Stoichiometry
L
What is it really?
• Cooking
• Method of finding
– how many cookies you can make given these
ingredients
– how many ingredients you had to make this many
cookies
All goes back to balanced equations
• First write the equation and balance it.
• You are observing the combustion of methane,
CH4. Your teacher asks you how many moles of
water are expected to form in the complete
combustion of 2.50 mol of methane.
CH4 + O2  CO2 + H2O
UNBALENCED
CH4 + 2O2  CO2 + 2H2O
BALENCED
Now the math
• Remember that the balanced equations shows
the relationship of how many moles of each
compound is used in this reaction
– Like the ingredients on a recipe (3 eggs + ½ cup
water = eggy water)
• So the coefficient of one compound to the
next is the mole ratio
Now the math (….continued)
You are observing the combustion of methane,
CH4. Your teacher asks you how many moles of
water are expected to form in the complete
combustion of 2.50 mol of methane.
CH4 + 2O2  CO2 + 2H2O
So if we had 1 mole of methane we would produce 2 moles of water
But since we have 2.50 mol CH4…
The math
You are observing the combustion of methane,
CH4. Your teacher asks you how many moles of
water are expected to form in the complete
combustion of 2.50 mol of methane.
Mole ratio
CH4 + 2O2  CO2 + 2H2O
5 mol H2O
Therefore we would expect to retrieve 5 mols of H2O
from the combustion of 2.5 mol of methane
mole to mole
• The synthesis reaction between nitrogen
monoxide and oxygen forms nitrogen dioxide.
– How many moles of nitrogen monoxide would be
needed to fully react with 3.6 moles of oxygen
– How many moles of nitrogen dioxide would be
produced from 3.6 moles of oxygen
– How many moles of oxygen would be needed to
produce 4.67 mol nitrogen dioxide?
– How many moles of nitrogen monoxide would be
needed to produce 4.67 mol nitrogen dioxide?
Building off that…
• If you can do mol to mol, the rest should be
child’s play
• Mass to mol
– Convert mass to mol, then mol to mol.
– The end
• Mass to mass
– Convert mass of given to mol, convert mols of given to
moles of wanted, convert mol to mass
– The end
• Particles to anything
– Use Avagotto’s number (6.022 x 1023 particles/mol)
Mass to mol
• In a reaction between the elements aluminum
and chlorine, aluminum chloride is produced.
How many grams of AlCl3 will be produced if
2.50 moles of Al react?
Al + Cl2  AlCl3
Unbalanced
2Al + 3Cl2  2AlCl3
Balanced
Mol to mass continued
2Al + 3Cl2  2AlCl3
How many grams of AlCl3 will be produced if 2.50
moles of Al react?
2.50
1 mol
mol
AlCl
AlCl
3 3
333.33 g AlCl3
Therefore 333.33 g of AlCl3 will be produced by the
reaction of 2.50 mol of Al
Practice
The ammonia (NH3) used to make fertilizers for
lawns and gardens is made by reacting nitrogen
and hydrogen.
– Determine the mass in grams of NH3 formed from
1.34 moles of nitrogen.
– What is the mass in grams of hydrogen required to
react with 1.34 moles of nitrogen?
– How many moles of nitrogen are required to produce
11.7 grams of NH3?
Mass to mass
• When nitrogen and hydrogen react, they form
ammonia gas, which has the formula NH3. If
56.0 g of nitrogen are used up in the reaction,
how many grams of ammonia will be
produced?
N2 + H2  NH3
Unbalanced
N2 + 3H2  2NH3
Balanced
When nitrogen and hydrogen react, they form ammonia gas, which has the
formula NH3. If 56.0 g of nitrogen are used up in the reaction, how many
grams of ammonia will be produced?
N2 + 3H2  2NH3
• Convert grams to mol
• Convert mol to gram
3.9971 mol NH3
68.11 g NH3
Therefore, 68.11 g of NH3 are produced from 56 g
of N2
Particles to Particles
• Hydrogen and Oxygen react to produce water.
How many particles of hydrogen would be
needed to react fully with 1.01 x 1012 particles
of oxygen?
H2+ O2  H2O
Unbalanced
2H2 + O2  2H2O
Balanced
Hydrogen and Oxygen react to produce water. How many
particles of hydrogen would be needed to react fully with 1.01 x
1012 particles of oxygen?
2H2 + O2  2H2O
• Convert particles to mol
• Convert mol to mol
• Convert mol to particles
1.6772 x 10-12 mol O2
2.14 x 1012
particles of H2
Therefore we expect to need 2.14 x 1012 particles of
H2 to react fully with 1.01 x 1012 particles of O2
Practice
• Packet!
Percent yield
• How much are you actually getting out of your
experiment?
% Yield = Actual yield/theoretical yield x 100
What is the total mass of H2O produced when 384g Cu is
completely consumed in the following reaction?
Cu + 4HNO3  Cu(NO3)2 + 2H2O + 2NO2
• Convert mass to mol
• Convert mol to mol
• Convert mol to mass
216g H2O
Therefore we expect to produce 216g of H2O.
What is the percent yield if this procedure was carried out in the lab and only
produced 200g of H2O
We expected to produce 216g of H2O.
What is the percent yield if this procedure was carried
out in the lab and only produced 200g of H2O
• What was the actual yield?
– 200g
• What was the theoretical yield?
– 216g
• Solve
% yield ==
92.6%
Actual
yield
200g
Theoretical
yield
216g
• Therefore, our % yield is 92.6%
Practice
• Ethanol (C2H5OH) is produced from • Lead (II) oxide is
the fermentation of sucrose in the
obtained by roasting
presence of enzymes:
galena, lead (II)
sulfide, in air:
C12H22O11(aq) + H2O(g)  C2H4OH(l) + CO2(g)
PbS(s) + O2(g)  PbO(s) + SO2(g)
• determine the theoretical and
• Determine the
percent yields of ethanol if 684g
theoretical and
sucrose undergoes fermentation
percent yields of PbO
and 349g ethanol is obtained
if 200.0g PbS is heated
and 170.0g PbO is
obtained
T.Y. = 369g
% yield = 94.6%
T.Y. = 186.6g PbO
% Yield = 91.10%
Limiting reagent
• The reactant that is
completely
consumed during a
rxn
• Reactants that
remain are excess
reactants
To find the limiting reagent
• Compare the actual mol:mol ratio to the
balanced formula’s mol:mol ratio
• Calculations must be based on given amounts
of limiting reactants
Fe + 2NiO(OH) + 2H2O  Fe(OH)2 + 2Ni(OH)2
• Determine the number of moles of Fe(OH)2
produced if 5.00 mol Fe and 8.00 mol of NiO(OH)
react. What is the limiting reactant? What
reactant is in excess and by how many moles?
• 1mol Fe::2mol NiO(OH). Therefore,5mol Fe would
need 10mol NiO(OH)
• Since we only have 8.00mol NiO(OH), this is our
limiting reagent.
Fe + 2NiO(OH) + 2H2O  Fe(OH)2 + 2Ni(OH)2
Determine the number of moles of Fe(OH)2 produced if 5.00 mol Fe and 8.00 mol of
NiO(OH) react. What is the limiting reactant? What reactant is in excess and by how
many moles?
• Since NiO(OH) is the limiting reagent, find how
many moles of Fe are needed to completely
react with the amount of it we do have
4mol Fe
• 4 Mols of Fe are needed
5.00mol Fe(provided) - 4mol Fe(needed) = 1mol Fe(excess)
• Therefore 1 mol of Fe is in excess
It gets complicated
• If you are given anything other then mol, you
need to first convert to mol for BOTH reagents
before you can determine the limiting factor!
Zn + 2MnO2 +H2O  Zn(OH)2 + Mn2O3
• Determine the limiting reactant if 25.0g Zn and 30.0g MnO2 are
used, Determine the mass of Zn(OH)2 produced.
• 0.77g moles of MnO2 required to react with 25.0g of Zn
• 0.17 moles of Zn required to react with 30.0g of Mn(OH)2
• Mn(OH)2 is the limiting reagent
– excess of 0.21mol Zn (0.38-0.17 = 0.21)
• Use 30.0g to find the mass of Zn(OH)2
• 17.07g Zn(OH)2