Writing Formulas for Simple
Binary Compunds
If the compound’s name ends in “ide”
chances are it’s a binary compound.
(There are a few exceptions which will be
covered later).
Find the 1st element on the periodic
table and determine its oxidation
number.
If there is more than 1 oxidation state
then go on to the section entitled
Writing formulas of Complex Binary
Compounds.
Oxidation numbers can be determined
from position in the periodic table
below.
1
2
3
2
2 3 4
4 5 6
-4 -3 -2
3
5
7
-1
Writing Formulas for Simple Binary
Compunds
Find the oxidation state of the 2nd element
remembering the first element is always
positive and the second is always negative.
Once both oxidation numbers
are determined the goal is to
combine the positives and the
negatives in ratios where the
total charge adds up to zero.
Example 1 - magnesium nitride.
2+
Mg
Since 3
particles have a charge
3of 6+ and 2 N particles have a
charge of 6- the total charge adds up
to zero.
The correct formula must be:
Mg2+
N3-
Mg2+
N3-
Mg2+
2 N 3- particles
3 Mg 2+ particles
Mg3N2
Writing Formulas for Simple Binary
Compunds
Example 2 - sodium phosphide.
Sodium is in group 1 so its oxidation number is
1+. Phosphide is short for phosphorus. Its
oxidation number must be negative since it’s the
2nd name so phosphide has a charge of 3-.
If the total charge is going to add
Na1+ P3up to zero 3 Na1+ and 1 P3- must be
used.
Na1+
Na1+
The final answer is:
Na3P
Writing Formulas for Complex Binary
Compounds - “ous,ic”
If the 1st name ends in “ous” or “ic” find the
element on the periodic table. Latin names are
sometimes used. Below is a list of the latin
names most frequently used:
ferrous, 2+ ferric - 3+
cuprous, 1+ cupric - 2+
stannous, 2+ stannic - 4+
plumbous, 2+plumbic - 4+
iron coppertin lead -
Fe
29Cu
50Sn
82Pb
26
Generally “ous” is used for the lowest
positive oxidation state and “ic” is the next
highest. There are exceptions to this rule.
Example - Stannous chloride
Cl1-
Sn2+ Cl1-
SnCl2
Example - plumbic sulfide
4+
Pb
2S -
2S -
PbS2
Writing Formulas for Complex
Binary Compounds
Roman Numeral System - IUPAC
When names for compounds have Roman
Numerals present, this number represents the
charge of the positive ion.
This charge can be used to determine the
number of negative particles needed to create a
combination of particles with an overall charge
of zero.
Example sulfur(VI) oxide
S6+ O2- O2- O2-
SO3
The Roman Numeral VI stands for 6. (V is 5
and I after it means add 1 to 5. The Roman
Numeral for 4 is IV)
If sulfur particles have a charge of 6+, 3 oxygen
particles, each with a charge of 2- are needed to
create a collection of particles where the total
charge adds up to zero.
Writing Formulas for Complex Binary
Compounds - Prefix System
Prefix
monoditritetrapentahexaheptaocta-
No. of Atoms When a compound ending in
“ide” also contains prefixes
1
like mono, di, tri, etc. the
2
formula can be written using
3
these prefixes instead of
4
using charges.
5 Example - dicarbon tetrahydride
6
2 carbons 4 hydrogens
7
8
C2H4
Writing Formulas of Oxyacids
The “ic” acids
O3
nitric
acid
H2 C
O3
carbonic
acid
H 1 Cl
O3
chloric
acid
H2 S
O4
sulfuric
acid
H3 P
O4
phosphoric
acid
H1 N
Writing Formulas of Oxyacids - “ous” and “ic”
Per_____ic
_____ ic
______ ous
HNO4
HNO3
HNO2
H2CO4
H2CO3 H2CO2
H2CO
HClO4
HClO3 HClO2
HClO
H2SO5
H2SO4 H2SO3
H2SO2
H3PO5
H3PO4 H3PO3
H3PO2
add 1 O
remove 1 O,
hypo ____ ous
HNO
remove 2 O’s
Family Substitutions
Since elements in the same
family have the same number
of valence shell electrons they
can sometimes be freely
substituted for one another.
Some examples of family
substitutions are:
F
P
S Cl
As
Se Br
Te I
Po
phosphate
arsenate
sulfite
selenite
PO43AsO43SO32SeO32-
hypochlorite
hypofluorite
hypobromite
ClO1FO1BrO1-
perchoric acid
HClO4
HBrO4
perbromic acid
Deriving Polyanions From the Oxyacids
Each of the oxyacids below is made up of positive H ions and
negative polyanions
HNO3 H1+ NO31-
nitrate
H2CO3 H1+ CO32- H1+
carbonate
HClO3 H1+ ClO 13
chlorate
H2SO4 H1+ SO42- H1+
sulfate
H3PO4 H1+ PO43- H1+ H1+ phosphate
Deriving “ites” from “ates”
per_____ate
_____ ates
______ ites
hypo ____ite
NO41-
NO31-
NO21-
NO1-
CO42-
CO32-
CO22-
CO2-
ClO41-
ClO31-
ClO21-
ClO1-
SO52-
SO42-
SO32-
SO22-
PO53-
PO43-
PO33-
PO23-
add 1 O
remove 1 O,
remove 2 O’s
Writing Formulas - “ates” & “ites”
Example 1 - aluminum carbonite
Determine the charges of each particle
Al3+ CO22Al3+
CO22-
CO22-
The answer is
Al2(CO
6- 2)3
6+
zero charge
Compile groups of particles where the sum of
positive charges equals the sum of negative
particles. This way the total charge is zero. 2
Al 3+ equals 6+ and 3 CO2 2- equals 6-.
Naming “ates” & “ites”
Example 1 - Fe(NO3)2
Since the positive ion has more than one oxidation
state, its specific oxidation state must be determined.
This is done by figuring out the charge on the negative particle.
Since the total negative charge
is 2- the positive charge on the
single particle of Fe must be 2+.
Remember oxyacid is HNO3
2+
Fe
Using the “ous” “ic” method, the
name is ferrous.
Using the IUPAC method the
name is iron(II).
NO31- is called nitrate.
NO31-
NO31-
2+
2-
zero charge
The answer is ferrous nitrate or iron(II) nitrate
Polyatomic Anions Containing H
H1+ ions can be added to any of the “ates”or “ites” with a
2- or 3- charges, reducing the overall charge of the newly
formed polyatomic negative ion by one. This is shown
below:
12H1+
CO2
HCO2
hydrogen carbonite
carbonite
or
bicarbonite
1+
If 2 H ions are added to any of the “ates”or “ites”
with a 3- charge the resulting particle is named:
H1+
PO43- H1+
phosphate
1-
H2PO4
dihydrogen phosphate
Polyatomic Anions Containing H
Write the formula for manganese (III) dihydrogen
phosphite
All formulae in this course consists of a positive particle
(cation) and a negative particle (anion). The positive
particle is Mn3+. This is determined from the Roman
Numeral in the name.
3- 1- H PO 1- H PO 1Mn3+ H
PO
PO
2 3 3
2
3
2
3
The negative particle is a dihydrogen phosphite. Phosphite i
After adding 2 hydrogens to the phosphite the charge
decreases to 1- and the anion becomes:
To balance charges 3 of these dihydrogen phosphites are needed.
The resulting answer is:
Mn(H2PO3)3
Water is
Special Polyatomic Ions
121+
1+
OH
H O H
In some instances one of these H1+ particles is
hydroxide
lost. The anion OH1- is formed. This particle is
called hydroxide.
Vinegar is a 5% solution of acetic acid. Its stucture is shown
below. Sometimes it loses a H1+ particle to water forming:
H
H
C
H
H
C O
O
H
H
C2H3O21-
HC2H3O2
C
H
C O
O
acetate
Special Polyatomic Ions
Sulfur is in group VIB and sulfate has the formula SO42-.
Chromium is in group VIA and chromate has the formula
CrO42chromate
O
O Cr O
O
If two of these chromate particles are combined one of the
oxygen atoms is lost forming:
O
O
O
O dichromate
Cr
2O
O Cr O CrO OCr Cr
OO
O
2 7
O
O
O
O
Special Polyatomic Ions
Cyanide salts used in execution chambers in the U.S. are
combined with sulfuric acid producing poisonous hydrogen
cyanide gas. The anion cyanide has the formula:
CN1cyanide
cyanate salts contain
the anion
CNO1cyanate
The prefix “thio” is used whenever an O particle is replaced by
a sulfur particle since both O and S are in the same family.
The anion thiocyanate has the formula:
CNO
S 1thiocyanate
Thiosulfate has the formula:
SO42-
S2O32-
Special Polyatomic Ions
Manganese is in group VIIA and chlorine is in group VIIB. This
allows some creative substitution. Chlorate has the formula
ClO31- perchlorate
chlorate
is
1- permanganate
ClO4
perchlorate
is
MnO41permanganate
A poisonous substance found in the leaves of rhubarb, potatoes,
tomatoes and countless other plants is called oxalic acid. Its
formula is
H2C2O4
If this acid loses two H1+ particles it creates the anion oxalate.
C2O42oxalate
Special Polyatomic Ions
The poisonous gas, nitrogen trihydride, NH3, has the common
name ammonia.
ammonia
When ammonia molecules
dissolve in water and collide with
water molecules they sometimes
form
Notice the water
molecule left a
H1+ particle
behind forming
N
H
H
H
H O
H O
HH O
H
ammonium - NH
N
H
H
H
4
1+
hydroxide
OH1-
cyanide
CN1-
oxalate
C2O42-
acetate
C2H3O21-
cyanate
CNO1-
oxalate
C2O42-
chromate
CrO41-
thiosulfate
S2O32-
ammonium
NH41+
dichromate
Cr2O72-
permanganate
MnO41-
Naming Chemical Compounds
Yes
Are there more than 2 elements?
It has polyatomic ions.
Does the negative ion come
from an oxyacid?
Yes
No
Use the HO, HO
table to determine
its name.
It must be a
special
polyatomic
ion.
No
It is a binary compound so the
name must end in ide.
Does the 1st element have
more than 1 oxidation state?
Yes
No
Determine the
oxidation state of the
1st element and use it
to name the
compound.
Name it.
Writing Formulas for Chemical Compounds
No
Does the name end in “ide”?
It’s made up of polyatomic ions.
Does the negative ion come
from an oxyacid?
Yes
Use the HO, HO
table to determine
its formula.
No
It must be a
special
polyatomic
ion.
Yes
It is probably a binary compound.
Does the 1st element have
more than 1 oxidation state?
Yes
No
Determine the
oxidation state of the
Determine
1st element and use it
the formula.
to determine the
formula.
Binary Acids
Hydrochloric acid HCl
Hydrobromic acid
HBr
Hydroiodic acid HI
Hydrofluoric acid HF
Hydrosulfuric acid
H2S
Hydroselenic acid H2Se