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Chapter 7
Chemical Formulas &
Chemical Compounds
Vocabulary
CHEMICAL FORMULA
IONIC
COVALENT
Formula
Unit
Molecular
Formula
NaCl
CO2
Vocabulary
COMPOUND
2 elements
Binary
Compound
NaCl
more than 2
elements
Ternary
Compound
NaNO3
Vocabulary
ION
1 atom
2 or more atoms
Monatomic
Ion
Polyatomic
Ion
+
Na
NO3
-
Ionic Nomenclature
Ionic Formulas

Write each ion, cation first. Don’t show
charges in the final formula.

Overall charge must equal zero.
 If
charges cancel, just write symbols.
 If not, use subscripts to balance charges.

Use parentheses to show more than one
polyatomic ion.

Stock System - Roman numerals indicate the
ion’s charge.
Ionic Nomenclature
Ionic Names

Write the names of both ions, cation first.

Change ending of monatomic ions to -ide.

Polyatomic ions have special names.

Stock System - Use Roman numerals to
show the ion’s charge if more than one is
possible. Overall charge must equal zero.
Ionic Nomenclature

Consider the following:
 Does
it contain a polyatomic ion?
2 elements  no
 -ate, -ite, 3+ elements  yes
 -ide,
 Does
it contain a Roman numeral?
 Check
 No
the table for metals not in Groups 1 or 2.
prefixes!
Ionic Nomenclature
Common Ion Charges
1+
0
2+
3+ NA 3- 2- 1-
Ionic Nomenclature

potassium chloride
 K+

Cl-
KCl

Mg(NO3)2

CuCl2
magnesium nitrate
 Mg2+


NO3-
copper(II) chloride
 Cu2+
Cl-
Ionic Nomenclature

NaBr
 sodium

Na2CO3
 sodium

bromide
carbonate
FeCl3
 iron(III)
chloride
Oxidation Numbers


Oxidation Numbers (aka Oxidation States)
are assigned to atoms composing a
compound or ion in order to indicate the
general distribution of electrons among the
bonded atoms.
Oxidation Numbers are very helpful in
naming compounds, writing formulas & in
balancing chemical equations.
Assigning Oxidation Numbers
1. The atoms in a pure element have an ox #
equal to ZERO. (ex. Na, O2, P4, S8)
2. The more electronegative element in a
binary molecular compound is assigned the
# equal to the negative charge it would have
as an anion. Likewise the less
electronegative element is assigned the
positive charge as the cation.
Assigning Oxidation Numbers (con’t)
3. Fluorine has an oxidation number of -1 in all of
its compounds. Why?
4. Oxygen has an oxidation number of -2 in most
compounds. Exceptions include the peroxides
(such as H2O2 => ox# = -1) and when
combined with halogens (ex. OF2=> ox# = +2)
Assigning Oxidation Numbers (con’t)
5. Hydrogen has an ox# = +1 in all compounds
containing elements that are more
electronegative than it. However it has an
ox# = -1 in compounds with metals.
6. The algebraic sum of the oxidation numbers
of all atoms in a neutral compound is equal
to zero.
Assigning Oxidation Numbers (con’t)
7.
The algebraic sum of the oxidation numbers
of all atoms in a polyatomic ion is equal to
the charge of the ion.
8. Although rules 1-7 apply to molecular
compounds, ox #’s can be assigned to
atoms in ionic compounds as well.
Practice: Assigning Oxidation Numbers
Try these now:
HCl
CF4
PCl3
SO2
HNO3
KH
P4O10
Practice: Assigning Oxidation Numbers
Try these now:
HClO3
N2O5
MgCl2
GeCl2
H2SO4
Mo2O3
Molecular Nomenclature

Prefix System (binary compounds)
1. Less e-neg atom
comes first.
2. Add prefixes to indicate # of atoms. Omit monoprefix on first element.
3. Change the ending of the
second element to -ide.
Molecular Nomenclature
PREFIX
monoditritetrapentahexaheptaoctanonadeca-
NUMBER
1
2
3
4
5
6
7
8
9
10
Molecular Nomenclature

CCl4
 carbon

tetrachloride
N 2O
 dinitrogen

monoxide
SF6
 sulfur
hexafluoride
Molecular Nomenclature

arsenic trichloride
 AsCl3

dinitrogen pentoxide
 N2O5

tetraphosphorus decoxide
 P4O10
Molecular Nomenclature

The Seven Diatomic Elements
Br2 I2 N2 Cl2 H2 O2 F2
H
N O F
Cl
Br
I
Definition

Acids
 Compounds
 Formulas

that form H+ in water.
usually begin with ‘H’.
Examples:
 HCl
– hydrochloric acid
 HNO3
– nitric acid
 H2SO4
– sulfuric acid
Acid Nomenclature
Anion
Ending
Acid Name
-ide
hydro-(stem)-ic acid
-ate
(stem)-ic acid
-ite
(stem)-ous acid
Acid Nomenclature
ACIDS
start with 'H'
2 elements
3 elements
hydro- prefix
-ic ending
no hydro- prefix
-ate ending
becomes
-ic ending
-ite ending
becomes
-ous ending
Acid Nomenclature

HBr
2


hydrobromic acid

carbonic acid

sulfurous acid
H2CO3
3

elements, -ide
elements, -ate
H2SO3
3
elements, -ite
Acid Nomenclature

hydrofluoric acid
2

 H+ F-
 HF
 H+ SO42-
 H2SO4
 H+ NO2-
 HNO2
sulfuric acid
3

elements
elements, -ic
nitrous acid
3
elements, -ous
Molar Conversions
Review of the Mole
What is the Mole?

A counting number (like a dozen)

Avogadro’s number (NA)

1 mol = 6.02  1023 items
A
large amount!!!!
What is the Mole?

1 mole of hockey pucks would equal
the mass of the moon!

1 mole of basketballs would fill a bag
the size of the earth!

1 mole of pennies would cover the
Earth 1/4 mile deep!
Molar Mass

Mass of 1 mole of an element or compound.

Atomic mass tells the...


atomic mass units per atom (amu)
grams per mole (g/mol)
Molar Mass Examples

carbon
12.01 g/mol

aluminum
26.98 g/mol

zinc
65.39 g/mol
Molar Mass Examples

water



H2O
2(1.01) + 16.00 = 18.02 g/mol
sodium chloride


NaCl
22.99 + 35.45 = 58.44 g/mol
Molar Mass Examples

sodium bicarbonate



NaHCO3
22.99 + 1.01 + 12.01 + 3(16.00)
= 84.01 g/mol
sucrose


C12H22O11
12(12.01) + 22(1.01) + 11(16.00)
= 342.34 g/mol
Molar Conversions
molar
mass
6.02  1023
MASS
NUMBER
MOLES
IN
GRAMS
OF
PARTICLES
(g/mol)
(particles/mol)
Molar Conversion Examples

How many moles of carbon are in
26 g of carbon?
26 g C
1 mol C
12.01 g C
= 2.2 mol C
Molar Conversion Examples

How many molecules are in 2.50
moles of C12H22O11?
2.50 mol
6.02  1023
molecules
1 mol
= 1.51  1024
molecules
C12H22O11
Molar Conversion Examples

Find the mass of 2.1  1024
molecules of NaHCO3.
2.1  1024
molecules
1 mol
6.02  1023
molecules
84.01 g
1 mol
= 290 g NaHCO3
Formula Calculations
Percentage Composition

the percentage by mass of each
element in a compound
mass of element
% composition 
 100
total mass
Percentage Composition

Find the % composition of Cu2S.
%Cu =
%S =
127.10 g Cu
159.17 g Cu2S
32.07 g S
159.17 g Cu2S
 100 =
79.852% Cu
 100 =
20.15% S
Percentage Composition

Find the percentage composition of a
sample that is 28 g Fe and 8.0 g O.
%Fe =
%O =
28 g
36 g
8.0 g
36 g
 100 = 78% Fe
 100 = 22% O
Percentage Composition

How many grams of copper are in a
38.0-gram sample of Cu2S?
Cu2S is 79.852% Cu
(38.0 g Cu2S)(0.79852) = 30.3 g Cu
Percentage Composition

Find the mass percentage of water in
calcium chloride dihydrate, CaCl2•2H2O?
%H2O =
36.04 g
147.02 g
 100 = 24.51%
H2O
Empirical Formula

Smallest whole number ratio of atoms in
a compound
C2H6
reduce subscripts
CH3
Empirical Formula
1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to find
subscripts.
4. When necessary, multiply subscripts
by 2, 3, or 4 to get whole #’s.
Empirical Formula

Find the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g 1 mol
= 1.85 mol N
14.01 g
1.85 mol
74.1 g 1 mol
16.00 g
= 4.63 mol O
1.85 mol
=1N
= 2.5 O
Empirical Formula
N1O2.5
Need to make the subscripts whole
numbers  multiply by 2
N2O5
Try Now: Empirical Formula
1. Quantitative analysis shows that a compound
contains 32.38% sodium, 22.65% sulfur, and
44.99% oxygen. Find the empirical formula of this
compound.
Try Now: Empirical Formula
2. Analysis of a 10.150g sample of a compound
known to contain only phosphorus and oxygen
indicates a phosphorus content of 4.433g. What is
the empirical formula of this compound?
Molecular Formula

“True Formula” - the actual number of
atoms in a compound
empirical
formula
CH3
?
molecular
formula
C2H6
Molecular Formula
1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the
empirical mass.
4. Multiply each subscript by the answer
from step 3.
MF mass
n
EF mass
EF n
Molecular Formula

The empirical formula for ethylene is
CH2. Find the molecular formula if the
molecular mass is
28.1 g/mol?
empirical mass = 14.03 g/mol
28.1 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4
Try Now: Molecular Formula
1. In a previous example we found the empirical
formula to be P2O5. Experimentation shows that
the molar mass of this compound is 283.89g/mol.
What is the compound’s molecular formula?
Try Now: Molecular Formula
2. Determine the molecular formula of the
compound with an empirical formula of CH and
a formula mass of 78.110amu.
Try Now: Molecular Formula
3. A sample of a compound with a formula mass of
34.00amu is found to consist of 0.44g hydrogen
and 6.92g oxygen. Find its molecular formula.