Elaine Teto
CHE 101-5
Due: 7/5/04
In chapter 5; Quantitative Relationships in Chemical Reactions, many
topics such as the concept of mole; atomic, formula, and molecular masses;
balanced chemical equations and stoichiometry; reactions in solutions; molar
concentration; and preparing dilute solutions from concentrated solutions are
discussed. In order to gain optimal knowledge on each individual topic they shall
be discussed and explained individually. In doing so, the reader will go from the
more basic concepts to those that require more detail. Therefore the first topic
that shall be discussed is the concept of the mole.
The mole, as with many other science and math related topics, is
equivalent to other numerical values. Firstly, the formula mass in grams of a
substance contains one mole of particles, and as previously stated that is not the
only way to identify a mole. A mole is also equivalent to Avogadro’s number
(6.022 x 1023.)
Now, one may ask what purpose a mole serves in chemistry. The
answer is actually quit simple, the mole allows us to weigh atoms in the lab and
most importantly it enables us to compare amounts of atoms in chemical
reactions. The basic formula for a mole is, 1 mole = 6.022 x 1023 atoms or
particles = formula mass in grams. Therefore, an example of a simple mole
calculation would be as follows:
What is the mass in grams of 0.452 mole of C6H12O6?
The formula would be: mole = n = mass (gr) / molar mass (gr/mole)
0.452 mole = x (gr) / 180.066 (gr/mole)
After multiplying the 0.452 and the 180.066, and canceling out the moles,
the answer is
81.36 grams of C6H12O6.
Returning back to what was previously stated a mole of anything has
Avogadro’s number of objects in it. For example, a mole of water (H2O) has
6.022 x 1023 water molecules. However, if the objects being discussed have
different masses, then the moles of these objects will also differ. Therefore a
mole of 12C atoms will weigh 12 grams, a mole of
197Au
atoms will weigh 197
grams. Now, before moving further in the chapter, through some basic
definitions, one can learn what the reasoning is for the explanations above.
Firstly, an atomic mass unit (amu) is defined by setting the mass of one
C atom equal to 12 amu. We can determine the amu experimentally to be 1.66 x
10-24 grams. Based on this, one can conclude that the atomic weight of a 12C
atom is 12 amu. These weights can be derived from the periodic table. After
learning what the atomic weight of a substance is, one can further expand on this
topic by calculating the formula weight, which is the sum of the atomic weight of
each atom in a chemical formula. For example, the formula weight for NaOH is
(22.99 + 15.99 + 1.0079) 39.988 amu. When chemical formulas equal the
molecular formula than the formula weight is also known as the molecular weight.
Finally, the molar mass, the mass in grams of 1 mole of a substance, can be
determined. The molar mass is numerically equal to the formula weight. For
example one mole of water (FW = 18 amu) has a molar mass of 18 grams.
Now that a basic understanding of the mole has been established the next
topic is balanced chemical equations and stoichiometry. Stoichiometry is the
math behind chemistry.
Given enough information, one can use stoichiometry to calculate masses,
moles, and percents within a chemical equation. In chemistry, we use symbols
to represent the various chemicals. For example, the symbol "C" represents an
atom of carbon. To represent a molecule of table salt, sodium chloride, we would
use the notation "NaCl", where "Na" represents sodium and "Cl" represents
chlorine. Therefore a chemical equation is a chemical process through which
reactants (substances being changed) produce products (newly formed
substances). An example of a chemical reaction is as follows:
AgNO3 (aq) + NaCl(aq) ---> AgCl (s) + NaNO 3 (aq)
Furthermore, coefficients are used in all chemical equations to show the
relative amounts of each substance present. However, in using these
coefficients we come to our next topic, balancing chemical reactions. One of the
most important laws to remember, when considering balancing equations is the
Law of Conservation of Mass, which states that matter can neither be created nor
destroyed. Based on this law we can conclude that certain equations, such as
Al + Fe3O4  Al2O3, could simply not exist. In short, one has to be sure that an
equal number of atoms are present on both sides of the equation. There are
many different ways and systems of doing this, but for all methods, it is important
to know how to count the number of atoms in an equation. For example:
2Fe3O4
In this substance we can see that there are two molecules of Fe 3O4.
Furthermore, in each molecule of this substance there are three Fe atoms, and
four O atoms. When actually balancing equations that have coefficients and
subscripts it is important to remember that the coefficient x the subscript = the
number of atoms. At this point one should be ready to try balancing a few
equations:
Example 1: Al + Fe3O4  Al2O3 + Fe

When balancing equations it is important to take
one atoms at a time and work with each
individually. Trying to balance an entire equation
at once can be difficult.
8 Al + 3 Fe3O4  4 Al2O3 + 9 Fe
Example 2: CH4 + 2O2  CO2 + H2O + Heat
CH4 +2O2  CO2 + 2H2O + Heat
After one has perfected balancing equations, it is easier to move forward
in stoichiometry. There are two types of compounds one may encounter, the
simpler being empirical formulas, and the other being molecular formulas. The
molecular formula is the form of the term as it would appear in a chemical
equation. The empirical formula and the molecular formula can be the same, or
the molecular formula can be any positive integer multiple of the empirical
formula.
Examples of empirical formulas: AgBr, Na2S, C6H10O5.
Examples of molecular formulas: P2, C2O4, C6H14S2, H2, C3H9.
One can calculate the empirical formula from the masses or percentage
composition of any compound. If we only have mass, all we are doing is
essentially eliminating the step of converting from percentage to mass.
Example: Calculate the empirical formula for a compound that has 43.7 g P
(phosphorus) and 56.3 grams of oxygen. First we convert to moles:
Next we divide the moles to try to get an even ratio.
When we divide, we did not get whole numbers so we must multiply by two (2).
The answer=P2O5
After one has calculated the empirical formula one can calculate the
molecular formula. By dividing the molecular mass of the compound by the mass
of the empirical formula, one would attain the molecular formula. It is also
possible to do this with one of the elements in the formula; by dividing the mass
of that element in one mole of compound by the mass of that element in the
empirical formula. The result should always be a natural number.
Example: The empirical formula of a compound is HCN, 2.016 grams of
hydrogen are necessary to make the compound, what is the molecular formula?
In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by
1.008 we see that the amount of hydrogen needed is twice as much. Therefore
the empirical formula needs to be increased by a factor of two (2). The answer is:
H2C2N2.
Progressing through the basics of stoichiometry, the next primary topic is
the concentration (strength) of solutions. The concentration of a solution is
typically given in molarity. Molarity is defined as the number of moles of solute
(what is actually dissolved in the solution) divided by the liters of solution (the
total volume of what is dissolved and what it has been dissolved in).
Example: If 5.00 grams of NaOH are dissolved in 5000 mL of water, what is the
molarity of the solution?
So the molarity (M) of the solution is 0.025 mol/L.
Molality is another common measurement of concentration. Molality is
defined as moles of solute divided by kilograms of solvent (the substance in
which it is dissolved, like water). Molality is sometimes used in place of molarity
at extreme temperatures because the volume can contract or expand.
Example: If the molality of a solution of C2H5OH dissolved in water is 1.5
and the weight of the water is 11.7 kg, figure out how much C2H5OH must have
been added in grams to the solution?
It is possible to convert between molarity and molality. The only information
needed is density.
Example: If the molarity of a solution is 0.30 M, calculate the molality of the
solution knowing that the density is 3.25 g/mL.
In summary, there are a few basic rules one can follow when practicing
stoichiometry. Firstly, complete a balanced equation. Secondly, convert mass
as given into respective moles. Thirdly, recognize the stoichiometry factor. And
finally, multiply each mole solved for by its molar mass to calculate mass sought.
Example: If you burn 1.6gr of CH
A.) Calculate the mass (g) of O required
B.) Calculate the mass (g) of CO required
C.) Calculate the mass (g) of H2O produced
1.) CH4 + 2O2  CO2 + 2H2O
2.) N = 1.6gr x 1 mole CH /16gr/mole = 0.1 moles of CH4
3.) N (O2) = N (CH4)X 2 moles of O2 /1 mol CH4 = 0.2 moles of O2
N (CO2) = 0.1 mol CH4 x 1 mol CO2 / 1mol CH4 = 0.1 moles of CO2
N (H2O) = 0.1 mol CH4 x 2 mol H2O / 1 mol CH4 = 0.2 moles of H2O
4.) m (g) O2 = 0.2 moles O2 x 32.0g/ 1 mole O2 = 6.4 gr
m (g) CO2 = 0.1 moles CO2 x 44 gr/ 1 mole CO2 = 4.4 gr
m (g) H2O = 0.2 moles H2O X 18.0 gr/ 1 mole H2O = 3.6 gr
Most chemical reactions take place in aqueous solution. However, before
beginning the explanation of this process it is important to understand a few key
concepts, which play a major role in this process. For instance, one should recall
the fact that one of the best solvents is water, because it is a polar molecule. An
example of what a polar molecule is and does is as follows; water dissolves an
ionic substance by attracting the cations and anions of salt so strongly that the
attractive forces in the ionic crystal are broken, causing the resulting ions to be
hydrated.
Electrolytes are substances whose aqueous solutions conduct electricity.
It is the presence of ions which can move through the solution that allows the
conduction of the electricity. Sometimes the ions are already there, and other
times they result from the reaction of a substance with the solvent, water. When
an ionic substance is broken apart into the solvated cations and anions, the
process is properly called dissociation. However, when the ions are formed by
the reaction of the substance with water, and didn't exist in the original
substance, then the process is properly called ionization.
Strong electrolytes are substances which are completely dissociated in
water or substances for which the ionization reaction goes to completion. In other
words, after the substance has dissolved in water, the original substance no
longer exists. Some examples of substances which are classified as strong
electrolytes are:
* strong inorganic acids (the common ones are: hydrochloric acid,
hydrobromic acid, hydroiodic acid, sulfuric acid, nitric acid, chloric acid, and
perchloric acid
* strong inorganic bases, e.g. NaOH.
Dissociation reaction: NaCl + H2O ---> Na+ (aq) + Cl- (aq)
Ionization Reaction: HCl(g) + H2O(l) ---> H3O+ (aq) + Cl- (aq)
Weak electrolytes only ionize partially in water. In other words, the
ionization reaction lies mostly to the left, producing only a small number of ions,
in solutions. Some examples substances which behave as weak electrolytes are:
* Weak inorganic acids (inorganic acids which are not listed above), e.g.
carbonic acid, nitrous acid, hypochlorous acid, etc.
* Organic acids (carboxylic acids, e.g. acetic acid)
* Organic amines, e.g. methylamine.
Finally, nonelectrolytes do not form ions in solution. Some examples of
these are, sugars and alcohols.
After gaining a basic understanding about the fundamentals in chemical
reactions, one can move forward in furthering their understanding about the
compositions of solutions. Once again, the mole plays a significant role. One
will use molarity (M), as the concentration unit. Molarity, is defined as the
number of moles of solute dissolved in a liter of solution. The equation is as
follows:
(Molarity) x (Volume [in liters]) = moles of solute
or
(Molarity) x (Volume [in milliliters]) = millimoles of solute
Another significant equation, in the composition of solutions deals with the
process of dilution. One fact should always be called to mind when dealing with
dilution equations, and that is, that more solvent is always added, in these
equations; but the number of moles of solute is the same in the diluted and
concentrated solutions. Due to this fact, the following equation applies:
MCVC = MDVD
However, this equation solely represents dilution, and is not to be used for
stoichiometric calculations. In the equation, the “c” represents concentration,
while the “d” represents dilution.
The primary types of reactions (at this point) are precipitation reactions,
acid/base reactions, and redox reactions. However, it is important to know a bit
about solubility, before furthering ones knowledge in this area. Generally, three
types of reactions may be written, when dealing with this area. The "molecular"
equation lists the whole formula for reactants and products, not taking into
account the form of the solute in solution. For example, NaCl doesn't exist as
NaCl in water solution; it exists as sodium and chloride ions. Nevertheless, the
formula NaCl would be used in a molecular equation. The ionic equation shows
all strong electrolytes, reactants and products, as separated ions. Therefore,
NaCl would be shown as Na+ and Cl- ions in an ionic equation. Sucrose (a
sugar), however, would be shown as the molecular substance C12H22O11, since it
is a nonelectrolyte. The net ionic equation results after common terms are
subtracted from the ionic equation. These common terms do not participate in the
reaction, and thus are called spectator ions.
Finally, in dealing with reactions, we reach one of the most significant
topics in this area. As discussed previously, an atom can gain or lose electrons
from another atom. Oxidation is the loss of electrons and reduction is the gain of
electrons. The method used for keeping track of the electrons is the oxidation
number. Oxidation states are not necessarily real charges. The only time that an
oxidation number is a real charge is for monatomic ions, like Cl- (oxidation
number -1) or aluminum ion, Al3+ (oxidation number +3). In hydrogen chloride,
the oxidation numbers of +1 for the hydrogen and -1 for the chlorine are clearly
not real; HCl is a covalent compound which doesn't contain ions. Also, it is
important to remember that charges on ions are written as n+ or n-, whereas
oxidation numbers are written as +n or -n. Although some simple redox reactions
can be balanced by inspection, most redox reactions must be balanced
systematically.
Chapter five contained many significant topics in the study of chemistry. It
is important to remember that, starting from the simplest form of a concept, such
as the mole, and than working toward a more elaborate concept, such as
stoichiometry, is the best way to master the topics. Numerous formulas have
been given, to aid in solving many of the equations one will encounter in this
area. Therefore, with a solid understanding of the key terms that apply, and
practice with equations, one can move forward in furthering there knowledge in
this field.
Works Sited
http://chemistry.about.com. Viewed on 6/13/04
http://richardbowels.tripod.com/chemistry/balance.htm#part0. Viewed on 6/18/04
www.chem.tamu.edu. Viewed on 6/18/04
www.iun.edu. Viewed on 6/15/04
www.howe.k12ok.us. Viewed on 6/13/04