The document provides an introduction to stoichiometry and the mole concept. It discusses key topics including: 1. The mole is a unit used to describe the amount of substance in chemistry and is equal to 6.022x1023 particles. 2. The molar mass of an element or compound is the mass in grams of one mole and can be used to calculate amounts in chemical reactions. 3. Conversions can be made between moles, particles, masses, and volumes using the molar mass and molar relationships like moles = mass/molar mass. 4. Solution concentration is expressed in molarity, which is the moles of solute per liter of solution. M

1. Introduction to Stoichiometry: The Mole Concept
Introduction: No class in chemistry would be complete without dealing with the “calculations of chemistry”. Not
only must chemists be able to make predictions about chemical interactions and understand the principles behind
those interactions, but chemists must be able to calculate the amounts of each chemical involved in a reaction. The
calculations related to the amounts of substances in chemical reactions is called stoichiometry.
Part 1 – The mole concept – What’s a mole?
Part 2 – Calculating molar mass – What does a mole weigh?
Part 3 – Conversions – Converting to and from moles, grams, atoms or molecules, liters or milliliters?
Part 4 – Concentration – How many moles are in a liter of solution? (molarity)
Part 5 – Moles and the balanced chemical equation – How many moles of this will react with moles of that?
Part 6 – Stoichiometry – How many grams or liters of this will react with grams or liters of that?
The Beginnings
Many chemistry historians mark the beginning of
modern chemistry with the discovery of oxygen by
Joseph Priestly and later by Antoine Lavoisier
(1778). Lavoisier actually named oxygen and
determined its characteristics. Lavoisier made a
number of contributions including the discovery of
hydrogen, the law of conservation of mass, and for
contributing to the development of the metric system.
In 1799 Joseph Proust gave us the law of definite
proportions which paved the way for John Dalton to
develop the theory of atoms. Dalton’s atomic theory
in 1803 had several important points.
1. All matter is composed of tiny, indestructible
particles called atoms.
2. Atoms of the same element are all identical,
including mass, and the atoms of different elements
are different from each other.
3. Atoms combine to form compounds in reactions
where the atoms are rearranged.
4. Atoms combine in simple whole-number ratios.
Dalton put Proust’s idea on a firm footing based on
the arrangement of atoms. Dalton also gave us the
law of multiple proportions which says that atoms
which combine in more than one way combine in
simple whole number ratios.
Dalton’s atomic theory was followed, in 1811, by
Avogadro’s hypothesis. Amedeo Avogadro said that
equal volumes of different gases at the same
temperature and pressure contain equal numbers
of molecules.
For any gas at a constant temperature and pressure,
the number of molecules is directly proportional to
the volume.
n ∝ V … or … n / V = k
Avogadro found that equal volumes of H2 and O2 are
in a mass ratio of 1:16. Therefore, if there are equal
numbers of molecules of each, a molecule of oxygen
is 16 times heavier than a molecule of hydrogen.
This paved the way for determining atomic weights.
The next question is how many molecules are in a
specific volume of gas at a specific temperature and
pressure? Avogadro didn’t actually determine that
number, but he is remembered for paving the way.
In honor of Avogadro, in 1909 the French physicist
Jean Perrin introduced the term “Avogadro’s
number”. He went on to receive the Nobel prize in
physics in 1926 for measuring Avogadro’s number in
several different ways.
Perrin proposed that Avogadro's number (N) to refer
to the number of molecules in exactly 32g of oxygen
gas, which is the molecular weight in grams of
oxygen. Today, Avogadro’s number is known to
even more precision: 6.02214129 x 1023
.
Part 1 – The Mole Concept
Avogadro’s number, therefore, is the number of
atoms in a mass of the element equal to its atomic
weight in grams. This amount of an element is called
the mole.
The mole is a convenient way to describe a large
number of atoms. Measuring the amount of an
H2 O2
Despite having different masses, the
number of H2 molecules is equal to
the number of O2 molecules.
CW stoichiometry intro 041112.doc

2. element or compound in moles avoids the
temperature dependence of volume, and allows the
direct comparison of the number of atoms or
molecules which can’t be done with mass.
According to some, the word mole comes from the
Latin “moles” which means “quantity” or “massive
heap”. More probably, the word mole comes the
German word mol coined by Wilhelm Ostwald who
developed the mole concept in 1902. The word mol
was an abbreviation for the German molekul, or for
molekulargewicht, the "molecular weight”.
If you understand the “dozen concept”, then you will
understand the “mole concept”. Dozen is a word
meaning 12. Mole is a word meaning 6.022×1023
.
We know that the number of particles (atoms or
molecules) in 1 mole of a substance is 6.022×1023
atoms or molecules.
The termination of the atomic weights of elements
has driven many of the discoveries in chemistry
including Dalton’s atomic theory, Mendeleev’s
periodic table and Avogadro’s studies of gases. We
now have very accurate atomic weights. They are
based on an isotope of carbon. The carbon-12
isotope is assigned a weight of exactly 12.0000
atomic mass units (or 12.0000 daltons).
1 amu = 1 dalton (µ) = 1.6605×10-24
grams
Avogadro’s number is then determined to be that
number of particles in the atomic weight of any
element or compound.
element atoms atomic wt moles
Li 6.02×1023
6.941 amu 1 mole
Al 6.02×1023
26.982 amu 1 mole
Fe 6.02×1023
55.847 amu 1 mole
Kr 6.02×1023
83.80 amu 1 mole
Part 2 – Calculating the Molar Mass of a Compound
We know that the molar mass of an element is the
mass in grams of one mole (6.022 x1023
) of that
element. We can also speak of the molar masses of
compounds.
One mole is defined as one molar mass
of an element or compound
The molar mass of a compound is the sum of the
molar masses of all of the atoms that make up the
compound. The best way to understand the
calculation of a compound’s molar mass is to use an
example. Consider the compound aluminum
carbonate: Al2(CO3)3
In exactly one mole of Al2(CO3)3 we have two moles
of aluminum atoms, 3 moles of carbon atoms, and 9
moles of oxygen atoms. The molar mass of
Al2(CO3)3 is the sum of the molar masses of all the
atoms:
27.0 + 27.0 + 12.0 + 12.0 + 12.0 + 16.0 + 16.0 + 16.0 + 16.0 + 16.0 + 16.0 +16.0 + 16.0 + 16.0 = 234.0 g/mol
A simpler approach for finding the molar mass of Al2(CO3)3 is:
2 Al atoms 2 mol Al x 27.0 g Al/mol = 54.0 g
3 C atoms 3 mol C x 12.0 g C/mol = 36.0 g
9 O atoms 9 mol O x 16.0 g O/mol = 144.0 g
molar mass = 234.0 g/mol
We can streamline the calculation to this: (2 x 27.0) + (3 x 12.0) + (9 x 16.0) = 234.0 g/mol
And now a word about precision: We are arbitrarily going to express molar masses to one decimal place. That
means that when we add up the individual molar masses to get the molar mass of the compound, the sum will be
expressed to one decimal place. The units of molar mass are g/mol. Always remember to include the units.
Sample Exercise 1 – Compute the molar mass of each of the following compounds. The answers are given in
italics. Note, the units for molar mass are “grams per mole”, and written “g/mol”.
a. BaCl2 b. CaCO3 c. (NH4)2C2O4 d. C12H22O11 e. MgSO4·12H2O
a. 208.3 g/mol b. 100.1 g/mol c. 124.0 g/mol d. 342.0 g/mol e. 336.4 g/mol

3. Practice Exercise 1 – Compute the molar masses of the following compounds.
a. FeBr3 b. Na3PO4 c. Al(NO3)3 d. CH3C6H6 e. Ni(ClO4)3·3H2O
Part 3 – Molar Conversions
1 mole = 1 molar mass of an element or compound. (The molecular weight)
1 mole = 6.022 x 1023
particles (Particles can be atoms, molecules, electrons, etc.)
1 mole = 22.4 L of any gas at standard temperature and pressure, STP (0C or 273K and 1.00 atm).
Sample Exercise 2 – Write the solution to the problems in the space provided.
1. How many moles of zinc are equivalent to 3.01 x 1025
atoms of zinc, Zn?
2. How many molecules of carbon dioxide are in 0.400 moles of carbon dioxide, CO2?
3. How many moles of fluorine gas (F2) are in 190.0 grams of fluorine, F2?
4. What is the mass of 0.0365 moles of calcium hydroxide, Ca(OH)2?
5. How liters of methane gas, CH4, are in 0.100 moles of the gas at STP?
6. What is the mass of 3.5L of nitrogen, N2, gas at STP?
7. How many atoms of chlorine, Cl, are in 0.250 L of carbon tetrachloride, CCl4, gas at STP?

4. Practice Exercise 2 – Answer the following in the space provided. Show your work. Answers must have the
correct number of significant digits.
1. What is the mass of 0.500 moles of sodium sulfate, Na2SO4?
2. How many moles of lithium nitrate, LiNO3, have a mass of 50.0 grams?
3. Find the number of moles in 4.53 x 1022
molecules of carbon monoxide, CO.
4. How many molecules are in 145.0 grams of ammonia, NH3?
5. How many liters of ozone (O3) gas at STP are in 0.400 moles of ozone?
6. How many moles of chlorine, Cl, are in 1022.5 grams of iron (III) chloride, FeCl3?
7. How many atoms of oxygen are in 3.55 liters of sulfur trioxide, SO3, at STP?
8. What is the mass of water in the hydrated compound magnesium sulfate heptahydrate, MgSO4·7H2O.

5. Part 4 – Concentration of solutes in solution (molarity)
Recall the classification of matter as
show in the concept map at the right. A
solution is a homogeneous mixture in
which a solute is dissolved in a solvent,
usually water. A mixture has no
definite composition as does a
compound. Because of that, there can be
a wide-ranging amount of solute
dissolved in a specific amount of
solution. The numerical measurement of
the amount of solute in a specific volume
of solution is the concentration.
The concentration of a solution is most
often described by its molarity.
Molarity is the ratio of the moles of
solute to the volume of solution in liters.
Molarity has units of mol/L (mol L-1
).
Molarity is abbreviated with a capital M.
L
moles
M
solutionofliters
soluteofmoles
Molarity ==
For example, consider the case where 30.0 grams of NaCl solid is
dissolved in enough water to make exactly 500.0 mL of solution. What
will be the concentration in moles per liter?
Molarity = moles of solute / L of solution
Molarity = M03.1
L500.0
)NaClg5.58/NaClmol1(gNaCl0.30
=
×
I. Use an equation. We can use the “molarity equation” above to solve for any one of the three variables.
Example: How many moles of sodium hydroxide are in 20.0 mL of 0.400 M solution?
L
mol
Molarity =
LMolaritymol ×=
L0.0200
L
mol
0.400mol ×=
NaOHmol0.00800mol =
Practice exercises:
1. What is the molarity of a solution made by dissolving 25.0 grams of Ba(NO3)2 in enough water to make
250.0 mL of solution?
500 mL of solution
30.0 g of NaCl

6. 2. How many grams of sodium hydroxide are needed to make 500.0 mL of 0.200 M solution?
3. Suppose you need 0.300 moles of silver nitrate. You have a 0.500 M solution of silver nitrate. How many
milliliters of the solution will contain 0.300 moles of silver nitrate?
II. Use molarity as a conversion factor. Since molarity is a ratio of two quantities (moles and liters) then we can
multiply by molarity and cancel out liters and get moles. We can also multiply by the reciprocal of molarity in
order to cancel out moles and get liters.
Example: How many moles of sodium hydroxide are in 20.0 mL of 0.400 M solution?
Note: Do not try to use the letter “M” in 0.400 M as a unit. The M is merely an abbreviation for mol/L. When
setting up a problem, you must use mol/L in place of M.
NaOHmol0.00800
mL1000
NaOHmol0.400
mL0.20 =×
Here’s a hint: When using volume in mL, use 1000 mL instead of 1 L
Practice exercises:
Set up the solution using conversion factors to the same practice problems above. Notice how you end up
doing the same math.
1. What is the molarity of a solution made by dissolving 25.0 grams of Ba(NO3)2 in enough water to make
250.0 mL of solution?
2. How many grams of sodium hydroxide are needed to make 500.0 mL of 0.200 M solution?
3. Suppose you need 0.300 moles of silver nitrate. You have a 0.500 M solution of silver nitrate. How many
milliliters of the solution will contain 0.300 moles of silver nitrate?