Chapter 5
•To accurately and concisely represent chemical reactions
we use the symbols of the elements and compounds in the
reaction.
•This is done with a chemical equation.
•But the equation must be balanced (because according to
the Law of Conservation of Mass, matter is never created or
destroyed and during a chemical reaction atoms are not
created nor destroyed, only rearranged).
•Therefore all equations must contain the same number of
each particular atom on both sides of the equation.
•This is accomplished using coefficients.
•We will practice balancing a few simple equations:
NOTE: When balancing you can only change
coefficients. (coefficients are the numbers in front
of the formulas of the elements or compounds)
1. C
+
O2 
CO
2. C
+ O2 
CO2
3. H2
+
Cl2 
HCl
4. Ba(NO3)2 + H2SO4  BaSO4 + HNO3
(Hint: treat polyatomic ions as one item)
Let’s do one more example and look at the balanced equation as a picture.
05_02.JPG
05_01.JPG
•Earlier we talked about atomic weights(AW) for the elements.
• A new value we need to understand now is molecular
weight (MW) or Molecular Mass for compounds (a more
general term is formula mass). This is determined by simply
adding the AW’s of every atom in the molecule.
•For example, 3 of the compounds we predicted formulas for
earlier have the following MW’s
NaCl  22.99 + 35.45 = 58.44
CaCl2  40.08 x1 + 35.45 x 2 = 110.98
MgSO4  24.31 + 32.06 + 16.00 x 4 = 120.37
Let’s look at what a balanced equation tells us
in more detail.
First, a new concept.
•An Italian chemist by the name of Amedeo Avogadro, in
the nineteenth century, realized a very important concept;
• Equal volumes of pure substance gases have equal
numbers of atoms or molecules.
•Amount equal to the atomic weight(for a monatomic
element) or the molecular weight for a compound, will
have the same volume in the gas phase.
•Therefore this # becomes very important and is known
as Avogadro’s #.
•Several experiments were developed to determine the value
of this #.
•As of today, we have extremely accurate instrumental ways
of determining this # and our best estimate today is that it’s
value is 6.02 X 1023.
•If we have a sample of any substance containing this # of
molecules (for a compound) or atoms (for a monatomic
element) we call that sample a MOLE, abbreviated mol.
6.02 X 1023 items = 1 mole
•Thus a mole is a specific # of particles. But this #
of particles will have a specific mass. For NaCl it
would be 58.44 grams, for CaCl2 it would be
110.98 g and for MgSO4 it would be 120.37 g.
•Important to be able to calculate grams from
moles and vice versa. Chemists use # of
molecules (too small to be seen) but measure
grams (large enough to see or handle).
•Use the molecular weight such as 58.44 for NaCl with the
proper units Grams/mol. This is essentially equal to 1
because 58.44 grams of NaCl = 1 mole of NaCl, therefore
dividing 58.44g by 1 mole or dividing 1 mole by 58.44 g for
NaCl = 1.00.
•Now I can convert moles of NaCl to grams or vice versa by
multiplying by the appropriate value of 1.00 (multiplying by 1
does not change the value), either 58.44 grams/mol or 1
mol/58.44 grams.
Example: Convert 1.65 moles of NaCl to equal
value in grams and then convert back to moles:
58.44 g of NaCl
1.65 moles of NaCl x
 96.426 g of NaCl
1 mol of NaCl
1 mol
96.426 g x
 1.65 moles
58.44 g
Let’s return to what we can learn from a balanced
chemical equation. Such as:
Cu2S + 2 Cu2O  6 Cu + SO2
1 mole
2moles
0.1 moles 0.2 moles
5 moles
10 moles
6 moles
1 mole
0.6 moles 0.1 moles
30 moles 5 moles
•Literally, this says that 1 molecule of Cu2S plus 2
molecules of Cu2O react together to form 6 atoms of
Cu plus 1 molecules of SO2.
•But this also means that 1 mole of Cu2S plus 2 moles of
Cu2O react together to form 6 moles of Cu plus 1 mole of
SO2, or any equivalent ratio, such as 0.10 mole of Cu2S plus
0.20 moles of Cu2O react together to form 0.60 moles of Cu
plus 0.10 mole of SO2.
•But if we know moles, we can also determine grams,
which is what we need to know in a lab, because we
measure grams, not moles or molecules in a typical lab
experiment.
•Earlier we discussed homogeneous and heterogeneous
mixtures and pointed out that homogeneous mixtures are
all called solutions.
•In solutions there is a solvent, which enables the solution
to form (the dissolver) and one or more solutes, (those
substances that are broken down to molecular level and
mixed in between the molecules of the solvent (the
dissolvees)
•The maximum amount of solute that can dissolve in a
particular solution (depends on the solute, the solvent, and
the temperature and sometimes the pressure (especially
when gases are dissolved in liquids) is called its Solubility.
•We frequently use the terms soluble or insoluble when
talking about a solid solute dissolving in a liquid solvent.
We use the terms miscible or immiscible when talking
about the solubility of a liquid solute in a liquid solvent
Concentrations of solutes in a water based solution
are measured in many ways. One of the most
common is a term called Molarity. It measures
how many moles of solute are present in every
liter of solution.
moles of solute
Molarity (M) 
liters of solution
Let’s practice a calculation using this equation.
See example 5.9 on page 141
WORKED EXAMPLE 5.9 Solution Concentration: Molarity and Moles
Calculate the molarity of a solution made by dissolving 3.50 mol of NaCl in enough water to produce 2.00 L of solution.
Solution
We read 1.75 M NaCl as “1.75 molar NaCl.”
Exercise 5.9A
Calculate the molarity of a solution that has 0.0500 mol of NH3 in 5.75 L of solution.
Exercise 5.9B
Calculate the molarity of a solution made by dissolving 0.750 mol of H3PO4 in enough water to produce 775 mL of solution.
Another common concentration term is %
concentration. This can be % by volume or % by
weight or mass (more common)
Volume of solute
% by volume 
x 100
Volume of solution
This is used for a solution between 2 liquids
mass of solute
% by mass 
x 100
mass of solution
This is used for any solutions, most commonly for
a solid solute in a liquid solvent. Let’s do example
5.14 on page 144.
WORKED EXAMPLE 5.14 Percent by Mass
What is the percent by mass of a solution of 25.5 g of NaCl dissolved in 425 g (425 mL) of water?
Solution
Use these values in the above percent-by-mass equation:
Exercise 5.14A
Hydrogen peroxide solutions for home use are 3.0% by mass solutions of H2O2 in water. What is the percent by mass of a solution
of 9.40 g of H2O2 dissolved in 335 g (335 mL) of water?
Exercise 5.14B
Sodium hydroxide (NaOH, lye) is used to make soap and is very soluble in water. What is the percent by mass of a solution that
contains 1.00 kg of NaOH dissolved in 950 mL of water?