STOICHIOMETRY
USING THE REACTION
EQUATION LIKE A RECIPE
Cool word, but it’s all Greek to me!
• Stoicheion = element
• Metron = measure
• STOICHIOMETRY…measuring and
calculating the chemical elements
• YOU WILL NEED YOUR CALCULATOR
EVERY DAY!
USING EQUATIONS
• Nearly everything we use is
manufactured from chemicals.
– Soaps, shampoos, conditioners, cd’s,
cosmetics, medications, and clothes.
• For a manufacturer to make a profit
the cost of making any of these items
can’t be more than the money paid for
them.
• Chemical processes carried out in
industry must be economical, this is
where balanced equations help.
USING EQUATIONS
• Equations are a chemist’s recipe.
– Eqs tell chemists what amounts of
reactants to mix and what amounts of
products to expect.
• When you know the quantity of one
substance in a rxn, you can calculate
the quantity of any other substance
consumed or created in the rxn.
– Quantity meaning the amount of a
substance in grams, liters, molecules,
formula units, or moles.
USING EQUATIONS
• The calculation of quantities in chemical reactions is called stoichiometry.
• Imagine you are in charge of manufacturing for Rugged Rider Bicycle
Company.
• The business plan for Rugged Rider
requires the production of 128
custom-made bikes each day.
• You are responsible for insuring that
there are enough parts at the start of
each day.
USING EQUATIONS
• Assume that the major components of
the bike are the frame (F), the seat
(S), the wheels (W), the handlebars
(H), and the pedals (P).
• The finished bike has a “formula” of
FSW2HP2.
• The balanced equation for the
production of 1 bike is.
F +S+2W+H+2P  FSW2HP2
USING EQUATIONS
• Now in a 5 day workweek, Rugged
Riders is scheduled to make 640 bikes.
How many wheels should be in the
plant on Monday morning to make
these bikes?
• What do we know?
– Number of bikes = 640 bikes
– 1 FSW2HP2=2W (balanced eqn)
• What is unknown?
–# of wheels = ? wheels
• The connection between wheels and
bikes is 2 wheels per bike. We can
use this information as a conversion
factor to do the calculation.
640 FSW2HP2
2W
1 FSW2HP2
.
= 1280
wheels
• We can make the same kinds of
connections from a chemical rxn eqn.
• N2(g) + 3H2(g)  2NH3(g)
• The key is the “coefficient ratio”.
COEFFICIENT RATIOS or
MOLE RATIOS
• N2(g) + 3H2(g)
• 1 mol N2
or
3 mol H2
• 1 mol N2
or
2 mol NH3
• 3 mol H2
or
2 mol NH3
 2NH3(g)
3 mol H2
1 mol N2
2 mol NH3
1 mol N2
2 mol NH3
3 mol H2
– The coefficients of the balanced
chemical equation indicate the
numbers of moles of reactants and
products in a chemical rxn.
• 1 mole of N2 reacts with 3 moles of H2
to produce 2 moles of NH3.
– N2 and H2 will always react to form
ammonia in this 1:3:2 ratio of moles.
• So if you started with 10 moles of N2 it
would take 30 moles of H2 and would
produce 20 moles of NH3
• Using the coefficients, from the balanced rxn equation to make connections
between reactants and products, is
the most important information that a
rxn equation provides.
– Using this information, you can calculate
the amounts of the reactants involved
and the amount of product you might
expect.
– Any calculation done with the next
process is a theoretical number,
the real world isn’t always perfect.
• Using the coefficients of balanced rxn
equations and our knowledge of mole
conversions we can perform powerful
calculations. A.K.A. stoichiometry.
• A balanced rxn equation is essential
for all calculations involving amounts
of reactants and products.
– If you know the number of moles of
1 substance, the balanced eqn allows
you to calc. the number of moles of all
other substances in a rxn equation.
MOLE – MOLE EXAMPLE
• The following rxn shows the synthesis
of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
• If you only had 1.8 mols of Al how
much product could you make?
Given: 1.8 moles of Al
Uknown: ____ moles of Al2O3
MOLE – MOLE EXAMPLE
• Solve for the unknown:
3O2(g) + 4Al(s)  2Al2O3(s)
1.8 mol Al
2 mol Al2O3
4 mol Al
Mole Ratio
= 0.90mol
Al2O3
MOLE – MOLE EXAMPLE 2
• The following rxn shows the synthesis
of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
• If you wanted to produce 24 mols of
product how many mols of each
reactant would you need?
Given: 24 moles of Al2O3
Uknown: ____ moles of Al
____ moles of O2
MOLE – MOLE EXAMPLE 2
• Solve for the unknowns:
3O2(g) + 4Al(s)  2Al2O3(s)
24 mol Al2O3
24 mol Al2O3
4 mol Al
2 mol Al2O3
3 mol O2
2 mol Al2O3
= 48 mol Al
= 36 mol O2
Cross the Mole-Mole Bridge
With thanks to Rossini, Willam Tell,
and the Lone Ranger
William Tell Overture Guitar
Hiho, Stoichiometry...Away!
Backwards
MASS – MASS CALCULAT’NS
• No lab balance measures moles
directly, generally mass is the unit of
choice.
• From the mass of 1 reactant or product, the mass of any other reactant or
product in a given chemical equation
can be calculated, provided you have
a balanced rxn equation.
• As in mole-mole calcs, the unknown
can be either a reactant or a product.
MASS – MASS CALCULAT’NS 1
Acetylene gas (C2H2) is
produced by adding water to
calcium carbide (CaC2).
CaC2 + 2H2O  C2H2 + Ca(OH)2
How many grams of C2H2 are produced
by adding water to 5.00 g CaC2?
MASS – MASS CALCULAT’NS 1
• What do we know?
– Given mass = 5.0 g CaC2
– Mole ratio: 1 mol CaC2 = 1 mol C2H2
– MM of CaC2 = 64 g CaC2
– MM of C2H2 = 26 g C2H2
• What are we asked for?
– grams of C2H2 produced
MASS – MASS CALCULAT’NS 1
mass A  moles A  moles B  mass B
5.0 g 1 mol CaC2 1 mol C2H2
CaC2
64 g CaC2 1mol CaC2
26 g C2H2
1mol C2H2
= 2.0 g C2H2
MASS – MASS CALCULAT’NS 2
You’ve recently learned that Copper will
replace silver ions out of solution. You’re
eyes light up with this money making
opportunity. However, you decide it might be
best if you did some preliminary calculations
to determine to the feasibility of this get rich
scheme. Copper is not very hard to find,
however the largest size of Silver nitrate
found in the Flinn Catalog is the 500 g size
and it costs $305.91. Currently Silver sells
for $9.00/ounce on the stock market. How
much money could you sell your
manufactured Silver for?
MASS – MASS CALCULAT’NS 2
Cu + 2AgNO3  2Ag + Cu(NO3)2
• What do we know?
–
–
–
–
–
–
Given mass = 500. g of AgNO3
Mole ratio: 2 mol AgNO3 = 2 mol Ag
MM of AgNO3: 169.84g = 1mol
MM of Ag: 107.87 g = 1mol
Price of Silver: $9.00 = 1 ounce
Conversion g to oz: 28.23g = 1 oz
MASS – MASS CALCULAT’NS 2
500. g 1mol AgNO3
2 mol Ag
AgNO3
169.84gAgNO3 2 mol AgNO3
107.87g Ag
1 oz
$9.00
1mol Ag
28.23 g
1 oz
= $101.26
=$101
• A balanced reaction equation indicates
the relative numbers of moles of
reactants and products.
• We can expand our stoichiometric
calculations to include any unit of
measure that is related to the mole.
• The given quantity can be expressed
in numbers of particles, units of mass,
or volumes of gases at STP.
• The problems can include massvolume, volume-volume, and particlemass calculations.
• In any of these problems, the given
quantity is first converted to moles.
• Then the mole ratio from the balanced
eqn is used to convert from the moles
of given to the number of moles of the
unknown
• Then the moles of the unknown are
converted to the units that the
problem requests.
• The next slide summarizes these steps
for all typical stoichiometric problems
MORE MOLE EXAMPLES
How many molecules of O2 are
produced when a sample of
29.2 g of H2O is decomposed by
electrolysis according to this
balanced equation:
2H2O  2H2 + O2
MORE MOLE EXAMPLES
• What do we know?
– Mass of H2O = 29.2 g H2O
– 2 mol H2O = 1 mol O2 (from balanced
equation)
– MM of H2O = 18.0 g H2O
– 1 mol O2 = 6.02x1023 molecules of O2
• What are we asked for?
– molecules of O2
mass A  mols A  mols B 
molecules B
29.2 g 1 mol H2O
1 mol O2
H2O
18.0 g H2O 2 mol H2O
6.02x1023
molecules O2
1 mol O2
= 4.88 x 1023
molecules O2
MORE MOLE EXAMPLES
The last step in the production of
nitric acid is the reaction of NO2
with H2O.
3NO2+H2O2HNO3+NO
How many liters of NO2 must react
with water to produce 5.0x1022
molecules of NO?
MORE MOLE EXAMPLES
• What do we know?
– Molecules NO = 5.0x1022 molecules NO
– 1 mol NO = 3 mol NO2 (from balanced
equation)
– 1 mol NO = 6.02x1023 molecules NO
– 1 mol NO2 = 22.4 L NO2
• What are we asked for?
– Liters of NO2
molecules A mols mols B volume B
5.0x1022 molecules NO
1 mol NO
3 mol NO2
6.02x1023 mol- 1 mol NO
ecules NO
22.4 L NO2
1 mol NO2
= 5.6 L NO2
Aspirin can be made from a chemical rxn
between the reactants salicylic acid and
acetic anhydride. The products of the rxn
are acetyl-salicylic acid (aspirin) and acetic
acid (vinegar). Our factory makes 125,000
100-count bottles of Bayer Aspirin/day.
Each bottle contains 100 tablets, and each
tablet contains 325mg of aspirin. How
much in kgs + 10% for production
problems, of each reactant must we have
in order to meet production?
C7H6O3 + C4H6O3  C9H8O4 + HC2H3O2
Salicylic
acid
Acetic
anhydride
aspirin
vinegar
• What do we know?
– Make 125,000 aspirin bottles/day
– 100 aspirin/bottle
– 325 mg aspirin/tablet
– Mole ratio of aspirin to salicylic acid
(1:1) and acetic anhydride (1:1)
– MM aspirin C9H8O4 = 180.11g
– MM SA C7H6O3 = 138.10g
– MM AA C4H6O3 = 102.06g
• What are we asked for?
– Mass of salicylic acid in kgs + 10%
– Mass of acetic anhydride in kgs + 10%
125,000 100 tablets 325mg asp.
bottles
1 bottle
1 tablet
1 g asp
1mol asp.
1000 mg asp
180.16g asp
= 22,549.4 mols aspirin
Salicylic Acid:
22,549.4 1 mol SA C7H6O3
mols
1 mol asp
aspirin
1 kg SA C7H6O3
1000 g SA C7H6O3
138.10g SA C7H6O3
1 mol SA C7H6O3
= 3068.97 kg salicylic
acid + (306.897 g)
= 3380 kg of salicylic acid
Acetic Anhydride:
22,549.4 1 mol AA C4H6O3
mols
1 mol asp
aspirin
1 kg AA C4H6O3
1000 g AA C4H6O3
102.06g AA C4H6O3
1 mol AA C4H6O3
= 2301.39 kg
Acetic anhydride
+ 230.139 kg
= 2530 kg Acetic anhydride