___________________________________ Slide 1 ___________________________________ ___________________________________ Chapter 8
___________________________________ Chemical Composition
___________________________________ ___________________________________ ___________________________________ Slide Chapter 8
2 Table of Contents
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
___________________________________ ___________________________________ Counting by Weighing
Atomic Masses: Counting Atoms by Weighing
The Mole
Learning to Solve Problems
Molar Mass
Percent Composition of Compounds
Formulas of Compounds
Calculation of Empirical Formulas
Calculation of Molecular Formulas
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Slide Section 8.1
3 Counting by Weighing
2
___________________________________ ___________________________________ •
Objects do not need to have identical masses
to be counted by weighing.
 All we need to know is the average mass of
the objects.
• To count the atoms in a sample of a given
element by weighing we must know the mass
of the sample and the average mass for that
element.
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___________________________________ Slide Section 8.1
4 Counting by Weighing
___________________________________ Averaging the Mass of Similar Objects
•
Example: What is the mass of 1000 jelly beans?
___________________________________ 1. Not all jelly beans have the same mass.
2. Suppose we weigh 10 jelly beans and find:
Bean
1
2
3
4
5
6
7
8
9
10
Mass
5.1 g
5.2 g
5.0 g
4.8 g
4.9 g
5.0 g
5.0 g
5.1 g
4.9 g
5.0 g
___________________________________ ___________________________________ 3. Now we can find the average mass of a bean.
___________________________________ ___________________________________ 4. Finally we can multiply to find the mass of 1000 beans!
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___________________________________ Slide Section 8.1
5 Counting by Weighing
___________________________________ Averaging the Mass of Different Objects
•
Two samples containing different types of
components (A and B), both contain the same
number of components if the ratio of the
sample masses is the same as the ratio of the
masses of the individual components.
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5
___________________________________ Slide Section 8.1
6 Counting by Weighing
___________________________________ Exercise
___________________________________ A pile of marbles weigh 394.80 g. 10 marbles
weigh 37.60 g. How many marbles are in the
pile?
Avg. Mass of 1 Marble =
___________________________________ ___________________________________ 37.60 g
= 3.76 g / marble
10 marbles
___________________________________ 394.80 g = 105 marbles
3.76 g
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6
Slide Section 8.2
7 Atomic Masses:
Masses Counting Atoms by Weighing
___________________________________ ___________________________________ •
Atoms have very tiny masses so scientists
made a unit to avoid using very small numbers.
___________________________________ ___________________________________ 1 atomic mass unit (amu) = 1.66 10–24 g
___________________________________ •
The average atomic mass for an element is the
weighted average of the masses of all the
isotopes of an element.
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Slide Section 8.2
8 Atomic Masses:
Masses Counting Atoms by Weighing
7
___________________________________ ___________________________________ Average Atomic Mass for Carbon
•
•
Even though natural carbon does not
contain a single atom with mass 12.01, for
our purposes, we can consider carbon to
be composed of only one type of atom with
a mass of 12.01.
This enables us to count atoms of natural
carbon by weighing a sample of carbon.
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Slide Section 8.2
9 Atomic Masses:
Masses Counting Atoms by Weighing
8
___________________________________ ___________________________________ Example Using Atomic Mass Units
Calculate the mass (in amu) of 431 atoms
of carbon.
___________________________________ The mass of 1 carbon atom = 12.01 amu.
___________________________________ Use the relationship as a conversion factor.
___________________________________ 12.01 amu
431 C atoms 
= 5176 amu
1 C atom
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___________________________________ Slide Section 8.2
10 Atomic Masses:
Masses Counting Atoms by Weighing
___________________________________ Exercise
___________________________________ Calculate the mass (in amu) of 75 atoms of
aluminum.
75 atoms Al 
___________________________________ 26.98 amu
=
1 Al atom
___________________________________ ___________________________________ 2024 amu
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Slide Section 8.3
11 The Mole
10
___________________________________ ___________________________________ •
•
•
The number equal to the number of carbon
atoms in 12.01 grams of carbon.
1 mole of anything = 6.022 x 1023 units of that
thing (Avogadro’s number).
1 mole C = 6.022 x 1023 C atoms = 12.01 g C
1 mol C
6.022  1023 C atoms
or
6.022  1023 C atoms
1 mol C
___________________________________ ___________________________________ ___________________________________ ___________________________________ 1 mol C
12.01 g C
or
12.01 g C
1 mol C
6.022  10 C atoms
12.01 g C
or
12.01 g C
6.022  1023 C atoms
23
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Slide Section 8.3
12 The Mole
11
___________________________________ ___________________________________ •
•
A sample of an element with a mass equal to that
element’s average atomic mass (expressed in g)
contains one mole of atoms (6.022 × 1023 atoms).
Comparison of 1-Mol Samples of Various Elements
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Slide Section 8.3
13 The Mole
___________________________________ ___________________________________ Concept Check
___________________________________ Determine the number of copper atoms in a
63.55 g sample of copper.
___________________________________ 6.022×1023 Cu atoms
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Slide Section 8.3
14 The Mole
13
___________________________________ ___________________________________ Concept Check
___________________________________ Which of the following is closest to the
average mass of one atom of copper?
a)
b)
c)
d)
e)
___________________________________ 63.55 g
52.00 g
58.93 g
65.38 g
1.055 x 10-22 g
1 Cu atom 
___________________________________ ___________________________________ ___________________________________ 1 mol Cu
63.55 g Cu
= 1.055  10 22 g Cu

1 mol Cu
6.022  1023 Cu atoms
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Slide Section 8.3
15 The Mole
14
___________________________________ ___________________________________ Concept Check
___________________________________ Calculate the number of iron atoms in a 4.48 mole
sample of iron.
___________________________________ 4.48 mol Fe 
6.022  10 atoms Fe
1 mol Fe
23
___________________________________ 2.70 × 1024 Fe atoms
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Slide Section 8.3
16 The Mole
___________________________________ ___________________________________ Concept Check
___________________________________ A sample of 26.98 grams of Al has the same
number of atoms as _______ grams of Au.
a)
b)
c)
d)
___________________________________ 26.98 g
13.49 g
197.0 g
256.5 g
___________________________________ ___________________________________ ___________________________________ Return to TOC
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Slide Section 8.3
17 The Mole
16
___________________________________ ___________________________________ Concept Check
___________________________________ Which of the following 100.0 g samples
contains the greatest number of atoms?
___________________________________ a) Magnesium
b) Zinc
c) Silver
___________________________________ ___________________________________ ___________________________________ Return to TOC
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Slide Section 8.3
18 The Mole
17
___________________________________ ___________________________________ Exercise
___________________________________ Rank the following according to number of
atoms (greatest to least):
___________________________________ a) 107.9 g of silver
b) 70.0 g of zinc
c) 21.0 g of magnesium
b)
a)
___________________________________ ___________________________________ c)
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___________________________________ Slide Section 8.4
19 Learning to Solve Problems
___________________________________ Conceptual Problem Solving
•
Where are we going?

•
___________________________________ How do we get there?

•
___________________________________ Read the problem and decide on the final
goal.
Work backwards from the final goal to decide
where to start.
___________________________________ Reality check.

___________________________________ Does my answer make sense? Is it
reasonable?
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Slide Section 8.5
20 Molar Mass
19
___________________________________ ___________________________________ •
Mass in grams of one mole of the substance:
___________________________________ Molar Mass of N = 14.01 g/mol
___________________________________ Molar Mass of H2O = 18.02 g/mol
(2 × 1.008 g) + 16.00 g
___________________________________ Molar Mass of Ba(NO3)2 = 261.35 g/mol
___________________________________ 137.33 g + (2 × 14.01 g) + (6 × 16.00 g)
___________________________________ Return to TOC
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Slide Section 8.5
21 Molar Mass
20
___________________________________ ___________________________________ Calculations Using Molar Mass
•
Moles of a compound =
mol = g 
•
___________________________________ mass of the sample (g)
molar mass of the compound (
g
)
mol
___________________________________ mol
g
___________________________________ Mass of a sample (g) = (moles of sample)(molar mass of compound)
g = mol 
g
mol
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21
Slide Section 8.5
22 Molar Mass
___________________________________ ___________________________________ Exercise
___________________________________ What is the molar mass of nickel(II)
carbonate?
a)
b)
c)
d)
118.7 g/mol
134.7 g/mol
178.71 g/mol
296.09 g/mol
___________________________________ The formula for nickel(II)
carbonate is NiCO3. Therefore
the molar mass is:
58.69 + 12.01 + 3(16.00) =
118.7 g/mol.
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Slide Section 8.5
23 Molar Mass
22
___________________________________ ___________________________________ Exercise
___________________________________ Consider equal mole samples of N2O,
Al(NO3)3, and KCN. Rank these from least to
most number of nitrogen atoms in each
sample.
a)
b)
c)
d)
___________________________________ ___________________________________ KCN, Al(NO3)3, N2O
Al(NO3)3, N2O, KCN
KCN, N2O, Al(NO3)3
Al(NO3)3, KCN, N2O
___________________________________ ___________________________________ Return to TOC
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Slide Section 8.5
24 Molar Mass
23
___________________________________ ___________________________________ Exercise
___________________________________ Consider separate 100.0 gram samples of
each of the following:
___________________________________ H2O, N2O, C3H6O2, CO2
___________________________________  Rank them from greatest to least number
of oxygen atoms.
___________________________________ ___________________________________ H2O, CO2, C3H6O2, N2O
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Slide Section 8.5
25 Molar Mass
___________________________________ ___________________________________ Exercise
___________________________________ How many grams of fluorine are contained
in one molecule of boron trifluoride?
a)
b)
c)
d)
___________________________________ 3.155 × 10–23 g
9.465 × 10–23 g
6.022 × 1023 g
3.433 × 1025 g
1 molecule BF3 
___________________________________ ___________________________________ 3 atoms F
1 mol F
19.00 g F


1 molecule BF3
1 mol F
6.022  1023 atoms F
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___________________________________ Slide Section 8.6
26 Percent Composition of Compounds
___________________________________ •
Mass percent of an element:
___________________________________ ___________________________________ •
For iron in iron(III) oxide, (Fe2O3):
___________________________________ ___________________________________ ___________________________________ Return to TOC
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___________________________________ Slide Section 8.6
27 Percent Composition of Compounds
___________________________________ Exercise
___________________________________ Morphine, derived from opium plants, has the
potential for use and abuse. It’s formula is
C17H19NO3. What percent, by mass, is the carbon
in this compound?
a)
b)
c)
d)
12.0 %
54.8 %
67.9 %
71.6 %
___________________________________ ___________________________________ ___________________________________ 17  12.01
17  12.01 + 19  1.008  + 14.01 +  3  16.00  
= 0.716  100 = 71.6% C
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___________________________________ Slide Section 8.6
28 Percent Composition of Compounds
___________________________________ Exercise
___________________________________ Consider separate 100.0 gram samples of
each of the following:
___________________________________ H2O, N2O, C3H6O2, CO2
___________________________________  Rank them from highest to lowest percent
oxygen by mass.
___________________________________ ___________________________________ H2O, CO2, C3H6O2, N2O
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Slide Section 8.7
29 Formulas of Compounds
28
___________________________________ ___________________________________ Empirical Formulas
•
•
The empirical formula of a compound is the
simplest whole number ratio of the atoms
present in the compound.
The empirical formula can be found from the
percent composition of the compound.
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Slide Section 8.7
30 Formulas of Compounds
29
___________________________________ ___________________________________ Formulas
•
Empirical formula = CH
 Simplest whole-number ratio
• Molecular formula = (empirical formula)n
[n = integer]
• Molecular formula = C6H6 = (CH)6
 Actual formula of the compound
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___________________________________ Slide Section 8.8
31 Calculation of Empirical Formulas
___________________________________ Steps for Determining the Empirical Formula of a Compound
A gaseous compound containing carbon and
hydrogen was analyzed and found to consist of
83.65% carbon by mass. Determine the empirical
formula of the compound.
•
___________________________________ ___________________________________ Obtain the mass of each element present (in
grams).
___________________________________ Assume you have 100 g of the compound.
___________________________________ 83.65% C = 83.65 g C
(100.00 – 83.65)
16.35% H = 16.35 g H
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___________________________________ Slide Section 8.8
32 Calculation of Empirical Formulas
___________________________________ Steps for Determining the Empirical Formula of a Compound
2. Determine the number of moles of each type
of atom present.
___________________________________ 83.65 g C 
1 mol C
= 6.965 mol C
12.01 g C
___________________________________ 16.35 g H 
1 mol H
= 16.22 mol H
1.008 g H
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___________________________________ Slide Section 8.8
33 Calculation of Empirical Formulas
___________________________________ Steps for Determining the Empirical Formula of a Compound
3. Divide the number of moles of each element
by the smallest number of moles to convert
the smallest number to 1. If all of the numbers
so obtained are integers, these are the
subscripts in the empirical formula. If one or
more of these numbers are not integers, go on
to step 4.
___________________________________ ___________________________________ ___________________________________ 6.965 mol C
=1
6.965 mol
16.22 mol H
= 2.33
6.965 mol
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___________________________________ Slide Section 8.8
34 Calculation of Empirical Formulas
___________________________________ Steps for Determining the Empirical Formula of a Compound
4. Multiply the numbers you derived in step 3 by
the smallest integer that will convert all of
them to whole numbers. This set of whole
numbers represents the subscripts in the
empirical formula.
___________________________________ ___________________________________ ___________________________________ C: 1  3 = 3
H: 2.33  3 = 7
___________________________________ The empirical formula is C H .
3
7
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___________________________________ Slide Section 8.8
35 Calculation of Empirical Formulas
___________________________________ Exercise
___________________________________ The composition of adipic acid is 49.3% C,
6.9% H, and 43.8% O (by mass). What is the
empirical formula?
___________________________________ ___________________________________ C3H5O2
___________________________________ ___________________________________ Return to TOC
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Slide Section 8.9
36 Calculation of Molecular Formulas
35
___________________________________ ___________________________________ •
•
The molecular formula is the exact
formula of the molecules present in a
substance.
The molecular formula is always an
integer multiple of the empirical formula.
___________________________________ ___________________________________ ___________________________________ Molecular formula = (empirical formula)n
___________________________________ where n is a whole number
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___________________________________ Slide Section 8.9
37 Calculation of Molecular Formulas
___________________________________ Continued Example…
A gaseous compound containing carbon and
hydrogen was analyzed and found to consist of
83.65% carbon by mass. The molar mass of the
compound is 86.2 g/mol. You determined the
empirical formula to be C3H7. What is the molecular
formula of the compound?
___________________________________ ___________________________________ ___________________________________ Molar mass of C3H7 = 43.086 g/mol
86.2 g/mol
=2
43.086 g/mol
___________________________________ ___________________________________ C3H7 × 2 = C6H14
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___________________________________ Slide Section 8.9
38 Calculation of Molecular Formulas
___________________________________ Exercise
___________________________________ The composition of adipic acid is 49.3% C,
6.9% H, and 43.8% O (by mass). The molar
mass of the compound is about 146 g/mol.
The empirical formula is C3H5O2. What is the
molecular formula?
___________________________________ ___________________________________ ___________________________________ C6H10O4
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Slide 39 38
___________________________________ Section 8.9
Chapter
8 Homework
Calculation
of Molecular Formulas
___________________________________ Homework
___________________________________ • Reading assignment
– Pages 205 through 238
___________________________________ • Homework Problems
– Questions and problems 7, 15, 17, 19, 21, 23, 29, 31,
33, 35, 37, 39, 41, 47, 49, 51, 55, 59, 61, 63, 65, 67,
69, 71, 73, 79, 81.
___________________________________ ___________________________________ • Due on
___________________________________ Return to TOC
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