Chapter 2:
Atoms and Molecules
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SYMBOLS AND FORMULAS
• A unique symbol is used to represent each element.
• Formulas are used to represent compounds.
ELEMENTAL SYMBOLS
• A symbol is assigned to each element. The symbol is
based on the name of the element and consists of one
capital letter or a capital letter followed by a lower case
letter.
• Some symbols are based on the Latin or German name of
the element.
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chlorine
bromine
iodine
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COMPOUND FORMULAS
• A compound formula consists of the symbols of the
elements found in the compound. Each elemental symbol
represents one atom of the element. If more than one
atom is represented, a subscript following the elemental
symbol is used.
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EXAMPLES OF COMPOUND FORMULAS
• Carbon monoxide, CO (one atom of C and one atom of O
are represented).
• Water, H2O (two atoms of H and one atom of O are
represented).
• Ammonia, NH3 (one atom of N and 3 atoms of H are
represented).
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THE STRUCTURE OF ATOMS
• Atoms are made up of three subatomic particles, protons,
neutrons, and electrons.
• The protons and neutrons are tightly bound together to
form the central portion of an atom called the nucleus.
• The electrons are located outside of the nucleus and
thought to move very rapidly throughout a relatively large
volume of space surrounding the small but very heavy
nucleus.
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PROTONS
• Protons are located in the nucleus of an atom. They carry
a +1 electrical charge and have a mass of 1 atomic mass
unit (u).
NEUTRONS
• Neutrons are located in the nucleus of an atom. They
carry no electrical charge and have a mass of 1 atomic
mass unit (u).
ELECTRONS
• Electrons are located outside the nucleus of an atom.
They carry a -1 electrical charge and have a mass of
1/1836 atomic mass unit (u). They move rapidly around
the heavy nucleus.
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CHARACTERISTICS OF SUBATOMIC PARTICLES
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ATOMIC NUMBER OF AN ATOM
• The atomic number of an atom is equal to the number of
protons in the nucleus of the atom.
• Atomic numbers are represented by the symbol Z.
MASS NUMBER OF AN ATOM
• The mass number of an atom is equal to the sum of the
number of protons and neutrons in the nucleus of the
atom.
• Mass numbers are represented by the symbol A.
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ISOTOPES
• Isotopes are atoms that have the same number of protons
in the nucleus but different numbers of neutrons. That is,
they have the same atomic number but different mass
numbers.
• Because they have the same number of protons in the
nucleus, all isotopes of the same element have the same
number of electrons outside the nucleus.
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SYMBOLS FOR ISOTOPES
• Isotopes are represented by the symbol
, where Z is
the atomic number, A is the mass number and E is the
elemental symbol.
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• An example of an isotope symbol is 28 Ni . This symbol
represents an isotope of nickel that contains 28 protons
and 32 neutrons in the nucleus.
• Isotopes are also represented by the notation: Name-A,
where Name is the name of the element and A is the mass
number of the isotope.
• An example of this isotope notation is magnesium-26.
This represents an isotope of magnesium that has a mass
number of 26.
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RELATIVE MASSES
• The extremely small size of atoms and molecules
makes it inconvenient to use their actual masses
for measurements or calculations. Relative
masses are used instead.
• Relative masses are comparisons of actual
masses to each other. For example if an object
had twice the mass of another object, their relative
masses would be 2 to 1.
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ATOMIC MASS UNIT (u)
• An atomic mass unit is a unit used to express the
relative masses of atoms. One atomic mass unit
is equal to 1/12 the mass of a carbon-12 atom.
• A carbon-12 atom has a relative mass of 12 u.
• An atom with a mass equal to 1/12 the mass of a
carbon-12 atom would have a relative mass of
1 u.
• An atom with a mass equal to twice the mass of a
carbon-12 atom would have a relative mass of
24 u.
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ATOMIC WEIGHT
• The atomic weight of an element is the relative mass of an
average atom of the element expressed in atomic mass
units.
• Atomic weights are the numbers given at the bottom of the
box containing the symbol of each element in the periodic
table.
• According to the periodic table, the atomic weight of
nitrogen atoms (N) is 14.0 u, and that of silicon atoms (Si)
is 28.1 u. This means that silicon atoms are very close to
twice as massive as nitrogen atoms. Put another way, it
means that two nitrogen atoms have a total mass very
close to the mass of a single silicon atom.
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MOLECULAR WEIGHT
• The relative mass of a molecule in atomic mass units is
called the molecular weight of the molecule.
• Because molecules are made up of atoms, the molecular
weight of a molecule is obtained by adding together the
atomic weights of all the atoms in the molecule.
• The formula for a molecule of water is H2O. This means
one molecule of water contains two atoms of hydrogen, H,
and one atom of oxygen, O. The molecular weight of
water is then the sum of two atomic weights of H and one
atomic weight of O:
MW = 2(at. wt. H) + 1(at. wt. O)
MW = 2(1.01 u) + 1(16.0 u) = 18.02 u
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ISOTOPES AND ATOMIC WEIGHTS
• Many elements occur naturally as a mixture of several
isotopes.
• The atomic weight of elements that occur as mixtures of
isotopes is the average mass of the atoms in the isotope
mixture.
• The average mass of a group of atoms is obtained by
dividing the total mass of the group by the number of
atoms in the group.
• A practical way of determining the average mass of a
group of isotopes is to assume the group consists of 100
atoms and use the percentage of each isotope to
represent the number of atoms of each isotope present in
the group.
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• The use of percentages and the mass of each isotope
leads to the following equation for calculating atomic
weights of elements that occur naturally as a mixture of
isotopes.
According to this equation, the atomic weight of an element is
calculated by multiplying the percentage of each isotope in the
element by the mass of the isotope, then adding the resulting
products together and dividing the resulting mass by 100.
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• A specific example of the use of the equation is shown
below for the element boron that consists of 19.78%
boron-10 with a mass of 10.01 u, and 80.22% boron-11
with a mass of 11.01u.
This calculated value is seen to agree with the value given in
the periodic table.
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THE MOLE CONCEPT APPLIED TO ELEMENTS
• The number of atoms in one mole of any element is called
Avogadro's number and is equal to 6.022x1023 .
• A one-mole sample of any element will contain the same
number of atoms as a one-mole sample of any other
element.
• One mole of any element is a sample of the element with a
mass in grams that is numerically equal to the atomic
weight of the element.
EXAMPLES OF THE MOLE CONCEPT
• 1 mole Na = 22.99 g Na = 6.022x1023 Na atoms
• 1 mole Ca = 40.08 g Ca = 6.022x1023 Ca atoms
• 1 mole S = 32.06 g S = 6.022x1023 S atoms
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THE MOLE CONCEPT APPLIED TO COMPOUNDS
• The number of molecules in one mole of any compound is
called Avogadro's number and is numerically equal to
6.022x1023.
• A one-mole sample of any compound will contain the same
number of molecules as a one-mole sample of any other
compound.
• One mole of any compound is a sample of the compound
with a mass in grams equal to the molecular weight of the
compound.
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EXAMPLES OF THE MOLE CONCEPT
• 1 mole H2O = 18.02 g H2O = 6.022x1023 H2O molecules
• 1 mole CO2 = 44.01 g CO2 = 6.022x1023 CO2 molecules
• 1 mole NH3 = 17.03 g NH3 = 6.022x1023 NH3 molecules
THE MOLE AND CHEMICAL CALCULATIONS
• The mole concept can be used to obtain factors that are
useful in chemical calculations involving both elements
and compounds.
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CALCULATIONS INVOLVING ELEMENTS
• The mole-based relationships given earlier as examples
for elements provide factors for solving problems.
• The relationships given earlier for calcium are:
1 mole Ca= 40.08 g Ca = 6.022x1023 Ca atoms.
• Any two of these quantities can be used to provide factors
for use in solving numerical problems.
• Examples of two of the six possible factors are:
and
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EXAMPLE PROBLEM
• Calculate the number of moles of Ca contained in
a 15.84 g sample of Ca.
• The solution to the problem is:
• We see in the solution that the g Ca units in the
denominator of the factor cancel the g Ca units in
the given quantity, leaving the correct units of
mole Ca for the answer.
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CALCULATIONS INVOLVING COMPOUNDS
• The mole concept applied earlier to molecules can be
applied to the individual atoms that are contained in the
molecules.
• An example of this for the compound CO2 is:
1 mole CO2 molecules = 1 mole C atoms + 2 moles O atoms
44.01 g CO2 = 12.01 g C + 32.00 g O
6.022x1023 CO2 molecules = 6.022x1023 C atoms +
(2) 6.022x1023 O atoms.
• Any two of these nine quantities can be used to provide
factors for use in solving numerical problems.
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• Example 1: How many moles of O atoms are contained in
11.57 g of CO2?
• Note that the factor used was obtained from two of the
nine quantities given on the previous slide.
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• Example 2: How many CO2 molecules are needed to
contain 50.00 g of C?
• Note that the factor used was obtained from two of the
nine quantities given on a previous slide.
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• Example 3: What is the mass percentage of C in CO2?
• The mass percentage is calculated using the
following equation:
• If a sample consisting of 1 mole of CO2 is used, the molebased relationships given earlier show that
1 mole CO2=44.01 g CO2=12.01 g C + 32.00 g O. Thus,
the mass of C in a specific mass of CO2 is known, and the
problem is solved as follows:
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• Example 4: What is the mass percentage of oxygen in
CO2?
• The mass percentage is calculated using the following
equation:
• Once again, a sample consisting of 1 mole of CO2 is used
to take advantage of the mole-based relationships given
earlier where:
1 mole CO2 = 44.01g CO2 = 12.01 g C + 32.00g O
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• Thus, the mass of O in a specific mass of CO2 is
known, and the problem is solved as follows:
• We see that the % C + % O = 100% , which
should be the case because C and O are the only
elements present in CO2.
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