Transcript CHEMISTRY 103 - University of Wisconsin–Eau Claire

CHEMISTRY 115
Fall 2012
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Why Study Chemistry?
Think about current issues in the news.
2
Why Study Chemistry?
Think about current issues in the news.
Energy sources
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Why Study Chemistry?
Think about current issues in the news.
Energy sources
Greenhouse effect
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Why Study Chemistry?
Think about current issues in the news.
Energy sources
Greenhouse effect
Pollution
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Why Study Chemistry?
Think about current issues in the news.
Energy sources
Greenhouse effect
Pollution
Ozone problem
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Why Study Chemistry?
Think about current issues in the news.
Energy sources
Greenhouse effect
Pollution
Ozone problem
Food additives
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Why Study Chemistry?
Think about current issues in the news.
Energy sources
Greenhouse effect
Pollution
Ozone problem
Food additives
Drugs
8
INTRODUCTION
CHEMISTRY:
A science that deals with the composition,
structure, and properties of substances and
of the changes they undergo.
There are six major subdivisions of Chemistry:
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1. Organic chemistry: Covers the compounds
of carbon and hydrogen (hydrocarbons). All
compounds derived from hydrocarbons.
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1. Organic chemistry: Covers the compounds
of carbon and hydrogen (hydrocarbons). All
compounds derived from hydrocarbons.
2. Inorganic chemistry: Covers all the elements,
and all compounds, except the hydrocarbons
and their derivatives.
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1. Organic chemistry: Covers the compounds
of carbon and hydrogen (hydrocarbons). All
compounds derived from hydrocarbons.
2. Inorganic chemistry: Covers all the elements,
and all compounds, except the hydrocarbons
and their derivatives.
3. Physical chemistry: Measurement of physical
properties. Interpretation of physical and
chemical properties.
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4. Biochemistry: Study of pure substances and
chemical reactions in living systems.
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4. Biochemistry: Study of pure substances and
chemical reactions in living systems.
5. Analytical chemistry: Measurement of the
amounts of substances. Measurement of
chemical composition of materials.
Separation of the components of mixtures
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4. Biochemistry: Study of pure substances and
chemical reactions in living systems.
5. Analytical chemistry: Measurement of the
amounts of substances. Measurement of
chemical composition of materials.
Separation of the components of mixtures
6. Theoretical chemistry: Mathematical
description of chemical structures and of
chemical changes.
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Scientific Method
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Scientific Method
A series of steps used to solve scientific
problems.
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Scientific Method
A series of steps used to solve scientific
problems.
Objective: Some problem to be solved, e.g.
how oxygen gas binds to the hemoglobin
molecule in our blood.
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Collection of data: Once the goal is defined, the
next step involves making careful observations
and collecting bits of information about the
system. The bits of information are called
data.
The word system here means that part of the
universe that is under investigation.
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The information obtained may be both
qualitative or quantitative.
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The information obtained may be both
qualitative or quantitative.
Qualitative: general and non-mathematical. E.g.
the object has a blue color.
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The information obtained may be both
qualitative or quantitative.
Qualitative: general and non-mathematical. E.g.
the object has a blue color.
Quantitative : numerical – related to
measurements. E.g. the density is 2.1 g/ml.
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Law: After a large amount of data has been
collected, it is often desirable to summarize
the information in a concise way. This
summarizing statement is called a law.
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Law: After a large amount of data has been
collected, it is often desirable to summarize
the information in a concise way. This
summarizing statement is called a law.
A law is a concise verbal or mathematical
statement of a relation between phenomena
that is always the same under the same
conditions.
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Hypothesis: Once enough information has
been gathered, a tentative explanation for the
observations can be formulated – this is the
hypothesis.
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Hypothesis: Once enough information has
been gathered, a tentative explanation for the
observations can be formulated – this is the
hypothesis.
Further experiments are devised to test the
validity of the hypothesis in as many ways as
possible.
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The hypothesis provides tentative
explanations that must be tested by many
experiments. If the hypothesis survives such
tests, the hypothesis develops into a theory.
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The hypothesis provides tentative
explanations that must be tested by many
experiments. If the hypothesis survives such
tests, the hypothesis develops into a theory.
Theory: A theory is a unifying principle that
explains a body of facts and those laws that
are based on them. Theories are constantly
being tested. If a theory is proved incorrect by
experiment, then it must be discarded or
modified, so that it becomes consistent with
experimental observations.
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Scientific progress is made by modifying old
laws and theories or replacing them with new
ones.
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Summary – the sequence
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Summary – the sequence
1. Objective
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Summary – the sequence
1. Objective
2. Data collection
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Summary – the sequence
1. Objective
2. Data collection
3. Summarize data, generalization, formulation
of law
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Summary – the sequence
1. Objective
2. Data collection
3. Summarize data, generalization, formulation
of law
4. Hypothesis (tentative explanation)
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Summary – the sequence
1. Objective
2. Data collection
3. Summarize data, generalization, formulation
of law
4. Hypothesis (tentative explanation)
5. Test hypothesis
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Summary – the sequence
1. Objective
2. Data collection
3. Summarize data, generalization, formulation
of law
4. Hypothesis (tentative explanation)
5. Test hypothesis
6. Formulate theory
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Summary – the sequence
1. Objective
2. Data collection
3. Summarize data, generalization, formulation
of law
4. Hypothesis (tentative explanation)
5. Test hypothesis
6. Formulate theory
7. Further testing
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Summary – the sequence
1. Objective
2. Data collection
3. Summarize data, generalization, formulation
of law
4. Hypothesis (tentative explanation)
5. Test hypothesis
6. Formulate theory
7. Further testing
8. Rejection or modification of theory as
required to account for new observations.
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Some basic definitions
Matter: Anything that occupies space and
possesses mass is called matter.
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Some basic definitions
Matter: Anything that occupies space and
possesses mass is called matter.
Mass: The mass of a body is a measure of the
quantity of matter contained in that body.
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Some basic definitions
Matter: Anything that occupies space and
possesses mass is called matter.
Mass: The mass of a body is a measure of the
quantity of matter contained in that body.
Weight: Refers to the force which gravity exerts
upon an object. Unfortunately, chemists very
frequently use the word “weight” when they
mean mass.
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Substance: A substance is a form of matter that
has a definite composition and distinct
properties. Examples: gold, water, oxygen.
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Substance: A substance is a form of matter that
has a definite composition and distinct
properties. Examples: gold, water, oxygen.
Mixture: A combination of two or more
substances in which the substances retain their
identities. Examples: air, a solution of table sugar
(sucrose) in water.
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Substance: A substance is a form of matter that
has a definite composition and distinct
properties. Examples: gold, water, oxygen.
Mixture: A combination of two or more
substances in which the substances retain their
identities. Examples: air, a solution of table sugar
(sucrose) in water.
Note: Mixtures do not have constant composition;
samples of air collected in Los Angles will have
different composition from samples collected in
Eau Claire.
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There are two types of mixtures: homogeneous and
heterogeneous.
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There are two types of mixtures: homogeneous and
heterogeneous.
Homogeneous mixture: The composition is the
same throughout. Example: a small amount of
sugar completely dissolved in water.
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There are two types of mixtures: homogeneous and
heterogeneous.
Homogeneous mixture: The composition is the
same throughout. Example: a small amount of
sugar completely dissolved in water.
Heterogeneous mixture: A mixture in which the
individual components remain physically separate
and can be seen as separate components.
Example: a mixture of sugar and sand.
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Any mixture, be it homogeneous or
heterogeneous, can be put together and then
separated into pure components without any
change in the identity of the components, by
physical means.
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For example: sugar can be removed from a
homogeneous sugar solution by evaporating
off the solvent water.
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For example: sugar can be removed from a
homogeneous sugar solution by evaporating
off the solvent water.
A sugar/sand mixture could be separated by
dissolving the sugar in water, drying the sand,
and reclaiming the sugar by evaporation of
the solution.
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For example: sugar can be removed from a
homogeneous sugar solution by evaporating
off the solvent water.
A sugar/sand mixture could be separated by
dissolving the sugar in water, drying the sand,
and reclaiming the sugar by evaporation of
the solution.
In the physical separation process, there has
been no change in the composition of each
substance making up the mixture.
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Physical property: Any property of a substance
that can be observed without permanently*
changing the substance to form some other
substance. Examples: color, density, melting
point.
* Some exceptions to this. E.g. some compounds
decompose at their melting point.
A physical property can be specified without
reference to any other substance.
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Chemical property: Any property of a
substance that cannot be studied without
resulting in a permanent change of the
substance to form some other substance.
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Chemical property: Any property of a
substance that cannot be studied without
resulting in a permanent change of the
substance to form some other substance.
Example: sodium metal is very reactive with
water.
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Chemical property: Any property of a
substance that cannot be studied without
resulting in a permanent change of the
substance to form some other substance.
Example: sodium metal is very reactive with
water.
Reactivity is a chemical property that refers to
the tendency of a substance to undergo a
particular chemical reaction.
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Element: An element is a pure substance that cannot
be separated into simpler substances by chemical
means.
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Element: An element is a pure substance that cannot
be separated into simpler substances by chemical
means.
Compound: A pure substance composed of two or
more elements chemically united in fixed
proportions.
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Element: An element is a pure substance that cannot
be separated into simpler substances by chemical
means.
Compound: A pure substance composed of two or
more elements chemically united in fixed
proportions.
Atom: The smallest particle of an element that retains
the chemical nature of the element.
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Element: An element is a pure substance that cannot
be separated into simpler substances by chemical
means.
Compound: A pure substance composed of two or
more elements chemically united in fixed
proportions.
Atom: The smallest particle of an element that retains
the chemical nature of the element.
Molecule: A structure consisting of two or more
atoms that are chemically bound together and
behave as an independent unit.
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Energy: Is the capacity to do work or produce
change.
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Energy: Is the capacity to do work or produce
change.
Potential Energy: Energy available because of
the position of an object.
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Energy: Is the capacity to do work or produce
change.
Potential Energy: Energy available because of
the position of an object.
Kinetic Energy: Energy available because of the
motion of an object.
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Energy: Is the capacity to do work or produce
change.
Potential Energy: Energy available because of
the position of an object.
Kinetic Energy: Energy available because of the
motion of an object.
Chemical Energy: The energy stored by
compounds.
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CHEMICAL SYMBOLS
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CHEMICAL SYMBOLS
Chemical symbols are shorthand notation for
the names of elements.
There are 117 known elements. The last
several have not been named.
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If an element has a single letter to represent it
– the letter must be capitalized.
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If an element has a single letter to represent it
– the letter must be capitalized.
If an element is represented by two letters –
the first letter must be capitalized and the
second letter must be lower case.
Examples: NO is not the symbol for Nobelium –
the correct symbol is No. CO is not the symbol
for cobalt – the correct symbol is Co.
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CHEMICAL FORMULAS
A chemical formula shows the chemical
composition of a compound. Three types of
chemical formulas will be of interest to us.
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CHEMICAL FORMULAS
A chemical formula shows the chemical
composition of a compound. Three types of
chemical formulas will be of interest to us.
1. Molecular formulas
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CHEMICAL FORMULAS
A chemical formula shows the chemical
composition of a compound. Three types of
chemical formulas will be of interest to us.
1. Molecular formulas
2. Empirical formulas
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CHEMICAL FORMULAS
A chemical formula shows the chemical
composition of a compound. Three types of
chemical formulas will be of interest to us.
1. Molecular formulas
2. Empirical formulas
3. Structural formulas
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Molecular formulas
An expression showing the exact numbers and
types of elements present in a molecule.
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Molecular formulas
An expression showing the exact numbers and
types of elements present in a molecule.
Example: The molecular formula for water is H2O
– this formula tells us that a molecule of water is
formed from
2 atoms of hydrogen (symbol H)
1 atom of oxygen (symbol O)
The subscript 2 tells how many atoms of the
element on the left of the subscript are present.
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When no subscripts are present, the implied
number is one.
When formulas like (CH3)2CO are encountered,
the parenthesis followed by the subscript 2
mean that there are two CH3 groups, so for
the given formula, we have:
3 atoms of carbon
6 atoms of hydrogen
1 atom of oxygen
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The compound Ni(CO)4 has one atom of Ni, four
atoms of carbon, and 4 atoms of oxygen.
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Empirical formulas
An expression showing types of elements present
and the ratio of the different kinds of atoms.
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Empirical formulas
An expression showing types of elements present
and the ratio of the different kinds of atoms.
The empirical formula is useful information when
the identity of an unknown compound is to be
determined.
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Empirical formulas
An expression showing types of elements present
and the ratio of the different kinds of atoms.
The empirical formula is useful information when
the identity of an unknown compound is to be
determined.
Note: Many different compounds may have the
same empirical formula.
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Examples: A molecule of glucose consists of 6
carbon atoms, 12 hydrogen atoms, and 6 oxygen
atoms – and so its molecular formula is C6H12O6.
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Examples: A molecule of glucose consists of 6
carbon atoms, 12 hydrogen atoms, and 6 oxygen
atoms – and so its molecular formula is C6H12O6.
From the formula we see that the proportion of
atoms for carbon : hydrogen : oxygen is
6 : 12 : 6
or
1:2:1
The empirical formula of glucose is CH2O.
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For many formulas, the empirical formula is the
same as the molecular formula, e.g. H2O.
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For many formulas, the empirical formula is the
same as the molecular formula, e.g. H2O.
Two examples with different molecular formulas
but the same empirical formula:
Benzene has the molecular formula C6H6 and its
empirical formula is CH.
Acetylene C2H2, has the empirical formula CH.
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For ionic compounds, e.g. sodium chloride, the
formula shows the ratio of elements that form
the compound.
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For ionic compounds, e.g. sodium chloride, the
formula shows the ratio of elements that form
the compound.
Solid sodium chloride consists of a collection of
positively charged sodium ions and negatively
charged chloride ions in a three-dimensional
structure. You cannot say which sodium ion is
associated with any particular chloride ion.
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For ionic compounds, e.g. sodium chloride, the
formula shows the ratio of elements that form
the compound.
Solid sodium chloride consists of a collection of
positively charged sodium ions and negatively
charged chloride ions in a three-dimensional
structure. You cannot say which sodium ion is
associated with any particular chloride ion.
The formula NaCl should be regarded as the
empirical formula.
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STRUCTURAL FORMULAS
An expression showing the exact numbers and
types of atoms present in a molecule, and
information about how the atoms are chemically
bonded to one another.
Examples:
O
H
water
H
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H
O
O
hydrogen peroxide
H
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H
O
O
hydrogen peroxide
H
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H
H
C
H
methane
H
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H
H
C
H
H
C
H
O
H
ethanol
(ethyl alcohol)
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Naming inorganic compounds
Two principal groups:
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Naming inorganic compounds
Two principal groups:
(1) metal with nonmetal (or nonmetal group)
use Stock system
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Naming inorganic compounds
Two principal groups:
(1) metal with nonmetal (or nonmetal group)
use Stock system
(2) nonmetal with nonmetal
use prefix system
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Stock system
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Stock system
Oxidation number: is the charge that an atom in a
compound would have if the electrons in each
bond belonged entirely to the more
electronegative atom.
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Stock system
Oxidation number: is the charge that an atom in a
compound would have if the electrons in each
bond belonged entirely to the more
electronegative atom.
Electronegativity: relative attraction of an atom
for electrons.
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Oxidation numbers:
In many cases the oxidation number of an
element in an ionic compound is the same as
the formal charge on the ions present.
Example: in the ionic compound KCl, which is
composed of K+ ions and Cl- ions, the oxidation
number of potassium is +1 and the oxidation
number of the chlorine is -1.
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For a neutral compound the sum of the
oxidation numbers of the elements = 0.
Example: for KMnO4 for which the oxidation
numbers are K (+1), Mn (+7), and O (-2),
the sum of the oxidation numbers
= 1 + 7 + 4x(-2) = 0.
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For a cation or an anion, the sum of the oxidation
numbers of the elements
= the charge on the species.
Example: For MnO4the sum of the oxidation numbers
= 7 + 4x(-2) = -1,
where -1 is the charge on the anion.
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1. The cation is named first.
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1. The cation is named first.
2. The oxidation number is given next (using
Roman numerals in parentheses) – if it is
needed.
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1. The cation is named first.
2. The oxidation number is given next (using
Roman numerals in parentheses) – if it is
needed.
The oxidation number is given only if the
element commonly has more than one
oxidation state.
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1. The cation is named first.
2. The oxidation number is given next (using
Roman numerals in parentheses) – if it is
needed.
The oxidation number is given only if the
element commonly has more than one
oxidation state.
3. The anion is named second.
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Examples: FeCl3
Iron forms two common cations, Fe2+ and Fe3+ ,
hence it will be necessary to specify the oxidation
number of the Fe.
The name is iron (III) chloride
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Examples: FeCl3
Iron forms two common cations, Fe2+ and Fe3+ ,
hence it will be necessary to specify the oxidation
number of the Fe.
The name is iron (III) chloride
CoPO4
Cobalt forms two common cations, Co2+ and Co3+ ,
hence it will be necessary to specify the oxidation
number of the Co.
The name is cobalt (III) phosphate
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LiClO3 The cation has only one common oxidation
state (which is +1), so it is not necessary to give
this as part of the name.
The name is lithium chlorate
You need to be able to name in both directions:
formula name
name formula
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Give the formula for the following:
Iron (II) phosphate
The two ions are Fe2+ and PO43- which can be
put together
Fe2+
PO43-
Fe3(PO4)2
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calcium sulfate
The two ions are Ca2+ and SO42- which can be
put together directly as
CaSO4
(Note: there are no subscripts of 2 on the
calcium and the sulfate ions in the final
formula. Keep in mind that we are working
with empirical formulas for ionic compounds.
The one odd exception to this that you will
encounter are the mercury (I) salts.)
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To use the Stock system, you need to know the
charges on the common cations and anions.
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For the anions, pay attention to the endings. Within a
given series, the name of the anion with more oxygen
atoms usually ends with an “ate” ending and that with
fewer oxygen atoms ends with an “ite” ending.
Examples: ClO-
hypochlorite
ClO2- chlorite
ClO3- chlorate
ClO4- perchlorate
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CHARGES OF SOME COMMON IONS
IA
II A
IB
II B
III A
IV A
VA
VI A
VII A
N-3
O-2
F-
P-3
S-2
Cl-
H+
Li+
Be+2
Na+
Mg+2
K+
Ca+2
Al+3
Cr+2
Cr+3
Rb+
Sr+2
Mn+2
Fe+2
Co+2
Fe+3
Co+3
Ni+2
Cu+1
Zn+2
Br-
Cu+2
Ag+1
Cd+2
Sn+2
I-
Sn+4
Cs+
Ba+2
Hg2+2
Pb+2
Bi+3
Hg+2
Group IA elements all form single positvely charged ions. Group IIA elements form doubly charged positive ions. Elements in the transition metal
group are frequently able to form cations with different charges. Note the unusual formula for the mercury (I) ion. Caions with a +4 charge are
very uncommon and are usually unstable in solution. The first four group VIA elements all form anions with a double negative charge. Group VIIA
elements all form anions with a single negative charge.
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(p.65)
112
(p.62)
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Prefix system
Use this for nonmetal
nonmetal compounds.
number
prefix
1
2
3
4
5
6
7
8
9
10
mon
di
tri
tetra
penta
hexa
hepta
octa
nona
deca
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Examples:
N2O
dinitrogen monoxide
(also called dinitrogen oxide)
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Examples:
N2O
dinitrogen monoxide
(also called dinitrogen oxide)
CO
carbon monoxide
116
Examples:
N2O
dinitrogen monoxide
(also called dinitrogen oxide)
CO
carbon monoxide
Cl2O7
dichlorine heptaoxide
117
Examples:
N2O
dinitrogen monoxide
(also called dinitrogen oxide)
CO
carbon monoxide
Cl2O7
dichlorine heptaoxide
P4O10
tetraphosphorous decaoxide
118
Examples:
N2O
dinitrogen monoxide
(also called dinitrogen oxide)
CO
carbon monoxide
Cl2O7
dichlorine heptaoxide
P4O10
tetraphosphorous decaoxide
PF5
phosphorous pentafluoride
119
Examples:
N2O
dinitrogen monoxide
(also called dinitrogen oxide)
CO
carbon monoxide
Cl2O7
dichlorine heptaoxide
P4O10
tetraphosphorous decaoxide
PF5
phosphorous pentafluoride
N2
dinitrogen
120
Chemical Reactions and Chemical
Equations – An introduction
Chemical reaction: The transformation of one or
more chemicals into different compounds.
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Examples:
2 Na + 2 H2O
(sodium) (water)
2 NaOH
+ H2
(sodium
dihydrogen
hydroxide)
122
Examples:
2 Na + 2 H2O
(sodium) (water)
2 NaOH
+ H2
(sodium
dihydrogen
hydroxide)
Cu + 4 HNO3
Cu(NO3)2
+ 2 NO2
(copper) (nitric acid) (copper (II) nitrate) (nitrogen
dioxide)
+ 2H2O
123
Examples:
2 Na + 2 H2O
(sodium) (water)
2 NaOH
+ H2
(sodium
dihydrogen
hydroxide)
Cu + 4 HNO3
Cu(NO3)2
+ 2 NO2
(copper) (nitric acid) (copper (II) nitrate) (nitrogen
dioxide)
+ 2H2O
Zn + 2HCl(aq)
ZnCl2 + H2
124
Reactants: The starting substances in a chemical
reaction. E.g. Na and H2O in the first reaction.
125
Reactants: The starting substances in a chemical
reaction. E.g. Na and H2O in the first reaction.
Products: Substances as a result of a chemical
reaction. E.g. Cu(NO3)2, NO2, and H2O in the
second reaction.
126
Reactants: The starting substances in a chemical
reaction. E.g. Na and H2O in the first reaction.
Products: Substances as a result of a chemical
reaction. E.g. Cu(NO3)2, NO2, and H2O in the
second reaction.
Balanced equation: The number of atoms of
each element (free or part of a compound) is
the same on both sides of the equation.
127
The three equations given are all balanced.
An example of a non-balanced equation is:
H2 + O2
H2O
The commonly employed definition of balanced
equation is the smallest set of integers that
leads to a balanced equation.
128
So,
H2 + ½ O2
H2O
and
4 H2 + 2 O2
4 H2O
Are not balanced equations. The balanced
equation is:
2 H2 + O2
2 H2O
129
The coefficients tell us how many molecules of
each species are reacting.
130
Measurements and Units
Most physical quantities we encounter in
chemistry have units.
Old system: The cgs system – centimeter,
gram, second were some of the key units in
use – hence the abbreviation cgs
131
The SI system:
132
The SI system:
Basic Quantity Name of Unit Symbol
133
The SI system:
Basic Quantity Name of Unit Symbol
length
meter
m
134
The SI system:
Basic Quantity Name of Unit Symbol
length
meter
m
mass
kilogram
kg
135
The SI system:
Basic Quantity Name of Unit Symbol
length
meter
m
mass
kilogram
kg
time
second
s
136
The SI system:
Basic Quantity Name of Unit Symbol
length
meter
m
mass
kilogram
kg
time
second
s
temperature kelvin
K
137
The SI system:
Basic Quantity Name of Unit Symbol
length
meter
m
mass
kilogram
kg
time
second
s
temperature kelvin
K
amount of
substance
mole
mol
138
Derived Units: These are obtained from the basic units
just given.
Examples: The SI unit of volume is derived as follows:
volume = length3 (think about the volume of a
cube).
Since the unit of length is m, then the unit of volume is
m3
139
Derived Units: These are obtained from the basic units
just given.
Examples: The SI unit of volume is derived as follows:
volume = length3 (think about the volume of a
cube).
Since the unit of length is m, then the unit of volume is
m3
The SI unit of density:
density  mass
volume
Hence the units of density are kg m-3
140
There are some non-SI units that are in common
use in chemistry. A common unit of volume is
the liter.
1 liter = 1 cubic decimeter (1 dm3)
= (10 cm)3
= 1000 cm3
and 1 liter = 1000 ml,
that is 1 ml = 1 cm3
141
The basic units are not always convenient for
reporting measurements. Various prefixes are
often employed.
Examples: 1 kg = 1000 g
1 km = 1000 m
1 nm = 0.000 000 001 m
1 cm = 0.01 m
142
The factor-dimensional method of
calculation
(also called the factor-label method)
143
The factor-dimensional method of
calculation
(also called the factor-label method)
Suppose we want to convert 45.6 m into
centimeters. From the conversion factor
100 cm = 1m
144
The factor-dimensional method of
calculation
(also called the factor-label method)
Suppose we want to convert 45.6 m into
centimeters. From the conversion factor
100 cm = 1m
we can write
1
1m
100 cm  1
or
100 cm
1m
These are unit conversion factors.
145
Multiplication of a quantity by a unit conversion
factor will change its units, and adjust the value
of the quantity for the new unit system used.
For the present problem we have:
Number of
cm
centimeters = (45.6m) 100
1m
= 4.56 x 103 cm
146
If we take the wrong conversion factor, the
result would be:
Number of
centimeters = (45.6m) 1 m
100 cm
= 4.56 x 10-1 m2 cm-1
147
If we take the wrong conversion factor, the
result would be:
Number of
centimeters = (45.6m) 1 m
100 cm
= 4.56 x 10-1 m2 cm-1
In this case we can see that the units on the
right-hand side do not match with what we
expect on the left-hand side of the expression.
148
Second example: convert 45.6 m to inches.
Conversion factors are 1 m = 100 cm and
1 inch = 2.54 cm (exactly)
149
Second example: convert 45.6 m to inches.
Conversion factors are 1 m = 100 cm and
1 inch = 2.54 cm (exactly)
Number of inches = (45.6m) 100 cm 1 inch
1m 2.54 cm
= 1.80 x 103 inches
150
Second example: convert 45.6 m to inches.
Conversion factors are 1 m = 100 cm and
1 inch = 2.54 cm (exactly)
Number of inches = (45.6m) 100 cm 1 inch
1m 2.54 cm
= 1.80 x 103 inches
In this factor label calculation, two conversion
factors are strung together. You can multiply any
number of conversion factors in a string.
151
Read the section in the book on:
Atomic view of matter (section 2.2)
Dalton’s atomic theory (section 2.3)
Nuclear atom model (section 2.4)
152
Atomic number, Mass number,
and Isotopes
Atomic Number: The number of protons in
the nucleus of each atom of an element.
The standard symbol for the atomic number
is Z.
153
Atomic number, Mass number,
and Isotopes
Atomic Number: The number of protons in
the nucleus of each atom of an element.
The standard symbol for the atomic number
is Z.
For a neutral atom, the atomic number also
indicates the number of electrons.
154
Example: The atomic number of C is 6. Thus,
carbon has 6 protons and 6 electrons.
155
Mass number: The total number of protons and
neutrons in the nucleus of each atom of an
element.
The symbol employed is A.
156
Mass number: The total number of protons and
neutrons in the nucleus of each atom of an
element.
The symbol employed is A.
The number of neutrons present in the nucleus
=A - Z
157
Mass number: The total number of protons and
neutrons in the nucleus of each atom of an
element.
The symbol employed is A.
The number of neutrons present in the nucleus
=A - Z
For example, the mass number of fluorine is 19
and the atomic number is 9, so the number of
neutrons present in the fluorine nucleus is
19 - 9 = 10.
158
Isotopes: Atoms having the same atomic
number, but different mass numbers.
159
Isotopes: Atoms having the same atomic
number, but different mass numbers.
The common symbol for denoting the atomic
number and mass number of element X is
A
ZX
160
Examples of isotopes: hydrogen has three
common isotopes
1H
1
hydrogen
2H
1
deuterium
3H
1
tritium
161
Examples of isotopes: hydrogen has three
common isotopes
1H
1
hydrogen
2H
1
deuterium
3H
1
tritium
Hydrogen is the only element for which special
symbols are used for different isotopes.
D for deuterium
T for tritium
162
Key Point: The chemical properties of an
element are primarily determined by the
number of protons and electrons, not by the
number of neutrons present. For this reason,
isotopes of the same element are chemically
similar.
163
Atomic mass scale
By international agreement, the reference adopted for
an atomic mass scale is the assignment of exactly a
mass of 12 for the most abundant isotope of carbon,
12C
6
164
Atomic mass scale
By international agreement, the reference adopted for
an atomic mass scale is the assignment of exactly a
mass of 12 for the most abundant isotope of carbon,
12C
6
The atomic mass unit (amu) is defined by
12C atom
mass
of
one
6
1 amu =
12
165
Most naturally occurring elements contain more than
one isotope – this means that when the atomic mass
of a naturally occurring element is determined, it is
an average quantity that is determined.
166
Most naturally occurring elements contain more than
one isotope – this means that when the atomic mass
of a naturally occurring element is determined, it is
an average quantity that is determined.
Example: the atomic masses of 12C and 13C
6
6
are 12.0000000 … amu (exact number) and 13.00335
amu and their natural abundances (relative amounts
of the isotopes present) are 98.89% and 1.11%
respectively.
167
The weighted average
atomic mass of carbon
 98.89 x 12  1.11 x 13.00335
1 00
1 00
= 12.011 amu
168
Molar mass
Chemists frequently use the term molecular
weight or for an atomic system, atomic
weight.
169
Molar mass
Chemists frequently use the term molecular
weight or for an atomic system, atomic
weight.
The molar mass is the mass of 1 mol of a
substance expressed in units of g/mol.
170
The mass of a single atom of carbon -12 is
determined from:
12C
molar
mass
of
6
mass of one 126C atom 
Avogadro's number
-1
12
g
mol

6.02x1023 mol-1
= 1.99 x 10-23 g
171
Molecules
The simplest molecules consist of only two
atoms – these are called diatomic molecules.
172
Molecules
The simplest molecules consist of only two
atoms – these are called diatomic molecules.
Examples: O2, H2, and Cl2
These are dioxygen, dihydrogen, and dichlorine.
Unfortunately, it is also very common to refer
to these as oxygen, hydrogen, and chlorine.
173
Molecules containing more than two atoms
are called polyatomic molecules.
174
Molecular mass
The molecular mass of a molecule is the sum
of the atomic masses of the atoms present in
the molecule.
Example: The molecular mass of H2O is given
by 2 x(1.00794) + 15.9994 = 18.0153 amu
175
The molar mass (frequently termed the
molecular weight) of a compound is the
molecular mass expressed in grams/mole
(units abbreviated as g/mol).
176
The molar mass (frequently termed the
molecular weight) of a compound is the
molecular mass expressed in grams/mole
(units abbreviated as g/mol).
Example: The molecular mass of H2O is 18.0153
amu and hence the molar mass of water is
18.0153 g/mol.
177
Laws of Chemical Combination
Law of Definite Proportions: In a given
compound, the elements are always
combined in the same proportion by mass.
178
Laws of Chemical Combination
Law of Definite Proportions: In a given
compound, the elements are always
combined in the same proportion by mass.
Example: water contains hydrogen and oxygen
in the proportion of 2.01588 g to 15.9994 g.
179
Law of Multiple Proportions: If two elements
combine to form more than one compound,
the amounts of one element that combine
with a fixed amount of the other element are
in the ratio of small whole numbers.
180
Law of Multiple Proportions: If two elements
combine to form more than one compound,
the amounts of one element that combine
with a fixed amount of the other element are
in the ratio of small whole numbers.
Example: Nitrogen and oxygen combine to form
several different oxides, of which three are
NO, N2O, and NO2. From experiment it is
known that 16 g of oxygen will combine with
14 g, 28 g, and 7 g of nitrogen, respectively,
for the three compounds. These masses are in
the ratio 14: 28: 7 or 2 : 4 : 1.
181
Percentage Composition
Determining a chemical formula of an
unknown compound begins with finding the
elements present.
182
Percentage Composition
Determining a chemical formula of an
unknown compound begins with finding the
elements present.
The next step is to determine the percentage
composition – which allows the empirical
formula to be determined.
183
Percentage Composition
Determining a chemical formula of an
unknown compound begins with finding the
elements present.
The next step is to determine the percentage
composition – which allows the empirical
formula to be determined.
Percentage composition: The percent by mass
of each element present in a compound.
184
Example: For H2O2 the molar mass is
34.0147 g/mol. There are 2.01588 g of H and
31.9988 g of O.
% H  2.01588 g x 100  5.92650%
34.0147 g
g x 100  94.0734%
% O  31.9988
34.0147 g
185
Example: For H2O2 the molar mass is
34.0147 g/mol. There are 2.01588 g of H and
31.9988 g of O.
% H  2.01588 g x 100  5.92650%
34.0147 g
g x 100  94.0734%
% O  31.9988
34.0147 g
The same percentage results would be
obtained if we started with the empirical
formula HO.
186
Example: Calculating % composition from
mass data.
187
Example: Calculating % composition from
mass data.
A sample of a liquid with a mass of 8.657 g
was decomposed into its elements to give
5.217 g of carbon, 0.9620 g of hydrogen, and
2.478 g of oxygen. What is the % by mass of
each element in the liquid?
188
Example: Calculating % composition from
mass data.
A sample of a liquid with a mass of 8.657 g
was decomposed into its elements to give
5.217 g of carbon, 0.9620 g of hydrogen, and
2.478 g of oxygen. What is the % by mass of
each element in the liquid?
Answer: Do a percentage calculation.
5.217 g x 100  60.26% C
8.657 g
189
Example: Calculating % composition from
mass data.
A sample of a liquid with a mass of 8.657 g
was decomposed into its elements to give
5.217 g of carbon, 0.9620 g of hydrogen, and
2.478 g of oxygen. What is the % by mass of
each element in the liquid?
Answer: Do a percentage calculation.
2.478 g x 100  28.62% O
5.217 g x 100  60.26% C
8.657 g
8.657 g
190
Example: Calculating % composition from
mass data.
A sample of a liquid with a mass of 8.657 g
was decomposed into its elements to give
5.217 g of carbon, 0.9620 g of hydrogen, and
2.478 g of oxygen. What is the % by mass of
each element in the liquid?
Answer: Do a percentage calculation.
2.478 g x 100  28.62% O
5.217 g x 100  60.26% C
8.657 g
8.657 g
0.9620 g x 100  11.11% H
8.657 g
191
Empirical Formulas
To determine the empirical formula using
percentage composition data – find the
number of moles of each element present.
192
Empirical Formulas
To determine the empirical formula using
percentage composition data – find the
number of moles of each element present.
Example 1: Ascorbic acid (vitamin C) is
composed of 40.92% C, 4.58% H and 54.50%
O. Determine the empirical formula.
193
Since we have % data, it is convenient to start
with 100 g of the substance. Therefore, in 100 g
of ascorbic acid, there are 40.92 g of C, 4.58 g of
H, and 54.50 g of O.
Now calculate the number of moles:
194
Since we have % data, it is convenient to start
with 100 g of the substance. Therefore, in 100 g
of ascorbic acid, there are 40.92 g of C, 4.58 g of
H, and 54.50 g of O.
Now calculate the number of moles:
moles of
1 mol C  3.406 mol C
40.92
g
C
carbon =
12.011 gC
195
Since we have % data, it is convenient to start
with 100 g of the substance. Therefore, in 100 g
of ascorbic acid, there are 40.92 g of C, 4.58 g of
H, and 54.50 g of O.
Now calculate the number of moles:
moles of
1 mol C  3.406 mol C
40.92
g
C
carbon =
12.011 gC
moles of
1 mol H  4.54 mol H
4.58
g
H
hydrogen =
1.008 gH
196
Since we have % data, it is convenient to start
with 100 g of the substance. Therefore, in 100 g
of ascorbic acid, there are 40.92 g of C, 4.58 g of
H, and 54.50 g of O.
Now calculate the number of moles:
moles of
1 mol C  3.406 mol C
40.92
g
C
carbon =
12.011 gC
moles of
1 mol H  4.54 mol H
4.58
g
H
hydrogen =
1.008 gH
moles of
1 mol O  3.406 mol O
54.50
g
O
oxygen =
15.9994 gO
197
Hence, the empirical formula is obtained as
follows:
C3.406H4.54O3.406
198
Hence, the empirical formula is obtained as
follows:
C3.406H4.54O3.406
Now divide by the smallest number, so that
C 3.406H 4.54 O 3.406
3.406
3.406
3.406
199
Hence, the empirical formula is obtained as
follows:
C3.406H4.54O3.406
Now divide by the smallest number, so that
C 3.406H 4.54 O 3.406
3.406
3.406
3.406
that is, C1.000H1.33O1.000 = C1H4/3O1
200
Hence, the empirical formula is obtained as
follows:
C3.406H4.54O3.406
Now divide by the smallest number, so that
C 3.406H 4.54 O 3.406
3.406
3.406
3.406
that is, C1.000H1.33O1.000 = C1H4/3O1
Now eliminate the fraction (multiply by 3) to
obtain the empirical formula as C3H4O3.
201
Example: A major air pollutant in coal burning
countries is a colorless pungent smelling gas
containing sulfur and oxygen. Chemical
analysis of a 1.078 g sample of this gas
showed that it contained 0.540 g of sulfur and
0.538 g of oxygen. What is the empirical
formula of this gas?
202
Example: A major air pollutant in coal burning
countries is a colorless pungent smelling gas
containing sulfur and oxygen. Chemical
analysis of a 1.078 g sample of this gas
showed that it contained 0.540 g of sulfur and
0.538 g of oxygen. What is the empirical
formula of this gas?
It is not necessary in this problem to calculate
the % composition – just calculate the number
of moles of each element present.
203
moles of sulfur = 0.540 gS 1 mol S  0.0168 mol S
32.064 gS
moles of oxygen = 0.538 gO 1 mol O  0.0336 mol O
15.999 gO
204
moles of sulfur = 0.540 gS 1 mol S  0.0168 mol S
32.064 gS
moles of oxygen = 0.538 gO 1 mol O  0.0336 mol O
15.999 gO
Hence,
S0.0168O0.0336
205
moles of sulfur = 0.540 gS 1 mol S  0.0168 mol S
32.064 gS
moles of oxygen = 0.538 gO 1 mol O  0.0336 mol O
15.999 gO
Hence,
S0.0168O0.0336
Now divide by the smallest number to obtain,
SO2
which is the empirical formula.
206
Molecular Formulas
A molecular formula gives the actual composition
of a molecule.
Example: Calculating a molecular formula from an
empirical formula and molar mass information.
207
Molecular Formulas
A molecular formula gives the actual composition
of a molecule.
Example: Calculating a molecular formula from an
empirical formula and molar mass information.
An oxide of nitrogen gave the following analysis:
1.52 g of nitrogen combined with 3.47 g of
oxygen. The molar mass of the compound was
found to be 92.0 g/mol. Determine the
molecular formula.
208
Approach 1: Determine the empirical formula and
then determine the molecular formula.
209
Approach 1: Determine the empirical formula and
then determine the molecular formula.
moles of oxygen = 3.47 g O 1 mol O  0.217 mol O
15.999 g O
moles of nitrogen = 1.52 g N 1 mol N  0.109 mol N
14.007 g N
210
Approach 1: Determine the empirical formula and
then determine the molecular formula.
moles of oxygen = 3.47 g O 1 mol O  0.217 mol O
15.999 g O
moles of nitrogen = 1.52 g N 1 mol N  0.109 mol N
14.007 g N
Empirical formula is N0.109O0.217
NO2
211
Approach 1: Determine the empirical formula and
then determine the molecular formula.
moles of oxygen = 3.47 g O 1 mol O  0.217 mol O
15.999 g O
moles of nitrogen = 1.52 g N 1 mol N  0.109 mol N
14.007 g N
Empirical formula is N0.109O0.217
NO2
The mass of the empirical formula unit is
14.0067 + 2 x 15.9994 = 46.0055 g
212
Approach 1: Determine the empirical formula and
then determine the molecular formula.
moles of oxygen = 3.47 g O 1 mol O  0.217 mol O
15.999 g O
moles of nitrogen = 1.52 g N 1 mol N  0.109 mol N
14.007 g N
Empirical formula is N0.109O0.217
NO2
The mass of the empirical formula unit is
14.0067 + 2 x 15.9994 = 46.0055 g
Since the mass of 1 mol is 92.0 g, the number
of empirical formula units in the compound is
213
equal to
92.0 g  2
46.0055 g
214
equal to
92.0 g  2
46.0055 g
Hence, the molecular formula is (NO2)2, that is
N2O4 (dinitrogen tetraoxide).
215
equal to
92.0 g  2
46.0055 g
Hence, the molecular formula is (NO2)2, that is
N2O4 (dinitrogen tetraoxide).
Example: A compound of carbon, hydrogen, and
oxygen has the following % composition (by mass):
C 42.10 %, H 6.479 %, O 51.421 %
The molar mass of the compound is 342.3 g/mol.
What is the molecular formula?
216
equal to
92.0 g  2
46.0055 g
Hence, the molecular formula is (NO2)2, that is
N2O4 (dinitrogen tetraoxide).
Example: A compound of carbon, hydrogen, and
oxygen has the following % composition (by mass):
C 42.10 %, H 6.479 %, O 51.421 %
The molar mass of the compound is 342.3 g/mol.
What is the molecular formula?
Approach 2: Work directly with the given molar
mass and find the moles of each element present.
217
moles of C =
42.10 x 342.3 g C 1 mol C  12.00 mol C
100
12.01 g C
moles of H =
6.479 x 342.3 g H 1 mol H  22.00 mol H
100
1.0079 g H
moles of O =
51.421x 342.3 g O 1 mol O  11.00 mol O
100
15.999 g O
Hence, the molecular formula is C12H22O11
218
Exercise: A compound of carbon, hydrogen, and
oxygen has the following composition:
C 40.00% and H 6.714 %.
The molar mass of the compound is 180.16
g/mol. What is the molecular formula?
Answer is C6H12O6
219