Chemical Formulas
and
Chemical Compounds
Heart cell rhythm depends on the opening and
closing of a complex series of valves on the cell
membrane, called ion channels. Some valves let
certain ions ike potassium (K+) flow out, others let
different ions like sodium (Na+) flow in. There
are also pumps that actively move ions one
direction or another.
Notes 7
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
Group 1: Lose 1 electron to form 1+ ions
H+
Li+ Na+
K+
Predicting Ionic Charges
Group 2: Loses 2 electrons to form 2+ ions
Be2+
Mg2+
Ca2+
Sr2+
Ba2+
Predicting Ionic Charges
B3+
Al3+
Ga3+
Group 13: Loses 3
electrons to form
3+ ions
Predicting Ionic Charges
Neither! Group 14
elements rarely form
ions.
Group 14: Lose 4
electrons or gain
4 electrons?
Predicting Ionic Charges
N3- Nitride
P3- Phosphide
As3- Arsenide
Group 15: Gains 3
electrons to form
3- ions
Predicting Ionic Charges
O2- Oxide
S2- Sulfide
Se2- Selenide
Group 16: Gains 2
electrons to form
2- ions
Predicting Ionic Charges
F1- Fluoride
Br1- Bromide
Cl1-Chloride
I1- Iodide
Group 17: Gains 1
electron to form
1- ions
Predicting Ionic Charges
Group 18: Stable
Noble gases do not
form ions!
Predicting Ionic Charges
Groups 3 - 12: Many transition elements
have more than one possible oxidation state.
Iron(II) = Fe2+
Iron(III) = Fe3+
Predicting Ionic Charges
Groups 3 - 12: Some transition elements
have only one possible oxidation state.
Zinc = Zn2+
Silver = Ag+
Binary Ionic Compounds
• Binary ionic compounds means two ions,
• one that is positive in charge (cation)
• one that is negative in charge (anion)
• that react to form a compound.
Naming Ionic Binary Compounds
1. The ion with the positive charge (cation) is
always written before the ion with the
negative charge (anion)
2. The first word is the name of the element of
which the cation originally came. (Example:
Na+ would be called Sodium)
3. The last word is the name of the element of
which the anion originally came. (Example: Clwould become chlorine, but the first part of
the word is used and -ide is added to the
end, so the last word would be Chloride.)
4. Put the two words together and that is the
name of the compound. (Example: Na+ and Clwould become Sodium Chloride.)
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
(continued)
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
Polyatomic Ions
• Polyatomic Ions are ions that
contain a number of ions.
• There is no way to learn how to
write their names, except to
commit them to memory
Ion
NH4+
NO2-
Common Polyatomic Ions (VIP)
Name
Ion
Name
Ammonium
O2-2
Peroxide
Nitrite
CrO4-2
Chromate
NO3SO3-2
SO4-2
Nitrate
Sulfite
Sulfate
Cr2O7-2
MnO4C2H3O2-
Dichromate
Permanganate
Acetate
HSO4OHCNPO4-3
Hydrogen Sulfate
Hydroxide
Cyanide
Phosphate
ClO4ClO3ClO2ClO-
Perchlorate
Chlorate
Chlorite
Hypochlorite
HCO3CO3-2
Hydrogen Carbonate
Carbonate
HPO4- Hydrogen Phosphate
H2PO4- Dihydrogen Phosphate
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.
( NH4+) SO42-
3. Balance charges , if necessary,
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
2
Not balanced!
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. Use parentheses
if you need more than one of a
polyatomic ion.
Fe3+ Cl-
3
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.
3+
Al
2
2S
3
Not balanced!
Writing Ionic Compound Formulas
Example: Magnesium carbonate
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges
are balanced.
Mg2+ CO32They 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.
2+
Zn
3. Balance charges , if necessary,
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
( OH- )2
Not balanced!
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.
2+
(
Ba NO3 ) 2
3. Balance charges , if necessary,
using subscripts. Use parentheses
if you need more than one of a
polyatomic ion.
Not balanced!
Writing Ionic Compound Formulas
Example: Aluminum phosphate
1. Write the formulas for the cation
and anion, including CHARGES!
2. Check to see if charges are
balanced.
3+
Al
PO4
3-
They ARE balanced!
Binary Covalent Compounds
• Binary covalent compounds are those
that do not involve metals or ions.
• An example of the difference between a
covalent compound is that CO2 is a
covalent compound, but NaCl is not,
because it contains Na+, a metal and an
ion.)
Naming Binary 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
CO2
CO
N2O
= diphosphorus pentoxide
= carbon dioxide
= carbon monoxide
= dinitrogen monoxide
Rules for naming a binary
covalent compound
1. The first element in the formula uses the
whole name of the element, much like binary
ionic compounds. (Example: In NO, the first
word would be Nitrogen.)
2. The second element in the formula only uses
the first half of the word and -ide is added
in place of the removed ending, much like
binary ionic compounds. (Example: In NO, the
first part of the second word would be oxide.)
3. So the person reading the name can determine
what the subscript is on each element, a
prefix is added to show how many of each
element are used. The first word of the
formula does not use momo, so CO (Carbon
Monoxide), NOT (Monocarbon Monoxide).
Prefixes in Chemical
Names
mono
1
di
tri
tetra
penta
2
3
4
5
hexa
hepta
octa
6
7
8
nona
9
deca
10
Calculating Formula Mass
Calculate the formula mass of magnesium carbonate,
MgCO3.
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
From previous slide:
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
 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
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
(continued)
Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas
(continued)
Formulas for molecular compounds MIGHT
be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Empirical Formula Determination
1. Base calculation on 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 
 6.85g 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
The Mole
1 dozen = 12
1 gross = 144
1 ream = 500
1 mole = 6.02 x 1023
There are exactly 12 grams of
carbon-12 in one mole of carbon-12.
Avogadro’s Number
6.02 x 1023 is called “Avogadro’s Number” in
honor of the Italian chemist Amadeo Avogadro
(1776-1855).
I didn’t discover it. Its
just named after me!
Amadeo Avogadro
Calculations with Moles:
Converting moles to grams
How many grams of lithium are in 3.50 moles of
lithium?
3.50 mol Li
6.94 g Li
1 mol Li
=
24.3
g Li
Calculations with Moles:
Converting grams to moles
How many moles of lithium are in 18.2 grams of
lithium?
18.2 g Li
1 mol Li
6.94 g Li
=
2.62
mol Li
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of
lithium?
3.50 mol Li 6.02 x 1023 atoms Li
1 mol Li
= 2.11 x 1024 atoms Li
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li 1 mol Li
6.94 g Li
6.02 x 1023 atoms Li
1 mol Li
(18.2)(6.02 x 1023)/6.94
= 1.58 x 1024 atoms Li