Chapter 4
!
Chemical Quantities
and
Aqueous Reactions
Stoichiometry
The study of the numerical relationship between
chemical 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.
Predicting Amounts from Stoichiometry
The amounts of any other substance produced or
consumed in a chemical reaction can be
determined from the amount of just one substance.
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
1 mol C6H12O6
6 mol H2O
6 mol water
1 mol glucose
conversion factors
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
Grams of
A
Grams of
B
The amounts of any other substance produced or
consumed in a chemical reaction can be
Molar Mass from the amount of justMolar
Mass
determined
one substance.
Moles of
A
Coefficients
Moles of
B
Avogadro’s Number
Avogadro’s Number
Particles of
Particles of
Estimate the mass of CO2 produced in 2007 by the
combustion of 3.5 x 1015 g of octane (C8H18).
C8H18(l) + O2(g) ➜ CO2(g) + H2O(g)
2 C8H18(l) + 25 O2(g) ➜ 16 CO2(g) + 18 H2O(g)
g C8H18
mol C8H18
mol CO2
g CO2
2 mol C8H18
16 mol CO2
conversion factors
16 mol CO2
2 mol C8H18
1 mol C8H18
114.22 g C8H18
conversion factors
114.22 g C8H18
1 mol C8H18
1 mol CO2
44.01 g CO2
conversion factors
44.01 g CO2
1 mol CO2
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
2 mol C8H18
16 mol CO2
mol C8H18
mol CO2
g CO2
conversion factors
16 mol CO2
2 mol C8H18
1 mol C8H18
114.22 g C8H18
conversion factors
114.22 g C8H18
1 mol C8H18
1 mol CO2
44.01 g CO2
conversion factors
44.01 g CO2
1 mol CO2
16 mol CO2! 44.01 g CO2!
3.5 x 1015 g C8H18 x 1 mol C8H18! x
x
!
114.22 g C8H18 2 mol C8H18
1 mol CO2
!
!
!
=
1.0789
x161016 g CO2
1.1 x 10
How many grams of glucose can be synthesized from 37.8 g
of CO2 in photosynthesis?
6 CO2 + 6 H2O ➜ C6H12O6 + 6 O2
g CO2
mol CO2
1 mol CO2
44.01 g CO2
mol C6H12O6
g C6H12O6
44.01 g CO2
1 mol CO2
conversion factors
1 mol C6H12O
6 mol CO2
conversion factors
6 mol CO2
1 mol C6H12O6
1 mol C6H12O6
180.2 g C6H12O6
conversion factors
180.2 g C6H12O6
1 mol C6H12O6
37.8 g CO2
!
!
!
x 1 mol CO2! x 1 mol C6H12O! x 180.2 g C6H12!O6!
1 mol C6H12O6
44.01 g CO2
6 mol CO2
=
25.796
g C6H12O6
25.8
Lead (IV) oxide decomposes to yield lead(II) oxide and
oxygen gas. 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)
g PbO2
1 mol PbO2
239.2 g PbO2
mol PbO2
mol O2
1 mol O2
2 mol PbO2
1 mol O2!
100.0 g PbO2 x 1 mol PbO2! x
239.2 g PbO2
2 mol PbO2
!
!
!
=
6.68896
g O2
6.689
g O2
32.00 g O2
1 mol O2
x
32.00 g O2! !
1 mol 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
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?
8 mol O2 x
1 mol CO2!
2 mol O2
= 4 mol of CO2
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.
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
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. The actual amount of product
made in a chemical reaction is called the
actual yield.
We can determine 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.
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?
g!
NH3
mol!
NH3
g
mol!
N2
mol
}
smaller amount is from
limiting reactant
mol
g!
CuO
mol
mol!
CuO
mol!
N2
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?
9.05 g NH3 x 1.00 mol NH3!x 1.00 mol N2! =
17.03 g NH3 2.00 mol NH3
0.2657 mol N2
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
Smaller # moles of N2
Limiting reactant
Theoretical yield
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
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
0.1894 mol N2 x 28.02 g N2! =
1.00 mol N2
0.189 mol N2 x 28.02 g N2! =
1.00 mol N2
4.61 g N2! x 100% = 87.0 %
5.30 g N2
percent yield
0.1894 mol N2
5.307 g N2
Theoretical yield
5.30 g N2
4.61 g N2! x 100% =
5.31 g N2
86.8 %
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
k
g
kg!
TiO2
g!
TiO2
mol!
C
mol!
Ti
g
mol
mol
mol
mol!
TiO2
}
smaller!
amount is!
from !
limiting!
reactant
mol!
Ti
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
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
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
smallest moles of Ti
Theoretical yield
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