Download Chemical Reactions and The Mole

Notes: Chemical Equations, Reaction Types, Atomic Mass, The Mole,
Solubility
Chemical Equations are defined as the symbolic representation of the chemicals involved in a
chemical reaction. The reactants are written first, each reactant is separated by an addition
sign. An arrow symbolizes the chemical conversion to new a substance or substances. On the
other side of the arrow the new substances, the products. are written and again separated by an
addition sign. The parts of a chemical equation are as follows:
1. Reactant chemical(s) and Product chemical(s)
2. States of each of the chemicals
3. Must obey Law of Conservation of Matter – Balanced
4. Arrow may show a catalyst or environmental condition such as pressure
5. Heat may be shown as a reactant or product
The states of matter will be represented by a subscript in parenthesis using the following major
four abbreviations:
gas (g)
solid (s)
liquid (l)
this is to be written as a lower case cursive ‘L’
aqueous (aq) only found in a water medium
Here follows the example of the reaction of methane gas and oxygen gas to yield carbon dioxide
gas and water gas.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
The 2 in front of the O2 and H2O actually means two of those molecules are needed or are
generated in the reaction. The reaction could be written as follows:
CH4(g) + O2(g) + O2(g)  CO2(g) + H2O(g) + H2O(g)
But it makes more sense to combine like terms, which is what has been done in the first
equation.
Another example is:
2 Na(s) + 2 H2O(l)  2 NaOH(s) + H2(g)
Balancing Equations (Stoichiometry):
The law of conservation of mass states that mass can neither be created nor destroyed, it may
only change form. When a chemical reaction occurs, sometimes it seems like mass has been
destroyed, but in fact it has not. When a tree burns, it seems as if the tree disappears, where
does all that mass or stuff go? It is converted into another form. These new forms are gasses.
There is a lot of carbon, oxygen and hydrogen in living organisms. When a living organism
burns, these atoms will now be found in the following molecules H2O and CO2. The remaining
soot is mostly carbon, plain carbon atoms.
C6H12O6(s) + O2(g)  CO2(g) + H2O(g)
C6H12O6 is the general formula for carbohydrates, which is the main component of a tree. The
problem with the above chemical equation is that it is not balanced. We need to have the same
number of each atom on each side of the equation. We must balance the equation.
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
The above equation is now balanced. There are 6 carbons on each side of the arrow. There
are 12 hydrogens on each side and there are 18 oxygens on each side of the arrow.
Dissolving:
The process of dissolving a solid in a liquid is very common, sugar in water, salt in soup are two
obvious examples, but what is the dissolving process? First, let’s define a few terms.
A solvent is the substance in a solution that is present in the greater amount. It is the major part
of a solution and determines the physical properties o f the mixture. The solute is the substance
that is present in the lesser amount. It is the smaller portion of a solution. A solution is a
homogeneous physical mixture of solvent and solute. The solute particle sizes are ions, atoms,
molecules or small combinations of these units.
Let’s say you added some salt to a cup of water. The salt seems to disappear as you stir. How
does this happen? The individual ions are said to be aqueous. An aqueous particle is
completely surrounded by water.
NaCl
Na+ and Cl- ions in
water
The negatively charged chlorine ions are attracted to the positive side of the water molecule, the
hydrogen side. The positively charged sodium ions are attracted to the negative side of the
water molecule, the oxygen side.
Note on the state of matter “aqueous”: This state only occurs for compounds that
have dissolved in water. Meaning the only substance that you can use the (aq)
label with are in water. So, for a substance like CuSO4, if you were to see CuSO4
in a container, it would be blue liquid. But this is not a liquid. The water is a liquid
but the CuSO4 is aqueous. It is dissolved, solvated, surrounded by water
molecules. What I am saying here is that even though the bottle says CuSO4 and
it is a liquid, the CuSO4 is not in liquid form. Most ionic compounds, like CuSO4,
require more than a thousand degrees to melt. Since we will never be using any
heating source to get us anywhere near that temperature, you should not write
any ionic compounds as liquids. One other point, if the substance is an ion, it has
a charge, then it will be aqueous.
Solubility:
 What compounds are soluble in water?
 Here is a nice little list for you.
 The higher rules supercede the lower rules.
1. All Group 1A metals and ammonium compounds are soluble.
2. Acids are soluble.
3. All acetate, perchlorate, chlorate and nitrate compounds are soluble.
4. Silver, lead and mercury(I) compounds are insoluble.
5. Chlorides, bromides and iodides are soluble
6. Carbonates, hydroxides, oxides, phosphates, silicates and sulfides are insoluble.
7. Sulfates are soluble except for calcium and barium.
Describing Reactions in Water:
There are three types of equations used to show you what is going on when a reaction occurs in
water.
Molecular Equation
LiF(aq) + AgNO3(aq)  AgF(s) + LiNO3(aq)
Full Ionic Equation
Li+(aq) + F-(aq) + Ag+(aq) + NO3-(aq)  AgF(s) + Li+(aq) + NO3-(aq)
Net Ionic Equation
Ag+(aq) + F-(aq)  AgF(s)
The molecular equation shows the ions of each chemical combined into a formula. The full ionic
equation shows the ions of each chemical separated into cations and anions, if that compound
is water soluble. The net ionic equation only shows the ions which actually participate in the
reaction. The other ions are simple watching and are named “Spectator Ions”. The spectator
ions are not included in the net ionic equation.
Mass of Atoms & Molecules:
Atomic Mass is the mass in AMUs of a particular atom, the average value of which will be found
on the periodic table. Formula or Molecular Mass is the sum of all the atoms in the molecule.
Some may refer to this as molecular weight, but this is not correct. For example, what if you are
an astronaut doing an experiment on the international space station? Weight does not apply but
mass does.
Calculating the molecular mass of a molecule is a simple process. Add the masses of each
atom in the molecule. Masses are found on the periodic
Chapter 4.2 uses the term weight, I will
table.
not. It says atomic weight in AMU, those
are different parameters (force and matter)
Examples:
H2O
1.008 x 2 + 16.00 x 1 = 18.02 AMU
CO2
12.01 x 1 + 16.00 x 2 = 44.01 AMU
C6H6
12.01 x 6 + 1.008 x 6 = 78.11 AMU
C2H6O
12.01 x 2 + 1.008 x 6 + 16.00 x 1 = 46.068 AMU
C8H18O
12.01 x 8 + 1.008 x 18 + 16.00 x 1 = 130.224 AMU
The Mole:
By definition an AMU is 1/12th the mass of a C-12 atom. The unit of mass needs a reference
point and a specific amount of matter to which all other matter can be referenced. This is a
standard, as C-12 is very abundant, but this mass is very small, too small to work with.
Generally, you will work with quantities of atoms greater than one. A new unit was developed
from work by the Italian Chemist Lorenzo Romano Amedeo Carlo Avogadro. The unit is named
the mole or Avogadro’s number.
This unit is nothing more than a number, a very big number. The mole works no differently than
the following units, a dozen equals 12, a baker’s dozen equals 13 or a gross equals 144.
The mole is simply a number equal to 6.02 x 1023. Written in decimal format it is equal to
602,000,000,000,000,000,000,000. Can you comprehend a number this big? Suppose the
entire population of the world, more than six billion people, were assigned to count the number
of atoms in one mole of gold atoms. If each person counted one atom per second and worked a
48-hour week, the task would take more than 8 million years. If you had a mole of pennies, you
would have enough money to pay all the expenses of the United States for the next billion
years. A mole of the larger marshmallows would cover the United States to a depth of more
than 600 miles.
Why would you need to use a number this big? The mass of atoms is so small that you need
many of them to have a mass significant enough for people to weigh. The mass of an atom is
measured in atomic mass units, AMU. So, scientists picked a number equal to a certain mass
of a certain element. The mass is 12.000 grams and the element is carbon-12. Carbon-12 has
6 protons and 6 neutrons. The masses of all other elements are standardized against the
carbon-12 atom.
The mole’s ease of use is that the mass of one atom of an element in AMUs is also equivalent
to the mass of one mole of that element in grams. So, looking at the periodic table the number
below the symbol has two meanings. Look at helium. The number below the He symbol is
4.00260. This means that the mass of one helium atom is 4.00260 AMU but it also means that
if we have 1 mole of helium atoms those helium atoms would weigh 4.00260 grams.
By definition a mole of carbon 12 atoms has a mass of 12 grams.
If you had 6.02214199 x 1023 carbon 12 atoms, they would have a mass of 12 grams.
Examples:
H2O
CO2
C6H6
C2H6O
C8H18O
1.008 x 2 + 16.00 x 1 = 18.02 g/mol
12.01 x 1 + 16.00 x 2 = 44.01 g/mol
12.01 x 6 + 1.008 x 6 = 78.11 g/mol
12.01 x 2 + 1.008 x 6 + 16.00 x 1 = 46.068 g/mol
12.01 x 8 + 1.008 x 18 + 16.00 x 1 = 130.224 g/mol
Below are few simple conversion examples:
If you are given two moles of iron, its mass would be what?
2 mol Fe
55.85 g Fe
1 mol Fe
= 2 x 55.85 =
1
111.7 g Fe
If you are given 50 grams of sulfur, how many moles would you have?
50 g S
1 mol S
28.09 g S
= 50 x 1
28.09
=
1.78 mol S
How many atoms are found in a 7 gram gold ring?
7 g Au
1 mol Au
197.97 g Au
6.022x1023 atoms Au
1 mol Au
=
7 x 1 x 6.022 x 1023
197.97 x 1
=
2.14 x 1022 atoms Au
Given 2.8 grams of NaCl, calculate the number of moles of NaCl present.
2.8 g NaCl
1 mol NaCl
58.44 g NaCl
= 2.8 x 1 =
58.44
0.048 mol NaCl
The mole, historically, has frustrated many a student. I offer this narrative in hopes of keeping
this frustration from occurring.
The phrase “Less of them” when discussing the amount of a substance is correct, not the
phrase “Less of it”. The difference is in the word “them”. That indicates a number of objects.
The word “it” connotes a size, as in a large object, not a lot of individual objects.
When using the mole the number matters, not the mass. Here is a real world example involving
the HUMV vehicle and its tires. The HUMV weighs 6550 pounds. So, when Chevrolet calls
Firestone and asks for tires they do not tell them the total weight of all the HUMVs they are
making they tell the how many HUMVs they are making. Chevy would not call them and say we
need tires for 32750000 lbs of HUMVs. They would tell them we have 5000 HUMVs. So, what
matters to Firestone is how many HUMVs are being made, that way they know how many tires
to send.
This is all fine and wonderful for Chevy, but in the lab we cannot count up all the atoms or
molecules we will be using in an experiment, we mass the substances out with a balance or
scale. We must then calculate the amount of objects, atoms or molecules, present. Because, in
a chemical reaction the atoms combine in simple whole number ratios to from compounds. That
was Dalton’s theory, and it is correct. When making water, you need 2 hydrogens and 1 oxygen,
that is all that matters. It does not matter how much each atom weighs. Firestone sends 4 tires
and a spare. They don’t care how much the total number of HUMVs weighs, just how many
there are so they know how many tires to send.
Reactions Types:
Decomposition – 1 chemical becomes 2 chemicals
AB+C
Combination – 2 chemicals become 1 chemical
A+BC
Single Replacement – 1 chemical replaces another chemical in a compound
A + BC  B + AC
Double Replacement – cations switch compounds
AB + CD  AD + CB
Combustion – a chemical is combined with oxygen in a rapid reaction.
A + O2  AxOy (+ Z…)
Redox – at least 2 chemicals in a reaction change their oxidation number/charge
 For every oxidation there must be a reduction
 All reactions except double replacement reactions are redox reactions.
Examples:
Decomposition
NH4NO2(s)  N2(g) + H2O(g)
Combination
P4(s) + O2(g)  P4O10(s)
Single Replacement
Zn(s) + CuCl2(aq)  Cu(s) + ZnCl2(aq)
Double Replacement
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
Combustion
C3H8(g) + O2  CO2(g) + H2O(g)
Redox
Ca(s) + S(s)  CaS(s)

Reasoning:
o The reactants Ca and S are both elements in their elemental state.
o This means their oxidation numbers are zero.
o If an element is in a compound it must have an oxidation number.
o So, if the products started with no oxidation number and now have an oxidation number, then this must
be a redox reaction.
o This works the same going the other direction. If two elements are combined in a compound, like H2O,
then are decomposed to H2 and O2. The hydrogen and oxygen used to have an oxidation number and
now do not, therefore a redox reaction occurred.
Stoichiometry Math:
The real usefulness of the mole is that you can calculate the number of grams of a substance
needed in a chemical reaction. The mole is needed because the mole represents a specific
number of atoms or molecules.
If we wanted to react sodium metal and chlorine gas together to make sodium chloride (a
combination reaction) we would want to use equal numbers of sodium atoms and chlorine
atoms; that does not mean equal masses of sodium metal and chlorine gas. We would not
react 10 grams of each as the chlorine is much more massive than the chlorine atom. 10 grams
of sodium would contain many billons more atoms than 10 grams of chlorine.
2 Na(s) + Cl2(g)  2 NaCl(s)
If we use the stoichiometry of the chemical equation and the molar masses of each of the
components in the reaction we can determine the amount of each reactant to add to the reaction
container.
If we start with 10 grams of sodium metal we would need to follow the next few steps to
determine how much chlorine to add to the sodium metal.
1 mol Na 1 mol Cl 2 70.906 g Cl 2
 15.42 g Cl 2
22.9898 g Na 2 mol Na 1 mol Cl 2
The use of mole in stoichiometry calculations follows a predictable pattern. You begin with a
mass in grams of a substance. You covert that substance’s mass into moles. You next use the
moles of the first substance and convert them into moles of the desired substance. Finally
convert the moles of the desired substance into grams of the desired substance.
10 g Na
The following can be used as a template for practically all stoichiometry calculations.
given mass in grams of A
1 mol A
Y mol B molar mass in grams of B
 mass in grams of B
molar mass in grams of A X mol A
1 mol B
If we use the above calculation with sodium and chloride, we begin with a given mass of A, 10
grams of sodium. We convert the mass of A into moles of A, 22.9898 g of Na per mole of Na.
Next we convert X moles of A into Y moles of B, 2 moles of Na into 1 mole of Cl 2. We finish by
converting moles of B into grams of B, 1 mole of Cl2 equals 70.906 grams of Cl2.
Limiting Reagent, and the like, Problems:
When two or more substances react, one of these substances will run out first. Just as when
you make a bunch of apple pies you are bound to run out of apples or pie crust, leaving leftover
apples or crust. You may be able to fudge this in the baking world but in the atomic world you
cannot.
There is no way around running out of one substance first. No matter how accurate you
measure, we do not have scales that have 23 decimal places, and even if we did, if you add one
grain to such a scale you would change the mass to the 6 place simply by adding a single grain
of sand.
So, whichever substance runs out first, that is your limiting reagent. These types of
stoichometric problems are very common to everyday life. If you are making steel you need to
know how much carbon to add to every ton of steel or your steel will not be as strong as you
need it to be or may be more brittle than you desire.
For instance, if you mix sodium metal and chlorine gas, you make table salt. You would not
want to have any sodium or chlorine left over, so you would want your measurements to be very
accurate.
There are three ways these problems work. Either you are given the amounts of each reactant
and asked how much product you could make. Or you are given one reactant and asked how
much of the other reactant(s) you need to complete the reaction. Or you are told you need to
make this much product, how much of each reactant will you need.
See the following examples:
Given 87.5 grams of silver and 75.0 grams of bromine liquid, how much silver bromide could be
made? Remember bromine is one of the 7 diatomics.
First you must write a balance chemical equation:
2 Ag(s) + Br2(l) → 2 AgBr(s)
87.5 g Ag
1 mol Ag
2 mol AgBr
107.87 g Ag 2 mol Ag
187.78 g AgBr
1 mol AgBr
= 152.31 g AgBr
This calculation must be completed twice, once for each reactant.
75.0 g Br2
1 mol Br2
159.8 g Br2
2 mol AgBr
1 mol Br2
187.78 g AgBr
1 mol AgBr
= 176.24 g AgBr
Given these two calculations, we can say the limiting reagent is silver and the most AgBr that
could be made with 87.5 g of Ag and 75.0 g of Br2 would be 152.31 grams. We could not make
176.24 grams because we would run out of silver once 152.31 grams were made.
Another example, if you wanted to make 58 grams of copper (II) oxide, how much copper and
oxygen would you need?
2 Cu(s) + O2(g) → 2 CuO(s)
58.0 g CuO
1 mol CuO
79.54 g CuO
2 mol Cu
2 mol CuO
63.54 g Cu
1 mol Cu
= 46.33 g Cu
58.0 g CuO
1 mol CuO
79.54 g CuO
1 mol O2
2 mol CuO
32 g O2
1 mol O2
= 11.67 g O2
So, to make 58 g of CuO you need 46.33 grams of Cu and 11.67 g of O2.
Predicting Products:
The ability to predict what products will result by mixing substances together is a very important
skill a chemist learns. We will discuss only the most basic of these predictions. For now, we will
only deal with single and double replacement reactions.
When dealing with a double replacement reaction, you simply swap our the cations, the
positively charge atoms, basically, the metals. But you must remember, double replacement
reactions are not redox reactions, so the oxidation number for each element must be the same
on each side of the equation.
Example:
MgCl2(aq) + AgNO3(aq) 
If you swap out the metals, Mg and Ag, you get the following
AgCl(s) + Mg(NO3)2(aq)
Notice the charges for all are the same, meaning not redox.
In the single replacement reactions, you will take the metal, sitting as an element, by itself and
replace it with the metal in the compound.
Example:
ZnCl2(aq) + Ca(s) →
If you swap out the metals, Zn and Ca, you get the following:
CaCl2(aq) + Zn(s)
Notice the charges for all are different, meaning this is a redox
reaction. Zn had a +2 charge and now it has a 0 charge. Ca
had a 0 charge and now has a +2 charge.