Chemistry 100(02) Fall 2009
Webpage: http://blackboard.latech.edu/
Dr. Upali Siriwardane
CTH 311 Phone 257-4941
Upali@chem.latech.edu
Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F 10:00
- 12:00.
Exams:
October 5,
2007 (Test 1): Chapter 1 & 2
October 22, 2007 (Test 3): Chapter 3 & 4
November 14, 2007 (Chapter 5 & 6)
November 15, 2007 (Make-up test) comprehensive:
Chapters 1 through 6 )
8:00-9:15 AM, CTH 328
CHEM 100, Fall 2009, LA TECH
4-1
Chapter 4. Quantities of Reactants and
Products
4.1 Chemical Equations
4.2 Patterns of Chemical Reactions
4.3 Balancing Chemical Equations
4.4 The Mole and Chemical Reactions: The MacroNano Connection
4.5 Reactions with One Reactant in Limited Supply
4.6 Evaluating the Success of a Synthesis:
Percent Yield
4.7 Percent Composition and Empirical Formulas
CHEM 100, Fall 2009, LA TECH
4-2
Chapter 4. Quantities of Reactants and
Products
•Chemical Equations
•Balanced chemical equation
•Patterns of Chemical Reactions
•Combination, decomposition and
displacement
•Stoichiometric coefficients
•Stoichiometric Conversion Factors
•Moles of Reactants
•Mole ratios of reactants and
products
•Converting moles of Reactants
to Products
• CHEM 100, Fall 2009, LA TECH
•Calculate grams of products from
grams of reactants
•Stiochiometric Reactions
•Reactions in One reactant in
Limited Supply
•Limiting reactant
•Evaluating Success of Synthesis
•Theoretical Yield
•Actual Yield
•Percent Yield
•Stoichiometric principles to find
the empirical formula
4-3
Types of Reactions
Synthesis reactions or combination reactions
Decomposition reactions
Displacement reactions
• Single
• Double(Exchange reactions)
Combustion Reactions
Formation Reactions
CHEM 100, Fall 2009, LA TECH
4-4
Types of Chemical Reactions
CHEM 100, Fall 2009, LA TECH
4-5
Reaction of H2 and I2
CHEM 100, Fall 2009, LA TECH
4-6
Synthesis or Combination Reactions
Formation of a compound from simpler
compounds or elements.
Decomposition reactions
Compound breaks up to from simpler compounds
or elements.
Displacement reactions
Single displacement: In a compound an element
is replaced by another element.
Exchange reactions
Double displacement: In a compound group or ion
is replaced by another group or ion in another
compound.
. CHEM 100, Fall 2009, LA TECH
4-7
Formation Reactions
Formation of a compound from elements at standard
state.
Combustion Reactions
Compound reacts with oxygen to produce oxides:
water and carbon dioxide for organic compounds
Acid/Base (Neutralization)Reactions
An acid and a base react to form water and salt
( most ionic compounds are salts)
Precipitation Reactions
when two aqueous salt solutions are mixed an
insoluble salt is formed when two aqueous salt
solutions are mixed.
CHEM 100, Fall 2009, LA TECH
4-8
Combination Reaction
CHEM 100, Fall 2009, LA TECH
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Decomposition Reactions
CHEM 100, Fall 2009, LA TECH
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Dynamite
CHEM 100, Fall 2009, LA TECH
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Electrolysis
Decomposition caused by an electric current
Anode
• electrode where oxidation occurs
Cathode
• electrode where reduction occurs
CHEM 100, Fall 2009, LA TECH
4-12
Electrolysis
CHEM 100, Fall 2009, LA TECH
4-13
Displacement Reactions
CHEM 100, Fall 2009, LA TECH
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Exchange Reactions
They are double displacement or exchange
reactions of ionic compounds where an insoluble
salt is formed (Precipitation Reactions) when two
aqueous salt solutions are mixed.
Ba(NO3)2(aq) +Na2SO4(aq)= BaSO4(s) + 2 NaNO3(aq)
CHEM 100, Fall 2009, LA TECH
4-15
Chemical Equation
P4O10 (s) + 6H2O (l)
=
4 H3PO4(l)
reactants enter into a reaction.
products are formed by the reaction.
Parantheses represent physical state
Stoichiometric Coefficients are numbers in front of
chemical formula formula
gives the amounts (moles) of each substance used
and each substance produced.
Equation Must be balanced!
CHEM 100, Fall 2009, LA TECH
4-16
Chemical Reaction
Could be described in words
Chemical equation:
Reactants?
Products?
Reaction conditions?
=, ---> , <==> or
?
Stoichiometric coefficients?
Number in front of substances representing moles,
atoms, molecules
CHEM 100, Fall 2009, LA TECH
4-17
Balanced Chemical Equation
representation of a chemical reaction which
uses coefficients (prefix numbers known as
stoichiometric coefficients) to represent the
relative amounts of reactants and products
CHEM 100, Fall 2009, LA TECH
4-18
Writing and Balancing
Chemical Equations
Write a word equation.
Convert word equation into formula equation.
Balance the formula equation by the use of prefixes
(coefficients) to balance the number of each type
of atom on the reactant and product sides of the
equation.
CHEM 100, Fall 2009, LA TECH
4-19
Balancing Chemical Equations
1. Check for Diatomic Molecules - H2 - N2- O2 - F2 - Cl2 - Br2 - I2
``If these elements appear
2. By Themselves in an equation, they Must be written with a
subscript of 2
3. Balance Metals
4. Balance Nonmetals
5. Balance Oxygen
6. Balance Hydrogen
7.Recount All Atoms
8. If EVERY coefficient will reduce, rewrite in the simplest
whole-number ratio.
CHEM 100, Fall 2009, LA TECH
4-20
Example
Hydrogen gas reacts with oxygen gas to
produce water.
Step 1.
hydrogen + oxygen  water
Step 2.
H2 + O2  H2O
Step 3.
2 H2 + O2  2 H2O
CHEM 100, Fall 2009, LA TECH
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Example
Iron(III) oxide reacts with carbon monoxide to
produce the iron oxide (Fe3O4) and carbon
dioxide.
iron(III) oxide + carbon monoxide  Fe3O4 + carbon
dioxide
Fe2O3 + CO  Fe3O4 + CO2
3 Fe2O3 + CO  2 Fe3O4 + CO2
CHEM 100, Fall 2009, LA TECH
4-22
Stoichiometry
stoi·chi·om·e·try noun
Calculations of the quantitative relationships
between reactants and products in a chemical
reaction.
CHEM 100, Fall 2009, LA TECH
4-23
Stoichiometric Relationships
CHEM 100, Fall 2009, LA TECH
4-24
Stoichiometric Converstion Factors
The large numbers in front of chemical formulas.
Coefficients represent the number of molecules of
the substance in the reaction. They provide the
convestion factor to conver moles of reactants to
products or vice versa.
4 NH3(g) + 5 O2(g) ------> 4 NO(g) + 6 H2O(g)
4 mol NH3 = 5 mol O2
5 mol O2 = 6 mol H2O
4 mol NH3 = 4 mol NO; 1mole NH3 = 1 mole NO
1 mol NH3 = 1 mol NO
CHEM 100, Fall 2009, LA TECH
4-25
Examples
Calculate the following using the chemical
equation given below:
4 NH3(g) + 5 O2(g) ----> 4 NO(g) + 6 H2O(g)
a) moles of NO(g) from 2 moles of NH3(g)
and excess O2(g).
b) moles of H2O(g) from 3 moles of O2(g) and
excess NH3(g).
CHEM 100, Fall 2009, LA TECH
4-26
The Mole and Chemical Reactions:
The Macro-Nano Connection
2 H2
2 H2 molecules
+ O2 
2 H 2O
1 O2 molecule
molecules
2 moles H2 molecules 1 mole O2 molecules
molecules
2 kmoles H2 molecules 1 kmole O2 molecules
molecules
2 mmoles H2 molecules 1 mmole O2 molecules
molecules
4 g H2
32 g O2
CHEM 100, Fall 2009, LA TECH
2 H 2O
2 moles H2O
2 kmoles H2O
2 mmoles H2O
36 g H2O
4-27
Examples
How many moles of H2O will be
produced by 0.80 mole of O2 with
excess H2 according to the
equation?
2H2(g) + O2(g) = 2 H2O(l)
CHEM 100, Fall 2009, LA TECH
4-28
Stoichiometric Reactions
Reactions where mole ratio of the products
and reactants are the same as mole ratio
from the stoichiometric coefficients in the
balanced chemical equation. All reactants
will be completely converted into products.
CHEM 100, Fall 2009, LA TECH
4-29
Limiting Reactant
CHEM 100, Fall 2009, LA TECH
4-30
Limiting Reactant
CHEM 100, Fall 2009, LA TECH
4-31
Analogy in Recipe : Making Cheese
Sandwiches
You were given 20 slices bread,
5 slices of cheese,
4 slices of ham If you want to make
sandwiches containing
two slices bread and one slice of cheese
and one slice of ham,
How many sandwiches you could make?
What is the limiting ingredient?
CHEM 100, Fall 2009, LA TECH
4-32
What is the limiting reagent?
Limiting reagent is the reactant, which
is used up first.
To find the limiting reactant you have
to compare the amounts of reactants
in moles.
CHEM 100, Fall 2009, LA TECH
4-33
Ways to find Limiting Reactant
There two ways to find the limiting ingredient:
1) Comparing mole ratio of the chemical
equation to mole ratio calculated from the
grams of reactants
2) Calculate moles of product and whichever
reactant produces lowest moles of product is
the limiting reactant
CHEM 100, Fall 2009, LA TECH
4-34
Examples
A 300.0 g sample of phosphorus
( M.W. 123.88 g/mol) burns in 500.0 g of
oxygen (M.W. 32.00 g/mol) according to
following equation:
P4(s) + 5O2(g) = P4O10(s)
What is the limiting reagent?
CHEM 100, Fall 2009, LA TECH
4-35
First Method
moles of
P4
and
O2
300
500
---------- = 2.42 mol P4;
----------= 15.63 mol O2
123.88
32.00
P4
and
O2
theoretical mole ratio 1
:
5
actually mole ratio
1
: 6.46
P4 is present in lower amount than required, it is used up first.
Therefore, P4 is the limiting reagent.
Second Method
2.42 mol P4
1 mol P4O10
1 mol P4
15.63 mole O2
1 mol P4O10
= 2.42 mole P4O10
= 3.13 mole P4O10
5 mole O2
CHEM 100, Fall 2009, LA TECH
4-36
EXAMPLE How much H2O, in moles
results from burning an excess of
H2 in 3.3 moles of O2?
2 H2 + O2  2 H2O
#mol H2O =
(3.3 mol O2) x (2 mol H2O)
(1 mol O2)
= 6.6 mol H2O
Mole ratio from
balanced chemical equation
CHEM 100, Fall 2009, LA TECH
4-37
Theoretical yield
Theoretical yield is the amount (grams)
of products formed according to
chemical equation.
Use the limiting reagent to calculate the
moles of the product and then convert
moles to grams.
CHEM 100, Fall 2009, LA TECH
4-38
Actual Yield
Actual yield is the grams of the product
obtained by an experiment.
Actual yield should be less than the
theoretical yield if the experiment was
carried out meticulously. If the
products are contaminated with
impurities or the formula of product
was wrong the actual yield could be
higher.
CHEM 100, Fall 2009, LA TECH
4-39
% Yield
actual yield
% yield = ------------------------ x 100
theoretical yield
If the products are contaminated with
impurities or the formula of product
was wrong the % yield could be higher
than 100%.
CHEM 100, Fall 2009, LA TECH
4-40
Question
Sulfur trioxide, SO3 , is made from the
oxidation of SO2 and the reaction is
represented by the equation
2SO2 + O2
-----> 2SO3
A 16.0-g sample of SO2 gives 18.0 g
of SO3. The percent yield of SO3
is
CHEM 100, Fall 2009, LA TECH
4-41
Examples
A 300.0 g sample of phosphorus ( M.W. 123.88
g/mol) burns in 500.0 g of oxygen (M.W. 32.00
g/mol) according to following equation:
P4(s) + 5O2(g) = P4O10(s)
a) What is the limiting reagent?
b) How many moles of P4O10 are produced
theoretically?
c) If 612 g of P4O10 is actually produced in this
reaction, calculate the percent yield.
CHEM 100, Fall 2009, LA TECH
4-42
Steps in Stoichiometric
Calculations
Check whether chemical equation is balanced
get the moles from grams of materials
find the limiting reactant
calculate moles of products from the limiting
reactant
convert moles of the products to grams
find the actual yield of the reaction
calculate % yield of the reaction
CHEM 100, Fall 2009, LA TECH
4-43
Question
How much hydrogen gas is
produced when 1 kg of sodium
reacts with water?
CHEM 100, Fall 2009, LA TECH
4-44
Question
2Al(s) + 6HCl(aq)--> 2AlCl3(aq)
+3H2(g) According to the equation
above, how many grams of
aluminum are needed to react with
0.582 mol of hydrochloric acid?
CHEM 100, Fall 2009, LA TECH
4-45
Answer
5.23 g Al
CHEM 100, Fall 2009, LA TECH
4-46
EXAMPLE How much iron(III) oxide
and aluminum powder are required to
field weld the ends of two rails
together?
Assume that the
rail is 132 lb/yard
and that one inch
of the rails will be
covered by an
additional 10%
mass of iron.
weld
Photo by Mike Condren
CHEM 100, Fall 2009, LA TECH
4-47
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field
weld the ends of two rails together?
Assume that the rail is 132 lb/yard and
that one inch of the rails will be covered
by an additional 10% mass of iron.
The mass of iron in 1 inch of this rail is:
#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)
= 1.67  103 g/in
The mass of iron in a 3weld adding 10% mass:
#g = (1.67  10 g) (0.10) = 167 g
CHEM 100, Fall 2009, LA TECH
4-48
EXAMPLE How much iron(III) oxide and aluminum
powder are required to field weld the ends of two
rails together? Assume that the rail is 132 lb/yard
and that one inch of the rails will be covered by an
additonal 10% mass of iron.
The mass of iron in a weld adding 10% mass:
#g = (1.67  103 g) (0.10) = 167 g
Balanced chemical equation:
Fe2O3 + 2 Al  2 Fe + Al2O3
What mass of Fe2O3 is required for the thermite process?
(1 mol Fe) (1 mol Fe2O3) (159.7 g Fe2O3)
#g Fe2O3 = (167 g Fe) 
(55.85 g Fe) (2 mol Fe) (1 mol Fe2O3)
= 238 g Fe2O3
CHEM 100, Fall 2009, LA TECH
4-49
EXAMPLE How much iron(III) oxide and aluminum
powder are required to field weld the ends of two rails
together? Assume that the rail is 132 lb/yard and that
one inch of the rails will be covered by an additonal
10% mass of iron.
The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe
Balanced chemical equation:
Fe2O3 + 2 Al  2 Fe + Al2O3
What mass of Fe2O3 is required for the thermite process?
#g Fe2O3 = 238 g Fe2O3
What mass of Al is required for the thermite process?
(1 mol Fe) (2 mol Al) (26.9815 g Al)
#g Al = (167 g Fe) 
(55.85 g Fe) (2 mol Fe) (1 mol Al)
= 80.6 g Al
CHEM 100, Fall 2009, LA TECH
4-50
EXAMPLE What is the number of moles of
CaSO4 (S) that can be produced by allowing 1.0
mol SO2, 2.0 mol CaCO3, and 3.0 mol O2 to react?
2SO2(g) + 2CaCO3(s) + O2(g)  2CaSO4(S) + 2CO2(g)
balanced equation relates:
2SO2(g)  2CaCO3(s)  O2(g)
have only:
1SO2(g)  2CaCO3(s)  3O2(g)
not enough SO2 to use all of the CaCO3 or the O2
not enough CaCO3 to use of the O2
SO2 is the limiting reactant
CHEM 100, Fall 2009, LA TECH
4-51
EXAMPLE What is the number of moles of
CaSO4 (S) that can be produced by allowing 1.0
mol SO2, 2.0 mol CaCO3, and 3.0 mol O2 to react?
2SO2(g) + 2CaCO3(s) + O2(g)  2CaSO4(S) + 2CO2(g)
have only:
1SO2(g)  2CaCO3(s)  3O2(g)
SO2 is the limiting reactant
if use all of SO2
#CaCO3 = (1 mol SO2)(2 mol CaSO4/2 mol SO2)
= 1 mol CaSO4
CHEM 100, Fall 2009, LA TECH
4-52
EXAMPLE A rocket fuel, hydrazine,
is produced by a reaction of Cl2 that is
reacted with excess NaOH and NH3.
What (a) theoretical yield can be
produced from 1.00 kg of Cl2?
2NaOH + Cl2 + 2NH3  N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for the overall process
#kg N2H4 =
(1.00 kg Cl2)
(1 kmol Cl2) (1 kmol N2H4)
(32.0 g N2H4)
(70.9 kg Cl2)
(1 mol N2H4)
molar mass
CHEM 100, Fall 2009, LA TECH
(1 kmol Cl2)
balanced
equation
= 0.451 kg N2H4
molar mass
4-53
EXAMPLE (b) What is the actual yield
if 0.299 kg of 98.0% N2H4 is produced
for every 1.00 kg of Cl2?
2NaOH + Cl2 + 2NH3  N2H4 + 2NaCl + 2H2O
(a) theoretical yield
#kg N2H4 = 0.451 kg N2H4
(b) actual yield
(0.299 kg product) (98.0 kg N2H4 )
= 0.293 kg N2H4
# kg N2H4 =
(100 kg product)
purity factor
CHEM 100, Fall 2009, LA TECH
4-54
EXAMPLE (c) What is the percent yield of
pure N2H4?
2NaOH + Cl2 + 2NH3  N2H4 + 2NaCl + 2H2O
(a) theoretical yield
#kg N2H4 = 0.451 kg N2H4
(b) actual yield
# kg N2H4 = 0.293 kg N2H4
(c) percent yield
0.293 kg
% yield =
 100 = 65.0 % yield
0.451kg
CHEM 100, Fall 2009, LA TECH
4-55
Combustion Analysis
CHEM 100, Fall 2009, LA TECH
4-56
Example Benzoic acid is known to contain
only C, H, and O. A 6.49-mg sample of benzoic acid
was burned completely in a C-H analyzer. The
increase in the mass of each absorption tube
showed that 16.4-mg of CO2 and 2.85-mg of H2O
formed. What is the empirical formula of benzoic
acid?
#mg C =
(16.4-mg of CO2 )(12.01-mg C)
= 4.48-mg C
(44.01-mg CO2)
%C =
4.48-mg C
 100 = 68.9% C
6.49-mg sample
CHEM 100, Fall 2009, LA TECH
4-57
Example Benzoic acid is known to contain
only C, H, and O. A 6.49-mg sample of benzoic acid
was burned completely in a C-H analyzer. The
increase in the mass of each absorption tube
showed that 16.4-mg of CO2 and 2.85-mg of H2O
formed. What is the empirical formula of benzoic
acid?
#mg H =
(2.85-mg of H2O )(2.02-mg H)
= 0.319-mg H
(18.02-mg H2O)
%C =
0.319-mg H
 100 = 4.92% H
6.49-mg sample
CHEM 100, Fall 2009, LA TECH
4-58
Example Benzoic acid is known to
contain only C, H, and O. A 6.49-mg sample
of benzoic acid was burned completely in a
C-H analyzer. The increase in the mass of
each absorption tube showed that 16.4-mg
of CO2 and 2.85-mg of H2O formed. What is
the empirical formula of benzoic acid?
68.9% C
4.92% H
% O = (100 - (68.9% C + 4.92% H) = 26.2% O
CHEM 100, Fall 2009, LA TECH
4-59
Example Benzoic acid is known to contain only C, H,
and O. A 6.49-mg sample of benzoic acid was burned
completely in a C-H analyzer. The increase in the mass of
each absorption tube showed that 16.4-mg of CO2 and
2.85-mg of H2O formed. What is the empirical formula of
benzoic acid?
Relative # Atoms
%
C 68.9
H 4.92
O 26.2
(%/gaw)
68.9/12.0 = 5.75
4.92/1.01 = 4.87
26.2/16.0 = 1.64
Divide by Smallest Multiply by Integer
5.75/1.64 = 3.51
3.51  2 = 7
4.87/1.64 = 2.97
2.97  2 = 6
1.64/1.64 = 1.00
1.00  2 = 2
C 7H 6O 2
CHEM 100, Fall 2009, LA TECH
4-60