Copyright Sautter 2003
STOICHIOMETRY
“Measuring elements”
Determining the Results of
A Chemical Reaction
PREDICTING HOW MUCH OF A SUBSTANCE CAN BE
MADE BY A CHEMICAL REACTION BEFORE IT IS
CARRIED OUT!!
• STEP I
• ALWAYS WRITE THE
EQUATION USING
CHEMICAL FORMULAE
AND BALANCE IT.
• (REMEMBER TO BE SURE
YOU USE THE CORRECT
SUBSCRIPTS FOR THE
FORMULAE AND CHANGE
ONLY THE COEFFICIENTS
WHEN BALANCING THE
EQUATION)
• EXAMPLE:
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CALCIUM CARBONATE
(LIMESTONE) WHEN HEATED
GIVES CALCIUM OXIDE (LIME)
AND CARBON DIOXIDE
CALCIUM = Ca (+2)
CARBONATE = CO3 (-2)
CALCIUM CARBONATE = CaCO3
OXIDE = O (-2)
CALCIUM OXIDE = CaO
CARBON DIOXIDE = CO2
CaCO3(S)  CaO(s) + CO2(g)
THERE IS ONE Ca ON EACH SIDE OF
THE EQUATION, ONE C ON EACH
SIDE AND THREE O ON EACH SIDE.
THE EQUATION IS BALANCED!
#1. In terms of Particles
An Element is made of atoms
A Molecular compound (made
of only nonmetals) is made up
of molecules (Don’t forget the diatomic elements)
Ionic Compounds (made of a
metal and nonmetal parts) are
made of formula units
Example: 2H2 + O2 → 2H2O
 Two
molecules of hydrogen and one
molecule of oxygen form two molecules of
water.
 Another example: 2Al2O3 Al + 3O2
2 formula units Al2O3 form 4 atoms Al
and 3 molecules O2
Now read this: 2Na + 2H2O  2NaOH + H2
#2. In terms of Moles
The
coefficients tell us how
many moles of each substance
2Al2O3 Al + 3O2
2Na + 2H2O  2NaOH + H2
Remember:
A balanced
equation is a Molar Ratio
#3. In terms of Mass
 The
Law of Conservation of Mass applies
 We can check mass by using moles.
2H2 + O2  2H2O
2 moles H2
1 mole O2
2.02 g H2
1 mole H2
32.00 g O2
1 mole O2
= 4.04 g H2
+
= 32.00 g O2
36.04 gg H
H22 ++ O2
36.04
reactants
In terms of Mass (for products)
2H2 + O2  2H2O
2 moles H2O
18.02 g H2O
= 36.04 g H2O
1 mole H2O
36.04 g H2 + O2 = 36.04 g H2O
36.04 grams reactant = 36.04 grams product
The mass of the reactants must
equal the mass of the products.
#4. In terms of Volume
At
STP, 1 mol of any gas = 22.4 L
2H2
+ O2
 2H2O
(2 x 22.4 L H2) + (1 x 22.4 L O2)  (2 x 22.4 L H2O)
67.2 Liters of reactant ≠ 44.8 Liters of product!
NOTE: mass and atoms are
ALWAYS conserved - however,
molecules, formula units, moles, and
volumes will not necessarily be
conserved!
Practice:
Show
that the following
equation follows the Law of
Conservation of Mass (show
the atoms balance, and the
mass on both sides is equal)
2Al2O3 Al + 3O2
Chemical Calculations
OBJECTIVES:
• Construct “mole ratios” from
balanced chemical
equations, and apply these
ratios in mole-mole
stoichiometric calculations.
Mole to Mole conversions
2Al2O3 Al + 3O2
• each time we use 2 moles of Al2O3
we will also make 3 moles of O2
2 moles Al2O3
3 mole O2
or
3 mole O2
2 moles Al2O3
These are the two possible conversion
factors to use in the solution of the problem.
Mole to Mole conversions
How
many moles of O2 are
produced when 3.34 moles of
Al2O3 decompose?
2Al2O3 Al + 3O2
3.34 mol Al2O3
3 mol O2
2 mol Al2O3
= 5.01 mol O2
Conversion factor from balanced equation
If you know the amount of ANY chemical in the reaction,
you can find the amount of ALL the other chemicals!
Practice:
2C2H2 + 5 O2  4CO2 + 2 H2O
• If 3.84 moles of C2H2 are burned, how
many moles of O2 are needed?(9.6 mol)
•How many moles of C2H2 are needed to
produce 8.95 mole of H2O? (8.95 mol)
•If 2.47 moles of C2H2 are burned, how
many moles of CO2 are formed? (4.94 mol)
How do you get good at this?
Steps to Calculate
Stoichiometric Problems
1. Correctly balance the equation.
2. Convert the given amount into
moles.
3. Set up mole ratios.
4. Use mole ratios to calculate moles
of desired chemical.
5. Convert moles back into final unit.
Mass-Mass Problem:
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4Al + 3O2  2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3 101.96 g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
(6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) =
= ? g Al2O3
12.3 g Al2O3
are formed
Another example:
If
10.1 g of Fe are added to a
solution of Copper (II) Sulfate, how
many grams of solid copper would
form?
2Fe + 3CuSO4  Fe2(SO4)3 + 3Cu
Answer = 17.2 g Cu
Volume-Volume Calculations:
 How
many liters of CH4 at STP are
required to completely react with 17.5 L
of O2 ?
CH4 + 2O2  CO2 + 2H2O
1 mol O2 1 mol CH4 22.4 L CH4
17.5 L O2
22.4 L O2 2 mol O2 1 mol CH4
= 8.75 L CH4
Notice anything relating these two steps?
Avogadro told us:
Equal
volumes of gas, at the same
temperature and pressure contain
the same number of particles.
Moles are numbers of particles
You can treat reactions as if they
happen liters at a time, as long as
you keep the temperature and
pressure the same. 1 mole = 22.4 L @ STP
Shortcut for Volume-Volume?
 How
many liters of CH4 at STP are
required to completely react with 17.5 L
of O2?
CH4 + 2O2  CO2 + 2H2O
17.5 L O2
1 L CH4
2 L O2
= 8.75 L CH4
Note: This only works for
Volume-Volume problems.
IN ORDER TO PREDICT THE RESULTS OF A CHEMICAL
REACTION WE MUST BE GIVEN THE QUANTITIES OF
MATERIALS THAT ARE TO BE USED IN THE REACTION
SUPPOSE WE ARE GIVEN 200 GRAMS OF CALCIUM CARBONATE
AND ASKED TO DECIDE HOW MUCH CALCIUM OXIDE AND
CARBON DIOXIDE CAN BE MADE??
WE ALREADY HAVE THE BALANCED EQUATION BUT WHAT DO
WE DO NEXT??
WELL, SINCE BALANCED EQUATIONS SHOW THE NUMBER OF
ATOMS AND MOLECULES INVOLVED, WE MUST WORK WITH
NUMBERS OF ATOMS AND MOLECULES.
REMEMBER WE COUNT ATOMS AND MOLECULES WITH MOLES.
(6.02 X 1023 = 1 MOLE)
**REMEMBER **
TO CONVERT GRAMS (MASS) TO MOLES WE DIVIDE
THE WEIGHT OF ONE MOLE FROM THE PERIODIC
TABLE INTO THE GIVEN NUMBER OF GRAMS
• FOR EXAMPLE:
• SINCE CaCO3 CONTAINS
• 1Ca FROM THE PERIODIC
TABLE WE USE 40grams x 1
• 1 C FROM THE PERIODIC
TABLE WE USE 12 grams x1
• 3 O FROM THE PERIODIC
•
•
TABLE WE USE 16 grams x 3
(1x 40) + (1x 12) + (3 x 16) = 100
THE MASS OF ONE MOLE OF
CALCIUM CARBONATE IS
100 grams
• IN OUR PROBLEM WE
ARE USING 200 grams of
CaCO3
• DIVIDING 200grams by
100grams per mole of CaCO3
WE FIND THAT WE HAVE
EXACTLY 2 MOLES OF
THE CALCIUM
CARBONATE REACTANT.
.
STEP II IN SOLVING STIOCHIOMETRY PROBLEMS
THEN IS TO CONVERT THE GIVEN NUMBER OF
GRAMS TO MOLES.
CaCO3(S)  CaO(s) + CO2(g)
• THE EQUATION SAYS THAT ONE CALCIUM CARBONATE
MAKES ONE CALCIUM OXIDE. HOW MANY CALCIUM
OXIDES WOULD TEN CALCIUM CARBONATES MAKES?
• HOW ABOUT TEN?
• WHAT ABOUT ONE HUNDRED CALCIUM CARBONATES?
• WOULDN’T THEY MAKE ONE HUNDRED CALCIUM
OXIDES?
• OF COURSE !!
• THEN WHAT ABOUT A MOLE OF CALCIUM CARBONATE?
WOULDN’T THEY BE EXPECTED TO MAKE A MOLE OF
CALCIUM OXIDE?
• AND OF COURSE THEN TWO MOLES WOULD MAKE TWO
MOLES!
• STEP III IS PROBABLY THE MOST DIFFICULT ONE!
• IT REQUIRES US TO PUT TOGETHER THE
BALANCED EQUATION FROM STEP I AND THE
MOLES THAT WE CALCULATED IN STEP II !!
• ** REMEMBER **
• HERE’S OUR BALANCED EQUATION
• CaCO3(S)  CaO(s) + CO2(g)
• AND HERE’S THE MOLES WE CALCULATED
• DIVIDING 200grams by 100grams per mole of CaCO3
WE FIND THAT WE HAVE EXACTLY 2 MOLES OF
THE CALCIUM CARBONATE REACTANT
CaCO3(S)  CaO(s) + CO2(g)
WHAT ABOUT THE CARBON DIOXIDE?
THE BALANCED EQUATION ALSO SAYS THAT ONE
CALCIUM CARBONATE MAKES ONE CARBON
DIOXIDE TOO.
• SO USING THE SAME LOGIC THAT WE APPLIED
TO THE CALCIUM OXIDE, IT IS OBVIOUS THAT
TWO MOLES OF CALCIUM CARBONATE WILL
PRODUCE EXACTLY TWO MOLES OF CARBON
DIOXIDE ALSO.
• STEP III – USING THE BALANCED EQUATION
RATIOS FOUND IN STEP I AND THE MOLES
DETERMINED IN STEP II, FIND THE MOLES OF
EACH PRODUCT MATERIAL FORMED.
STEP IV – CONVERT THE MOLES FOUND IN STEP III
TO GRAMS (MASS)
• ** REMEMBER** TO CONVERT MOLES TO GRAMS, MULTIPLY THE
MASS OF ONE MOLE FROM THE PERIODIC TABLE BY THE NUMBER
OF MOLES.
• EXAMPLE:
• CaO CONTAINS 1 Ca (1 X 40grams from the Periodic Table) and 1 O (1 x 16
from the Periodic Table)
• THE MASS OF ONE MOLE OF CaO IS (1 x 40) + (1 x 16) = 56 grams per
mole
• TWO MOLES OF CaO ARE FORMED THEREFORE, 2 MOLES x 56 grams
per mole = 112 gram of CaO ARE FORMED IN THE REACTION
• CO2 CONTAINS 1 C (1 x 12grams from the Periodic Table) and 2 O (2 x 16
grams from the Periodic Table)
• THE MASS OF ONE MOLE OF CO2 IS (1 x 12) + (2 x 16) = 44 grams per
moles
• TWO MOLES OF CO2 ARE FORMED THEREFORE, 2 MOLES x 44 grams
per mole = 88 grams of CO2 ARE FORMED IN THE REACTION
CaCO3(S)  CaO(s) +
200 grams
2 moles
0 grams
0 moles
CO2(g)
0 grams
0 moles
before reaction occurs
0 grams
0 moles
112 grams
2moles
after reaction occurs
88 grams
2moles
STARTING TOTAL MASS = 200 +0 +0 = 200 GRAMS
FINAL TOTAL MASS = 0 + 112 + 88 = 200 GRAMS
(CONSERVATION OF MASS)
• QUESTIONS ?????
NOW FOR A HARDER ONE!
• HYDROGEN GAS REACTS WITH OXYGEN GAS TO
FORM WATER. HOW MUCH HYDROGEN AND OXYGEN
MUST BE COMBINED TO MAKE 45 GRAMS OF WATER?
•
•
•
•
•
•
STEP I – WRITE AND BALANCE THE EQUATION
HYDROGEN = H2 (DIATOMIC ELEMENT)
OXYGEN = O2 (DIATOMIC ELEMENT)
WATER = H2O
2 H2(g)
+ 1 O2(g)  2 H2O(g)
THERE ARE 4 ATOMS OF HYDROGEN ON EACH SIDE
AND TWO ATOMS OF OXYGEN ON EACH SIDE.
• THE EQUATION IS BALANCED !!
STEP II – CONVERT THE GIVEN GRAMS TO MOLES
• WATER HAS A MOLAR MASS (MASS OF ONE
MOLE) FROM THE PERIODIC TABLE IS 18 GRAMS.
• { (2 x 1) FOR HYDROGEN AND (1 x 16) FOR
OXYGEN }
• 45 grams of water DIVIDED BY 18 grams per mole = 2.5
moles
• We want to make 2.5 moles of water in our reaction.
STEP III – USE THE BALANCED EQUATION TO FIND
THE MOLES OF REACTANT OR PRODUCT REQUIRED
• Here’s our balanced equation from STEP I
• 2 H2(g)
+
1 O2(g)  2 H2O(g)
• TWO MOLES OF WATER ARE MADE FROM TWO
MOLES OF HYDROGEN. HOW MANY MOLES OF
HYDROGEN WOULD BE NEEDED TO MAKE 2.5
MOLES OF WATER? (A ONE FOR ONE RATIO)
• HOW ABOUT 2.5 ??
• NOW FOR THE OXYGEN !
• IT TAKES ONLY 1 MOLE OF OXYGEN TO MAKE 2
MOLES OF WATER, HALF AS MUCH ! (A TWO FOR
ONE RATIO)
• THEREFORE TO MAKE 2.5 MOLES OF WATER
NEEDS ONLY ½ OF 2.5 MOLES OF OXYGEN!
THEREFORE 1.25 MOLES OF OXYGEN IS
REQUIRED.
STEP IV – CONVERT MOLES OF UNKNOWN
TO GRAMS
• SINCE WE NOW KNOW THAT 2.5 MOLES OF
HYDROGEN IS REQUIRED, WE CAN MULTIPLY 2.5
TIMES 2.0 GRAMS, THE MOLAR MASS OF
HYDROGEN TO GET 5.0 GRAMS OF H2(g) NEEDED!
• THE MOLAR MASS OF OXYGEN IS 32 GRAMS.
MULTIPLYING 32 TIMES 1.25 MOLES GIVES 40
GRAMS OF OXYGEN ARE NEEDED.
THEN, IN ORDER TO MAKE 45.0 GRAMS OF
WATER, USING HYDROGEN GAS AND
OXYGEN GAS WE HAVE CALCULATED
THAT:
• THE MOLES OF HYDROGEN NEEDED ARE 2.5
MOLES OR 5.0 GRAMS AND
• THE MOLES OF OXYGEN NEEDED ARE 1.25 MOLES
OR 40.0 GRAMS AND
• ALL OF THIS HAS BEEN DETERMINED WITHOUT
EVER PICKING UP A TEST TUBE BY USING
STOICHIOMETRY!!!
References
Walt Sautter