Dr Inam Ul Haq Jazbi 10. Inorganic Section
1.22. Rules for writing Chemical Formulae of Compounds
To write chemical formulae of compounds, following steps are involved:
1. Every compound is supposed to consist of two types of radicals namely cation and anion. First of all
name of types of ions or radicals are identified in the given compounds. Then symbols or formulae of
both ions or radicals are written side by side (showing cation first then anion) with their real or formal
charges (oxidation numbers) or valencies in superscripts (at the top or at the upper side each ion or
radical).
2. Now the correct combination of ions is used to produce a compound with a net charge of zero. To do
so, the real or formal charges or valencies of corresponding cation and anion are cross multiplied. If
the charges are same then they are cancelled out otherwise they become the subscripts of ions or
radicals showing the ratios of atoms.
Solved Examples
Example # 1. Write down chemical formula of calcium oxide.
Step One
Write the symbols for the elements in the compound. Note that the ending Calcium = Ca , Oxide = O
"ide" is used for oxide to show that it is a negative ion of oxygen.
Step Two
Write down the charge or oxidation numbers of the elements or radicals
involved as superscripts to the right of the elemental symbols. (Note that Calcium = Ca2+, Oxide = O2ˉ
when no number accompanies a charge symbol, as in the case of fluoride
below, they charge value is understood to be "1").
Step Three
Use the correct combination of ions to produce a compound with a net charge
of zero. In this case, (2+) + (-2) = 0. So, one oxide ion will cancel out one
calcium ion. Since it would take one oxide ion (each with a charge of negative
Ca2+ O2ˉ CaO
two) to cancel out one barium ion (with a charge of plus two) we do not use a
subscript of one after the symbol for calcium and oxygen to show the ratio as
when no number accompanies in subscript of elemental symbol the atomic
ratio is understood to be "1:1").
Example # 2 Write the correct formula for Barium Fluoride.
Step One
Write the symbols for the elements in the compound. Note that the ending Barium = Ba , Fluoride = F
"ide" is used for fluoride to show that it is a negative ion of fluorine.
Step Two
Look up the oxidation numbers of the elements involved and write them as
superscripts to the right of the elemental symbols. Note that when no number Barium = Ba2+ , Fluoride = Fˉ
accompanies a charge symbol, as in the case of fluoride below, they charge
value is understood to be "1".
Step Three
Use the correct combination of ions to produce a compound with a net charge
of zero. In this case, (2+) + 2(-1) = 0. So, two fluoride ions will cancel out one
barium ion. Since it would take two fluoride ions (each with a charge of Ba2+ F ˉ
negative one) to cancel out one barium ion (with a charge of plus two) we use
a subscript of two after the symbol for fluorine to show the ratio. Ba2+ 2F ˉ
BaF2
Conceptual Chemistry IX and X -1-
, Dr Inam Ul Haq Jazbi 10. Inorganic Section
Example # 3;Write the proper formula for the ionic compound lithium bromide.
Step One
Write the symbols for the elements in the compound. Note that the ending Lithium = Li Bromide = Br
"ide" is used for bromide to show that it is a negative ion of bromine.
Step Two
Look up the oxidation numbers of the elements involved (in ion chart), and
write them as superscripts to the right of the elemental symbols. Note that Lithium = Li+ Bromide = Brˉ
when no number accompanies a charge symbol, as in the case of fluoride
below, they charge value is understood to be "1".
Step Three
Use the correct combination of ions to produce a compound with a net charge Li+ Brˉ
of zero. In this case, (+1) + (-1) = 0. So, one lithium ion will cancel out the Li+ 2Brˉ
charge of one bromide ion. This means that the two elements will combine in LiBr2
a one to one ratio, and no subscripts will be needed
Example # 4; Show the correct formula for Zinc Nitrate.
Step One
Write the symbols for the monatomic and polyatomic ions in the compound. Zinc = Zn Nitrate = NO3
Step Two
Write down the charge or oxidation numbers of the elements or radicals
involved as superscripts to the right of the elemental or radical symbols. (Note Zinc = Zn2+ Nitrate = NO3ˉ
that when no number accompanies a charge symbol, as in the case of nitrate
below, they charge value is understood to be "1").
Step Three
Use the correct combination of ions to produce a compound with a net charge
of zero. In this case, (2+) + 2(-1) = 0. So, two nitrate ions will cancel out one
zinc ion. Since it would take two nitrate ions (each with a charge of negative Zn2+ NO3ˉ
one) to cancel out one zinc ion (with a charge of plus two) we use a subscript Zn2+ 2NO3ˉ
of two after the symbol for nitrate ion to show the ratio. Parenthesis must be Zn(NO3)2
used if you need more than one of a polyatomic ion. We need to show two
nitrate ions in our formula. The subscript is put on the outside of the
parenthesis to show that the entire polyatomic ion is doubled. The correct use
of parenthesis will seem hard at first, but you must master this skill with
practice!
Example # 5; Show the correct formula for ferric sulphate.
Step One
Write the symbols for the monatomic and polyatomic ions in the compound. Ferric = Fe , Sulphate = SO4
Step Two
Write down the charge or oxidation numbers of the elements or radicals
involved as superscripts to the right of the elemental or radical symbols. Ferric = Fe3+ , Sulphate = SO42ˉ
Step Three
Use the correct combination of ions to produce a compound with a net charge
of zero. In this case, 2(3+) + 3(2-) = 0. So, three sulphate ions will cancel out
the charge of two ferric ions. Since it would take three sulphate ions (each
with a charge of negative two) to cancel out two ferric ions (with a charge of
plus three) we use a subscript of three after the symbol for sulphate ion and a Fe3+ SO42ˉ
subscript of two after the symbol of ferric ion to show the ratio. Parenthesis 2Fe3+ 3SO42ˉ
must be used if you need more than one of a polyatomic ion. We need to
Fe2(SO4)3
show three sulphate ions in our formula. The subscript is put on the outside
of the parenthesis to show that the entire polyatomic ion is tripled. The correct
use of parenthesis will seem hard at first, but you must master this skill with
practice!
Conceptual Chemistry IX and X -2-
1.22. Rules for writing Chemical Formulae of Compounds
To write chemical formulae of compounds, following steps are involved:
1. Every compound is supposed to consist of two types of radicals namely cation and anion. First of all
name of types of ions or radicals are identified in the given compounds. Then symbols or formulae of
both ions or radicals are written side by side (showing cation first then anion) with their real or formal
charges (oxidation numbers) or valencies in superscripts (at the top or at the upper side each ion or
radical).
2. Now the correct combination of ions is used to produce a compound with a net charge of zero. To do
so, the real or formal charges or valencies of corresponding cation and anion are cross multiplied. If
the charges are same then they are cancelled out otherwise they become the subscripts of ions or
radicals showing the ratios of atoms.
Solved Examples
Example # 1. Write down chemical formula of calcium oxide.
Step One
Write the symbols for the elements in the compound. Note that the ending Calcium = Ca , Oxide = O
"ide" is used for oxide to show that it is a negative ion of oxygen.
Step Two
Write down the charge or oxidation numbers of the elements or radicals
involved as superscripts to the right of the elemental symbols. (Note that Calcium = Ca2+, Oxide = O2ˉ
when no number accompanies a charge symbol, as in the case of fluoride
below, they charge value is understood to be "1").
Step Three
Use the correct combination of ions to produce a compound with a net charge
of zero. In this case, (2+) + (-2) = 0. So, one oxide ion will cancel out one
calcium ion. Since it would take one oxide ion (each with a charge of negative
Ca2+ O2ˉ CaO
two) to cancel out one barium ion (with a charge of plus two) we do not use a
subscript of one after the symbol for calcium and oxygen to show the ratio as
when no number accompanies in subscript of elemental symbol the atomic
ratio is understood to be "1:1").
Example # 2 Write the correct formula for Barium Fluoride.
Step One
Write the symbols for the elements in the compound. Note that the ending Barium = Ba , Fluoride = F
"ide" is used for fluoride to show that it is a negative ion of fluorine.
Step Two
Look up the oxidation numbers of the elements involved and write them as
superscripts to the right of the elemental symbols. Note that when no number Barium = Ba2+ , Fluoride = Fˉ
accompanies a charge symbol, as in the case of fluoride below, they charge
value is understood to be "1".
Step Three
Use the correct combination of ions to produce a compound with a net charge
of zero. In this case, (2+) + 2(-1) = 0. So, two fluoride ions will cancel out one
barium ion. Since it would take two fluoride ions (each with a charge of Ba2+ F ˉ
negative one) to cancel out one barium ion (with a charge of plus two) we use
a subscript of two after the symbol for fluorine to show the ratio. Ba2+ 2F ˉ
BaF2
Conceptual Chemistry IX and X -1-
, Dr Inam Ul Haq Jazbi 10. Inorganic Section
Example # 3;Write the proper formula for the ionic compound lithium bromide.
Step One
Write the symbols for the elements in the compound. Note that the ending Lithium = Li Bromide = Br
"ide" is used for bromide to show that it is a negative ion of bromine.
Step Two
Look up the oxidation numbers of the elements involved (in ion chart), and
write them as superscripts to the right of the elemental symbols. Note that Lithium = Li+ Bromide = Brˉ
when no number accompanies a charge symbol, as in the case of fluoride
below, they charge value is understood to be "1".
Step Three
Use the correct combination of ions to produce a compound with a net charge Li+ Brˉ
of zero. In this case, (+1) + (-1) = 0. So, one lithium ion will cancel out the Li+ 2Brˉ
charge of one bromide ion. This means that the two elements will combine in LiBr2
a one to one ratio, and no subscripts will be needed
Example # 4; Show the correct formula for Zinc Nitrate.
Step One
Write the symbols for the monatomic and polyatomic ions in the compound. Zinc = Zn Nitrate = NO3
Step Two
Write down the charge or oxidation numbers of the elements or radicals
involved as superscripts to the right of the elemental or radical symbols. (Note Zinc = Zn2+ Nitrate = NO3ˉ
that when no number accompanies a charge symbol, as in the case of nitrate
below, they charge value is understood to be "1").
Step Three
Use the correct combination of ions to produce a compound with a net charge
of zero. In this case, (2+) + 2(-1) = 0. So, two nitrate ions will cancel out one
zinc ion. Since it would take two nitrate ions (each with a charge of negative Zn2+ NO3ˉ
one) to cancel out one zinc ion (with a charge of plus two) we use a subscript Zn2+ 2NO3ˉ
of two after the symbol for nitrate ion to show the ratio. Parenthesis must be Zn(NO3)2
used if you need more than one of a polyatomic ion. We need to show two
nitrate ions in our formula. The subscript is put on the outside of the
parenthesis to show that the entire polyatomic ion is doubled. The correct use
of parenthesis will seem hard at first, but you must master this skill with
practice!
Example # 5; Show the correct formula for ferric sulphate.
Step One
Write the symbols for the monatomic and polyatomic ions in the compound. Ferric = Fe , Sulphate = SO4
Step Two
Write down the charge or oxidation numbers of the elements or radicals
involved as superscripts to the right of the elemental or radical symbols. Ferric = Fe3+ , Sulphate = SO42ˉ
Step Three
Use the correct combination of ions to produce a compound with a net charge
of zero. In this case, 2(3+) + 3(2-) = 0. So, three sulphate ions will cancel out
the charge of two ferric ions. Since it would take three sulphate ions (each
with a charge of negative two) to cancel out two ferric ions (with a charge of
plus three) we use a subscript of three after the symbol for sulphate ion and a Fe3+ SO42ˉ
subscript of two after the symbol of ferric ion to show the ratio. Parenthesis 2Fe3+ 3SO42ˉ
must be used if you need more than one of a polyatomic ion. We need to
Fe2(SO4)3
show three sulphate ions in our formula. The subscript is put on the outside
of the parenthesis to show that the entire polyatomic ion is tripled. The correct
use of parenthesis will seem hard at first, but you must master this skill with
practice!
Conceptual Chemistry IX and X -2-
