Mole Concept and Stoichiometry
The mole concept and the closely related concept of stoichiometry are the hardest concepts for most students to
fully understand.
The Mole
The “mole” is the concept of measuring the amount of a substance and it is a fundamental unit in the système
international d'unités (International System of Units – the modern form of the metric system). It is defined as the
number of representative particles in exactly 12.0 grams of C-12. So by this definition, the mass of one mole
of carbon-12 must be exactly 12.0 grams.
Representative Particles
The term “representative particles” is used to describe the particles or entities that represent the smallest unit that
has all of the properties of that “group” – examples would be:
Large Group
geese
mice
people
VW Beetle
Representative particle
goose
mouse
person
a single car
When counting VW Beetles, you do not count wheels, or windows, because they do not have the same
“properties” of a whole VW Beetle. You must count a whole car, same with counting geese, you do not count
feathers or beaks – you must count the whole bird (and that is a goose).
For chemistry, there are three representative particles that we normally use in mole conversions and one other
unit that we use in electrochemistry and oxidation-reduction problems. Again, the representative particles depend
on what is being “counted” and you have to be absolutely sure as to the term used – it helps with the AP questions.
 for elements the representative particles  atoms
 for ionic compounds the representative particles  formula units
 for covalent compounds (also called molecular compounds) the representative particles  molecules
When dealing with electrochemistry and oxidation-reduction problems, the moles of electrons are used in the
calculations; the representative particle for electrons is (drum roll please) an electron.
Substance
Iron
Water
Table salt
Oxygen gas
Mole and Stoichiometry Packet
Representative particles for various substances
Representative particle
Atom
Molecule
Formula unit (and a formula unit is one grouping of NaCl)
Molecule (remember that oxygen is diatomic and is O2, therefore
must be a molecule and not an atom.
Page 1
The number of representative particles in exactly 12.0 grams of carbon-12 (in this case, the representative particles
are atoms, since carbon is an element) is 6.0221415 x 1023 (on the AP Chemistry formula charts, it is written as
6.022 x 1023). That number is referred to as Avogadro’s Number (NA) in honor of the nineteenth century Italian
scientist Amedeo Avogadro. It was Avogadro, in the early 1800’s that first proposed that the volume of a gas is
proportional to the number of particles of gas (either atoms or molecules – again that idea of representative
particles and what you are counting). Since gas volume changes with temperature and pressure, the gases have
to be at the same pressure and temperature for the comparison to be made.
Molar Mass
There are two definitions that are closely related in chemistry, atomic mass units and that of a mole. They are
related so that the following connections can be made. The two definitions are:
 Atomic mass unit is exactly 1/12th the mass of a carbon-12 nucleus.
 A mole is the number of particles in exactly 12.0 grams of C-12.
If you want to see how they are related, read the table below. If not, skip below the table to the paragraph
at the top of the next page. Let’s see how they are related.
Atomic Mass Units
Definition: An atomic mass unit is exactly
1/12th the mass of a carbon-12 nucleus
An atom of carbon-12 has:
 6 protons
 6 neutrons
 6 electrons
The atomic mass of C-12 is 12 amu.
Atomic mass = #p+ + #no (since electrons have
relatively no mass)
So: 12 amu = 6p+ + 6no
Since the mass of a proton and the mass of a
neutron are the same, we can substitute protons
for neutrons (or vice versa) in the equation
above and get:
12 amu = 6p+ + 6p+
1 amu = 1 p+
Mole and Stoichiometry Packet
Moles
Definition: A mole is the number of particles
in exactly 12.0 grams of C-12
An atom of carbon-12 has:
 6 protons
 6 neutrons
 6 electrons
A mole of carbon-12 has:
 6 moles protons
 6 moles neutrons
 6 moles electrons
What is the mass of one mole of C-12? By the
definition, one mole of carbon-12 has to have a
mass of 12 grams.
Molar mass = #p+ + #no (since electrons have
relatively no mass)
So: 12 grams = 6 moles p+ + 6 moles no
Since the mass of a proton and the mass of a
neutron are the same, we can substitute protons
for neutrons (or vice versa) in the equation
above and get:
12 grams = 6 moles p+ + 6 moles p+
The mass of a proton is 1 amu, so
12 grams = 6 mol (1 amu) + 6 mol (1 amu)
12 grams = 12 moles of amu’s
1 gram = 1 mole of amu’s
Page 2
So what does this all mean? The definitions of atomic mass units (amu) and moles are related, so as to allow
the periodic table to show both the average atomic mass and the molar mass of every element. The average
atomic mass of chlorine is 35.45 amu, while the molar mass is 35.45 grams. Everything that is done with
isotopes and atomic masses, applies to moles and molar masses.
The definition of molar mass is the mass of one mole of a substance.
 Units are g/mol
 Numerically equal to the average atomic weight in atomic mass units.
 Specific names depend on the representative particles.
 atoms = gram atomic mass
 ionic compounds = gram formula mass
 covalent compounds = gram molecular mass
It is common for AP free response questions to ask for the students to find molar mass of an unknown
compound. This is especially true with the lab based question.
Year
1983
1978
1989
1979
1973
How was problem stated?
The molecular weight of a monoprotic acid HX was to
be determined.
Calculate the molecular weight of the acid HA.
In an experiment to determine the molecular weight and
the ionization constant for ascorbic acid (vitamin C)
In a laboratory determination of the atomic weight
What minimum data are needed to determine the
molecular weight
What does this tell you?
Molecular weight =
covalent compound
Molecular weight =
covalent compound
Molecular weight =
covalent compound
Atomic weight = element
Molecular weight =
covalent compound
Molar Volume
As stated above, it was Avogadro which first proposed that the volume of a gas is proportional to the number
of particles of gas at constant temperature and pressure. Since gas volume changes with temperature and
pressure, the pressure and temperature have to be stated for the comparison to be made. The temperature and
pressure that has been agreed upon is called Standard Temperature and Pressure (STP). Standard temperature
is 0OC (273.15 K) and 1atmosphere of pressure (1 atm). If and only if you are at those conditions can
you use the value for molar volume! Molar volume is the volume occupied by one mole of ANY gas
at STP; that volume is 22.4 liters.
Avogadro used his law with equal volumes of gases under the same conditions and equal numbers of
molecules to determine how many particles were in a mole.
Students often see “volume” or “liters” in a problem and jump to using 22.4 liters! You must have two
things in order to use molar volume.
1. Must have a gas!
2. Must be at STP!
Mole and Stoichiometry Packet
Molarity and molar does not
mean a gas! You cannot use
22.4 liters in the problem!
Page 3
Mole Diagram
The mole diagram is designed to help students visualize how to do the conversions. If you are asked to go
from grams to moles, that is a one step conversion – going from the “mass in grams” to the “moles” in the
middle. If you are asked to convert liters @STP to atoms of He, then that is a two step problem: first is to
convert from liters (at the top) to moles in the middle, then convert from the moles to the representative
particles at the bottom.
Mole and Stoichiometry Packet
Page 4
Examples
Find the number of moles in 35.78 grams of (NH4)2SO4.
35.78 grams (NH 4 ) 2 SO 4 x
1 mol (NH 4 ) 2 SO 4
 2.707x10 1 moles (NH 4 ) 2 SO 4
132.17 grams (NH 4 ) 2 SO 4
To do this problem, you need to first find the molar mass of (NH4)2SO4:
Element in
the formula
Nitrogen
Hydrogen
Sulfur
Oxygen
Number of moles of the
element in the formula
2
8
1
4
Molar Mass
(periodic table)
14.01
1.01
32.07
16
Add them up
Final value
28.02
8.08
32.07
64
132.17 grams
Find the volume of 4.65 x 1024 atoms of He @ STP.
4.65 x 10 24 atoms He 
1 mol of He
22.4 liters

 1.73 x 10 2 liters
23
6.02 x 10 atoms of He 1 mol He
Find the number of representative particles in 2.76 moles of HCl.
2.76 mol HCl 
6.02 x 10 23 formula units HCl
 1.66 x 10 24 formula units
1 mol HCl
Find the moles of 38.9 grams of oxygen gas.
38.9 grams O 2 
1 mole O 2
 1.22 x 10 0 mol O 2
32.00 grams O 2
Hardest problem: Find the number of atoms of oxygen in 343 grams of ammonium oxalate.
343 grams (NH 4 ) 2 C 2 O 4 
1 mol (NH 4 ) 2 C 2 O 4
6.02 x 10 23 formula units (NH 4 ) 2 C 2 O 4
4 atoms of oxygen

x
124.12 grams (NH 4 ) 2 C 2 O 4
1 mol (NH 4 ) 2 C 2 O 4
1 formula unit (NH 4 ) 2 C 2 O 4
 6.65 x10 24 atoms of oxygen
Whenever you need to break up a formula into smaller parts, always place the
number “1” in front of the most complicated formula and then determine the other
coefficient.
Mole and Stoichiometry Packet
Page 5
Determination of the formula of a compound
Percent Composition
Percent Compositio n 
mass of desired element (or ion)
x 100
total mass
Percent composition is a very useful tool to determine the amount of an element in a compound. Since it is
based on the total mass (and we use the molar mass for the total mass) the percent composition of an element
in a compound is independent of the amount of that compound.
Example: Find the percent composition for nitrogen in (NH4)2SO4.
To do this problem, you need to first find the molar mass of (NH4)2SO4:
Element in
the formula
Nitrogen
Hydrogen
Sulfur
Oxygen
Number of moles of the
element in the formula
2
8
1
4
Percent Compositio n 
Molar Mass
(periodic table)
14.01
1.01
32.07
16
Add them up
Final value
28.02
8.08
32.07
64
132.17 grams
28.02 grams
x 100  21.20%
132.17 grams
If the problem asked for the amount of grams of nitrogen in 402 grams of (NH4)2SO4 you can do that
problem two ways. The first way is using mole conversions.
402 grams (NH 4 ) 2 SO 4 
1 mol (NH 4 ) 2 SO 4
2 mol N
14.01 grams N


 85.2 grams N
132.17 grams (NH 4 ) 2 SO 4 1 mol (NH 4 ) 2 SO 4
1 mol N
This problem can also be done using percent composition. 21.20% of (NH4)2SO4 is nitrogen, so 21.20% of
402 grams of (NH4)2SO4 must give you grams of nitrogen.
21.20% x 402 grams (NH4)2SO4 = 0.2120 x 402 = 85.2 grams
Notice that the numbers are exactly the same if you look at how the 21.20% was calculated:
Percent Compositio n 
2(14.01) grams
x 100  21.20%
132.17 grams
Look at the mole conversion, all of the numbers are located in the exact same place, 402, 2 and 14.01 on the
top and the 132.17 on the bottom.
Mole and Stoichiometry Packet
Page 6
Empirical and Molecular Formulas
There are two formulas that are asked for on the AP exam, empirical and molecular formulas. They are related
to each other and sometimes the problems ask for one or the other or both. There are two ways to determine
the molecular formulas and only one way to get the empirical formula. Since a formula, such as (NH4)2SO4
relates the number of moles of each element present, the molecular and empirical formulas deal with the moles
of each element in the compound.
 empirical formula
 the lowest whole number ratio of atoms in a formula
 generally used to determine ratio of elements in a compound
 can be used to determine the molecular formula
 two common problems for determining the data needed to find the empirical formula
 from percent composition
 combustion analysis
 Molecular formula
 the formula showing the actual ratio of atoms in a formula
 usually found from the empirical formula
 the molar mass of the actual compound is usually determined by laboratory data
 Find the grams of the compound (given to you in the problem) and use the other information
in the problem to do stoichiometry to determine the moles of the compound.
 Molar mass is grams per mole so take the grams and divide by the moles.
 There are three common methods in the laboratory to determine the molar mass:
 colligative properties
 titration (usually acid-base)
 gas laws
 Finding empirical and molecular formulas:
 find empirical formulas
 if given grams of each element, start at step three of the percentage problems
 for percentage problems
 step one: assume a 100 gram sample
 step two: for each element multiply the percentage by the mass (100g) to give you grams
 step three: convert all grams into moles
 step four: divide by smallest number of moles
 step five: clear fractions by multiplication each moles by the denominator of the fraction
(if needed)
 find molecular formulas
 If given the molecular formula’s molar mass
 Use the empirical formula’s molar mass and see how many times it goes into the
molecular formula’s molar mass. Must be a whole number of times (2x, 3x etc.)
 Can use the molecular formula’s molar mass to find molecular formula directly.
 May have to determine the molecular formula’s molar mass from laboratory data
There are two common ways to find the molecular formula…find it directly or use the empirical formula.
Many times on the AP free response questions, you will have to solve a problem to determine the molecular
Mole and Stoichiometry Packet
Page 7
formula’s molar mass. If you want to solve this first and then find the molecular formula that is fine or you
can find the empirical formula and then find the molecular formula.
Examples from AP Free Response Questions
Problem involving combustion analysis:
(a) A 0.7549 g sample of the compound burns in O2(g) to produce 1.9061 g of CO2(g) and 0.3370 g of
H2O(g).
(i)
Calculate the individual masses of C, H, and O in the 0.7549 g sample.
(ii)
Determine the empirical formula for the compound.
Answer:
You start out with an equation:
CxHyOz + O2  CO2 + H2O
Our job is to find the x, y and z. Using the Law of Conservation of Mass, we know that all of the carbon in
CO2 came from the original compound and that all of the H in H2O came from the original compound. So
we need to find the C in the CO2 and the H in the H2O and we can do this by either mole conversions or by
percent composition. As good review, we are going to find the amount of H by mole conversion and the
amount of C by percent composition.
(a i) 0.3370 g H 2 O 
1 mole H 2 O 2 moles H 1.01 g H


 0.03744 g H
18.02 g H 2 O 1 mole H 2 O 1 mole H
Percent Compositio n 
12.01 grams
x 100  27.28%
44.01 grams
27.28% of 1.9061 g CO2  0.5198 g of C
0.7549 g sample – (0.5198 g C + 0.0374 g H) = 0.1977 g of Oxygen
1 mole C
 0.04322 moles C
12.01 g C
1 mole H
 0.0370 moles H
(a ii) 0.0374 g H 
1.01 g H
1 mol O
0.1977 g O 
 0.01236 mol O
16.00 grams O
0.5198 g C 
0.04322 mole C
Divide by
the smallest
number of
moles
0.01236
 3.5 moles C
 3 moles H
0.01236
0.01236 mol O
 1 moles O
0.01236
0.0370 mol H
Since 3.5 moles of C is not a whole number, take the denominator of the fraction (0.5 = ½) and multiple each
moles by 2 and get 7 C : 6 H : 2 O, thus an empirical formula of C7H6O2. Let’s say that the problem said
that the molecular formula’s molar mass was determined to be 244 g/mol (more likely is that you are going to
have to solve for this). The empirical formula has to be a “reduced” version of the molecular formula, so we
can go back and put in the factor that was taken out to get the “reduced” empirical formula.
C7H6O2 = EF = 122 g/mol
X2
X2
C?H?O? = EF = 244 g/mol
Mole and Stoichiometry Packet
First find the empirical formula’s molar mass (in this case 122 g/mol).
Next, how many times does the EF’s mass go into the MF’s mass?
Answer here is two times. So multiple the EF by two for the number of
atoms and get the molecular formula to be C14H12O4.
Page 8
Problem involving percentages:
An unknown compound contains only the three elements C, H, and O. A pure sample of the compound is
analyzed and found to be 65.60 percent C and 9.44 percent H by mass. Determine the empirical formula of
the compound.
Answer:
All three percentages have to add up to be a 100, so 65.60 %C + 9.44 %H + X %O = 100, so 24.96% O.
Step one: Assume a 100 gram sample.
Step two: Then 65.60% of 100 grams = 65.60 grams of C (likewise you have 24.96 g O and 9.44 g H)
Step Three: Convert to moles
1 mole C
65.60 g C 
 5.462 moles C
12.01 g C
1 mole H
9.44 g H 
 9.37 moles H
1.01 g H
1 mol O
24.96 g O 
 1.56 mol O
16.00 g O
Step Four: divide by smallest number of moles
5.462 mole C
 3.5 moles C
1.56
9.37 mol H
 6 moles H
1.56
1.56 mol O
 1 moles O
1.56
Step Five: clear fractions by multiplication of denominator of fraction (if needed)
3.5 moles C  2  7 moles C
6 moles H  2  12 moles H
1 moles O  2  2 moles O
Empirical formula = C7H12O2
Finding the molecular formula
We are going to use the information above to illustrate the two methods of finding the molecular formula.
Again, we will assume that the problem said that the molecular formula’s molar mass was determined to be
256 g/mol. So to find it from the empirical formula:
C7H12O2 = EF = 128 g/mol
X2
X2
C?H?O? = EF = 256 g/mol
First find the empirical formula’s molar mass (in this case 128 g/mol).
Next, how many times does the EF’s mass go into the MF’s mass?
Answer here is two times. So multiple the EF by two for the number of
atoms and get the molecular formula to be C14H24O4.
NOTE: If the EF’s mass does not go evenly into the MF’s mass, you made a
mistake.
Mole and Stoichiometry Packet
Page 9
To find the molecular formula directly you use the same process you did for finding the empirical formula.
However, to find the molecular formula directly, you will have to do more work at the beginning of the problem
to get the grams of each element, but there will be no work at the end of the problem: you will not get a fraction
and you do not have to divide by the smallest number.
The original problem for the last example was a four part problem…only the first part was shown above. Part
(a) and part (b) are below.
An unknown compound contains only the three elements C, H, and O. A pure sample of the compound
is analyzed and found to be 65.60 percent C and 9.44 percent H by mass.
(a) Determine the empirical formula of the compound.
(b) A solution of 1.570 grams of the compound in 16.08 grams of camphor is observed to freeze
at a temperature 15.2 Celsius degrees below the normal freezing point of pure camphor.
Determine the molar mass and apparent molecular formula of the compound. (The molal
freezing-point depression constant, Kâ, for camphor is 40.0 kg•K•mol-1.)
We have already done the work for part (a) above and shown how to find part (b) from part (a) by given you
the molar mass that would be found from part (b).
However, on the AP exam, you can answer the question in any order you would like, so the second method to
finding the molecular formula involves doing part (b) first (finding the molar mass), then using the percentages
to find the molecular formula and finally using that to find part (a).
Again, the answer for the molar mass from part (b) is 256 grams/mol.
We are going to assume that we have a one mole sample of this compound – that means that we have a sample
mass of 256 grams. So using that information, we need to find the amount of each element in that compound.
65.60% of 256 grams = 167.9g C
9.44% of 256 grams = 24.2g H
24.96% of 256 grams = 63.9g O
Now convert the grams into moles.
1 mole C
 13.98 moles C
12.01 g C
1 mole H
24.2 g H 
 23.96 moles H
1.01 g H
1 mol O
63.9 g O 
 3.99 mol O
16.00 g O
167.9 g C 
}
This will give you the molecular formula automatically,
so there is no dividing by the smallest number. The 13.98
moles of C is 14 moles, 23.96 moles H becomes 24 moles
and the 3.99 moles of O becomes 4 moles. So the
molecular formula is C14H24O4.
Once you have the molecular formula, C14H24O4, just reduce it to lowest terms and get the empirical formula.
Since the numbers are 14, 24 and 4, we have a common factor of 2, so they would become 7, 12 and 2 and the
empirical formula becomes C7H12O2.
Mole and Stoichiometry Packet
Page 10
HYDRATES
A hydrate is a compound that has water attached to it. It is common for the AP exam to ask questions about
hydrates, usually it is to determine the amount of water that the hydrate has. It is nothing more than an
empirical formula problem. A dot is used to state how many water molecules are attached (but not chemically
combined) to the crystal. A formula might look like CuSO45H2O.
 Terms
o How to name them
 uses prefixes to state number of water then put it in the form of compound’ name
followed by the prefix for the amount of water attached to the word “hydrate”
 copper(II) sulfate pentahydrate
o anhydrous – without water
o water of hydration – amount of water attached to the hydrate
Usually the problem will give enough information to determine the grams of the water of hydration and the
grams of the anhydrous. Once you have the grams of each one, follow the steps to find the empirical
formulas. The only difference is that you will be using formulas and not elements.
Example:
When 1.357 grams of CoCl2x H2O is heated, 1.062 grams of cobalt(II) chloride is left behind. Find the
value for “x”.
1.357 grams of CoCl2x H2O – 1.062 grams CoCl2= 0.295 grams H2O.
0.295 g H 2 O x
1 mol H 2 O
 0.0164 mol H 2 O
18.02 g H 2 O
1.062 g CoCl2 x
1 mol CoCl2
 0.00818 mol CoCl2
129.83 g CoCl2
Divide by
the smallest
number of
moles
0.0164 mol H 2 O
2
0.00818
0.00818 mol CoCl2
1
0.00818
Empirical formula is CoCl22 H2O
CONCENTRATION
Reactions that take part in solutions are very common in chemistry and will be found on the AP exam. For
solutions, it is important to remember that there are two components: solute and solvent. The solute is the
substance that is dissolved and the solvent does the dissolving. The solute is always in the smaller amount.
There are several methods to describe the amount of solute that is dissolved in a solution. One method is to
use the amount of solute in the solvent expressed as a percentage. Since the amount of solute and solvent can
be expressed as either mass or volume, you have four common percentages that are listed below. In each case
it is proper to have a notation as to how the calculations where done. This is written as “m/m”, “m/v”, “v/v”
and “v/m” with the first letter telling you how the solute was measured (either mass of volume) and the second
letter indicates how the solvent was measured. The percentages are listed below:
Mole and Stoichiometry Packet
Page 11

percent by mass (so common that any percent is assumed to be percent by mass if nothing is
indicated, but to be perfectly clear, the notation of “m/m” should be written)
grams of solute
x100
100
grams
of
solution
o Defined as

percent by volume (written “v/v” after the percentage)
milliliter s of solute
x100
o Defined as 100 milliliter s of solution
percent mass – volume (written “m/v” after the percentage)
grams of solute
x100
o Defined 100 milliliter of solution

If the percentages are very small, then they can be expressed as “ppm”, “ppb” or “ppt” – parts per million,
billion or trillion, respectively. You usually see this with pollution and other environmental hazards.
Other common methods are mole fraction, mole ratio and molarity (also called molar concentration). Each is
defined below.

mole fraction (written as Xi)
Xi 


o Defined as
mole ratio (written as ri)
o Defined as
molarity
ri 
moles of substance " i"
total number of moles of everything
moles of subtance " i"
total number of moles of everything ELSE (don' t count " i" here)
moles of solute
o Defined as liter of solution
Molarity is the most common unit for concentration encountered on the AP chemistry exam. It is so often
used that chemists have a shorthand notation for it. Anything that is written inside square brackets [ ] means
the molar concentration (or molarity). The concept is called “molarity”, the unit is “M” and “M” is pronounced
as “molar”, so “3.0 M HCl” is read as “three molar hydrochloric acid”. So [MgCl2] = 0.5 M is read as “the
molar concentration of magnesium chloride is 0.5 molar” and can be thought of as:
(a) [MgCl2] = 0.5 M
(b) [Mg2+] = 0.5 M
(c) [Cl-1] = 1.0 M
(i) since MgCl2 has two Cl- ions,
0.5 moles of MgCl 2
2 moles of Cl1
1.0 moles of Cl1


1 moles of MgCl 2
liter of solution
(ii) [MgCl ] = 0.5 M  liter of solution
2
This is very common on the
multiple choice part of the
AP exam. Simple numbers
– remember no
calculators!!!
Mole and Stoichiometry Packet
Page 12
STOICHIOMETRY
Stoichiometry is the study of the quantitative relationship between the amount of reactants and products. There
is a simple graphical approach to setting up the information in the problem and planning your approach to
solving the problem. First you must start with a balanced chemical equation.
Graphical approach to solving stoichiometry problems
Remember that the coefficients
 Place the information in the problem.
of a balanced equation relate the
 Balance equation on first line.
number of moles of each
substance that reacts together
 Moles go on the second line.
and the moles of the products
 Anything that can be changed into moles on the third line.
that are formed. It is important
 Grams
that you MUST start with a
 molarity
balanced equation.
 representative particles
 liters @ STP
 Use a question mark (?) to identify the unknown.
 Plan how to solve the problem.
 Use arrows to plan to do the problem and to guide you in your conversions.
 Draw from you starting information
 Must be in moles to go from side to side in a balance equation.
 Draw arrows down if need be.
Balanced
+ O2 
2 H2
equation
here
2 H2O
Moles are placed on this line.
Anything that can be changed into moles goes here.
A typical problem should look like this:
Problem: When 150.0 grams of TiO2 is reacted with BrF3, find the number of liters of oxygen gas at STP, the
number of molecules of bromine, the number of moles of titanium(IV) fluoride and the grams of BrF3 required
for complete reaction.
Your setup should look like this:
TiO2
+
BrF3

TiF4
+
Br2
+
O2
?
150.0 g
?g
?
molecules
?
L@STP
Draw the arrows to help you with the problem, we will draw only one set at a time, however, on your problems,
you can draw all of the sets at the same time. The only time that you can go side to side is when you are in
moles. A balanced chemical equation relates the moles of each species in the equation. DO NOT FORGET
Mole and Stoichiometry Packet
Page 13
TO BALANCE THE EQUATION.
3 TiO2
+
4 BrF3

3 TiF4
+
2 Br2
+
3 O2
?
150.0 g
?g
?
molecules
?
L@STP
Anytime that you are going up or down in the diagram, the number in front of moles will ALWAYS be one.
You are using the three definitions of (i) mole, (ii) molar mass and (iii) molar volume. Since all three of these
definitions are based on one mole, the number in front of moles has to be one.
Your problem would be solved like this:
150.0 grams TiO 2 x
3 TiO2
+
4 BrF3
1 mole TiO2
2 moles Br2
6.02 x10 23 molecules Br2

x
x
79.9 grams TiO 2 3 moles TiO 2
1mole Br2

3 TiF4
+
2 Br2
+
7.534 x 1023
3 O2
?
150.0 g
?g
150.0 grams TiO 2 x
3 TiO2
+
?
molecules
?
L@STP
4 moles BrF3 136.90 grams BrF3
1 mole TiO 2
x
x
 342.7grams BrF3
79.9 grams TiO 2 3 mole TiO 2
1 moles BrF3
4 BrF3

3 TiF4
+
2 Br2
+
3 O2
?
150.0 g
?g
150.0 grams TiO 2 x
?
molecules
?
L@STP
1 mole TiO 2
3 moles TiF 4
x
 1.877moles TiF 4
79.9 grams TiO 2 3 mole TiO 2
Mole and Stoichiometry Packet
Page 14
3 TiO2
+

4 BrF3
3 TiF4
+
2 Br2
+
3 O2
?
150.0 g
?g
150.0 grams TiO 2 x
?
molecules
?
L@STP
1 mole TiO 2
3 moles O 2
22.4 liters O 2
x
x
 42.05 liters O 2
79.9 grams TiO 2 3 mole TiO 2
1 moles O 2
Practice Problem #1
How many grams of water can be made from 6.0 grams of hydrogen, and how many moles of oxygen are
needed for the reaction to go to completion?
H2
+
O2 
H2O
Now…set up the problem and then solve the problem – without a calculator (the numbers are easy!).
Mole and Stoichiometry Packet
Page 15
LIMITING REAGENTS
 limiting reagent
 The limiting reagent determines or limits the amount of product formed.
 At least two amounts are given in the problem (may be more than two, but on the AP test, it has only
been two – does not mean that it could not be more than two on the AP exam).
 Once identified, ALL calculations MUST be based on this reactant.
 excess reagent
 The reagent that is left over…have more than enough of it.
 Uses up all of the limiting reagent.
 Easiest way to determine this type of problem is to change each starting amount to moles (or grams –
depends on what the problem is asking) of the product that you are looking for.
 Smaller amount comes from the LIMITING REAGENT and is your answer.
 MAKE SURE THAT YOU ANSWER THE QUESTION CORRECTLY…THE PRODUCT IS
NOT THE LIMITING REAGENT.
 Read the problem before you start and see what product they are asking about…and then go find that
product.
You can always tell this type of problem by setting up the problems with the three lines. Once you have two
starting amounts on the diagram…you know that it is a limiting reagent problem – see example below.
4 Fe(s) + 3O2(g)  2 Fe2O3(s)
42 g
75 g
?g
The two starting amounts tell you that it is a limiting reagent problem. To solve this problem, talk both starting
amounts and find the grams of the Fe2O3. The one with the smallest answer will be the correct value for the
amount of Fe2O3.
75 g O 2 
1 mol O 2 2 mol Fe 2 O 3 159.70 grams Fe 2 O 3


 250 grams Fe 2 O 3
32.0 g O 2
3 mol O 2
1 mol Fe 2 O 3
42 g Fe 
1 mol Fe 2 mol Fe 2 O 3 159.70 grams Fe 2 O 3


 60. grams Fe 2 O 3
55.85 g Fe
4 mol Fe
1 mol Fe 2 O 3
Since the 60. grams is smaller than the 250 grams, the 60. grams of Fe2O3 is the amount that is produced.
Since the 60. grams of Fe2O3 is produced from the 42 grams of Fe, the Fe must be the limiting reagent and
the 75 grams of oxygen gas must be the excess reagent.
Mole and Stoichiometry Packet
Page 16
Part of an AP question – Practice Problem #2
CH4(g) + 2 Cl2(g)  CH2Cl2(g) + 2 HCl(g)
Methane gas reacts with chlorine gas to form dichloromethane and hydrogen chloride, as represented by the
equation above.
(a) A 25.0 g sample of methane is placed in a reaction vessel containing 2.58 mol of Cl2(g).
(i) Identify the limiting reactant when the methane and chlorine gases are combined. Justify your answer
with a calculation.
(ii) Calculate the total number of moles of CH2Cl2(g) in the container after the limiting reactant has been
totally consumed.
CH4 + 2 Cl2  CH2Cl2 + 2 HCl
2.58
?????
25.0 g
NOTE: You can so the questions in any order that you want so this approach answer
part (ii) first and then you answer part (i).
Mole and Stoichiometry Packet
Page 17
Part of an AP question – Practice Problem #3
4 Fe(s) + 3O2(g)  2 Fe2O3(s)
Iron reacts with oxygen to produce iron(III) oxide as represented above. A 75.0 g sample of Fe(s) is mixed
with 40.0 grams of oxygen gas.
(a) Calculate the number of moles of each of the following before the reaction occurs.
(i) Fe(s)
(ii) O2(g)
(b) Identify the limiting reactant when the mixture is heated to produce Fe2O3. Support your answer with
calculations.
(c) Calculate the number of moles of Fe2O3 produced when the reaction proceeds to completion.
Mole and Stoichiometry Packet
Page 18
Percent Yield
Percent yield is a measurement of the efficiency of a chemical reaction.
 theoretical yield
 This is what you calculate from stoichiometry.
 This is fantasy land! Most reactions will not give you 100% of the product that you expect.
 actual yield
 What you get in the experiment or laboratory investigation (what happens in the real world).
 Remember, no chemical process is 100% efficient.
 Therefore no yield will be 100% (useful for lab reports!)
 if more than 100%  contamination of some kind
Percent yield =
actual yield (real)
theoretical yield (ideal)
x 100
You can use a magic circle if it helps you.
AY
%Y TY
Laboratory free response questions on the AP exam may give you the information in a data table. The data
table should give you information that will allow you to determine the actual results and the starting amount
of the limiting reactant – usually in the lab it is clearly stated which reagent is the limiting reagent. From that
information, you can do the stoichiometry to determine the theoretical yield (the maximum amount possible)
and then determine the percent yield.
Reagent versus reactant
Although most people consider them interchangeable (and in most cases they are) there is a very subtle
difference between the two. The AP exam does not care which one you use.
Reactant: Any substance that participates in a chemical reaction, especially one that is present at the initiation
of that reaction.
Reagent: A chemical that generally is added to a substance to produce a specific chemical change or to detect
a specific chemical.
Mole and Stoichiometry Packet
Page 19
Acid-base and Redox Titrations Terms



Titration:
o A procedure in which one solution is added to another solution until the chemical reaction
between the two reactants (solutions) is complete; the concentration of one solution is known
and that of the other is unknown.
Titrant:
o the solution in a titration that is added from a burette to a measured quantity of another solution.
You are trying to find the concentration of the titrant.
Standardized:
o A solution in a titration that is of known concentration.
o Primary Standard:
 A substance of known purity that can be weighed directly and used to standardize other
solutions.
o Secondary Standard:
 A solution whose concentration has been determined by reaction with a primary
standard.
o indicator
 For acid-base titrations, a substance that exhibit different colors in solutions of different
acidities; used to determine the point at which reaction between two solutes is complete.
 For redox titrations, a substance that exhibits a different color when either the oxidizing
agent or the reducing agent is consumed.
o End Point
 The point at which an indicator changes color and a titration is stopped.
o Equivalence Point
 The point at which the reactants in a titration are stoichiometrically equal.
 For acid-base reactions, when the acid and the base are stoichiometrically equal.
 For redox reactions, when the oxidizing agent and the reducing agent are
stoichiometrically equal.
o Neutralize (neutralization)
 The process, by titration, of reacting an acid with a base to form a salt (and usually
water).
 It is another term to describe that you have reached the equivalence point.
 You have made the acid harmless by reacting with a base and in most cases you only
have a salt and water at the end.
o The indicator visually indicates the end point (color change) and the end point should be as
close to the equivalence point as possible.
 Equipment: pipet or buret is used to determine the volume of one of the solutions and a
buret is used to determine the volume of the unknown solution.
In titration problems, you are usually given the molarity and the volume of one substance (hence the moles)
and the volume of the other and are asked calculate the molarity of the titrant by doing stoichiometry to
determine the moles of the titrant, then you can calculate the molarity.
Mole and Stoichiometry Packet
Page 20
Answers to Practice Problems
Problem 1
Equation is: 2 H2 + O2  2 H2O
6 g H2 
1 mol H 2 2 mol H 2 O 18 grams H 2 O


 54 grams H 2 O
2.0 g H 2 2 mol H 2
1 mol H 2 O
6 g H2 
1 mol H 2 1 mol O 2

 1.5 grams O 2
2.0 g H 2 2 mol H 2
Problem 2
25.0 g CH 4 
1 mol CH 4 1 mol CH 2 Cl2

 1.55 moles CH 2 Cl2
16.04 g CH 4
1 mol CH 4
2.58 mol Cl2 
1 mol CH 2 Cl2
 1.29 moles CH 2 Cl2
2 mol Cl2
Since 1.29 moles is the smallest answer, that is the amount of CH2Cl2 that is formed and the limiting reagent
has to be the Cl2.
Problem 3
75.0 g Fe 
1 mol Fe
 1.34 moles Fe
55.85 g Fe
40.0 g O 2 
1 mol O 2
 1.25 moles O 2
32.0 g O 2
(a)
1.34 mol Fe 
2 mol Fe2O3
 0.667 moles Fe2 O3
4 mol Fe
1.25 mol O 2 
2 mol Fe2 O3
 0.833 moles Fe2 O3
3 mol O 2
(b and c)
Since the 0.667 moles of Fe2O3 is the smallest amount – that must be the amount produced [answer to part
(c)]. Since the Fe produces that amount, the Fe must be the limiting reagent [the answer to part (b)].
Mole and Stoichiometry Packet
Page 21
Problems
Directions: Put all of your work on separate sheets of paper. DO NOT DO YOUR WORK ON THESE
PAGES.
Find the number of moles in each of the following:
1)
2)
3)
4)
1.204 x 1024 atoms of He
24.6 grams of Sr
5004 grams of Na2CO3
456 liters of O2 (g) @ STP
Find the number of grams in each of the following
5)
6)
7)
4.57 moles of NH3
2.35 moles of FeCO3
2.20 moles of Sn
Find the number of representative particles in each of the following
8)
9)
10)
11)
3.45 moles of Ar
23.4 moles of Fe2(CO3)3
6.35 grams of Cu
34.5 liters of H2O(g) @STP
Make the following conversions:
12)
13)
14)
15)
16)
17)
18)
19)
20)
5.75 grams of H2 gas to liters @ STP
6.35 moles of oxygen gas to grams
2.35 x 1027 atoms of Np to grams
3.54 x 1022 molecules of H2O to grams
45.7 liters of H2 (g) @ STP to atoms of H (be careful on this one!!)
4.5 moles of Ne to grams
6.75 x 1026 molecules of CH4 (g) to liters @ STP
245 grams of SrCO3 to formula units
24.2 liters of chlorine gas to grams @ STP
21)
Calculate the empirical formula if a compound is 27.59% C, 1.15% H, 16.09% N, and 55.17% O.
22)
Find the empirical and molecular formulas of an unknown compound. Its percent composition is
58.8% C, 9.8% H, and 31.4% oxygen. The molecular compound’s molar mass is 306 grams/mol.
23)
What is the molar mass of Fe2(CO3)3?
24)
What is the percent composition of O in Fe2(CO3)3?
25)
What is the grams of oxygen in 867 grams of Fe2(CO3)3?
26)
Convert 23.4 moles of Fe2(CO3)3 to grams.
Mole and Stoichiometry Packet
Page 22
27)
Convert 437 grams of Fe2(CO3)3 to atoms of carbon.
28)
Styrene is a compound contains only C and H. If 0.438 grams of styrene is burned in oxygen, 1.481
grams of CO2 and 0.303 grams of water are produced, what is the E.F.
29)
A hydrate of magnesium iodide has the formula MgI2 • X H2O. To determine a value for X , the
hydrate sample is heated until all the water of hydration is removed. A 1.628 g sample is heated,
leaving 1.072 grams of magnesium iodide. What is the value of X ?
30)
2 NH3
+
5 F2

N2F4
??????
45.75 grams
31) 3 (NH4)2PtCl6 
???????
??? Liters @STP
3 Pt +
?
758 g
+ 6 HF
2 NH4Cl +
2 N2 +
16 HCl
?
? L @STP
? grams
32)
When 2.56 grams of barium chloride reacts with silver nitrate, how many grams of each product are
produced?
33)
5.86 x 1024 atoms of aluminum react with oxygen. How many grams of oxygen are reacted and how
many grams of product are produced? If only 354 grams of product is produced, what is the percent
yield?
34)
What is the molarity if 83.75 grams of (NH4)2SO4 is dissolved in 745 mL of water. What is the
molarity of each ion present?
35)
It is found that 38.25 mL of a 0.158 M solution of sulfuric acid is required to neutralize 64.35 mL of
aluminum hydroxide. What is the molarity of the aluminum hydroxide solution?
36)
In order to reach the equivalence point, 20.0 mL of 0.125 M solution of potassium hydroxide is reacted
with 50.0 mL of hydrofluoric acid. What is the molarity of the hydrofluoric acid? This is a typical AP
chemistry exam question that can be found on the multiple choice problems. The numbers are easy
enough that the calculations in this problem can be done without a calculator.
Mole and Stoichiometry Packet
Page 23