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Unit 2
Moles, Chemical Reactions,
and Stoichiometry
The Mole
• A mole (mol) is the
amount of a substance
that contains as many
particles as there are
atoms in exactly 12 g
of carbon-12.
• It is a counting unit, similar to a dozen. In a
dozen, there are 12 things. In a mole, there
are 6.02 x 1023 things.
Visual Concept
Molar Mass
• Molar Mass is the mass of one mole of a pure
substance.
• Molar mass units
are g/mol.
• The molar mass of
an element is the
same number as
its atomic mass,
only the units are
different.
• Examples: Molar mass of H = 1.0 g/mol
Molar mass of O = 16.0 g/mol
Formula Mass
• The formula mass of any compound is the sum
of the masses of all the atoms in its formula.
Example:
Formula mass of water, H2O:
H2 = 1.0 amu x 2 = 2.0 amu.
O=
+ 16.0 amu.
18.0 amu
• A compound’s molar mass is numerically equal
to its formula mass. Only the units are
different. (Ex: Molar mass of H2O = 18.0 g.)
Molar Masses
Sample Problem
Determine the molar mass of each of the
following compounds:
Solution:
Al2 = 27.0 x 2 =
54.0 g
a. Al2S3
S3 = 32.1 x 3 = + 96.3 g
150.3 g
=
b. Ba(OH)2 Ba
O2 = 16.0 x 2 =
H2 = 1.0 x 2 =
137.3 g
32.0 g
+ 2.0 g
171.3 g
Percent Composition
• The percentage by
mass of each element
in a compound is
known as the
percent composition
of the compound.
% of element = mass of element in compound x 100
molar mass of compound
Percent Composition
Sample Problem
Find the percentage composition of copper(I)
sulfide, Cu2S.
Cu2 = 63.5 x 2 = 127.0 g
Solution:
S
= + 32.1 g
1. Find the molar
mass of Cu2S:
159.1 g
2. Find the percentage by mass of each element:
127.0 g
% Cu =
x 100 = 79.80% Cu
159.1 g
32.1 g
%S=
x 100 = 20.2% S
159.1 g
Molar Mass as a Conversion Factor
• The molar mass of a compound can be used as
a conversion factor to convert between moles
and grams for a given substance.
Example:
• What is the mass of 2.5 moles of H2O?
molar mass of H2O = 18.0 g/mol
given
conversion factor
2.5 mol H2O x 18.0 g H2O = 45 g H2O
1 mol H2O
Molar Mass as a Conversion Factor
Sample Problem
Calculate the moles in 1170 g of copper (II) nitrate.
Solution:
2+
1. Determine the correct formula: Cu (NO3) 2 Cu(NO3)2
= 63.5 g
2. Calculate the molar mass: Cu
N2 = 14.0 x 2 = 28.0 g
3. Convert from g to mol:
O6 = 16.0 x 6 = + 96.0 g
Given
1170 g Cu(NO3)2 x
Conversion factor
187.5 g
1 mol Cu(NO3)2 = 6.24 mol
Cu(NO3)2
187.5 g Cu(NO3)2
Avogadro’s Number
• Avogadro’s number:
6.022 1415 × 1023— the
number of particles in exactly
one mole of a pure substance.
• Named for
nineteenth-century
Italian scientist
Amedeo Avogadro.
Visual Concept
Conversions with Avogadro’s Number
• Avogadro’s number can be used as a conversion
factor between number of particles (atoms or
molecules) and moles.
Example:
•How many moles of silver, Ag, are in 3.01  1023
atoms of silver?
Given
3.01 x 1023 atoms Ag x
Conversion factor
1 mol Ag
= 0.500 mol Ag
6.02 x 1023 atoms Ag
Combining Conversion Factors
Sample Problem
What is the mass in grams of 1.20  108 atoms
of copper, Cu?
nd
2
Solution:
st
Given
1.20 x 108 atoms Cu x
1
Conversion factor
Conversion
factor
1 mol Cu
x 63.5 g Cu
6.02 x 1023atoms Cu
1 mol Cu
= 1.27 x 10-14 g Cu
Standard Molar Volume
• Standard Temperature and Pressure (STP) is
0oC and 1 atm.
• The Standard Molar Volume of a gas is the
volume occupied by one
mole of a gas at STP.
It has been found to
be 22.4 L.
Molar Volume Conversion Factor
• Standard Molar Volume can be used as a
conversion factor to convert from the
number of moles of a gas at STP to volume
(L), or vice versa.
Molar Volume Conversion
Sample Problem
a. What quantity of gas, in moles, is contained
in 5.00 L at STP?
5.00 L x
1 mol
22.4 L
= 0.223 mol
b. What volume does 0.768 moles of a gas
occupy at STP?
0.768 mol x
22.4 L
1 mol
= 17.2 L
The Mole Map
• You can now convert between number of particles,
mass (g), and volume (L) by going through moles.
Chemical Reactions
• A chemical reaction is the process by which one
or more substances are changed into one or more
different substances.
• Signs that a chemical reaction is taking place:
1. Release of energy as heat and/or light.
2. Production of a gas.
3. Formation of a precipitate - a solid
that separates from a liquid solution.
4. Color change.
The Law of Conservation of Mass
• In any chemical reaction, the original
substances are called reactants and the
resulting substances are called products.
• According to the law of conservation of
mass, the total mass of reactants must equal
the total mass of
products for any
given chemical
reaction.
Chemical Equations
• A chemical equation represents a chemical
reaction using symbols and formulas.
Example:
2H2O(l)
2H2(g) + O2(g)
Reactant
Products
Diatomic Molecules
• Oxygen gas (O2) is an example of an element that
normally exists as a diatomic molecule. You need
to memorize all seven:
Balancing Equations
• The final step in writing correct chemical
equations is to make sure the law of conservation
of mass is satisfied.
– The numbers and types of atoms on both sides
of the equation must be the same – this is
called balancing an equation.
– Equations are balanced by inserting
coefficients - whole numbers that appear in
front of formulas in a chemical equation.
Balancing Equations
Sample Problem A
Balance the following equation:
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Solution:
• Start with the easiest element…carbon.
– Carbon is already balanced.
• Next count the hydrogen atoms.
– Two more hydrogen atoms are needed on the right.
• Finally, count oxygen atoms.
– There are 4 oxygens on the right side of the equation, but
only two on the left.
– Add a coefficient 2 in front of the O2 on the left.
Balancing Equations
Sample Problem B
Balance the following equation:
Al4C3(s) + 12 H2O(l)
3 CH4(g) + 4 Al(OH)3(s)
Solution:
• Let’s start with aluminum.
– Add a coefficient 4 to Al(OH)3 on the right.
• Next count the carbon atoms.
– Add a coefficient 3 to CH4 on the right.
• Balance the oxygen atoms.
– Add a 12 to the H2O on the left.
• Lastly, count the hydrogen atoms.
– Hydrogen is already balanced.
Writing Chemical Equations
Sample Problem
Write an equation for the reaction that occurs when solid
copper metal reacts with aqueous silver nitrate to produce
solid silver metal and aqueous copper(II) nitrate.
Solution:
• First, use correct formulas and symbols to write a chemical
equation.
• Then, balance your equation.
Cu(s) + 2 AgNO3(aq)
2 Ag(s) + Cu(NO3)2(aq)
Types of Chemical Reactions
• There are 5 basic types of chemical reactions:
1.
2.
3.
4.
5.
Synthesis
Decomposition
Single-Displacement
Double-Displacement
Combustion
Synthesis Reactions
• In a synthesis reaction (also called a composition
reaction) two or more substances combine to
form a new compound.
• This type of reaction is represented by the
following general equation:
A+X
AX
Decomposition Reactions
• In a decomposition reaction, a single compound
breaks apart to form 2 or
more simpler substances.
• Decomposition is the opposite of synthesis.
• This type of reaction is represented by the following
general equation:
AX
A+X
Single-Displacement Reactions
• In a single-displacement reaction (also called
single-replacement) one element replaces a similar
element in a compound.
• They often take place in aqueous solution.
• This type of reaction is represented by the following
general equation:
A + BX
AX + B
Double-Displacement Reactions
• In double-displacement reactions, the ions
of 2 compounds exchange places in an
aqueous solution to form 2 new compounds.
• One of the compounds formed is usually either a
precipitate, a gas, or water.
• Represented by the following general equation:
AX + BY
AY + BX
Combustion Reactions
• In a combustion reaction,
a hydrocarbon fuel combines with
oxygen, releasing a large amount of
energy in the form of light and heat.
• Products of combustion reactions are always
carbon dioxide and water vapor.
• Example: Combustion of propane
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
Types of Reactions
Sample Problem
Classify each of the following reactions as a
synthesis, decomposition, single-displacement,
double-displacement, or combustion reaction.
a. N2(g) + 3H2(g) → 2NH3(g)
synthesis
b. 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)
single-displacement
c. 2NaNO3(s) → 2NaNO2(s) + O2(g)
decomposition
d. 2C6H14(l) + 19O2(g) → 12CO2(g) + 14H2O(l)
combustion
What is Stoichiometry?
• Stoichiometry is the branch
of chemistry that deals with
the mass relationships of
elements in a chemical
reaction.
• Stoichiometry calculations
always start with a balanced
chemical equation.
Mole Ratio
• A mole ratio is a conversion factor that relates
the amounts in moles of any two substances
involved in a chemical reaction
Example:
2Al2O3(l) → 4Al(s) + 3O2(g)
Mole Ratios:
2 mol Al2O3
4 mol Al
2 mol Al2O3
3 mol O2
4 mol Al
3 mol O2
Mole to Mole Conversions
• Using the mole ratio, you can convert from moles
of one substance to moles of any other substance
in a chemical reaction.
Example:
• For the reaction N2 + 3H2 → 2NH3, how many moles
of H2 are required to produce 12 moles of NH3?
This is what
we’re given:
Use mole ratio as
a conversion factor
12 mol NH3 x
3 mol H2
2 mol NH3
= 18 mol H2
Mole to Mass Conversions
• To convert from moles of one substance to grams
of another, you need 2 conversion factors:
1. mole ratio.
2. molar mass of the unknown.
• To set up your conversion factor, always put the
units you have on the bottom and the units you
need on the top.
Mole to Mass Conversions (continued)
Example:
Given the equation: 2Mg(s) + O2(g)→ 2MgO(s),
Calculate the mass in grams of magnesium oxide
which is produced from 2.00 mol of magnesium.
Given
2.00 mol Mg x
1st C.F.
Mole Ratio
2nd C.F.
Molar Mass
of the Unknown
2 mol MgO x 40.3 g MgO
= 80.6 g MgO
2 mol Mg
1 mol MgO
Mass to Mole Conversions
• To convert from grams of one substance to
moles of another, you need 2 conversion
factors:
1. molar mass of the given.
2. mole ratio.
• Mass to mole conversion
factors are the inverse
of mole to mass
conversion factors.
Mass to Mole Conversions (continued)
Example:
Given the equation: 2HgO(s) → 2Hg(s) + O2(g),
How many moles of HgO are needed to produce
125g of O2?
Given
125 g O2
1st C.F.
Molar Mass
of the Given
x 1 mol O2
32.0 g O2
2nd C.F.
Mole Ratio
x
2 mol HgO
1 mol O2
= 7.81 mol HgO
Mass to Mass Conversions
• To convert from grams of one substance to grams
of another, you need 3 conversion factors:
1. molar mass of the given.
2. mole ratio.
3. molar mass of the unknown.
Mass to Mass Conversions (continued)
Example:
Given the equation: 2HgO(s) → 2Hg(s) + O2(g),
How many grams of HgO are needed to produce
45.0 g of O2?
1st
Given
C.F.
Molar Mass
of the Given
1
mol O2
x
45.0 g O2
32.0 g O2
3rd C.F.
Molar Mass of
2nd C.F.
Mole Ratio the Unknown
x 2 mol HgO x 216.6 g HgO = 609 g
1 mol O2
1 mol HgO HgO
Volume Ratios
• You can use the volume ratios as conversion
factors just like you would use mole ratios.
2CO(g)
+ O2(g)
→
2 molecules 1 molecule
2 mole
1 mole
2 volumes
1 volume
2CO2(g)
2 molecules
2 mol
2 volumes
• Example: What volume of O2 is needed to react
completely with 0.626 L of CO to form CO2?
0.626 L CO x
1 L O2
2 L CO
= 0.313 L O2
Gas Stoichiometry
Sample Problem
Assume that 5.61 L H2 at STP reacts with excess CuO
according to the following equation:
CuO(s) + H2(g) → Cu(s) + H2O(g)
a. How many moles of H2 react?
5.61 L H2 x 1 mol H2 = 0.250 mol H2
22.4 L H2
b. How many grams of Cu are produced?
5.61 L H2 x 1 mol H2 x 1 mol Cu x 63.5 g Cu = 15.9 g
22.4 L H2 1 mol H2 1 mol Cu
Cu
Limiting Reactants
• When combining 2 or more different things to
make a product, you have to stop when one of
the things is used up.
• For example, no matter how many tires there
are, if there are only 8 car bodies, then only 8
cars can be made.
Limiting Reactants (continued)
• The limiting reactant is the reactant that
limits the amount of product formed.
• The excess reactant is the substance that is
not used up completely.
• Once the limiting reactant is used up, a
chemical reaction will stop.
Limiting Reactants (continued)
Example:
Silicon dioxide reacts with hydrogen fluoride according
to the following equation:
The limiting
SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)
reactant
If 6.0 mol HF is added to 4.5 mol SiO2, which makes the
is the limiting reactant? Mole ratio C.F.
least product.
Set up 2 6.0 mol HF x
equations,
one for each
given:
4.5 mol SiO2 x
1 mol SiF4 = 1.5 mol SiF44
4 mol HF Limiting Reactant is HF
1 mol SiF4 = 4.5 mol SiF4
1 mol SiO2
Percentage Yield
• The theoretical yield is the maximum amount of
product that can be produced from a given
amount of reactant (found with stoichiometry).
• The actual yield of a product is the measured
amount of that product obtained from a reaction
(given in problem).
• The percentage yield
is the ratio of the
actual yield to the
theoretical yield, multiplied by 100.
Percentage Yield
Sample Problem A
Given the following equation:
C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g)
When 36.8 g C6H6 react with excess Cl2, the actual
yield of C6H5Cl is 38.8 g. What is the % yield of C6H5Cl?
Set up a mass to mass conversion
This is the
Theoretical Yield
36.8 g C6H6 x 1 mol C6H6 x 1 mol C6H5Cl x 112.5 g C6H5Cl = 53.1 g
78.0 g C6H6 1 mol C6H6
1 mol C6H5Cl C6H5Cl
Actual x 100 38.8 g C6H5Cl x 100
Percent
= 73.1%
Now use
=
=
Yield
Theoretical
53.1 g C6H5Cl
the formula
Percentage Yield
Sample Problem B
According to the following reaction :
CO(g) + 2H2(g) → CH3OH(l)
If the typical yield is 80.0%, what mass of CH3OH should
be expected if 75.0 g of CO reacts with excess H2 gas?
Set up a mass to mass conversion
This is the
Theoretical Yield
75.0 g CO x 1 mol CO x 1 mol CH3OH x 32.0 g CH3OH = 85.7 g
28.0 g CO
1 mol CO
1 mol CH3OH
CH3OH
Multiply theoretical
yield by percentage
to get actual yield
85.7 g CH3OH x 80.0% =
68.6 g CH3OH