Chapter 3: Stoichiometry
Emily Scheerer
Section 3.1–Chemical Equations
Section 3.2–Patterns of Chemical Reactivity
Section 3.3–Atomic and Molecular Weights
Section 3.4–The Mole
Section 3.5–Empirical Formulas from Analyses
Section 3.6–Quantitative Information from Balanced
Equations
Section 3.7 – Limiting Reactants
3.1Chemical Equations
~ Law of conservation of mass: total mass of all
substances after a chemical reaction is the same as the
total mass before the reaction.
~ Balancing Chemical Equations (a few tips)
Δ above the arrow means the addition of heat
Never change subscripts, only coefficients
Use the symbols (g), (l), (s), (aq)
3.2 Patterns of Chemical Reactivity
•All alkali metals react with water to form hydrogen and
hydroxide compounds.
2M(s) + 2H2O(l)  2MOH(aq) + H2(g)
•Combustion reactions are rapid reactions that produce flame,
generally involve O2.
Example: C3H8+5O2 3CO2+4H2O
•In combination reactions, two or more substances form one
product.
Example: 2Mg(s) + O2(g)2MgO(s)
In decomposition reactions one substance produces multiple
other substances.
Example: CaCO3(s)  CaO(s) + CO2
3.3 Atomic and Molecular Weights
We can measure the masses of individual atoms with
the atomic mass unit, amu. The amu is defined by
assigning the mass of exactly 12 amu to the 12C
isotope of carbon.
1 amu = 1.66054 x 10-24g and 1g = 6.02214 x 1023amu
Most elements are a mixture of isotopes. The average
atomic mass of each element (expressed in amu’s) is
known as it’s atomic weight. The atomic weights are
in the periodic table.
3.3 Atomic and Molecular Weights
The formula weight of a substance is the sum of the atomic weights of each
atom in its chemical formula.
If the chemical formula is merely the chemical symbol of an element, such
as Na, then the formula weight equals the atomic weight of the element. If
the chemical formula is that of a molecule, then the formula weight is also
called the molecular weight.
Percentage Composition from Formulas
Occasionally we must calculate the percentage composition of a compound
(that is, the percentage by mass contributed by each element in the
substance). Here’s how:
3.4 The Mole
Mole - The amount of matter that contains as many objects as as the number
of atoms in 12g of 12C.
This number is 6.02214199 x 1023 and is called Avogadro’s number. (use 6.
02 x 1023)
Molar Mass : The mass in grams of 1 mole is called the molar mass (in
grams) and is numerically equal to it’s formula weight (in amu)
Example: What is the mass in grams of 1.000 mol of glucose, C6H12O6?
Solution
Analyze: We are given the chemical formula for glucose and asked to determine
its molar mass.
Plan: The molar mass of a substance is found by adding the atomic weights of its
component atoms.
Solve:
Because glucose has a formula weight of 180.0 amu, 1 mol of this substance has a
3.4 The Mole
Calculate the number of H atoms in 0.350 mol of C6H12O6
Solution
Analyze: We are given both the amount of the substance (0.350 mol) and
its chemical formula (C6H12O6). The unknown is the number of H atoms.
Plan: Avogadro's number provides the conversion factor between the
number of moles of and the number of molecules of C6H12O6. Once we
know the number of molecules of we can use the chemical formula, which
tells us that each molecule of contains 12 H atoms. Thus, we convert moles
of to molecules of and then determine the number of atoms of H from the
number of molecules of C6H12O6
Solve:
3.4 The Mole
Interconverting Masses, Moles, and
Numbers of Particles (also called
dimensional analysis)
Calculate the number of moles of
glucose(C6H12O6) in 5.380 g of
C6H12O6.
Solution
Analyze: We are given the number of grams of C6H12O6 and asked to calculate
the number of moles.
Plan: The molar mass of a substance provides the conversion factor for
converting grams to moles. The molar mass of C6H12O6
is 180.0
Solve: Using 1 mol C6H12O6 to write the appropriate conversion factor, we have
3.5 Empirical Formulas from Analyses
The empirical formula tells the relative number of atoms
of each element it contains. The general procedure for
finding empirical formula is seen below:
3.5 Empirical Formulas from Analyses
Remember that the formula from percent composition is always empirical. To transfer to
molecular, you need to know the molecular weight. *The subscripts in the molecular
formula are always whole number multiples of the subscripts in the empirical formula.
Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical
formula of C3H4. The experimentally determined molecular weight of this substance is 121
amu. What is the molecular formula of mesitylene?
Solution
Analyze: We are given the empirical formula and molecular weight of mesitylene and
asked to determine its molecular formula.
Plan: The subscripts in a molecular formula are whole-number multiples of the subscripts
in its empirical formula. To find the appropriate multiple, we must compare the molecular
weight with the formula weight of the empirical formula.
Solve: First we calculate the formula weight of the empirical formula, C3H4
Next we divide the molecular weight by the empirical formula weight to obtain the factor
used to multiply the subscripts in C3H4
Only whole-number ratios make physical sense because we must be dealing with whole
atoms. We therefore multiply each subscript in the empirical formula by 3 to give the
molecular formula: C9H12
3.6 Quantitative information from
balanced equations
~ The coefficients in a balanced equation can be
interpreeted as the relative number of molecules and as
the relative number of moles involved in the reaction.
consider the combustion of butane (C4H10), the fuel in
disposable cigarette lighters:
the conversion sequence is:
combined in a single sequence of factors:
3.6 Quantitative information from
balanced equations
Outline of the procedure used to calculate the number of
grams of a reactant consumed or of a product formed in a
reaction, starting with the number of grams of one of the
other reactants or products.
3.7 Limiting Reactants
The reactant that is completely consumed in in a reaction
is called the limiting reactant.
Real World Example:
You have 10 pieces of bread and 7 pieces of cheese. You
want to make grilled cheese sandwiches with two pieces
of bread and 1 piece of cheese, which will run out first?
The bread will run out first, because even though there is
less cheese, you need more bread for each sandwich,
making bread the limiting reactant.
3.7 Limiting Reactants
The quantity of the product that is calculated to form
when all of the limiting reactant has reacted is the
theoretical yield. The amount of product actually obtained
is the actual yield. The percent yield relates the actual
yield to the theoretical.
3.7 Limiting Reactants
Example:
Adipic acid, H2C6H8O4 is used to produce nylon. It is made commercially by a
controlled reaction between cyclohexane (C6H12) and O2.
(a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane, and that
cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid?
(b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of
adipic acid?
Solution
Plan: (a) The theoretical yield is the calculated quantity of adipic acid formed in the
reaction. We carry out the following conversions: g C6H12  mol C6H12  mol
H2C6H8O4  g H2C6H8O4. (b) Once we have calculated the theoretical yield, we use
Equation 3.13 to calculate the percent yield.
Solve: (a)
(b)
Key things to remember
~ How to balance equations
~ common chemical reactions (like decomposition,
combustion, combination, etc)
~ amu’s, molecular weight, percent composition
~ Avagadro’s number and how to use it
~ dimensional analysis
~ Empirical formulas: how to get them, how to get
molecular formulas from them, how to get theoretical
yield from them.
~ Limiting reactants and percent yield formula