 mol C
=
4.5 mol H
=
2.22 mol O
12.0 g C
4.5 g H
x
1 mol H
1.01 g H
35.5 g O x
1mol O
16.0 g O
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?_____
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 the results whole numbers?_____
Multiply everything x 4
C: 2.25 mol C
x 4 = 9 mol C
H: 2.0 mol H
x 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
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.
(1/2)
(1/3)
(1/4)
(2/3)
(3/4)
0.5
0.333
0.25
0.667
0.75
x2
x3
x4
x3
x4
=
=
=
=
=
1
1
1
2
3
Molecular Formula
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?
Solution
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
Molecular formula = S3N3Cl6
351/117.1 = 3