Reading Notes
for Chapter 4
These are Dr. Bodwin's reading notes for Chapter 4 of "Chemistry
2e" from OpenStax.
I am using a local .pdf copy that was downloaded in May 2020.
Chapter
Summary:
Stoichiometry (stow-ick-ee-AHM-it-tree) is a sometimes cumbersome word,
but it is probably the biggest fundamental concept in all of chemistry.
The etymology of the word is from the Greek and it means the study of
or measuring of elements. We've already seen a little glimmer of it in
the previous chapters when we were writing balanced chemical formulas,
and "balance" is really the key here. Nature does not allow random
dangling bits. When writing balanced chemical formulas, that means that
every positive charge must have a corresponding negative charge to
balance it. In this chapter, we are going to move on to chemical
reactions, and once again, we need to avoid the random dangling bits!
One way we can frame stoichiometry is by the Law of Conservation of
Matter... one way to say this is "matter cannot be created or
destroyed", but another way to state this that is more specific to
stoichiometry is "everything that goes in to a reaction, must come out,
and everything that comes out must have gone in".It might seem like a
pretty simple concept, but "simple" doesn't always mean "easy".
Balancing chemical equations and working through stoichiometry take
practice. I like to focus on using a consistent process because even
though every problem is just a little bit different, the process to solve those problems is
usually very close to the same.
Writing
Balanced Chemical Equations:
There is some basic vocabulary on page 176 (reactants, products,
coefficients) that we will be using constantly. Make sure you read over
these definitions and cement them into your vocabulary. You will be
seeing and hearing them often enough that they will become "normal"
words, but I want to make sure everyone has these in place.
Back to the Conservation of Matter concept, every element that appears
on the reactant side of an equation (the "left" side) must appear in
equal number on the product side of an equation (the "right" side).
NOTE - In a nuclear reaction, the rules
change a little bit, but at this point we are not studying nuclear equations, so
let's keep it simple. Nuclear reactions are important, but the VAST
majority of what we call "chemical reactions" take place in the
electron cloud outside the
nucleus, so at this point we can consider the nucleus to be constant
and unchanging.
A chemical equation connects the microscopic and the macroscopic world,
and can be used to describe both, depending upon context. For example,
the balanced chemical equation:
2 Na + Cl2 --> 2 NaCl
tells us that two sodium atoms react with one chlorine molecule (under
"normal" conditions, chlorine and the other halogens exist as diatomic
species) to produce 2 sodium chlorides. But we could also say that 20
sodium atoms react with 10 chlorine molecules to form 20 sodium
chlorides. The balanced chemical equation shows us the smallest
whole-number ratio of reactants and products, but it's a ratio, so as
long as the realtive relationship is maintained, it is still valid.
This is where the mole concept is important... remember a "mole" is
just a grouping unit, so we can say that 2 moles of sodium atoms react
with 1 mole of chlorine molecules to form 2 moles of sodium chlorides.
"Atoms" and "molecules" are descriptions in the microscopic world, "moles" are
descriptions in the macroscopic
world. The ratios established by the balanced chemical equation are
equally correct in both
worlds!
When we are working in the macroscopic domain, it is usually considered
good practice to include state labels in our balanced chemical
equations. Unless you have reason to believe otherwise, always assume
that you are under "normal" conditions of temperature and pressue
(around room temperature and around 1 atmosphere of pressue). Under
these "normal" conditions, sodium is a metallic solid, chlorine is a
gas, and sodium chloride is a solid. Adding these extra bits of
information into the above equation...
2 Na(s) + Cl2(g) --> 2 NaCl(s)
There are also liquids, (l), and aqueous solutions, (aq).
Writing correct balanced chemical equations takes practice. Take every
opportunity to look at balanced equations and practice balancing them
yourself. If someone handed you a bassoon and asked you to play an F
major scale, you wouldn't expect to be able to do it without
significant practice, and the F major scale isn't even a song. {Shout
out to any bassoon players!} Balancing chemical equations is similar...
you have to practice to become proficient.
Reaction
Classes:
There are a few different ways to classify chemical reactions to make
them easier to recognize and understand. One method looks at what type
of reactants and products the reaction involves.
Combination
- two (or more) elements combine to form an ionic or molecular product
Decomposition
- The opposite of "combination"; an ionic or molecular reactant breaks
apart into elements
Displacement/Replacement
- An element reacts with (usually) an ionic compound to product a
different element and ionic compound
Double
displacement/Double replacement/Metathesis - Two ionic or
molecular reactants "switch partners" to produce two new ionic or
molecular products.
These are one way to start thinking about chemical reactions, and I
include them because their terminology comes up in a number of other
places. There are also a number of reaction types that also help see
commonalities between chemical reactions...
Reaction
Type - Precipitation:
Precipitation reactions are those in which 2 or more soluble reactants
combine to form products, at least one of which is not soluble. For convenience at
this point, we will limit ourselves to aqueous solutions so water is
our solvent. Precipitation reactions are metathesis reactions.
Key review topics - writing balanced chemical formulas, molarity and
concentration
Table 4.1 summarizes the solubility rules we will use. Pair these up
with your polyatomic ion memorization practice. If you use physical
flashcards for the polyatomic ions, you can even add some additional
notes to those flashcards with solubility information.
These solubility rules are a good example of why you will see nitrate
so often in sample problems (and acetate, chlorate, and perchlorate)...
nitrate salts are soluble. When I'm writing a question and I need a
soluble salt, I go with nitrate because there are no significant
exceptions to the "All nitrate salts are soluble" rule.
Net Ionic
Equations:
A chemical reaction is a shorthand way to describe chemical change. It
can also be a "recipe" for a chemical reaction. If we are most
interested in describing chemical change, then the reaction can be
simplified by eliminating anything that is not undergoing chemical change.
Let's start with a precipitation reaction, the reaction of calcium
nitrate solution with sodium phosphate solution:
3 Ca(NO3)2(aq) + 2 Na3PO4(aq)
--> Ca3(PO4)2(s) + 6 NaNO3(aq)
This is the "full" or "molecular" equation. It contains all the species
present in the reaction, and it would be a very good version of the
reaction if you were trying to prepare these reagents from bottles of
solids that you might find on a shelf. If we want to explore the actual
chemical changes that are happening in this reaction, we might first
want to more correctly represent what is present in solution. "Ca(NO3)2(aq)"
is not actually intact calcium nitrate units floating around in
solution, it is actually calcium ions in solution and nitrate ions in
solution. The same is true of the other species in this reaction that
are labelled "aq". The "Ca3(PO4)2(s)"
actually is intact calcium
phospahte units, now as part of a solid designated by the "(s)". With
this in mind, we can re-write the balanced equation as a "full ionic"
or "complete ionic" equation:
3 Ca+2(aq) + 6 NO3-1(aq) + 6 Na+1(aq)
+ 2 PO4-3(aq)
--> Ca3(PO4)2(s) + 6 Na+1(aq)
+ 6 NO3-1(aq)
Looking at that reaction, there are a number of ions that appear to
just be sitting around watching the chemical change take place. They
are spectators. If we want a
version of the balanced chemical reaction that only describes the chemical change
that is taking place, then we can eliminate those spectators, and we
are left with the "net ionic" equation:
3 Ca+2(aq) + 2 PO4-3(aq)
--> Ca3(PO4)2(s)
This net ionic equation describes the chemical change that is taking
place. For any molecular equation, there is only one net ionic
equation, but what about the reverse of that statement? Think about
solubility rules... I need a soluble sourse of Ca+2(aq) ions
for my reaction, so I could choose a few different options, for
example, calcium chlorate. I also need a soluble source of PO4-3(aq)
ions; that could be potassium phosphate or lithium phosphate. Any
combination of a soluble calcium salt with a soluble phosphate salt
should give the exact same net ionic equation because the net ionic
equation is only describing
the chemical change that is taking place in the system.
Reaction
Type - Acid-Base:
Acid-base reactions are usually metathesis reactions.
The easiest way to recognize and acid-base reaction is to be able to
recognize common acids and bases. There are a number of definitions of
acids and bases, let's start with these:
NOTE: Just because a reaction has as
acid or a base doesn't make it an
Acid-Base reaction.
Acids - substances which when dissolved in water increase the
concentration of H+1 {or H3O+1}. An H+1-donor.
Bases - substances which when dissolved in water increase the
concentration of OH-1. An OH-1-donor.
For an acid or base to be considered "strong", it must "completely
dissociate in water". That means it breaks up into its component ions
completely. Weak acids and bases tend to "stick" together to varying
extent.
For our purposes, the acids in Table 4.2 are "strong". If an acid is
not in that table (and you aren't specifically told otherwise), then
you can assume that the acid is weak.
For our purposes, soluble hydroxides (as noted in the "exceptions"
column of Table 4.1) are "strong". Unless you are specifically told
otherwise, all other bases are "weak".
For strong acids and strong hydroxide bases, the net
ionic equation is always the same... H+1(aq) + OH-1(aq)
-->
H2O(l)
If either the acid or base is weak,
then that species has to appear in the net ionic equation because it is
not completely dissociated.
Reaction
Type - Redox:
This topic has its own chapter later on, so let's try to focus on the
basics here.
Let's add one more "number" to our toolbox of numbers that keep track
of subatomic particles. "Oxidation number" is the number of protons
minus the number of electrons. Yes, I know, that sounds just like
charge, BUT oxidation number works for individual ions as well as atoms
that are part of a polyatomic ion or molecule.
Redox reactions are reactions in which reduction and oxidation are both
occuring. If you have reduction, you MUST have oxidation and vice-versa.
Oxidation Is Losing electrons (OIL)
Reduction Is Gaining electrons (RIG)
So OIL-RIG might help you remember. If that doesn't click with you...
Losing Electrons is Oxidation (LEO)
Gaining Electrons is Reductions (GER)
So LEO the lion says GER might work better for you. Either way, they
both work.
For monoatomic ions, oxidation number of equal to charge.
There are a couple other "rules" that can help us at this point.
1. For uncombined,
uncharged elements, the Oxidation Number os zero
2. For monoatomic ions, the Oxidation Number is equal to
the charge
3. The sum of the Oxidation Numbers for a polyatomic ion or
molecule is equal to the charge on the polyatomic ion or is equal to
zero for a neutral molecule
4. Oxygen is almost always Oxidation Numebr -2, unless it
is molecular oxygen (Ox# = 0) or a peroxide (Ox# = -1). For our
purposes, a peroxide will have contain either "peroxide" or "peroxo" in
its name.
5. Hydrogen is almsot always Oxidation Number +1, unless it
is molecular hydrogen (Ox# = 0) or a hydride (Ox# = -1). For our
purposes, a hydride will have the word "hydride" in its name.
Redox reactions can be single
displacements, combination,
decomposition, and
might even look like metathesis
reactions. One way to quickly recognize redox reactions is to look for
uncombined elements as either reactants or products or both; not every
redox reaction has an uncombined element, but every reaction that
includes an uncombined element is a redox reaction. {Well, that may not
be absolutely true, but it's a pretty reliable "rule" for now...}
Your textbook goes through a very nice process for balancing redox
reactions using the half-cell method. Take a look at it, but don't
worry too much about this process at this point. Many (most?) of the
redox reactions we will use in this course can be balanced by
inspection rather than the half-cell method. Just remember, similar to
atoms in the Law of Conservation of Matter, charge must balance in a
correctly balanced chemical equation. Nature doesn't like random
charges dangling about.
Balancing by inspection:
Zinc metal reacts with copper(II) ions to form copper metal and
zinc(II) ions
Zn(s) + Cu+2(aq) --> Cu(s) + Zn+2(aq)
Magic! It's already balanced! One zinc on each side, one copper on each
side, +2 total charge on both sides... balanced!!
Zinc metal reacts with iron(III) ions to form iron metal and zinc(II)
ions
Zn(s) + Fe+3(aq) --> Fe(s) + Zn+2(aq)
One zinc on each side, one iron on each side... um, but there's a
charge problem. +3 vs +2... We can get +3 up to +6 by doubling it, and +2 up to +6 by tripling it, so
3 Zn(s) + 2 Fe+3(aq) --> 2 Fe(s) + 3
Zn+2(aq)
Reaction
Stoichiometry:
Stoichiometry problems all have the same 4 steps and they're all about
the moles:
- Write a balanced chemical equation.
- Find the number of moles of whatever reactant(s) or product(s)
for which you have enough information to do so.
- Use the mole-to-mole ratio(s) in the balanced chemical equation
to find the moles of whatever you're interested in.
- Once you have moles of what you're looking for, find whatever
quantity you're interested in finding.
This is nicely shown in the flowchart in Figure 4.11.
There are certainly shortcuts, but don't take them. Figure out how to
go through the 4-step process, and just do it. Explicitly write out
mathematical expressions and keep track of the units, make sure the
units cancel. Example 4.9 has a nice example of writing out an
expression and cancelling units. So does Example 4.10. And Example 4.11.
To start thinking about reaction stoichiometry, think about what you
would have to do to plan a chemical reaction. If we want to use 22.99g
of sodium metal to react with water to form sodium hydroxide and
hydrogen gas, how much water do we need to use?
2 Na(s) + 2 H2O(l) --> 2 NaOH(s) + H2(g)
22.99g of sodium metal is 1 mol of Na(s). 2 moles of Na(s) react with 2
moles of water... or 1 mol of Na(s) reacts with 1 mol of water. We just
did stoichiometry.
Yield:
Whenever we perform a chemical reaction, it's nice to evaluate how
successful that reaction was. This is "yield". Yield is how much we
made.
Theoretical yield is the amount of product that could be made if all of
the limiting reactant reacted perfectly to produce product.
To find the limiting reactant, do a stoichiometry problem. If I use all of reactant A to make product, how much product can I make? If I use all
of reactand B to make product, how much product can I make? Whichever
reactant makes less product is the limiting reactant, and the amount of
product you just calculated for the limiting reactant is the
theoretical yield.
"Limiting reactant" and "limting reagent" are slightly different terms for the exact same thing. They are fully interchangeable.
Don't make percent yield into something harder than it is. You have
enough cookie dough to make 28 cookies. You only make 23 cookies
because you accidentally ate some of the cookie dough while you were
baking. Your percent yield is (23/28)*100% = 82%.
Application
- Titration:
Titration is a specific type of stoichiometry problem. Treat it like a
stoichiometry problem. Don't make it into a whole new and different
thing.
Memorization
Notes:
Polyatomic ions
Diatomic elements - nitrogen, oxygen, halogens
Solubility Rules
Common acids & bases
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