Homework
Chapter 3
Chemical Equations and
Stoichiometry
(This is a VERY important chapter)
Chapter 3
„
27, 29, 31, 39, 45, 47, 49, 55, 57, 65, 71, 75,
77, 81, 87, 91, 95, 99, 101, 111, 117, 121
1
2
The Mole
Introduction
When two chemicals react with each other, we
want to know how many atoms or molecules of
each are used do that a formula of the product
can be established.
This means we need a way to count atoms, but
HOW?
The solution to this problem is to define a
convenient unit of matter that contains a know
number of particles.
1 moles = 6.022136736×1023 particles
Up until now we have been talking about
compounds and elements and how to write and
understand what they mean.
Chemistry in more than just that. When we think
of chemistry, we think of chemical reactions.
In order to understand chemical reaction, we
need to study stoichiometry and the chemical
equation.
„
„
Stoichiometry: the quantitative study of chemical
reactions.
Chemical equation: a written representation of a
chemical reaction, showing the reactants and
products, their physical states, and the direction in
which the reaction proceeds.
„
Remember that one mole always contains the
same number of particles, no matter what the
substance.
3
„
„
Thinking about moles with respect to
molecules and compounds is not all that
different than with atoms.
If I have 1 mole of CH4:
1 mole of sodium weighs 22.99 g
To determine the molecular mass of a molecule, we add
up the atomic masses of the elements that make up the
molecule.
Example:
„
CCl4
We have 1 carbon and 4 chlorines
12.001 g/mol + 4(35.4527 g/mol) = 153.81 g/mol
Therefore, if we have 1 mole of CCl4 it would weigh 153.81
g.
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4
Molecules, Compounds, and The
Mole
Molar Mass
The mass of one mole of atoms of any element is the
molar mass, which is numerically equal to the atomic
mass, but in grams.
Example:
This not only true for atoms, but also for molecules,
but we use the molecular mass instead of the atomic
mass.
This value is commonly known has Avogadro’s
number.
1.
How many CH4 molecules do I have?
2.
How much does it weigh?
3.
How many carbon atoms do I have?
4.
How many hydrogen atoms do I have?
I have 6.022 ×1023 molecules of CH4
16.032 g
6.022 ×1023
4 × 6.022 ×1023 = 2.409×1024
6
1
EXAMPLES
MOLES are the CENTER
# of
particles
Vol. of an
“ideal” gas @ STP
e
b
At first these concepts are difficult to understand, but
working problems is the best way to become comfortable
with these concepts.
What is the mass, in
grams, of 1.5 mols of
silicon?
42.1283 g
How many moles are
represented by 454 g of
sulfur? How many
atoms?
14.2 mols
8.53×1024 atoms
f
a
# of moles of X
a) Multiply by Avogadro’s
Number
e) Multiply by 22.4 L / mol
b) Divide by Avogadro’s
Number
f) Divide by 22.4 L / mol
c
d
c) Multiply by the
molecular weight of X
d) Divide by the molecular
weight of X
Mass
7
Percent Composition
8
Empirical and Molecular Formulas
Percent Composition
„
Calculate the molar mass
of CaCO3 (limestone).
100.08 g/mol
If you have 454 g of
CaCO3, how many moles
do you have? How many
molecules? How many
atoms of oxygen?
4.54 mols
2.73×1023 molecules
8.20×1024 atom of O
We have discussed molecular formulas
Empirical formulas
The percentage of the mass of a compound represented by each
of its constituent elements.
According to the law of constant composition, any sample
of a pure compound always consists of the same elements
combined in the same proportion by mass.
Example: What is the percent composition of N and H in
ammonia (NH3)?
Mass Percent N in NH3 =
Mass of N in 1 mol of NH3
14.01 g N
=
× 100 = 82.27%
Mass of 1 mol of NH3
17.03 g NH3
Mass Percent H in NH3 =
Mass of H in 1 mol of NH3 3(1.008) g H
=
× 100 = 17.76%
Mass of 1 mol of NH3
17.03 g NH3
„
A molecular formula showing the simplest
possible ratio of atoms in a molecule.
From percent compositions we can
determine empirical and molecular
formulas by the following procedure.
% A → g A → x mols A
% B → g B → x mols B
Find mole
ratio
x mol A
y mol B
9
Ratio gives
formula
AxBy
10
Example
Example
Eugenol is the active
component of oil of cloves. It
has a molar mass of 164.2
g/mol and is 73.14% C and
7.37% H; the remainder is
oxygen. What are empirical and
molecular formulas for eugenol?
Where to start?!
Eugenol is the active component of oil of cloves. It has a molar mass of 164.2
g/mol and is 73.14% C and 7.37% H; the remainder is oxygen. What are
empirical and molecular formulas for eugenol?
Where to start?!
1.
What is the % of oxygen?
2.
How many grams of C, H, and O?
100% - (73.14% + 7.37%) = 19.49%
Assume you have 100 g of the compound therefore your percentages are the number of
grams of C, H, and O.
„
„
„
3.
73.14 g of C
7.37 g of H
19.49 g of O
Change the grams into moles.
73.14 g of C × 1 mole of C / 12.011g of C = 6.09 mols of C
7.37 g of H × 1 mole of H / 1.0079 g of H = 7.31 mols of H
19.49 g of O × 1 mole of O / 15.9994 g of O = 1.22 mols of O
4.
Find the mole ratios (divide the large amount(s) of moles by the smallest amount of
moles).
6.09 mols of C / 1.22 mols of O = 5 mols of C to 1 mol of O
7.37 mols of H / 1.22 mols of O = 6 mols of H to 1 mol of O
5.
So what does this mean?
6.
How to get the molecular formula?
C5H6O is the empirical formula
Molecular weight of the empirical formula.
„
82 g/mol
Divide the molecular weight of eugenol by the molecular weight empirical formula.
„
11
164.2 g/mol / 82 g/mol = 2
So there are 2 units of the empirical formula.
„
(C5H6O)2 = C10H12O2
12
2
Hydrated Compounds
Determining the Units of Hydration
If ionic compounds are prepared water solution
and then isolated as solids, the crystals often
have molecules of water trapped in the lattice.
Compounds in which molecules of water are
associated with the ions of the compound are
called hydrated compounds.
Examples:
„
„
Example: The following reaction takes place.
1.023 g of CuSO4 · x H2O + heat → 0.654 g CuSO4 + ? g H2O
Find the units of hydration.
Step 1: determine the mass of water
1.023 g of CuSO4 · x H2O – 0.654 g of CuSO4 = 0.369 g of water
Step 2: determine the number of moles
0.369 g of water × 1 mol of water / 18.02 g of water = 0.0205 mols of water
0.654 g CuSO4 × 1 mols CuSO4 / 159.6 g CuSO4 = 0.00410 mols CuSO4
NiCl2 • 6H2O
CaSO4 • 2H2O
Step 3: determine the molar ratio
But what about the molar mass? (Do I add the
water?)
„
„
0.0205 mol H2 O
5.00 mol H2 O
=
0.00410 mol CuSO 4 1.00 mol CuSO 4
So, we know that we started with 1.023 g of
CuSO4 · 5 H2O
YES!
NiCl2 • 6H2O → 237.7 g/mol
129.6 g/mol from NiCl2
108.1 g/mol from 6H2O
„
CaSO4 • 2H2O → 172.14 g/mol
13
Balancing Chemical Equations
Chemical Equation
A written representation of a chemical reaction, showing
the reactants and products, their physical states, and the
direction in which the reaction proceeds.
Example:
„
2 Al(s) + 3 Br2(l) → Al2Br6 (s)
So it takes:
„
„
2 atoms (or moles) of solid aluminum
3 molecules (or moles) of liquid bromine
reactants
To make:
product(s)
„
1 molecule (or mole) of solid Al2Br6
The given reaction is an example of a balanced equation, which is
our next topic.
We have seen what a balanced chemical
equation looks like.
Balancing chemical equations is an
application of both the modern atomic
theory and the law of conservation of
matter.
„
Example:
„
14
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
What are the stoichiometric coefficients?
If you used 8000 atoms of Fe, how many molecules of O2 are
required to consume the iron completely?
8000 atoms of Fe ×
3 molecules of O 2
= 6000 molecules of O 2
4 atoms of Fe
15
16
Example Balancing Chemical Equations
More Examples
Example: write the balanced equation for the complete combustion of propane
(C3H8).
„
„
„
Combustion is the process of burning a compound in the presence of oxygen (O2) to form
CO2 gas and H2O liquid.
This is the unbalanced chemical equation.
Balance the number of C atoms
How many on the left?
„
3
1
„
Change the stoichiometric coefficient.
Balance the number of H atoms
How many on the left?
„
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l)
Combustion of pentane (C5H12).
So, we need to add three C atoms to the right, how?
„
Combustion of butane (C4H10).
„
How many of the right?
„
The number of on kind of atom on the
reactants side must be equal to the number of
the same kind of atom on the product side.
„
8
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l)
How many on the right?
„
2
So, we need to add 6 H atoms to the right, by changing the stoichiometric coefficient, but end up
with a total of 8 H atoms on the right.
„
Balance the number of O atoms
How many on the left?
„
2
How many on the right?
„
10
So, we need to add 8 O atoms to the left and end up with a total of 10 on the left.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
17
18
3
The POWER of a Balanced
Chemical Equation
Figure 3.12: Hydrogen and nitrogen react
to form ammonia according to the equation
N2 + 3H2
2NH3.
With a balance chemical equation, we can do many things.
„
„
Determine how much (moles & grams) of our products we will make
with a known amount of reactants.
Determine how much reactants (moles & grams) we will need to
create a known amount of product.
For example we want to make 64 g of H2O(l) from the
combustion of methane (CH4), how much methane and
oxygen will we need?
„
„
„
„
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
64 g of H2O = 3.56 mols of H2O
3.56 mols of H2O × 1 mol of CH4 / 2 mols of H2O = 1.78 mols of CH4
3.56 mols of H2O × 2 mol of O2 / 2 mols of H2O = 3.56 mols of O2
How did I do that?
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20
Using the Balanced Equation
From Chapter 3
MOLES are the CENTER
# of
particles
Vol. of an
“ideal” gas @ STP
e
b
f
f
a
b) Divide by Avogadro’s
Number
f) Divide by 22.4 L / mol
d
c) Multiply by the
molecular weight of X
d) Divide by the molecular
weight of X
Mass
21
„
„
„
„
b
h
Multiply by
Avogadro’s Number
b)
Divide by
Avogadro’s Number
c)
Multiply by the
molecular weight of
X or Y
d)
Divide by the
molecular weight of
X or Y
g)
cx
dx
h)
Mass
Multiply by the
stoichiometric
coefficient of Y and
divide by the
stoichiometric
coefficient of X.
Multiply by the
stoichiometric
coefficient of X and
divide by the
stoichiometric
coefficient of Y.
cy
# of
particles
dy
Mass
e)
Multiply by 22.4 L / mol
f)
Divide by 22.4 L / mol
22
Limiting Reactants
If 454 g of propane (C3H8) is burned, what mass
of oxygen (in grams) is required for complete
combustion and what masses of carbon dioxide
and water (in grams) are formed?
„
a
# of moles of Y
a
a)
Example
e
g
# of moles of X
e) Multiply by 22.4 L / mol
c
f
e
b
# of
particles
# of moles of X
a) Multiply by Avogadro’s
Number
Vol. of and
“ideal gas” @ STP
Vol. of an
“ideal gas” @ STP
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
454 g of C3H8 × 1 mol of C3H8 / 44.10 g of C3H8= 10.3
mols of C3H8
10.3 mols of C3H8 × 5 mols of O2 / 1 mol of C3H8 =
51.5 mols of O2 × 32.0 g/mol = 1648 g of O2
10.3 mols of C3H8 × 3 mols of CO2 / 1 mol of C3H8 =
30.9 mols of CO2 × 44.0 g/mol = 1360 g of CO2
10.3 mols of C3H8 × 4 mols of H2O / 1 mol of C3H8 =
41.2 mols of H2O × 18.0 g/mol = 742 g of H2O
23
Limiting Reactant
„
„
The reactant present in limited supply that determines the
amount of product formed.
In other words, one of our reactants in a chemical reaction will
run out first and this dictates the amount products formed.
Example: 225 g of SiCl4 is mixed with 225 g of Mg, which
compound is the limiting reactant and how much Si (s)
will be produced.
„
„
SiCl4(l) + 2 Mg(s) → Si(s) + 2 MgCl2(s)
225 g of SiCl4 × 1 mol of SiCl4 / 169.90 g of SiCl4 = 1.32 mols of
SiCl4
1.32 mols of SiCl4 ×1 mol of Si(s) / 1 mol of SiCl4(l) = 1.32 mols of
Si(s)
„
225 g of Mg × 1 mol of Mg / 24.3050 g of Mg = 9.26 mols of Mg
9.26 mols of Mg ×1 mol of Si(s) / 2 mol of Mg = 4.63 mols of Si(s)
„
„
So, SiCl4(l) is the limiting reactant because it only will form 1.32
mols of Si(s).
1.32 mols of Si(s) × 28.0855 g / 1 mol = 37.1 g of Si(s).
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4
Percent Yield
Percent Yield
„
The actual yield of a chemical reaction as a percentage of its
theoretical yield.
Percent yeild =
actual yield
× 100%
theoretical yield
Example: methanol (CH3OH) decomposes to form
hydrogen and carbon monoxide gas. If 125 g of
methanol decomposes what is the theoretical yield of
hydrogen gas? If you obtained only 13.6 g of hydrogen,
what is the percent yield of the gas?
CH3OH(l) → 2 H2(g) + CO(g)
„
„
„
„
125 g of CH3OH × 1 mol of CH3OH / 32.042 g of CH3OH = 3.90
mols of CH3OH(l)
3.90 mol of CH3OH(l) × 2 mols of H2(g) / 1 mol of CH3OH(l) =
7.8 mols of H2(g)
7.8 mols of H2(g) × 2.0158 g of H2(g) / 1 mol of H2(g) = 15.72 g of
H2(g)
25
13.6 / 15.72 × 100 = 86.51 % yield
5