Stoichiometry
The study of the numerical relationship between
quantities in a chemical reaction
Making Pizza
The number of pizzas you can make depends
on the amount of the ingredients you use.
1 crust + 5 oz. tomato sauce + 2 cu cheese ➜ 1 pizza
This relationship can be expressed mathematically
1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza
If you want to make more than one pizza, you can use the
amount of cheese you have to determine the number of
pizzas you can make.
Grams of Amounts from Stoichiometry
Grams of
Predicting
A
B
The amounts of any other substance produced or
consumed in a chemical reaction can be
determined from the amount of just one substance.
Molar Mass
Molar Mass
Moles of
A
Avogadro’s Number
Coefficients
Moles of
B
Avogadro’s Number
1. According to the following equation, how many moles of water
are made in the combustion of 0.10 moles of glucose?
C 6 H 12 O 6 + 6 O 2 ➜ 6 CO 2 + 6 H 2 O
glucose + oxygen gas ➜ carbon dioxide + water
1 mol glucose
6 mol water
6 mol water
1 mol glucose
conversion factors
1 mol C6H12O6
6 mol H2O
6 mol H2O
1 mol C6H12O6
conversion factors
mol
C6H12O6
0.10 mol C6H12O6 x
mol H2O
6 mol H2O
1 mol C6H12O6
= 0.60 mol H2O
Stoichiometry Road Map
Grams of
A
Grams of
B
Molar Mass
Moles of
A
Mole to Mole
Ratio from
balanced
equation
Moles of
B
Avogadro’s Number
Particles of
A
Particles of
B
2. Estimate the mass of CO2 produced in 2007 by the
combustion of 3.5 x 1015 g of octane (C8H18).
2 C8H18(l) + 25 O2(g) ➜ 16 CO2(g) + 18 H2O(g)
g C8H18
mol C8H18
mol CO2
g CO2
1 mol C8H18
114.22 g C8H18
conversion factors
114.22 g C8H18
1 mol C8H18
2 mol C8H18
16 mol CO2
conversion factors
16 mol CO2
2 mol C8H18
1 mol CO2
44.01 g CO2
conversion factors
44.01 g CO2
1 mol CO2
3.5 x 1015 g C8H18 x
1 mol C8H18 x 16 mol CO2 x 44.01 g CO2
114.22 g C8H18 2 mol C8H18
1 mol CO2
=
1.0789
x161016 g CO2
1.1 x 10
3. How many grams of glucose can be synthesized from
37.8 g of CO2 in photosynthesis?
6 CO2 + 6 H2O ➜ C6H12O6 + 6 O2
mol CO2
g CO2
1 mol CO2
44.01 g CO2
1 mol C6H12O6
180.2 g C6H12O6
37.8 g CO2
44.01 g CO2
1 mol CO2
conversion factors
1 mol C6H12O6
6 mol CO2
g C6H12O6
mol C6H12O6
conversion factors
6 mol CO2
1 mol C6H12O6
conversion factors
180.2 g C6H12O6
1 mol C6H12O6
180.2 g C6H12O6
x 1 mol CO2 x 1 mol C6H12O6 x
1 mol C6H12O6
44.01 g CO2
6 mol CO2
=
25.796
g C6H12O6
25.8
4. How many grams of O2 can be made from the
decomposition of 100.0 g of PbO2?
2 PbO2(s) → 2 PbO(s) + O2(g)
(PbO2 = 239.2, O2 = 32.00)
mol PbO2
g PbO2
1 mol PbO2
239.2 g PbO2
conversion factors
2 mol PbO2
1 mol O2
conversion factors
1 mol O2
32.oo g O2
100.0 g PbO2
g O2
mol O2
conversion factors
239.2 g PbO2
1 mol PbO2
1 mol O2
2 mol PbO2
32.00 g O2
1 mol O2
x 1.000 mol PbO2 x 1.000 mol O2 x 32.00 g O2
1.000 mol O2
239.2 g PbO2
2.000 mol PbO2
=
6.689 g O2
Stoichiometry Road Map
Grams of
A
Grams of
B
Molar Mass
Moles of
A
Mole to Mole
Ratio from
balanced
equation
Moles of
B
Avogadro’s Number
Particles of
A
Particles of
B
More Making Pizzas
1 crust + 5 oz. tomato sauce + 2 cu cheese ➜1 pizza
What would happen if we had 4 crusts,
15 oz. tomato sauce, and 10 cu cheese?
Limiting
reagent
Theoretical
yield
The Limiting Reactant
For reactions with multiple reactants, it is
likely that one of the reactants will be
completely used before the others.
When this reactant is used up, the reaction
stops and no more product is made.
Limiting Reactant
The reactant that is completely consumed in a chemical reaction
Theoretical Yield
The amount of product that can be made in a chemical reaction based on the
amount of limiting reactant
Actual Yield
The amount of product actually produced by a chemical reaction
Actual Yield
Percent Yield = --------------------------------- x 100%
Theoretical Yield
Stoichiometry Road Map
A+B
Grams of
A
C
Molar Mass
Moles of
A
Moles of
B
Molar Mass
Grams of
B
Mole to
Mole
Ratio fr
om bala
n
equation ced
ole
Mole to M
alanced
b
m
o
r
f
o
i
Rat
equation
Molar Mass
Moles of
C
Compare the
results of
A and B
Grams of
C
Use the smaller
result to get a
theoretical yield
Limiting and Excess Reactants in the
Combustion of Methane
CH4(g) + O2(g) ➜ CO2(g) + H2O(g)
CH4(g) + 2 O2(g) ➜ CO2(g) + 2 H2O(g)
➜
➜
If we have five molecules of CH4 and eight molecules of O2,
which is the limiting reactant?
5 molecules CH4 x 1 molecule CO2 =
1 molecule CH4
8 molecules O2 x
8 mol O2 x
1 molecule CO2 =
2 molecules O2
1 mol CO2
2 mol O2
5 molecules of CO2
4 molecules of CO2
= 4 mol of CO2
What happens when you mix five molecules of CH4
and eight molecules of O2?
➜
5. How many moles of Si3N4 can be made from
1.20 moles of Si and 1.00 mole of N2 in the reaction:
3 Si + 2 N2 ➜ Si3N4 ?
Limiting
reactant
1.20 mol Si x
1.00 mol Si3N4
3.00 mol Si
=
0.400 mol Si3N4
Theoretical
yield
1.00 mol N2 x
1.00 mol Si3N4
2.00 mol N2
=
0.500 mol Si3N4
6. How many grams of N2(g) can be made from 9.05 g of NH3
reacting with 45.2 g of copper(II) oxide?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
g
NH3
mol
NH3
g
mol
N2
mol
}
smaller amount is from
limiting reactant
mol
g
CuO
mol
mol
CuO
mol
N2
6. How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g
of copper(II) oxide?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
9.05 g NH3 x 1.00 mol NH3 x 1.00 mol N2 =
17.03 g NH3 2.00 mol NH3
45.2 g CuO x 1.00 mol CuO x 1.00 mol N2 =
79.55 g CuO 3.00 mol CuO
0.1894 mol N2
Theoretical yield
Limiting reactant
0.189 mol N2 x
0.2657 mol N2
28.02 g N2 =
1.00 mol N2
5.30 g N2
More Making Pizzas
Let’s now assume that as we are making pizzas, we
burn a pizza, drop one on the floor, or other
uncontrollable events happen so that we only make
two pizzas. This is the actual yield.
We can calculate the efficiency of making pizzas by
calculating the percentage of the maximum number of
pizzas we actually make. In chemical reactions, we
call this the percent yield.
Actual Yield x 100 % = Percent Yield
Theoretical Yield
2 pizzas x 100 % = 67%
3 pizzas
6. How many grams of N2(g) can be made from 9.05 g of NH3
reacting with 45.2 g of copper(II) oxide?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
If 4.61 g of N2 are made, what is the percent yield?
smaller
mol
N2
g
N2
Theoretical Yield
Actual Yield
=
Theoretical Yield
% Yield
6. How many grams of N2(g) can be made from 9.05 g of NH3
reacting with 45.2 g of copper(II) oxide?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
If 4.61 g of N2 are isolated, what is the percent yield?
45.2 g CuO x 1.00 mol CuO x 1.00 mol N2 =
79.55 g CuO 3.00 mol CuO
0.189 mol N2 x
28.02 g N2 =
1.00 mol N2
5.30 g N2
0.1894 mol N2
Theoretical yield
4.61 g N2 x 100% = 87.0 %
5.30 g N2
percent yield
7. When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of
Ti are obtained. Find the limiting reactant, theoretical yield,
and percent yield.
TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)
kg
C
g
C
kg
g
kg
TiO2
g
TiO2
mol
C
mol
Ti
g
mol
mol
mol
mol
TiO2
}
smaller
amount is
from
limiting
reactant
mol
Ti
7. When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of
Ti are obtained. Find the limiting reactant, theoretical yield,
and percent yield.
TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)
smaller
mol
Ti
g
Ti
kg
Ti
Theoretical Yield
Actual Yield
=
Theoretical Yield
% Yield
7. When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of
Ti are obtained. Find the limiting reactant, theoretical yield,
and percent yield.
TiO2(s) + 2 C(s)➜
Ti(s) + 2 CO(g)
Collect needed relationships:
1000 g = 1 kg
Molar Mass Ti = 47.87 g/mol
Molar Mass C = 12.01 g/mol Molar Mass TiO2 = 79.87 g/mol
1 mole TiO2 : 1 mol Ti
2 mole C : 1 mol Ti
7. When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained.
Find the limiting reactant, theoretical yield, and percent yield.
TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)
28.6 kg C x
1000 g C
1.00 mol C
1.00 mol Ti
x
x
1 kg C
12.01 g C
2.00 mol C
3 3 mol Ti
1.19 xx10
= 1.1907
10
88.2 kg TiO2 x 1000 g TiO2 x 1.00 mol TiO2 x 1.00 mol Ti
1 kg TiO2
1.00 mol TiO2
79.87 g TiO2
3 3 mol Ti
1.10 xx10
= 1.1043
10
limiting reactant
Theoretical yield
7. When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of
Ti are obtained. Find the limiting reactant, theoretical yield,
and percent yield.
TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)
1.10 x 103 mol Ti x 47.87 g Ti x
1 mol Ti
1.00 kg Ti
1000 g Ti
=
52.9 kg Ti
theoretical yield
percent yield
Enthalpy
A Measure of the Heat Evolved
or Absorbed in a Reaction
Enthalpy
Exothermic Reactions emit thermal energy
when they occur.
Endothermic Reactions absorb thermal
energy when they occur.
Enthalpy - The amount of thermal energy
emitted or absorbed by a chemical reaction, under
conditions of constant pressure.
We can only measure the change in enthalpy,
therefore the important quantity is ΔH.
Molecular View of Exothermic Reactions
The products of the reaction
have less chemical potential
energy than the reactants.
The difference in energy is
released as heat to the
surroundings.
REACTANTS
Enthalpy
The temperature of the
surroundings rises due to
release of thermal energy by
the reaction.
Hi
decrease in enthalpy
PRODUCTS
Reaction coordinate
ΔH = Hf -Hi
ΔH is “-’’
Hf
Molecular View of Endothermic Reactions
The products of the reaction
have more chemical potential
energy than the reactants.
The difference in energy is
absorbed and becomes part of
the chemical potential energy
of the products.
PRODUCTS
Enthalpy
The temperature of the
surroundings decreases due to
absorption of thermal energy
by the reaction.
Hf
increase in enthalpy
REACTANTS
Reaction coordinate
ΔH = Hf -Hi
ΔH is “+’’
Hi
Sign of ΔHrxn
Combustion of CH4, the main component in natural gas:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
ΔHrxn = -802.3 kJ
This reaction is exothermic and therefore has a
negative enthalpy of reaction.
The sign and magnitude of ΔHrxn tell us that 802.3 kJ of
heat are emitted when 1 mol CH4 reacts with 2 mol O2.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) + 802.3 kJ
Sign of ΔHrxn
Reaction between nitrogen and oxygen gas to form nitrogen monoxide:
N2(g) + O2(g) → 2 NO(g)
ΔHrxn = +182.6 kJ
This reaction is endothermic and therefore has a
positive enthalpy of reaction.
The sign and magnitude of ΔHrxn tell us that 182.6 kJ of heat are
absorbed from the surroundings when 1 mol N2 reacts with 1 mol O2.
182.6 kJ + N2(g) + O2(g) → 2 NO(g)
8. How much heat is evolved in the complete
combustion of 13.2 kg of C3H8(g)?
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
ΔH = −2044 kJ
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
g C 3H 8
kg C3H8
+
2044 kJ
kJ
mol C3H8
1000
g
1 mol
C 3H 8
kg
44.09 g
C 3H 8
2044
kJ
1 mol
C 3H 8
1000 g C3H8
1 mol C3H8
2044 kJ
13.2 kg C3H8 x
x
x
1.00 kg C3H8
44.09 g C3H8
1 mol C3H8
5 kJ
5 kJ
= 6.12
6.1195
x 10
x 10