Chapter 5.1, 5.3, 5.5,
5.8, 5.9 and 5.10
Homework from the book
1, 11, 35, 37, 39, 41, 43, 45, 45, 47,
67, 69, 75, 77, 79, 83, 85, 87, 89, 91,
97, 103, 107, 109, 115, 119, 121, 127,
129 and 131
Molecules and Compounds
5.1
A molecule is an aggregate of two or more atoms in a definite arrangement
held together by chemical bonds
A compound is a substance composed of two or more elements
combined in a specific ratio and held together by chemical bonds.
A molecule is formed when two or more atoms join together chemically. A
compound is a molecule that contains at least two different elements. All
compounds are molecules but not all molecules are compounds
Which of the following is a molecule and NOT a compound?
Answer:
H2
H2O
NH3
CH4
H2
There are 7 elements that occur in nature as a diatomic molecule
One way to remember these
elements is:
Mr. BrINClHOF
Empirical and Molecular formulas
A molecular formula shows the exact number of atoms of
each element in the smallest unit of a substance
An empirical formula shows the simplest
whole-number ratio of the atoms in a substance
molecular
empirical
H2O
H2O
C6H12O6
CH2O
O3
O
N2H4
NH2
Ions
Atoms are neutral—meaning that the number of protons
is equal to the number of electrons
If an atom loses or gains electrons the atom is
No longer neutral but has a charge.
Why do Ions form?
Cations form from metals because they have a low Ionization
energy and will readily give up electrons to obtain the
electron configuration of a noble gas
Na
11 protons
+ energy
11 electrons
Na+
11 protons
10 electrons
+ e-
Anions form from non-metals because they have low Electron
affinity and will readily accept electrons to obtain the
electron configuration of a noble gas
Cl
17 protons + e17 electrons
Cl-
17 protons
18 electrons
Electrons are transferred from the Cation to the Anion and the
charged ions attract each other
+ energy
Ionic Bonds
Occur between a metal and a non-metal. Ionic bonds form from the
Attraction of positive and negative ions that form because of a transfer
Of electrons.
Lattice Energy
The amount that the potential energy
of the system decreases
When the ions in one mole of the
compound are brought from
A gaseous state to the positions the
ions occupy in a crystal of the
Compound. This is the reason Ionic
Compounds form
Electrostatic (Lattice) Energy
Lattice energy (E) is the energy required to completely separate
one mole of a solid ionic compound into gaseous ions.
Q+Q E=k
r
Q+ is the charge on the cation
Q- is the charge on the anion
r is the distance between the ions
Lattice energy (E) increases
as Q increases and/or
as r decreases.
cmpd
MgF2
MgO
LiF
LiCl
lattice energy
2957 Q= +2,-1
3938 Q= +2,-2
1036
853
r F < r Cl
The Octet Rule
Atoms tend to gain or lose electrons until they have achieved an
outer shell that contains an octet (8) of electrons
Exceptions to the Octet Rule
Atoms that have valence electrons in the d and f orbitals
And all atoms before Carbon in the periodic table
Recognizing Ionic Compounds
A compound is ionic if it contains a metal from group 1
Or group 2 or one of the polyatomic ions. Binary metal
Oxides and sulfides also have ionic character
Fig 2-23
Pg 59
Courtesy Ken Karp
Pure water(left)
and a solution
of sugar(right)
do not conduct
electricity
because they
contain virtually
no ions. A
solution of salt
(center)
conducts
electricity well
because it
contain mobile
cations and
anions.
When Atoms Combine to make Molecules
Atoms contain both positive and negative charges. When they come
Together they arrange themselves so that the attractive forges of opposite
Charges is greater than the repulsive forces of like charges
Bond Length and Bond Energy
Bond Length (Bond Distance)
The distance in picometers between the nucleii of two
Bound atoms in a molecule
Bond Energy
The strength of the bond between atoms in a molecule.
The amount of energy (J) required to break the bond.
Fig 8-3
Pg 330
The interaction energy of a pair of hydrogen atoms varies with
internuclear separation.
Change in electron
density as two hydrogen
atoms approach each
other.
A covalent bond is a chemical bond in which two or more
electrons are shared by two atoms.
Why should two atoms share electrons?
F
+
7e-
F
F F
7e-
8e- 8e-
Lewis structure of F2
single covalent bond
lone pairs
F
F
single covalent bond
lone pairs
F F
lone pairs
lone pairs
The Octet Rule
Atoms tend to share electrons until they have achieved an
outer shell that contains an octet (8) of electrons
Exceptions to the Octet Rule
Atoms that have valence electrons in the d and f orbitals
And all atoms before Carbon in the periodic table
Lewis structure of water
H
+
O +
H
single covalent bonds
H O H
or
H
O
H
2e-8e-2eDouble bond – two atoms share two pairs of electrons
O C O
or
O
O
C
double bonds
- 8edouble
8e- 8ebonds
Triple bond – two atoms share three pairs of electrons
N N
triple
bond
8e-8e
or
N
N
triple bond
5.10
Molar Mass
The molar mass (MM ) of a substance is the mass in grams of one
mole of the substance.
The molar mass of an element is numerically equal to its atomic
mass.
1 mol C = 12.01 g
1 C atom = 12.01 amu
The molar mass of a compound is the sum of molar masses of the
elements it contains.
1 mol H2O = 2 ×1.008 g + 16.00 g = 18.02 g
1 mol NaCl = 22.99 g + 35.45 g = 58.44 g
5.8
Molecular and Formula Mass
The molecular mass is the mass in atomic mass units (amu) of an
individual molecule.
To calculate molecular mass, multiply the atomic mass for each
element in a molecule by the number of atoms of that element and
then total the masses
Molecular mass of H2O = 2(atomic mass of H) + atomic mass of O
= 2(1.008 amu) + 16.00 amu = 18.02 amu
Because the atomic masses on the periodic table are average atomic
masses, the result of such a determination is an average molecular
mass, sometimes referred to as the molecular weight.
Molar mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams/mole)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
Worked Example 5.12
Calculate the molecular mass or the formula mass, as appropriate, for each of the
following corresponds: (a) propane, C3H8, (b) lithium hydroxide, LiOH, and (c)
barium acetate, Ba(C2H3O2)2.
Strategy Determine the molecular mass (for each molecular compound) or
formula mass (for each ionic compound) by summing all the atomic masses.
Solution
ForAbout
each compound,
multiply
theyou
number
atomsthe
by number
the atomic mass
Think
It Double-check
that
have of
counted
of eachofelement
and then for
sumeach
the calculated
atoms correctly
compound values.
and that you have used the
proper atomic masses from the periodic table.
(a) The molecular mass of propane is 3(12.01 amu) + 8(1.008 amu) = 44.09 amu
(b) The formula mass of lithium hydroxide is 6.941 amu + 16.00 amu + 1.008
amu = 23.95 amu.
(c) The formula mass of barium acetate is 137.3 amu + 4(12.01 amu) + 6(1.008
amu) + 4(16.00 amu) = 255.4 amu.
Interconverting Mass, Moles, and
Numbers atoms or molecules
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x
x
x
=
1 mol C3H8O
1 mol H atoms
60 g C3H8O
5.82 x 1024 atoms H
Worked Example 5.14
Determine (a) the number of moles of CO2 in 10.00 g of carbon dioxide and
(b) the mass of 0.905 mole of sodium chloride.
Strategy Use molar mass to convert from mass to moles and to convert from
moles to mass. The molar mass of carbon dioxide (CO2) is 44.01 g/mol and the
molar mass of sodium chloride (NaCl) is 58.44 g/mol.
Solution
1 mol CO2
(a) 10.00 g CO2 × 44.01 g CO = 0.2272 mol CO2
2
(b) 0.905 mol NaCl ×
58.44 g NaCl
= 52.9 g NaCl
1 mol NaCl
Think About It Always double-check unit cancellations in problems such as
these–errors are common when molar mass is used as a conversion factor. Also
make sure that the results make sense. In both cases, a mass smaller than the
molar mass corresponds to less than a mole of substance.
Worked Example 5.15
(a) Determine the number of water molecules and the numbers of H and O atoms
in 3.26 g of water.
(b) Determine the mass of 7.92×1019 carbon dioxide molecules.
Strategy Use molar mass and Avogadro’s number to convert from mass to
molecules, and vice versa. Use the molecular formula of water to determine the
numbers of H and O atoms.
Solution
1 mol H O
6.022×1023 H O molecules
2
(a) 3.26 g H2O× 18.02 g H O ×
2
2
1 mol H2O
= 1.09×1023 H2O molecules
Using the molecular formula, we can determine the number of H and O atoms in 3.26 g of
H2O as follows:
2 H atoms
1.09×1023 H2O molecules × 1 H O molecule = 2.18×1023 H atoms
2
1 O atom
1.09×1023 H2O molecules × 1 H O molecule = 1.09×1023 H atoms
2
Solution
1 mol CO2
44.01 g CO2
(b) 7.92×1019 CO2 molecules × 6.022×1023 CO molecules × 1 mol CO
2
2
= 5.79×10-3 g C
Think About It Again, check the cancellation of units carefully and make sure that the magnitudes
of your results are reasonable.
5.9
Percent Composition of Compounds
A list of the percent by mass of each element in a compound is
known as the compound’s percent composition by mass.
percent mass of an element =
n  atomic mass of element
 100%
molecular or formula mass of compound
where n is the number of atoms of the element in a molecule or
formula unit of the compound
Percent composition of an element in a compound
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
%C =
%H =
%O =
C2H6O
2 x (12.01 g)
46.07 g
6 x (1.008 g)
46.07 g
1 x (16.00 g)
46.07 g
x 100% = 52.14%
x 100% = 13.13%
x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
3.5
Percent Composition of Compounds
For a molecule of H2O2:
2  1.008 amu H
%H =
 100% = 5.926%
34.02 amu H2O2
2  16.00 amu O
%O =
 100% = 94.06%
34.02 amu H2O2
Percent Composition of Compounds
We could also have used the empirical formula of hydrogen peroxide
(HO) for the calculation.
In this case, we could have used the empirical formula mass, the
mass in amu of one empirical formula, in place of the molecular
formula.
The empirical formula mass of H2O2 (the mass of HO) is 17.01 amu.
%H =
1.008 amu H × 100% = 5.926%
17.01 amu H2O2
%O =
16.00 amu O × 100% = 94.06%
17.01 amu H2O2
Worked Example 5.13
Lithium carbonate, Li2CO3, was the first “mood-stabilizing” drug approved by the
FDA for the treatment of mania and manic-depressive illness, also known as
bipolar disorder. Calculate the percent composition by mass of lithium carbonate.
Strategy Use Equation 5.1 to determine the percent by mass contributed by
each element in the compound.
Think About It Make sure that the percent composition results for
Solution
For eachsum
element,
multiply the100.
number
of atoms
atomic
a compound
to approximately
(In this
case, by
thethe
results
summass,
divide to
byexactly
the formula
mass, and multiply
by 100 percent.
100 percent––18.79%
+ 16.25%
+ 64.96% = 100.00%––
but remember
that because
2×6.941
amu Li of rounding, the percentages may sum to
%Li
=
×100%
= 18.79%
very
slightly
more
slightly
less.
73.89
amuorLivery
CO
2
3
12.01 amu C
%C = 73.89 amu Li CO ×100% = 16.25%
2
3
%O =
3×16.00 amu O
×100% = 64.96%
73.89 amu Li2CO3
Worked Example 5.16
Determine the empirical formula of a compound that is 30.45 percent nitrogen
and 69.55 percent oxygen by mass. Given that the molar mass of the compound is
approximately 92 g/mol, determine the molecular formula of the compound.
Strategy Assume a 100-g sample so that the mass percentages of nitrogen and
oxygen given in the problem statement correspond to the masses of N and O in
the compound. Then, using the appropriate molar masses, convert the grams of
each element to moles. Use the resulting numbers as subscripts in the empirical
formula, reducing them to the lowest possible whole numbers for the final
answer. To calculate the molecular formula, first divide the molar mass given in
the problem statement by the empirical formula mass. Then, multiply the
subscripts in the empirical formula by the resulting number to obtain the
subscripts in the molecular formula.
The molar masses of N and O are 14.01 and 16.00 g/mol, respectively. One
hundred grams of a compound that is 30.45 percent nitrogen and 69.35 percent
oxygen by contains 30.45 g N and 69.55 g O.
Worked Example 5.16 (cont.)
Solution
1 mol N
= 2.173 mol N
14.01 g N
1 mol O
69.55 g O ×
= 4.347 mol O
16.00 g O
30.45 g N ×
The gives a formula of N2.173O4.347. Dividing both subscripts by the smaller of the
two to get the smallest possible whole numbers (2.173/2.173 = 1, 4.347/2.173 ≈
2) gives an empirical formula of NO2.
Finally, dividing the approximate molar mass (92 g/mol) by the empirical formula
mass [14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol] gives 92/46.01 ≈ 2. Then,
multiplying both subscripts in the empirical formula by 2 gives the molecular
formula, N2O4.
Think About It Use the method described in Worked Example 5.13 to calculate
the percent composition of the molecular formula N2O4 and verify that it is the
same as that given in this problem.
Empirical Formula
The listing of the mass of each element present in
100 g of a compound (elemental analysis)
Empirical formula: contains the smallest set of wholenumber subscripts that match the elemental analysis
What is the empirical formula for a compound with
The following elemental analysis?
49.5% C 5.2% H 28.8%N 16.5% O
Finding Empirical Formula from percentages………
1. Assume you have a 100 g sample so the mass of each
element is the same as the percentage
2. Divide by the molar mass of each element to get moles
49.5 g C x
1 mol C = 4.125 mol C ÷ 1.03 =
12.00 g
4
5.20 g H x
1 mol H =
1.00 g
5.2 mol H
÷ 1.03 =
5
28.8 g N x 1 mol N = 2.05 mol N
14.00 g
÷ 1.03 =
2
16.5 g O x
÷ 1.03 =
1
1 mol O = 1.03 mol O
16.00 g
3. Divide the moles of each element by the smallest (O) to get
ratios of the atoms in the molecule
C4H5N2O
Finding Molecular Formula……………
Molecular formula: contains the real set of wholenumber subscripts that are found in the molecule
Need: Empirical Formula and
Molar Mass of the compound……….
For the preceeding problem where the empirical formula was
found to be C4H5N2O with a molar mass of 291 g/ mol
1. Find the molar mass of the empirical formula
(97g/ mol)
2. Divide the Molecular Formula MM by the
empirical formula MM giving us the “multiplying factor”
291 ÷ 97 = 3 therefore 3 is the multiplying factor
3. Multiply all subscripts by the multiplying factor to get the
molecular formula
C12H15N6O3