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.

1

2 3

24-4

35-3

246-2

357-1

Oxidation numbers can be determined from position in the periodic table below.

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.

Mg2+ N3-

Since 3 Mg2+ particles have a charge of 6+ and 2 N3- particles have a charge of 6- the total charge adds up to zero. The correct formula must be:

Mg2+

Mg2+

N3-

Mg3N22 N 3- particles

3 Mg 2+ particles

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-.

Na1+ P3-

If the total charge is going to add up 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, ferric - iron - 26 Fecuprous, cupric - copper- 29Custannous, stannic - tin - 50Snplumbous, plumbic - lead - 82Pb

2+ 3+1+ 2+

2+ 4+2+ 4+

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

Sn2+ Cl1- SnCl2Example - plumbic sulfide

Pb4+S2-- PbS2

Cl1-

S2--

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

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)

S6+

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.

O2- O2- O2-

SO3

Writing Formulas for Complex Binary Compounds - Prefix System

Prefix No. of Atoms

mono- 1di- 2 tri- 3tetra- 4 penta- 5hexa- 6 hepta- 7octa- 8

When a compound ending in “ide” also contains prefixes like mono, di, tri, etc. the formula can be written using these prefixes instead of using charges.

Example - dicarbon tetrahydride

2 carbons 4 hydrogens

C2H4

Writing Formulas of Oxyacids The “ic” acids

H O

H O

H O

H O

H O

N

C

Cl

S

P

1

2

1

2

3

3

3

3

4

4

nitric acid

carbonic acid

chloric acid

sulfuric acid

phosphoric acid

Writing Formulas of Oxyacids - “ous” and “ic”

HNO3

H2CO3

HClO3

H2SO4

H3PO4 remove 1 O,

HNO2

H2CO2

HClO2

H2SO3

H3PO3 remove 2 O’s

HNO

H2CO

HClO

H2SO2

H3PO2add 1 O

HNO4

H2CO4

HClO4

H2SO5

H3PO5

Per_____ic _____ ic ______ ous hypo ____ ous

Family SubstitutionsSince 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

Cl

Br

I

S

Se

Te

Po

P

As

phosphatearsenate

PO43-

AsO43-

sulfiteselenite

SO32-

SeO32-

hypochloritehypofluoritehypobromite

ClO1-

FO1-

BrO1-

perchoric acid

perbromic acid

HClO4

HBrO4

HNO3

H2CO3

HClO3

H2SO4

H3PO4

Deriving Polyanions From the OxyacidsEach of the oxyacids below is made up of positive H ions and negative polyanions

H1+ NO31-

H1+ H1+CO32-

H1+ ClO31-

H1+ H1+SO42-

H1+ H1+ H1+PO43-

nitrate

carbonate

chlorate

sulfate

phosphate

Deriving “ites” from “ates”

NO31-

CO32-

ClO31-

SO42-

PO43-

remove 1 O,

NO21-

CO22-

ClO21-

SO32-

PO33-

remove 2 O’s

NO1-

CO2-

ClO1-

SO22-

PO23-

add 1 O

NO41-

CO42-

ClO41-

SO52-

PO53-

per_____ate _____ ates ______ ites hypo ____ite

Writing Formulas - “ates” & “ites”

Example 1 - aluminum carbonite

Determine the charges of each particle

Al3+ CO22-

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-.

Al3+

CO22-

CO22-

6+ 6-zero charge

The answer is

Al2(CO2)3

Naming “ates” & “ites”Example 1 - Fe(NO3)2Since 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+.

NO31- NO3

1-

Remember oxyacid is HNO3

Fe2+

2+ 2-zero charge

Using the “ous” “ic” method, the name is ferrous.Using the IUPAC method the name is iron(II).NO3

1- is called nitrate.The answer is ferrous nitrate or iron(II) nitrate

Polyatomic Anions Containing HH1+ 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:

CO22-

carbonite

H1+ HCO21-

hydrogen carboniteor

bicarboniteIf 2 H1+ ions are added to any of the “ates”or “ites” with a 3- charge the resulting particle is named:

PO43-H1+ H1+

phosphate dihydrogen phosphate

H2PO41-

Polyatomic Anions Containing HWrite the formula for manganese (III) dihydrogen phosphiteAll 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.

Mn3+

The negative particle is a dihydrogen phosphite. Phosphite is:

PO33-

After adding 2 hydrogens to the phosphite the charge decreases to 1- and the anion becomes:

H2PO31-

To balance charges 3 of these dihydrogen phosphites are needed.

H2PO31- H2PO3

1-

The resulting answer is: Mn(H2PO3)3

Special Polyatomic IonsWater is H1+ H1+O2-

In some instances one of these H1+ particles is lost. The anion OH1- is formed. This particle is called hydroxide.

OH1-

hydroxide

Vinegar is a 5% solution of acetic acid. Its stucture is shown below. Sometimes it loses a H1+ particle to water forming:

C

H

H

H C O

O

HHC2H3O2

C

H

H

H C O

O

C2H3O21-

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

CrO42-

OCrO OO

chromateIf two of these chromate particles are combined one of theoxygen atoms is lost forming:

OCrO OO

OCrO OO

OCrO OO

OCrO OO

dichromate Cr2O7

2-

Special Polyatomic IonsCyanide 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:

CN1-

cyanide

cyanate salts contain the anion CNO1-

cyanate

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:

CNO1-Sthiocyanate

Thiosulfate has the formula:

SO42- S2O3

2-

Special Polyatomic IonsManganese is in group VIIA and chlorine is in group VIIB. This allows some creative substitution. Chlorate has the formula

ClO31-

chlorateperchlorate

isClO4

1-

perchlorate

permanganateis

MnO41-

permanganate

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.

C2O42-

oxalate

Special Polyatomic IonsThe poisonous gas, nitrogen trihydride, NH3, has the common name ammonia.

HN

H H

ammonia

When ammonia molecules dissolve in water and collide with water molecules they sometimes form

HN

H H

H O

H H O

H

H O

H

Notice the water molecule left a H1+ particle behind forming

ammonium - NH41+

hydroxideOH1-

acetateC2H3O2

1-

chromateCrO4

1-

dichromateCr2O7

2-

cyanideCN1-

cyanateCNO1-

thiosulfateS2O3

2-

permanganateMnO4

1-

oxalateC2O4

2-

oxalateC2O4

2-

ammoniumNH4

1+

Naming Chemical Compounds

Are there more than 2 elements?Yes No

It is a binary compound so the name must end in ide.

It has polyatomic ions.

Does the negative ion comefrom an oxyacid?

Yes No

Use the HO, HO table to determineits name.

It must be a specialpolyatomic ion.

Does the 1st element have more than 1 oxidation state?

Yes No

Name it.

Determine the oxidation state of the 1st element and use it to name the compound.

Writing Formulas for Chemical Compounds

Does the name end in “ide”? YesNo

It is probably a binary compound. It’s made up of polyatomic ions.

Does the negative ion comefrom an oxyacid?

Yes No

Use the HO, HO table to determineits formula.

It must be a specialpolyatomic ion.

Does the 1st element have more than 1 oxidation state?

Yes No

Determinethe formula.

Determine the oxidation state of the 1st element and use it to determine the formula.

Binary AcidsHydrochloric acid HClHydrobromic acid HBrHydroiodic acid HIHydrofluoric acid HFHydrosulfuric acid H2SHydroselenic acid H2Se