STOICHIOMETRY
Objectives:
1. Describe
the types of
relationships indicated by a
balanced chemical equation
2. State the mole ratios from a
balanced chemical equation
 Defintion:
Stoichiometry is the calculation of
quantities in chemical equations
 The quantities are usually amount of a
substance expressed in grams or moles, but
can be liters, molecules, etc.
 Definition: a mole ratio is the ratio between
the number of moles of any two of the
substances in a balanced chemical equation
 Find the mole ratios for the reaction
2K(s) + Br2(l) ------ 2KBr(s)
 A: 2 mol K/1 mol Br2
2 mol K/2mol KBr
1mol Br2/2 mol K 1 mol Br2/2 mol KBr
2 mol KBr/2 mol K 2 mol KBr/1 mol Br2
A
balanced chemical equation
tells you
• The relative amount of molecules
• The relative amounts of moles
Additional information you can
calculate is
• The mass of any product or
reactant
• The volume of any gas present
 The
coefficients in an equation can be
interpreted on the micro or macro
levels
• Micro: how many representative units
are reacting or produced
• Macro: how many moles of each
substance are present
 The coefficients in front of each
chemical equals how many moles & how
many molecules of each chemical there
are
 A balanced chemical equation gives the
ratio of moles of one chemical to moles
of any other chemical in the equation
 The
number of moles can be used to make
conversion factors.
 For example in the equation
N2(g) + 3H2(g) → 2NH3(g)
1
mol of nitrogen reacts with three moles of
hydrogen to form two moles of ammonia
 Mass: you can calculate the mass of each
product or reactant by multiplying the
number of moles by the chemical’s
molecular mass
• Mass is conserved, so the mass of the
products should equal the mass of the
reactants
The
volume of any gas present, if you
assume STP conditions, can be found
by using the molar volume definition
• 1 mol = 22.4L
The
volume is simply the number of
moles multiplied by (22.4L / 1mol)
Let’s try a problem:
Interpret the combustion of propane in
terms of representative particles,
moles and conservation of mass
 Note: most
stoichiometric calculations require a
balanced chemical equation
 Always check that the equation you are working
with is balanced
 C3H8(g) + 5O2(g) -----
3CO2(g) + 4H2O(g)
 1 mol C3H8 x 44.09 g C3H8 /mol C3H8 = 44.09 g C3H8
 5 mol O2 x 32.00 g O2 /mol O2= 160.0 g O2
 3 mol CO2 x 44.01 g CO2 /mol CO2 = 132.0 g CO2
 4 mol H2O x 18.02 g H2O /mol H2O= 72.08 g H2O
 44.09 g C3H8 + 160 g O2= 204.1 g reactants
 132.0 g CO2 +72.08 g H2O = 204.1 g products
 The coefficients in the equation indicate the number
of molecules as well as the number of moles:
 1 molecule (mole) C3H8 +5 molecules (moles) 5O2
------ 3 molecules (moles) CO2 + 4 molecules
(moles) H2O
Objectives:
1. List
the sequence of steps
used in solving stoichiometric
problems
2. Solve stoichiometric
problems
Mole
ratios can then be used to find
unknown quantities of reactants or
products
The general approach to solving for
the unknown is to use the given
amount of moles from the
information in the problem,
multiply it by the mole ratio and
then calculate
 Example: 2
S + 3 O2-------> 2 SO3 How
many moles of sulfur react with 9 moles
of O2?
 Set up your ratio like this:
9 mol O2 x 2 mol S
3 mol O2 = 6 mol sulfur
 Notice the cross cancellation for the
moles of O2
 You can go grams to moles and in more
complex problems you go from grams
of one compound to grams of the other
compound by using the four steps of
gram, mol, mol, gram
 4Al(s)
+ 3O2(g) → 2Al2O3(s)
 How many moles of aluminum are needed to
form 3.7 moles of Al2O3(s)?
 The amount of substance is usually
determined by finding its mass in grams
 Knowing the mass of one reactant or product
in a balanced equation will let you determine
the mass of any other reactant or product in
the equation
 You cannot directly convert between products
and reactants
 Use the gram, mol, mol, gram calculations that
follow this slide
 An
amount given in grams is first converted
to moles using the molar mass of the given
substance
 Then using the molar ratios from the
balanced equation, convert the moles of
given substance (which you just calculated)
to the number of moles of the unknown
using the principles of cross-cancellation
 Finally, convert to grams using the molar
mass of the unknown
 The unknown can be either a reactant or a
product
 To go from the mass of the given to the
unknown use the four steps that follow
Steps
to solving gram to gram
problems:
1. Write down what you know (the
givens, including proper units)
2. Convert to moles using molar
mass(always include proper units)
3. Use the balanced equation to
convert to moles of the unknown
quantity (the desired quantitykeep using units)
4. convert back to grams using
molar mass
Acetylene
as (C2H2) is produced by
adding water to calcium carbide (CaC2)
CaC2(s) + 2H2O ------> C2H2(g) +
Ca(OH)2(aq)
How many grams of acetylene are
produced by adding water to 5.00 g
CaC2?
Using the same equation determine how
many moles of CaC2 are needed to react
completely with 49.0 g of H2O?
A
balanced chemical equation indicates the
relative number of molecules and moles
 Any unit of measurement can be found from
this relationship
• These include: number of particles, units of
mass, volumes of gases (@ STP)
 The problems based on these are varied and
can be: mass-volume, volume-volume,
particle-mass; any combination
 (1)The first step in any problem is convert to
moles
 Change the measured quantity (what you
know) to moles
(2) The
second step is to use the
mole ratio (from the balanced
eqn) of wanted to given
substances
(3) Change moles of the wanted
substance to a measured quantity
(what you are looking for)
How many molecules of oxygen
are produced by the
decomposition of 6.54g of
potassium chlorate?
2KClO3 (s) ------> 2KCl (s) + 3O2 (g)
 Objectives:
 1. Identify
the limiting reactant in a
chemical equation
 2. Identify the excess reactant, and
calculate the remaining after the reaction
is complete
 3. Calculate the mass of a product when
the amount of more than one reactant are
given
 Chemicals
in a reaction are required according to
the coefficients in the balanced chemical
equation
 Definition: a limiting reactant limits the extent of
the reaction and therefore, the amount of product
formed
 The chemical that gets used up first is called the
limiting reactant or reagent
 It limits the amount of products that can be
formed
 When the reaction stops, the limiting reactant will
be all used up
 Definition: excess reactant are left over when a
reaction stops
 The
known amount of one of the reactant
is multiplied by the mole ratio (from the
balanced equation) to calculate the
required amount of the other reactant
 The required amount is compared to the
given amount to see if it is limiting or not
 Sodium chloride can be prepared by the
reaction of sodium metal with chlorine
gas. Suppose 6.7 mol Na reacts with 2.3
mol Cl2. What is the limiting reagent?
How many moles of product are
produced?
 2Na + Cl2 --- 2NaCl
 The
reaction between solid white
phosphorus (P4) and oxygen produces
tetraphosphorus decoxide (P4O10).
 A. Determine the mass of P4O10 formed if 25
g of P4 and 50 g of oxygen are combined/
 B. How much excess reactant remains
afterward?
 1. givens: mass of 25 g of P4 and 50 g of
oxygen
 2. write the equation: P4(s) + 5O2(g) -
P4O10(s)
 3. What are you looking for?
 4. How are you going to get there? Solve it!!
Objectives:
1. Calculate
the theoretical
yield of a chemical reaction
from data
2. Determine the percent
yield for a chemical reaction
 Usually
chemical reactions do not produce
the quantity of product you would expect
 Definition: theoretical yield is the maximum
amount of product that can produced from a
given amount of reactant
 The amount of product expected is referred
to as the theoretical yield
 Its value comes from the balanced chemical
equation
 Definition: the amount of product that is
formed from performing the reaction (the
experiment) is the actual yield
 The
measure of how close the actual yield is
to the theoretical yield is the percent yield
 The equation: percent yield equals actual
yield divided by the theoretical yield times
100
 Percent yield is always a number less than
100
 Percent yield is a measure of how efficient a
reaction is in producing the desired product
 Reasons for low percent yield include:
competing side reactions, impure reactants,
improper procedures, improper
measurements
 1. Calcium
carbonate decomposes by the
reaction below. What is the theoretical yield
of CaO if 24.8 g of CaCO3 is heated?
 2. What is the percent yield if 13.1 g of CaO
is formed? CaCO3 --- CaO + CO2
 3. Determine the theoretical yield of silver
chromate if 0.5 g of sliver nitrate reacts from
the following equation:
 2AgNO3(aq) + K2CrO4(aq) - Ag2CrO4(s) +
2KNO3(aq)
 4. What is the percent yield if the reaction
yields 0.455 g Ag2CrO4?