Chapter 7
Chemical Formulas &
Chemical Compounds
7.1 Chemical Names &
Formulas
Ions
• Cation: A positive ion
• Mg2+, NH4+
• Anion: A negative ion
• Cl-, SO42• Ionic Bonding: Force of attraction
between oppositely charged ions.
Predicting Ionic Charges
Groups 3 - 12:
Iron (II) = Fe2+
Iron (III) = Fe3+
Many transition elements have
more than one possible
oxidation state.
Predicting Ionic Charges
Groups 3 - 12:
Zinc = Zn2+
Silver = Ag1+
Some transition elements
have only one possible
oxidation state.
Formula Writing for Binary Ionic
Compounds
criss-cross the oxidation numbers to balance
out the charge.
Magnesium Bromide
Mg+ 2
Br – 1
Calcium Sulfide
Ca + 2
S –2
Mg1Br2
Ca2S2
MgBr2
CaS
Writing Ionic Compound Formulas
Example: Iron (III) chloride
1. Write the formulas for the cation and
anion, including CHARGES!
2. Check to see if charges
are balanced.
3. Balance charges, if necessary,
using subscripts.
Fe3+ Cl-
3
Not balanced!
Naming Ionic Compounds
1. Cation first, then anion
2. Monatomic cation = name of the
element
• Ca2+ = calcium ion
3. Monatomic anion = root + -ide
• Cl- = chloride
• CaCl2 = calcium chloride
Naming Ionic Compounds
Metals with multiple oxidation states
• some metal forms more than one cation
• use Roman numeral in name
• PbCl2
• Pb2+ is cation
• PbCl2 = lead (II) chloride
Elements with Multiple Oxidation Numbers
Copper I
Copper II
Iron II
Iron III
Mercury I
Mercury II
Lead II
Lead IV
Tin II
Tin IV
Chromium II
Chromium III
Chromium VI
Cu+1
Cu+2
Fe+2
Fe+3
Hg+1
Hg+2
Pb+2
Pb+4
Sn+2
Sn+4
Cr+2
Cr+3
Cr+6
Manganese II
Manganese III
Manganese VII
Cobalt II
Cobalt III
Gold I
Gold III
Nickel II
Nickel III
Nickel IV
**Silver
Ag+1
**Zinc
Zn+2
**Cadmium Cd+2
Mn+2
Mn+3
Mn+7
Co+2
Co+3
Au+1
Au+3
Ni+2
Ni+3
Ni+4
♥ Poly Atomic Ions to Know and Love ♥
Name
Formula
Name
Formula
Hypochlorite
ClO-1
Acetate
C2H3O2-1
(CH3COO -1)
Dichromate
Cr2O7-2
Chlorite
ClO2-1
Ammonium
NH4+1
Chlorate
ClO3-1
Nitrate
NO3-1
Perchlorate
ClO4-1
Nitrite
NO2-1
Cyanide
CN-1
Hydroxide
OH-1
Carbonate
CO3-2
Phosphate
PO4-3
Chromate
CrO4-2
♥ More Poly Atomic Ions to Know and Love
♥
Name
Formula
Name
Formula
Sulfite
SO3-2
Hydrogen
HCO3-1
Sulfate
SO4-2
Hydrogen
Sulfite
Permanganate
HSO3-1
MnO4-1
Carbonate
Hydrogen
Phosphate
Hydrogen
Sulfate
Oxalate
Hydronium
H3O+
Silicate
SiO3-2
Peroxide
O2-2
Phosphite
PO3-3
Bromate
BrO3-1
Arsenate
AsO4-2
HPO4-2
HSO4-1
C2O4-2
Naming Compounds with Polyatomic Ions
•
•
•
•
Formula
(NH4)2SO4
ZnCO3
NH4Br
Li2CO3
Name
ammonium sulfate
zinc carbonate
ammonium bromide
lithium carbonate
* Polyatomic & monatomic cation names remain
the same, monatomic anions change their
ending to –ide.
Writing Ionic Compound Formulas
Example: Barium nitrate
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges are
balanced.
3. Balance charges , if necessary,
using subscripts. Use
parentheses if you need more
than one of a polyatomic ion.
Ba2+ ( NO3-)
Not
balanced!
2
Writing Ionic Compound Formulas
Example: Ammonium sulfate
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges
are balanced.
3. Balance charges , if necessary,
using subscripts. Use
parentheses if you need more
than one of a polyatomic ion.
( NH4+)
2
SO42-
Not
balanced!
Writing Ionic Compound Formulas
Example: Aluminum sulfide
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges
are balanced.
3. Balance charges , if necessary,
using subscripts. Use
parentheses if you need more
than one of a polyatomic ion.
Al3+2
S2- 3
Not
balanced!
Writing Ionic Compound Formulas
Example: Magnesium carbonate
Mg2+
CO32-
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges
are balanced.
3. Simplify to a formula unit.
They are
balanced!
Writing Ionic Compound Formulas
Example: Zinc hydroxide
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges are
balanced.
3. Balance charges , if necessary,
using subscripts. Use
parentheses if you need more
than one of a polyatomic ion.
Zn2+ ( OH- ) 2
Not
balanced!
Writing Ionic Compound Formulas
Example: Aluminum phosphate
Al3+
PO43-
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges are
balanced.
They ARE
balanced!
More Examples…
Chemical Formula
1.
2.
3.
4.
5.
6.
7.
Cr2O3
Cr2O
CuSO4
Ni(OH)2
Cr2(C2O4)3
Cu2S
CuS
Chemical Name
1.
2.
3.
4.
5.
6.
7.
chromium (III) oxide
chromium (I) oxide
copper (II) sulfate
nickel (II) hydroxide
chromium (III) oxalate
copper (I) sulfide
copper (II) sulfide
Hydrates
• Hydrate – when a water molecule (s) are
chemically bonded to the ionic compound.
• Normal ionic naming protocol are used,
then followed by the word “hydrate.”
• Prefixes are added to indicate the number
of water molecules when naming hydrates.
Hydrate Prefixes
# of water
molecules
1
prefix
mono-
# of water
molecules
6
prefix
hexa-
2
di-
7
hepta-
3
tri-
8
octa-
4
tetra-
9
nona-
5
penta-
10
deca-
Hydrates
• Example: MgBr2 ∙ 6H2O
Magnesium bromide hexahydrate
• The “ ∙ ” means “loosely bonded”
• Hygroscopic - easily absorb water molecules
from the air.
• Deliquescent- very hygroscopic; takes out
water from the air to dissolve completely to
form a liquid solution.
• Anhydrous – when all of the water has been
removed.
Naming Binary Covalent Compounds
•
•
•
•
•
Compounds between two nonmetals
First element in the formula is named first.
Second element is named as if it were an anion.
Use prefixes
Only use mono on second element
P2O5 = diphosphorus pentoxide
CO2 = carbon dioxide
CO = carbon monoxide
N2O = dinitrogen monoxide
Acids
• Acids always begin with Hydrogen
Anion
Formula
Name
Cl-1
HCl
Hydrochloric Acid
Br-1
HBr
Hydrobromic Acid
SO4-2
H2SO4
Sulfuric Acid
SO3-2
H2SO3
Sulfurous Acid
NO3-1
HNO3
Nitric Acid
CN-1
HCN
Hydrocyanic Acid
PO4-3
H3PO4
Phosphoric Acid
Bases
Cation
Formula
Name
Na+1
NaOH
Sodium Hydroxide
K+1
KOH
Potassium Hydroxide
NH4+1
NH3
Ammonia
Organic Compounds
• Organic compounds are named using a
different set of rules.
• The simplest group is the hydrocarbons.
These compounds are composed solely of
the elements carbon and hydrogen.
• Carbon atoms can link to each other in
chains and in rings.
Naming Hydrocarbons
• The stem of the compound name is then
chosen from the following table:
# of carbon
atoms
1
prefix
meth-
# of carbon
atoms
6
prefix
hexa-
2
eth-
7
hepta-
3
prop-
8
octa-
4
but-
9
nona-
5
penta-
10
deca-
Hydrocarbons: Alkanes
• These molecules have the generic formula:
CnH2n+2
• They contain all single bonds.
CH4
C2H6
C3H8
C4H10
C5H12
C6H14
methane
ethane
propane
butane
pentane
hexane
Hydrocarbons: Alkenes
• These molecules have the generic formula:
CnH2n
• They contain double bonds between carbon
atoms.
C2H4
C3H6
C4H8
C5H10
C6H12
ethene
propene
butene
pentene
hexene
Hydrocarbons: Alkynes
• These molecules have the generic formula:
Cn Hn
• They contain triple bonds between carbon
atoms.
C2H2
C3H3
C4H4
C5H5
C6H6
ethyne
propyne
butyne
pentyne
hexyne
Chapter 7
Chemical Formulas &
Chemical Compounds
7.2 Oxidation Numbers
Oxidation Numbers
• Oxidation Number – numbers assigned to
atoms composing a compound or ion that
indicate the general distribution of
electrons among bonded atoms
Chapter 7
Chemical Formulas &
Chemical Compounds
7.3 Using Chemical
Formulas
Molar Mass
• The mass of 1 mole of a pure substance is called
its Molar Mass.
• Ex: Molar mass of Iron is 55.847 g/mol
What is the molar mass of Platinum?
195.08 g/mol
Molar Mass
• The molar mass depends on the particles that
compose the compound. If your element exists as a
molecule, i.e. BrINClHOF, one mole of these
particles contains 2 moles of the element as an
atom.
• Determine the molar mass of oxygen molecules (O2)
(16.00 g/mol) x (2 atoms) = 32.00 g/mol
The molar mass of oxygen molecules (O2) is twice the
molar mass of oxygen atoms!
Formula Mass
• The molar mass of a compound is the
mass of the atomic mass units of one
molecule.
• This takes into consideration the number
of atoms of each element in a compound.
• Formula Mass is calculated the same way
as molar mass except it is measured in
amu, instead of g/mol.
Calculating Formula Mass
Calculate the formula mass of magnesium carbonate, MgCO3.
24.31 + 12.01 + 3(16.00) =
84.32 amu
Steps for Calculating Molar Mass
for Compounds
1. List the elements
2. Determine how many atoms of each
3. Identify the atomic masses from the periodic
table
4. Multiply how many atoms by the respective
atomic mass
5. Add up the totals for the Molar Mass
Practice
• H2O
• NaCl
H
2 x 1.008 = 2.016
Na 1 x 22.9 = 22.9
O
1 x 15.99 = 15.99
Cl
1 x 35.45 = 35.45
18.006 g/mol
58.35
• C6H12O6
C
6 x 12.01 = 72.06
H 12 x 1.008 = 12.096
O
6 x 15.99 = 95.94
180.096
g/mol
• K2O
K
2 x 39.1 = 78.2
O
1 x 15.99 = 15.99
94.19
g/mol
g/mol
Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
24.31 + 12.01 + 3(16.00) = 84.32 amu
 24.31 
Mg  
  100  28.83%
 84.32 
 12.01 
C 
  100  14.24%
 84.32 
 48.00 
O
  100  56.93%
 84.32 
100.00
Mass Percent
• So…. In one mole of H2O, how many grams of
Hydrogen are there?
2 mol H x 1.008g H = 2.016 g H in 1 mol H2O
1 mol H
• What % of Hydrogen, by mass, is in H2O?
2.016 g H
18 g H20
x 100 = 11.2 % H
*Must also find molar mass of H2O
What % of Oxygen, by mass is in H2O?
Formulas
Empirical formula: the lowest whole number ratio
of atoms in a compound.
Molecular formula: the true number of atoms
of each element in the formula of a
compound.
 molecular formula = (empirical formula)n
[n = integer]
 molecular formula = C6H6 = (CH)6
 empirical formula = CH
Formulas
Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio). Often,
these are called formula units.
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas
Formulas for molecular compounds MIGHT
be empirical (lowest whole number ratio).
Molecular:
H2 O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Chapter 7
Chemical Formulas &
Chemical Compounds
7.4 Determining
Chemical Formulas
Empirical Formula Determination
1. Base calculation on assumption of 100 grams
of compound.
2. Determine moles of each element in 100
grams of compound.
3. Divide each value of moles by the smallest of
the values.
4. Multiply each number by an integer to obtain
all whole numbers.
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical
formula of adipic acid?
 49.32 g C 1 mol C   4.107 mol C
12.01 g C 
 6.85 g H 1 mol H   6.78 mol H
1.01 g H 
 43.84 g O 1 mol O   2.74 mol O
16.00 g O 
Empirical Formula Determination
(part 2)
Divide each value of moles by the smallest of the values.
4.107
mol
C
Carbon:
 1.50
2.74 mol O
6.78 mol H
Hydrogen:
 2.47
2.74 mol O
2.74 mol O
Oxygen:
 1.00
2.74 mol O
Empirical Formula Determination
(part 3)
Multiply each number by an integer to obtain all whole
numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Oxygen: 1.00
x 2
2
Empirical formula: C3H5O2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid is
146 g/mol. What is the molecular formula of
adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid is
146 g/mol. What is the molecular formula of
adipic acid?
2. Divide the molecular mass by the
mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid is
146 g/mol. What is the molecular formula of
adipic acid?
3. Multiply the empirical formula by this
number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
(C3H5O2) x 2 =
C6H10O4
Determining Chemical Formulas from Mass Percents
A sample has been analyzed, here are the
results:
18.8 % Na
29.0 % Cl
52.2 % O
• How can you determine the chemical formula?
• Step 1: Assume a 100 g sample.
Then, your percent quantities become gram (mass)
quantities.
18.8 g Na , 29.0 g Cl & 52.2 g O
• Step 2: Convert those masses to moles.
18.8 g Na x 1 mol Na = 0.817 mol Na
23 g Na
29.0 g Cl x 1 mol Cl = 0.817 mol Cl
35.5 g Cl
52.2 g O x 1 mol O = 3.26 mol O
16 g O
• Step 3: Since your empirical formula is in
small, whole number ratios, divide your mole
amounts by the smallest mole quantity.
0.817 mol Na / 0.817 = 1.00 mol Na
0.817 mol Cl / 0.817 = 1.00 mol Cl
3.26 mol O / 0.817 = 3.99 ≈ 4.00 mol O
• Step 4: Use these values as subscripts in
your formula
Na1Cl1O4 ≈ NaClO4
• Step 5: In the event the chemical formula is
not the same as the empirical formula, you
need the molar mass of the desired
compound and you must compare it to the
molar mass of the empirical formula.
• Step 6: Divide the given molar mass by the
empirical molar mass to get the multiple
quantity.
• Step 7: Multiply each subscript in the formula
by that multiple quantity.
• Ex: MM of molecular formula = 180 g/mol
Using steps 1-4, you found that the empirical
formula is CH2O.
Find the molar mass of the empirical formula:
MM EF = 30 g/mol
Divide MM MF / MM EF to get a whole number.
Ex: 180 / 30 = 6
C1x6 H2x6 O1x6
C6H12O6
Practice Problems:
• A sample has been analyzed to be 10.04 % C, 0.84
% H & 89.12% Cl. Find the Empirical Formula.
• A compound’s empirical formula has been
determined to be HF. The compound’s molar mass
is 40 g/mol. What is its chemical formula?
• A compound’s empirical formula has been
determined to be CH2. The compound’s molar mass
is 42 g/mol. What is its chemical formula?