

AP 1.3 draft
Presentation
•
Chemistry
•
11th Grade
•
Practice Problem
•
Hard
Peter Spiessbach
FREE Resource
12 Slides • 0 Questions
1
1.3 Composition of pure
substances
John D. Bookstaver
St. Charles Community
College
Cottleville, MO
© 2009, Prentice-Hall, Inc.
Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E.
Bursten
2
Stoichiometry
Formula Weight (FW)
© 2009, Prentice-Hall, Inc.
●A formula weight is the sum of the atomic weights
for the atoms in a chemical formula.
●So, the formula weight of calcium chloride, CaCl2,
would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
●Formula weights are generally reported for ionic
compounds.
© 2009, Prentice-Hall, Inc.
3
Stoichiometry
Molecular Weight (MW)
© 2009, Prentice-Hall, Inc.
●A molecular weight is the sum of the atomic
weights of the atoms in a molecule.
●For the molecule ethane, C2H6, the molecular
weight would be
C: 2(12.0 amu)
30.0 amu
+ H: 6(1.0 amu)
© 2009, Prentice-Hall, Inc.
4
Stoichiometry
Percent Composition
© 2009, Prentice-Hall, Inc.
One can find the percentage of the mass of a
compound that comes from each of the elements in
the compound by using this equation:
% element =
(number of atoms)(atomic weight)
(FW of the compound)
x 100
© 2009, Prentice-Hall, Inc.
5
Stoichiometry
Percent Composition
© 2009, Prentice-Hall, Inc.
So the percentage of carbon in ethane (C2H6) is…
%C =
(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu
=
x 100
= 80.0%
© 2009, Prentice-Hall, Inc.
6
Stoichiometry
Calculating Empirical Formulas
One can calculate the empirical formula from
the percent composition.
© 2009,
Prentice-Hall, Inc.
The compound para-aminobenzoic acid (PABA used to be included in
sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen
(10.21%), and oxygen (23.33%). Find the empirical formula of PABA.
© 2009,
Prentice-Hall,
Inc.
7
Stoichiometry
Calculating Empirical Formulas
© 2009, Prentice-Hall, Inc.
Assuming 100.00 g of para-aminobenzoic acid,
C:
61.31 g x
= 5.105 mol C
H:
5.14 g x
= 5.09 mol H
N:
10.21 g x
= 0.7288 mol N
O:
23.33 g x
= 1.456 mol O
1 mol
12.01 g
1 mol
14.01 g
1 mol
1.01 g
1 mol
16.00 g
© 2009, Prentice-Hall, Inc.
8
Stoichiometry
Calculating Empirical Formulas
© 2009, Prentice-Hall, Inc.
Calculate the mole ratio by dividing by the smallest number of moles:
© 2009, Prentice-Hall, Inc.
C:
= 7.005 ≈ 7
H:
= 6.984 ≈ 7
N:
= 1.000
O:
= 2.001 ≈ 2
5.105 mol
0.7288 mol
5.09 mol
0.7288 mol
0.7288 mol
0.7288 mol
1.458 mol
0.7288 mol
9
Stoichiometry
Calculating Empirical Formulas
© 2009, Prentice-Hall, Inc.
These are the subscripts for the empirical formula:
C7H7NO2
© 2009, Prentice-Hall, Inc.
10
Stoichiometry
Copyright © Houghton
Mifflin Company. All rights
reserved.
3a–*
Figure 3.6: Examples of substances whose
empirical and molecular formulas differ. Notice that
molecular formula = (empirical formula)n, where n
is an integer.
© 2009, Prentice-Hall, Inc.
11
Stoichiometry
© 2009, Prentice-Hall, Inc.
12
Stoichiometry
© 2009, Prentice-Hall, Inc.
1.3 Composition of pure
substances
John D. Bookstaver
St. Charles Community
College
Cottleville, MO
© 2009, Prentice-Hall, Inc.
Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E.
Bursten
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