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					               Writing Chemical Formulas


A chemical formula is a combination of elemental symbols and subscript numbers that
is used to show the composition of a compound. Depending of the type of compound
that the formula represents, the information that it provides will vary slightly. Before we
go about learning how to write chemical formulas, it is important that you clearly
understand the difference between molecular compounds and ionic compounds.

    Ionic compounds are composed of charged ions that are held together by electrostatic
forces. A typical type of ionic compound, called a binary compound because it is made
up of two elements, will be composed of metallic positive ions (cations) and nonmetal
negative ions (anions). Another type of ionic compound, called a ternary compound as
it contain three elements, is composed of monatomic ions and polyatomic ions. When
dealing with ionic formulas it is very important to remember that the formula does not
show how the compound actually exists in nature. It only shows the ratio by which the
individual ions combine. For example, the ionic formula for calcium chloride is CaCl2.
Since calcium chloride is an ionic compound, this formula does not mean that there are
actually two chlorine atoms floating around attached to one calcium atom. Ionic
compounds are actually continuous, lacking the discrete units that make up a sample of a
molecular substance. Rather, the formula shows that a sample of calcium chloride
contains twice as many chlorine atoms as calcium atoms. Remember that ionic
compounds are not molecules, so the formula CaCl2 is said to represent one formula
unit of calcium chloride.

   Molecular compounds are held together by covalent bonds, or shared pairs of
electrons. Molecular formulas do show these molecules as they actually exist as discrete
units in nature. When we say that the molecular formula of water is H2O, we can see that
the molecules of water are made up of three atoms, two hydrogen atoms are covalently
bonded to each oxygen atom. A special type of chemical formula, called an empirical
formula, shows the composition of a molecule not as it actually exists, but in a simple
whole number ratio. The difference between empirical and molecular formulas will be
explained in lesson 5-5.

   This lesson will concentrate on writing simple chemical formulas when given a
formula name. In learning how to write chemical formulas, you will make use of the
oxidation numbers that you learned about in lesson 5-2. For your convenience, print out
the tables from lesson 5-2 before you continue with this lesson, as they will be referred to
from time to time.

Writing Ionic Formulas
I. Binary Compounds - Binary compounds are compounds that are composed of only
two elements. When you write the formulas for binary compounds, they will consist of
two elemental symbols, and they may also have one or two subscript numbers, if the
elements don't combine in a one to one ratio. You are probably familiar with the formula
NaCl for table salt. This formula shows no subscripts because one ion of Na will be
present for each ion of Cl, in any sample of table salt.

You will be given the name of a binary compound and you will be expected to be able to
write the proper formula for the compound. There will be two sources of information for
writing the correct formula. The compounds name will give you the elements that make
up the compound. The oxidation numbers of the ions involved will show you the ratio by
which they combine. Let's go through an example;

Example 1. Write the correct formula for Barium Fluoride.

Step one - Write the symbols for the elements in the compound. If you need to review
the elemental symbols, see lesson 5-1. Note that the ending "ide" is used for fluoride to
show that it is a negative ion of fluorine.

                           Barium = Ba            Fluoride = F

Step two - Look up the oxidation numbers of the elements involved (in table 5-2b or
some similar table), and write them as superscripts to the right of the elemental symbols.
Note that when no number accompanies a charge symbol, as in the case of fluoride
below, they charge value is understood to be "1".

                          Barium = Ba2+           Fluoride = F-

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 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.

                                           BaF2

               If this seems confusing to you, it will get simpler over time.

Example 2. 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
"ide" is used for bromide to show that it is a negative ion of bromine.

                            Lithium = Li        Bromide = Br
Step two - Look up the oxidation numbers of the elements involved (in table 5-2b or
some similar table), and write them as superscripts to the right of the elemental symbols.
Note that when no number accompanies a charge symbol, as in the case of fluoride
below, they charge value is understood to be "1".

                           Lithium = Li+        Bromide = Br-

Step three - Use the correct combination of ions to produce a compound with a net charge
of zero. In this case, (+1) + (-1) = 0. so, one lithium ion will cancel out the charge of one
bromide ion. This means that the two elements will combine in a one to one ratio, and
know subscripts will be needed.

                                           LiBr



II. Ternary Compounds - Ternary compounds are composed of three different
elements. The most common types of ternary compounds consist of a metallic cation
(positive ion) and a polyatomic anion (negative ion). The only common polyatomic ion
with a positive charge is the ammonium ion. At any rate, To write these formulas you
will want to have reference tables with the information provided on tables 5-2b and 5-2d.

Example 1. Write the proper chemical formula for potassium hydroxide.

Step one - Write the symbols for the monatomic and polyatomic ions in the compound.
You will find the symbol potassium on table 5-2b. Hydroxide is a polyatomic ion, which
will be found on table 5-2d. Eventually you will recognize the name of a polyatomic ion,
but for now if you can't find an ion on one table, look on the other.

                           Potassium = K       Hydroxide = OH

Step two - Look up the oxidation numbers of the ions involved (in table 5-2b and 5-2d, or
some similar tables), and write them as superscripts to the right of the elemental symbols.

                          Potassium = K+      Hydroxide = OH-

Step three - Use the correct combination of ions to produce a compound with a net charge
of zero. Parenthesis must be used if you need more than one of a polyatomic ion. In this
case, (+1) + (-1) = 0. So, only one of each ion is used. No subscripts are necessary. If
you needed more than one hydroxide ion, it would be put in parenthesis with the
subscript on the outside.

                                           KOH

                       Note the importance of upper and lower case
Example 2. Show the correct formula for Calcium Nitrate.

Step one - Write the symbols for the monatomic and polyatomic ions in the compound.

                            Calcium = Ca        Nitrate = NO3

Step two - Look up the oxidation numbers of the ions involved (in table 5-2b and 5-2d, or
some similar tables), and write them as superscripts to the right of the elemental symbols.

                           Calcium = Ca2+       Nitrate = NO3-

Step three - Use the correct combination of ions to produce a compound with a net charge
of zero. Parenthesis must be used if you need more than one of a polyatomic ion. In
this case (+2) + 2(-1) = 0. 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.

                                         Ca(NO3)2

The correct use of parenthesis will seem hard at first, but you must master this skill with
                                        practice!



III. The Stock System - Some elements, like iron and lead, have more than one
oxidation number. If you were given a compound name like lead chloride, you would not
know if you should used an oxidation number of +2 or +4 for the lead. The stock system
is used to specify which form of an element, that shows multiple oxidation numbers, is
used in a particular compound. A roman numeral is shown after the name of the positive
ion (cation) to indicate the oxidation number of the positive ion.

Example 1. Show the correct formula for lead(IV) nitrate.

Step one - Write the symbols for the ions in the compound.

                             Lead = Pb         Nitrate = NO3

Step two - Look up the oxidation number of the negative ion involved (in table 5-2b and
5-2d, or some similar tables). The positive ion will have a positive oxidation number
equal to the roman numeral. Write the numbers as superscripts to the right of the
elemental symbols.

                            Lead = Pb4+        Nitrate = NO3-

Step three - Use the correct combination of ions to produce a compound with a net charge
of zero. Parenthesis must be used if you need more than one of a polyatomic ion.
                                      Pb(NO3)4



Example 2. Show the correct formula for Copper(II) Fluoride

Step one - Write the symbols for the ions in the compound.

                            Copper = Cu       Fluoride = F

Step two - Look up the oxidation number of the negative ion involved (in table 5-2b and
5-2d, or some similar tables). The positive ion will have a positive oxidation number
equal to the roman numeral. Write the numbers as superscripts to the right of the
elemental symbols.

                           Copper = Cu2+        Fluoride = F-

Step three - Use the correct combination of ions to produce a compound with a net charge
of zero. Parenthesis must be used if you need more than one of a polyatomic ion.

                                         CuF2

Writing Molecular Formulas

I. Binary Molecular Compounds - The standard method for naming binary molecular
compounds has changed over the years. Currently, the stock system is commonly used
for naming molecular compounds. Names like "carbon dioxide", "carbon monoxide",
and "dinitrogen pentoxide" are really remnants of an older system that used prefixes to
identify the number of elements involved. When you are writing the formula for a
molecular compound using the stock system, you will not really notice any difference
from the methods described above, until you study bonding. You should be aware that
you are not dealing with ions when you are working with molecular formulas, rather you
are looking up what might be called the apparent charge on each atom.

Example 1. Write the correct formula for nitrogen(IV) oxide.

Step one - Write the symbols for the elements involved.

                              Nitrogen = N      Oxide = O

Step two - Use the roman numeral as the apparent charge of the first element. Find the
apparent chart of the second element by looking on reference tables such as 5-2a.

                            Nitrogen = N4+      Oxide = O2-
Step three - Determine the ratio by which the elements will bond to show a net charge of
zero. Use subscripts to indicate the number of atoms of each element present. In this
case, (+4) + 2(-2) = 0.

                                          NO2

Example 2. Write the correct formula for nitrogen(III) oxide.



Step one - Write the symbols for the elements involved.

                              Nitrogen = N      Oxide = O



Step two - Use the roman numeral as the apparent charge of the first element. Find the
apparent charge of the second element by looking on reference tables such as 5-2a.

                            Nitrogen = N3+      Oxide = O2-

Step three - Determine the ratio by which the elements will bond to show a net charge of
zero. Use subscripts to indicate the number of atoms of each element present. In this
case, 2(+3) + 3(-2) = 0.

                                         N2O3
         Using Coefficients with Formulas
You have learned that the subscript numbers in a chemical formula represent the number
of atoms in one molecule or in one formula unit of an ionic compound. Now you will
learn about the other numbers, called coefficients, that you often see to the left of a
chemical formula. While a subscript number acts as a multiplier for a single element
(unless there are parenthesis), a coefficient number acts as a multiplier for all of the
atoms in the entire compound. As with subscripts, when no number is present then "1" is
understood. Look at the example below:


                                         CO2
   Here we have one molecule of carbon dioxide. The subscript 2 in the formula above
only pertains to the oxygen in the compound. The total number of atoms in the
compound is 3.

   Now let us put a coefficient in front of the molecule and see how that changes things.


                                       5 CO2
   The coefficient 5 refers to the entire molecule. It shows that there are 5 molecules of
carbon dioxide. Since each molecule is made up of 3 atoms, the total number of atoms is
now 15. There are 5 carbon atoms and 10 oxygen atoms.



Now, for an example with parenthesis;


                                    Ba(NO3)2
   Here we have one formula unit of the ionic compound, barium nitrate. We say
"formula unit" instead of "molecule" because ionic compounds don't form molecules.
The subscript 3 pertains to the oxygen, showing 3 oxygen atoms for each polyatomic ion
of nitrate. The subscript 2 is a multiplier for everything in the parenthesis, because it is
showing that there are two nitrate ions for every barium ion. The total number of atoms
for each formula unit of barium nitrate is 9. There are; 1 barium atom, 2 nitrogen atoms,
and 6 oxygen atoms.

    Now let's put a coefficient in front of the formula unit and see how it changes the
tally:
                                 3 Ba(NO3)2
   Now we have 3 formula units of barium nitrate. The 3 coefficient acts as a multiplier
for the entire compound. If there are 9 atoms in one formula unit of barium nitrate, then
there must be 27 atoms in three formula units. There are: 3 barium atoms, 6 nitrogen
atoms and 18 oxygen atoms.

Table 5-7a will show you some more examples of coefficients and formulas. Study the
table to make sure that you understand each atomic tally.

                                    Table 5-7a
     Atomic Tallies for Specific Quantities of Molecules and Ionic Compounds
        Given              2 NaNO3               6 C2H6             3 (NH4)2S
                       2 atoms Na            12 atoms C            6 atoms N

                       2 atoms N             36 atoms H            24 atoms H
     Atomic Tally
                       6 atoms O                                   3 atoms S

                       Total Atoms 10        Total Atoms 48        Total atoms 33
                       Naming Compounds

As first mentioned in an early lesson, there are two main types of compounds, ionic and
molecular. Some of the compounds that you will learn about this year will require
special systems for naming, and we will learn about them at a later time. For example,
we will learn how to correctly name the various types of acids and bases when we study
the chapters on acids and bases. In this lesson you will learn enough to name most of the
compounds that you will come in contact with in your laboratory activities this year.
References will be made to the tables from lesson 5-2, so it would be wise to have them
handy as you go over this material.

I. Binary Compounds. As you know, binary compounds consist of only two elements.
The formula for a binary compound may contain more than two letters, but it will contain
only two capital letters. When naming a binary compound, regardless of whether it is
ionic or molecular, follow the following steps:

1. Write the name of the element represented by the first symbol in the formula.

2. Write the name of the element represented by the second symbol in the formula, but
change the ending of the element's name to "ide".

3. Check a reference table to determine the number of positive oxidation numbers that
the first element forms. If it only forms one then you are done.

4. If the first element shows more than one oxidation number, than use the stock system.
Determine the oxidation number that the first element is showing and write that roman
numeral in-between the two elemental names.

Example 1. What is the correct name for the compound AlBr3 ?

Step 1. Write the name of the element represented by the first symbol in the formula.

                                       aluminum

Step 2. Write the name of the element represented by the second symbol in the formula,
but change the ending of the element's name to "ide". In this case, bromine becomes
bromide.
                                   aluminum bromide

Step 3. Check a reference table to determine the number of positive oxidation numbers
that the first element forms. If it only forms one then you are done. Aluminum always
has an oxidation number of +3, therefore there is no need for a roman numeral. Our
answer is;

                                   aluminum bromide



Example 2. What is the correct name for the element NiS ?

Step 1. Write the name of the element represented by the first symbol in the formula.

                                          nickel

Step 2. Write the name of the element represented by the second symbol in the formula,
but change the ending of the element's name to "ide". In this case, sulfur becomes
sulfide.

                                      nickel sulfide

Step 3. Check a reference table to determine the number of positive oxidation numbers
that the first element forms. If it only forms one then you are done. Nickel forms
oxidation numbers of +2, +3 and +4, so we must go to the next step.

Step 4. If the first element shows more than one oxidation number, than use the stock
system. Determine the oxidation number that the first element is showing and write that
roman numeral in-between the two elemental names. We check the oxidation number of
sulfide and find that it is -2. If one nickel is canceling out one sulfur than the apparent
charge on the nickel must be +2. (+2) + (-2) = 0.

                                     nickel(II) sulfide
Example 3. What is the correct name for the compound Fe2O3 ?

Step 1. Write the name of the element represented by the first symbol in the formula.

                                           iron

Step 2. Write the name of the element represented by the second symbol in the formula,
but change the ending of the element's name to "ide". So, oxygen becomes oxide.

                                        iron oxide

Step 3. Check a reference table to determine the number of positive oxidation numbers
that the first element forms. If it only forms one then you are done. Iron can be +2 or +3,
so we must go on to step 4.

Step 4. If the first element shows more than one oxidation number, than use the stock
system. Determine the oxidation number that the first element is showing and write that
roman numeral in-between the two elemental names. We know that oxygen is -2 in this
case. Since we have 3 atoms of oxygen, each with a charge of -2, then the total negative
charge is -6. We must have +6 to balance out the -6. Since there are two iron atoms to
make up a total of +6, each must be +3.
2(+3) + 3(-2) = 0.

                                      iron(III) oxide



II. Ternary Compounds - Ternary compounds contain three elements. The only type of
ternary compounds that we will learn how to name in this chapter are those that consist of
one polyatomic ion and one monatomic ion. The vast majority of these types of
compounds consist of a positive monatomic ion and a negative polyatomic ion. For this
type of compound you follow the steps below:

Step 1. Write the name of the element represented by the first symbol in the compound.

Step 2. Write the name of the polyatomic ion, without changing the ending.

Step 3. Check a reference table to determine the number of positive oxidation numbers
that the first element forms. If it only forms one then you are done.

Step 4. If the first element shows more than one oxidation number, than use the stock
system. Determine the oxidation number that the first element is showing and write that
roman numeral in-between the two names.
Example 1. Name the compound Ca(CN)2 ?

Step 1. Write the name of the element represented by the first symbol in the compound.

                                         Calcium

Step 2. Write the name of the polyatomic ion, without changing the ending.

                                    Calcium Cyanide

Step 3. Check a reference table to determine the number of positive oxidation numbers
that the first element forms. If it only forms one then you are done. Calcium is always
+2, so the final answer is as below:

                                    Calcium Cyanide

Example 2. What is the name of the compound Fe(NO3)2 ?

Step 1. Write the name of the element represented by the first symbol in the compound.

                                           iron

Step 2. Write the name of the polyatomic ion, without changing the ending.

                                       iron nitrate

Step 3. Check a reference table to determine the number of positive oxidation numbers
that the first element forms. If it only forms one then you are done. Iron can be +2 or +3,
so we must go on to step 4.

Step 4. If the first element shows more than one oxidation number, than use the stock
system. Determine the oxidation number that the first element is showing and write that
roman numeral in-between the two names. Nitrate shows an oxidation number of -1.
Since there are two nitrate ions in the compound, the total negative charge is -2.
Therefore, the iron must be +2. (+2) + 2(-1) = 0.

                                     iron(II) nitrate

Special Exception: The Ammonium ion (NH4 +) is a positive polyatomic ion. When it
combines with a negative monatomic ion, you change the ending of the negative ion to
"ide". When it combines with a negative polyatomic ion, you just name both ions.

                           (NH4)2S is called ammonium sulfide

                         NH4OH is called ammonium hydroxide

				
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