Chapter #6
Chemical Composition
Chapter Contents
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6-1 Chemical Composition
6-2 Counting Units
6-3 Counting Atoms By the Gram
6-4 Counting Molecules by the Gram
6-5 Mole Conversions
6-6 Mass Percent Composition
6-8 Empirical Formula Calculation
6-9 Molecular Formula Calculation
6-1 The Chemical Package
About Packages
• The baker uses a package called the dozen.
All dozen packages contain 12 objects.
• The stationary store uses a package called a
ream, which contains 500 sheets of paper.
• So what is the chemistry package?
6-2 The Chemical Package
About Packages
• The baker uses a package called the dozen.
All dozen packages contain 12 objects.
• The stationary store uses a package called a
ream, which contains 500 sheets of paper.
• So what is the chemistry package? Well, it is
called the mole (Latin for heap).
Each of the above packages contain a number of
objects that are convenient to work with, for that
particular discipline.
6-3 The Mole
A mole contains 6.022X1023 particles, which is the number
of carbon-12 atoms that will give a mass of 12.00 grams,
which is a convenient number of atoms to work with in the
chemistry laboratory.
The atomic weights listed on the periodic chart are the
weights of a mole of atoms. For example a mole of
hydrogen atoms weighs 1.00797 g and a mole of carbon
atoms weighs 12.01 g.
6-4 Moles of Objects
Suppose we order a mole of marshmallows for a
chemistry party. How much space here at Central
would be required to store the marshmallows?
6-4 Moles of Objects
Suppose we order a mole of marshmallows for a
chemistry party. How much space here at Central
would be required to store the marshmallows?
Would cover the entire 50 states 60 miles deep
6-4 Moles of Objects
Suppose we order a mole of marshmallows for a
chemistry party. How much space here at Central
would be required to store the marshmallows?
Would cover the entire 50 states 60 miles deep
How about a mole of computer paper instead
of a ream of computer paper, how far would
that stretch?
6-4 Moles of Objects
Suppose we order a mole of marshmallows for a
chemistry party. How much space here at Central
would be required to store the marshmallows?
Would cover the entire 50 states 60 miles deep
How about a mole of computer paper instead
of a ream of computer paper, how far would
that stretch? Way past the planet Pluto!
6-5 Formula Weight Calculation
To calculate the molar mass of a compound we sum
together the atomic weights of the atoms that make up
the formula of the compound. This is called the formula
weight (MW, M).
Formula weights are the sum of atomic weights of
atoms making up the formula.
The following outlines how to find the formula weight of water
symbol
H
O
number
weight
= 2.02
1.01 X
2
1
= 16.0
X
16.0
18.0 g/mole
6-7 Percent Composition
Find the formula weight and the percent composition of
glucose (C6H12O6)
symbol
C
H
O
%C =
weight
number
12.0 x 6 = 72.0
1.01 x 12 = 12.12
16.0 x 6 = 96.0
180.1 g/mole
72.0
X 100 = 40.0 %C
180.1
%H = 12.12
180.1
96.0
%O =
180.1
X 100 = 6.73 %H
X 100 = 53.3 %O
6-7 Percent Conversions
Seawater contains approximately 3.5% NaCl by
mass and has a density of 1.02 g/mL. What
volume of seawater would be required to produce
1.0 g of NaCl when evaporated?
6-7 Percent Conversions
Seawater contains approximately 3.5% NaCl by
mass and has a density of 1.02 g/mL. What
volume of seawater would be required to produce
1.0 g of NaCl when evaporated?
mL seawater
1.02 g seawater
6-7 Percent Conversions
Seawater contains approximately 3.5% NaCl by
mass and has a density of 1.02 g/mL. What
volume of seawater would be required to produce
1.0 g of NaCl when evaporated?
mL seawater
100 g seawater
1.02 g seawater 3.5 g NaCl
6-7 Percent Conversions
Seawater contains approximately 3.5% NaCl by
mass and has a density of 1.02 g/mL. What
volume of seawater would be required to produce
1.0 g of NaCl when evaporated?
mL seawater
100 g seawater 1.00 g NaCl
1.02 g seawater 3.5 g NaCl
6-7 Percent Conversions
Seawater contains approximately 3.5% NaCl by
mass and has a density of 1.02 g/mL. What
volume of seawater would be required to produce
1.0 g of NaCl when evaporated?
mL seawater
100 g seawater 1.00 g NaCl
= 28 mL
1.02 g seawater 3.5 g NaCl
6-8 Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of
glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022
X 1023 particles then a mole of glucose must contain 6 moles of C
atoms.
How many moles of hydrogen atoms are contained in a mole of
glucose?
In 5 moles of H2SO4 how many moles of oxygen atoms is there?
6-8 Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of
glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022
X 1023 particles then a mole of glucose must contain 6 moles of C
atoms.
How many moles of hydrogen atoms are contained in a mole of
glucose? 12 Moles of hydrogen.
How many moles of oxygen and hydrogen are in one mole of
H2O contains:
6-8 Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of
glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022
X 1023 particles then a mole of glucose must contain 6 moles of C
atoms.
How many moles of hydrogen atoms are contained in a mole of
glucose? 12 Moles of hydrogen.
How many moles of oxygen and hydrogen are in one mole of
H2O contains:
One mole of oxygen atoms
Two moles of hydrogen atoms
6-8 Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of
glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022
X 1023 particles then a mole of glucose must contain 6 moles of C
atoms.
How many moles of hydrogen atoms are contained in a mole of
glucose? 12 Moles of hydrogen.
How many moles of oxygen and hydrogen are in one mole of
H2O contains:
One mole of oxygen atoms
Two moles of hydrogen atoms
In 5 moles of H2SO4 how many moles of oxygen atoms are there?
6-8 Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of
glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022
X 1023 particles then a mole of glucose must contain 6 moles of C
atoms.
How many moles of hydrogen atoms are contained in a mole of
glucose? 12 Moles of hydrogen.
How many moles of oxygen and hydrogen are in one mole of
H2O contains:
One mole of oxygen atoms
Two moles of hydrogen atoms
In 5 moles of H2SO4 how many moles of oxygen atoms is there?
20 moles of O atoms.
6-8 Mole Conversions
In 50.0g of H2SO4 how many moles of sulfuric acid are there?
50.0g of H2SO4
6-8 Mole Conversions
In 50.0g of H2SO4 how many moles of sulfuric acid are there?
50.0g of H2SO4 mole H2SO4
98.0g of H2SO4
=
6-8 Mole Conversions
In 50.0g of H2SO4 how many moles of sulfuric acid are there?
50.0g of H2SO4 mole H2SO4
98.0g of H2SO4
= 0.510 mole H2SO4
6-8 Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
6-8 Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
50.0g of H2SO4
=
6-8 Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
50.0g of H2SO4 mole H2SO4
98.0g of H2SO4
=
6-8 Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
50.0g of H2SO4 mole H2SO4
4mole O
98.0g of H2SO4 mole H2SO4
=
6-8 Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
50.0g of H2SO4 mole H2SO4
4mole O
98.0g of H2SO4 mole H2SO4
= 2.04 mole O
6-8 Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are present?
6-8 Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are present?
5 moles H2SO4
=
6-8 Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are present?
5 moles H2SO4 4 mole O
mole H2SO4
6-8 Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are present?
5 moles H2SO4 4 mole O
6.02 x 1023 atoms O
=
mole
O
mole H2SO4
1.20 x 1025 atoms
6-8 Empirical Formulas
Empirical formula is the smallest whole number
ratio between atoms and can be calculated from
the percent composition.
Molecular formulas happen to be the exact
number of atoms making up a molecule, and may
or may no be the simplest whole number ratio.
Molecular formulas are whole number multiples of
the empirical formula.
6-8 Empirical Formula Steps
1.
2.
3.
4.
Assume 100 g of compound.
Convert percent to a mass number.
Convert the mass to moles.
Divide each mole number by the smallest mole
number.
5. Rounding:
a. If the decimal is ≤ 0.1, then drop the decimals
b. If the decimal is ≥0.9, then round up.
c. All other decimal need to be multiplied by a whole
number until roundable.
6-8 Empirical Formula Example
A compound is composed of 75.0% C and 25.0%
H. Find its empirical formula.
Step #1 Assume 100 g of compound
75.0 g C
25.0 g H
6-8 Empirical Formula Example
A compound is composed of 75.0% C and 15.0%
H. Find its empirical formula.
Step #2 Convert grams to moles.
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole H
1.008 g H
6-8 Empirical Formula Example
A compound is composed of 75.0% C and 15.0%
H. Find its empirical formula.
Step #3 Divide each mole number by the smallest.
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole H
1.008 g H
6.225
= 1.00
6.225
24.802
6.225
= 3.98
6-8 Empirical Formula Example
A compound is composed of 75.0% C and 15.0%
H. Find its empirical formula.
Step #4 Rounding; Decimal ≤ 0.1, drop decimals
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole H
1.008 g H
6.225
= 1.00
6.225
24.802
6.225
= 3.98
6-8 Empirical Formula Example
A compound is composed of 75.0% C and 15.0%
H. Find its empirical formula.
Step #4 Rounding; Decimal ≤ 0.1, drop decimals
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole H
1.008 g H
6.225
=1C
6.225
24.802
6.225
= 3.98
6-8 Empirical Formula Example
A compound is composed of 75.0% C and 15.0%
H. Find its empirical formula.
Step #4 Rounding; Decimal ≥ 0.9, round up
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole H
1.008 g H
6.225
=1C
6.225
24.802
6.225
= 3.98
6-8 Empirical Formula Example
A compound is composed of 75.0% C and 15.0%
H. Find its empirical formula.
Step #4 Rounding; Decimal ≥ 0.9, round up
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole H
1.008 g H
6.225
=1C
6.225
24.802
6.225
= 3.98
6-8 Empirical Formula Example
A compound is composed of 75.0% C and 15.0%
H. Find its empirical formula.
Step #4 Rounding; Decimal ≥ 0.9, round up
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole H
1.008 g H
6.225
=1C
6.225
24.802
6.225
=4H
6-8 Empirical Formula Example
A compound is composed of 75.0% C and 15.0%
H. Find its empirical formula.
Step #4 Rounding; Decimal ≥ 0.9, round up
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole H
1.008 g H
6.225
=1C
6.225
24.802
6.225
Empirical Formula = CH4
=4H
6-9 Molecular Formulas
Empirical formula, is the smallest ratio
between atoms in a molecular or formula
unit.
Molecular formula, is the exact number of
atoms in a molecule; a whole number
multiple of an empirical formula
6-9 Possible Molecular
Formulas
Assume an empirical formula of C3H5O
Empirical formula
Integer
C3H5O
1
C3H5O
2
C3H5O
3
C3H5O
4
C3H5O
5
Molecular Formula
C3H5O
6-9 Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula
Integer
Molecular Formula
C3H5O
1
C3H5O
C3H5O
2
C6H10O2
C3H5O
3
C3H5O
4
C3H5O
5
6-9 Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula
Integer
Molecular Formula
C3H5O
1
C3H5O
C3H5O
2
C6H10O2
C3H5O
3
C9H15O3
C3H5O
4
C3H5O
5
6-9 Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula
Integer
Molecular Formula
C3H5O
1
C3H5O
C3H5O
2
C6H10O2
C3H5O
3
C9H15O3
C3H5O
4
C12H20O4
C3H5O
5
C15H25O5
6-9 Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
6-9 Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #1 Assume 100g of compound
83.6 g C
16.3 g H
6-9 Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #2 Convert grams to moles
83.6 g C mole
12.01 g C
16.3 g H mole
1.008 g H
6-9 Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #2 Convert grams to moles
83.6 g C mole
12.01 g C
16.3 g H mole
1.008 g H
= 6.961 mole
= 16.17 mole
6-9 Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #3 Divide each mole number by the smallest.
83.6 g C mole
12.01 g C
16.3 g H mole
1.008 g H
= 6.961 mole
= 16.17 mole
6-9 Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #3 Divide each mole number by the smallest.
83.6 g C mole
12.01 g C
16.3 g H mole
1.008 g H
6.961 = 1.00
= 6.961 mole
6.961
= 16.17 mole
= 2.32
6-9 Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #4 Round if---Not Roundable
6.961 = 1.00
83.6 g C mole
= 6.961 mole
6.961
12.01 g C
16.3 g H mole
= 16.17 mole
= 2.32
1.008 g H
Step #4, Multiply by an integer until roundable
1.00 X 3 = 3
Empirical formula C3H7
2.32 X 3 = 7
6-9 Molecular Formula Integer
Divide empirical weight into molecular weight
3x12 + 7x1 =43
43
2
86
Now multiply the empirical formula by 2
6-9 Molecular Formula Integer
Divide empirical weight into molecular weight
3x12 + 7x1 =43
43
2
86
Now multiply the empirical formula by 2
Molecular Formula is C6 H14
The End