Spencer L. Seager
Michael R. Slabaugh
www.cengage.com/chemistry/seager
Chapter 2:
Atoms and Molecules
Jennifer P. Harris
SYMBOLS & 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.
CHEMICAL ELEMENTS & THEIR SYMBOLS
CHEMICAL ELEMENTS & THEIR SYMBOLS
(continued)
Elements from Group 7A
chlorine
bromine
iodine
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.
COMPOUND FORMULAS EXAMPLES
• Carbon monoxide, CO
• one atom of C
• one atom of O
• Water, H2O
• two atoms of H
• one atom of O
• Ammonia, NH3
• one atom of N
• 3 atoms of H
COMPOUND FORMULAS PRACTICE
• The clear liquid is carbon disulfide. It is composed of carbon
(left) and sulfur (right). If carbon disulfide contains one atom
of carbon for every two atoms of sulfur, what is the chemical
formula for carbon disulfide?
• Answer: CS2
ATOMIC STRUCTURE
• 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.
SUBATOMIC PARTICLES
• 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 are located in the
nucleus of an atom. They carry
no electrical charge and have a
mass of 1 atomic mass unit (u).
• 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.
SUBATOMIC PARTICLE CHARACTERISTICS
ATOMIC STRUCTURE REVIEW
• Which subatomic particles are
represented by the pink spheres?
• Answer: electrons
• Which subatomic particles are
represented by the yellow and
blue spheres?
• Answer: protons and neutrons
• What structure do the yellow and
blue spheres form?
• Answer: the nucleus
ATOMIC & MASS NUMBERS
• 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.
A
Z
E
• MASS NUMBER OF AN ATOM
• The mass number of an atom is equal to the sum of the
number of protons & neutrons in the nucleus of the atom.
• Mass numbers are represented by the symbol A.
ATOMIC & MASS NUMBERS APPLICATION
• Based on the information given above, what is the atomic
number of fluorine?
• Answer: The atomic number of fluorine is 9.
• On the periodic table, the atomic number is written as a
whole number above the symbol F.
• In the written description, fluorine is said to have 9 protons
(the atomic number is the number of protons).
• In the symbol, the number 9 is written in the atomic
number or Z (lower left) position.
ATOMIC & MASS NUMBERS APPLICATION
• Based on the information given above, what is the mass
number of fluorine?
• Answer: The mass number of fluorine is 19.
• In the written description, fluorine is said to have 9 protons
and 10 neutrons (the mass number is the sum of the
numbers of protons and neutrons).
• In the symbol, the number 19 in written in the mass
number or A (upper left) position.
• Note: The periodic table does not show the mass number
for an individual atom. It lists an average mass number
for a collection of atoms!
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.
ISOTOPE SYMBOLS
A
• Isotopes are represented by the symbol Z
E
, where Z is the
atomic number, A is the mass number, and E is the
elemental symbol.
60
• 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.
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.
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.
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.
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(16.00 u) = 18.02 u
MOLECULAR WEIGHT PRACTICE
• The clear liquid is carbon disulfide, CS2. It is composed of
carbon (left) and sulfur (right). What is the molecular weight
for carbon disulfide?
• Answer: MW = 1(atomic weight C) + 2(atomic weight S)
12.01 u + 2(32.07 u) = 76.15 u
ISOTOPES & 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.
ISOTOPES & ATOMIC WEIGHTS
(continued)
• 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.
isotope %isotope mass 
Atomic wei ght 
100
• 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.
ISOTOPES & ATOMIC WEIGHTS EXAMPLE
• 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.

19.78% 10.01 u  80.22%11.01 u)
AW 
100
198.0 u  883.2 u

 10.81 u
100
• This calculated value is seen to agree with the value given in
the periodic table.
THE MOLE CONCEPT
• 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 = 32.07 g S = 6.022x1023 S atoms
THE MOLE CONCEPT (continued)
• 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.
• 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 CONCEPT (continued)
• 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.
One mole quantities of six
metals; top row (left to
right): Cu beads (63.5 g), Al
foil (27.0 g), and Pb shot
(207.2 g); bottom row (left
to right): S powder (32.1 g),
Cr chunks (52.0 g), and Mg
shavings (24.4 g).
One mole quantities of four
compounds: H2O (18.0 g);
small beaker NaCl (58.4 g);
large beaker aspirin,
C9H8O4, (180.2 g); green
(NiCl2 · 6H2O) (237.7 g).
MOLE CALCULATIONS
• 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:
1 mole Ca
40.08 g Ca
and
40.08 g Ca
23
6.022  10 Ca atoms
MOLE CALCULATION EXAMPLE
• Calculate the number of moles of Ca contained in a 15.84 g
sample of Ca.
• The solution to the problem is:
1mole Ca
15.84 g Ca 
 0.3952 moles Ca
40.08 g Ca
• 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.
MOLE CALCULATIONS (continued)
• 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.
MOLE CALCULATION EXAMPLES
• Example 1: How many moles of O atoms are contained in
11.57 g of CO2?
2 moles O atoms
11.57 g CO2 
 0.5258 moles O atoms
44.01 g CO2
• Note that the factor used was obtained from two of the nine
quantities given on the previous slide.
MORE MOLE CALCULATION EXAMPLES
• Example 2: How many CO2 molecules are needed to contain
50.00 g of C?
6.022  10 CO2 molecules
50.00 g C 
12.01 g C
23
 2.507  10
24
CO2 molecules
• Note that the factor used was obtained from two of the nine
quantities given on a previous slide.
MORE MOLE CALCULATION EXAMPLES
• Example 3: What is the mass percentage of C in CO2?
• The mass percentage is calculated using the
following equation:
mass of C
%C 
 100
mass of CO 2
• 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
MORE MOLE CALCULATION EXAMPLES
(continued)
• Thus, the mass of C in a specific mass of CO2 is known, and
the problem is solved as follows:
12.01 g C
%C 
 100  27.29%
44.01 g CO2
MORE MOLE CALCULATION EXAMPLES
• Example 4: What is the mass percentage of oxygen in CO2?
• The mass percentage is calculated using the following
equation:
mass of O
%O 
 100
mass of CO2
• 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
MORE MOLE CALCULATION EXAMPLES
(continued)
• Thus, the mass of O in a specific mass of CO2 is known, and
the problem is solved as follows:
32.00 g O
%O 
 100  72.71%
44.01 g CO2
• We see that the % C + % O = 100% , which should be the
case because C and O are the only elements present in CO2.
MOLE CALCULATIONS HELP
LABORATORY APPLICATION
• The mass of an empty crucible is 20.057 g. Copper sulfate
pentahydrate is added to the crucible. The combined mass is
45.551 g. What is the mass of the copper sulfate
pentahydrate?
45.551 g crucible + copper sulfate pentahydrate
−20.057 g crucible
25.494 g copper sulfate pentahyrdate
LABORATORY APPLICATION (continued)
• The crucible is heated and allowed to cool. The mass of the
crucible and anhydrous solid after heating is 38.547 g. What
is the mass of the anhydrous solid?
38.547 g crucible + anhydrous copper sulfate
−20.057 g crucible
18.490 g anhydrous copper sulfate
• What is the mass of the water that was removed by heating?
25.494 g copper sulfate pentahydrate
−18.490 g anhydrous copper sulfate
7.004 g water
LABORATORY APPLICATION (continued)
• How many moles of water (H2O) were removed by heating?
 1 mole H2 O 
  0.3887 moles H2 O
7.004 g H2 O
 18.02 g H2 O 
• How many moles of anhydrous copper sulfate (CuSO4)
remained after heating?
 1 mole CuSO 4 
  0.11584 moles CuSO 4
18.490 g CuSO 4 
 159.62 g CuSO 4 
• What is the ratio of moles anhydrous copper sulfate to moles
of water?
0.11584 moles CuSO4 : 0.3887 moles H2O
1 mole CuSO4 : 3.355 moles H2O
LABORATORY APPLICATION (continued)
• The formula for a hydrate is written as CuSO4·x H2O, where x
is the number of moles of water associated with each mole of
CuSO4. Based on the calculated mole ratio, what is the
experimentally determined formula for copper sulfate
pentahydrate?
CuSO4·3.355 H2O
• The actual formula for copper sulfate pentahydrate is
CuSO4·5 H2O. Why is there a difference between the actual
formula and the experimentally determined formula?
• Answer: The copper sulfate pentahydrate was not heated
thoroughly enough. Some of the water remained in the
“anhydrous” copper sulfate, which caused the number of
moles of water removed by heating to be smaller than the
actual value.
LABORATORY APPLICATION (continued)
• What is the mass percentage of H2O in CuSO4·5 H2O?
 18.02 g H2 O 

5 moles H2 O
 1 mole H2 O   100  36.08% H O
2
249.72 g CuSO 4  5 H2 O
• What mass of water could have been removed from the
25.494 g copper sulfate pentahydrate, if the sample had been
heated completely?


90.10 g H2 O

  9.1983 g H2 O
25.494 g CuSO 4  5 H2 O
 249.72 g CuSO 4  5 H2 O 
or


36.08 g H2 O
  9.198 g H2 O
25.494 g CuSO 4  5 H2 O
 100 g CuSO 4  5 H2 O 
LABORATORY APPLICATION (continued)
• What percentage of the total amount of water was removed
during the experiment?
7.004 g H2 O
 100  76.14%
9.1983 g H2 O
• Had all the water been removed, what would the mass of the
anhydrous copper sulfate been after heating? Would this
have been the mass displayed on the balance? Why or why
not?
25.494 g copper sulfate pentahydrate
−9.1983 g water
16.296 g anhydrous copper sulfate
• Answer: No, the mass displayed on the balance would have
been 36.353 g because the anhydrous copper sulfate was in
the 20.057 g crucible.