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CLASS NOTES FOR CHEMISTRY 1411
4.1 Climate Change and the Combustion of Fossil Fuels
4.2 Reaction Stoichiometry: How Much Carbon Dioxide?
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) [This is fossil fuel combustion.]
The coefficients in a chemical equation specify the relative amounts in moles of each of the
substances involved in the reaction. The numerical relationships between chemical amounts in
a balanced chemical equation are called reaction stoichiometry.
Making Pizza: The Relationships among Ingredients
1 crust + 5 ounces tomato sauce + 2 cups of cheese  1 pizza
If we have 6 cups of cheese and enough of everything else how many pizzas could we make?
6 cups of cheese X
1 pizza
2 cups of cheese
= 3 pizzas
Making Molecules: Mole-to-Mole Conversions
Looking at the equation above for fossil fuel combustion:
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)
2 mol C8H18(l) : 16 mol CO2
So if we have 22.0 mol C8H18(l) how many moles of CO2 would be produced?
Plan: moles of octane  moles of CO2
22 mol C8H18 X
16 mol CO2
2 mol C8 H18
=
176 mol of CO2
Making Molecules: Mass-to-Mass Conversions
Plan: mass of A to moles of A to moles of B to mass of B
Molar mass of C8H18(l), octane, is 114.22 g/mol
Molar mass of CO2(g) is 44.01 g/mol
In 2010 the world burned approximately 3.5 x 1015 g of octane (gasoline). How many g of
carbon dioxide are released into the atmosphere when that much gasoline is burned?
Plan: g of octane  moles octane  moles of CO2  grams of CO2
3.5 x 1015 g octane X
1 mol octane
114.22 g octane
X
16 mol CO2
2 mol octane
X
44.01 g CO2
1 mol CO2
= 1.1 x 1016 g CO2
Volcanos emit 2.0 x 1014 g of CO2 per year or 1.8% as much.
Example 4.1 In photosynthesis, plants convert carbon dioxide and water into glucose,
C6H12O6 according to the equation:
6 CO2(g) + 6 H2O(l)  6 O2(g) + 1 C6H12O6(aq)
Suppose you determine that a particular plant consumes 37.8 g of CO2 and there is more
than enough water present to react with all the CO2, what mass of glucose in grams can the
plant synthesize from CO2?
PLAN: g of CO2  moles of CO2 moles of glucose  g of glucose
Molar mass of CO2 = 44.01 g/mol
Molar mass of C6H12O6 = 180.2 g/mol
37.8 g CO2 X
1 mol CO2
44.01 g CO2
X
1 mol C6 H12 O6
6 mol CO2
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X
180.2 g C6 H12 O6
1 mol C6 H12 O6
=
25.8 g
Example 4.2 Sulfuric acid, H2SO4 is a component of acid rain that forms when SO2, a
pollutant, reacts with oxygen and water according to the simplified reaction:
2 SO2 + O2(g) + 2 H2O(l)  2 H2SO4(aq)
The generation of electricity used by a medium-sized home produces about 25 kg of of SO2
per year. Assuming that there is more than enough O2 and H2O, what mass of sulfuric acid in
kg can form from this much sulfur dioxide?
Plan: kg SO2  g SO2  moles SO2  moles of H2SO4  g of H2SO4  kg H2SO4
Molar mass of SO2 = 64.07 g/mol
Molar mass of H2SO4 = 98.09 g/mol
25 kg SO2 X
1000 g
1 kg
X
1 mol SO2
64.07 g
x
2 mol H2 SO4
2 mol SO2
X
98.09 g H2 SO4
1 mol H2 SO4
X
1 kg
1000 g
= 38 kg
4.3 Limiting Reactant, Theoretical Yield, and Percent Yield
1 crust + 5 ounces tomato sauce + 2 cups of cheese  1 pizza
Suppose we have 4 crusts, 10 cups of cheese and 15 ounces of tomato sauce. How many pizzas
can we make?
We have enough crusts to make: 4 crusts X
1 pizza
1 crust
We have enough cheese to make: 10 cups cheese X
= 4 pizzas
1 pizza
2 cups cheese
= 5 pizzas
We have enough tomato sauce to make:
15 ounces tomato sauce X
1 pizza
5 ounces of tom.sauce
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= 3 pizzas
We can only make 3 pizzas because we don’t have enough tomato sauce to make more than
that. It is the limiting reactant. Three pizzas is the theoretical yield—the maximum we could
make.
What if two pizzas were the actual yield? Maybe we burned one or dropped one.
% yield =
actual yield
theoretical yield
X 100
IMPORTANT:
The limiting reactant (or limiting reagent) is the reactant that is completely consumed in a
chemical reaction and limits the amount of product.
The reactant is excess (or excess reagent) is any reactant that occurs in a quantity greater than
is required to completely react with the limiting reactant. Or what’s left over.
The theoretical yield is the amount of product that can be made in a chemical reaction based
on the amount of limiting reactant.
The actual yield is the amount of product actually produced by a chemical reaction.
The percent yield is calculated as
actual yield
theoretical yield
X 100
Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Reactant Masses
2 Mg(s) + O2(g)  2 MgO(s)
A reaction mixture contains 42.5 g Mg and 33.8 g O2, what is the limiting reactant and the
theoretical yield?
Plan: (a) g of Mg  to moles of Mg moles of MgO if all Mg is used up. (There is enough
oxygen for that amount of Mg.)
(b) Then g of O2  to moles of O2 moles of MgO if all O2 is used up. (There is enough Mg
for that amount of O2.
(c) Compare which gives us less moles of MgO. Use the lesser amount and calculate g of MgO.
42.5 g Mg X
33.8 g O2 X
1 mol Mg
24.31 g Mg
1 mol O2
32.00g O2
X
X
2 mol MgO
2 mol Mg
2 mol MgO
1 mol O2
= 1.7481 mol MgO LIMITING REACTANT
= 2.1125 mol MgO
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EXCESS REACTANT
We could go on to grams for each of the equations above and then compare the number of
grams obtained. The smaller of the two would be what we get.
40.31 g MgO
Use the smaller one: 1.7481 mol Mg X
1 mol MgO
Is the actual yield is 55.9 g, what is the percent yield?
= 70.5 g is the theoretical yield
55.9 g MgO
70.5 gMgO
X 100 = 79.3 %
Example 4.3 Ammonia, NH3, can be synthesized by the reaction:
2NO(g) + 5 H2(g)  2 NH3(g) + 2 H2O(g)
Starting with 86.3 g of NO and 25.6 g H2, find the theoretical yield of ammonia in grams.
Plan: g of NO  mol NO  mol NH3
g of H2  mol H2  mol NH3
From lesser number of mol NH3  g of NH3
86.3 g NO X
25.6 g H2 X
1 mol NO
30.01 g NO
1 mol H2
2.016 g H2
2.876 mol NH3 X
X
X
2 mol NH3
2 mol NO
2 mol NH3
5 mol H2
17.03 g NH3
1 mol NH3
number to go on to grams.
= 2.876 mol of NH3
NO IS LIMITING REACTANT
= 5.079 mol of NH3
= 49.0 g of NH3
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We use the amount of NO, the smaller
Example 4.4 We can obtain titanium metal from the oxide according to the following
balanced equation:
TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti is produced. Find the limiting
reactant, theoretical yield and % yield.
Plan kg C  g C mol C mol Ti
kg TiO2  g TiO2  mol TiO2  mol Ti
From lesser number of mol Ti g of Ti
g of Ti  % yield
28.6 kg C X
1000 g
1 kg
88.2 kg TiO2 C X
1 mol C
12.01 g C
1000 g
1 kg
1.104 x 103 mol Ti X
Per cent yield =
X
X
2 mol C
79.87 g TiO2
1 mol Ti
52.8 kg
1 mol Ti
1 mol TiO2
47.87 g Ti
42.8 kg
X
X
1 kg
1000 g
X
= 1.191 x 103 mol Ti
1 mol Ti
1 mol TiO2
= 1.104 x 103 mol Ti
LIMITING
= 52.8 kg THEORETICAL YIELD [in text 52.9 kg]
X 100 = 81.1 % [in text 80.9%]
4.4 Solution Concentration and Solution Stoichiometry
Solution = homogeneous mixture
Solvent = the larger component in the solution
Solute = the lesser component in the solution
The terms dilute and concentrated are relative amounts and not everyone agrees.
Concentration units to be used: Molarity =
mol solute
liters of solution
Notice it says liters of solution and not liters of solvent.
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Example 4.5 If you dissolve 25.5 g KBr in enough water to make 1.75 L of solution, what is the
molarity of the solution?
Plan: g of KBr  moles of KBr to molarity of solution
Molar mass of KBr = 119.00 g/mol
25.5 g x
1 mol
119.00 g
Molarity =
= 0.21428 mol of KBr
mol solute
liters of solution
=
0.21429 mol
1.75 L
= 0.122 M
Example 4.6 How many liters of a 0.125 M NaOH solution contain 0.255 mol of NaOH?
0.125 M =
0.125 mol
0.255 mol X
We can use this as a conversion factor.
1 Liter
1 liter
0.125 mol
= 2.04 L
Dilution –to dilute a solution, we add additional solvent; therefore, the amount of solute does
not change.
M1V1 = M2V2 This is saying that the moles of the solute at first equals the moles of solute in the
final solution.
Suppose a laboratory procedure calls for 3.00 L of a 0.500 M CaCl2 solution. If we have 10.0 M
solution in stock, how do we prepare the new solution?
M1V1 = M2V2 or
V1 =
M2 V2
0.500 𝑥 3.00
M1
10.0
= 0.150 L
We would take 0.150 L of the 10.0 M solution and dilute to 3.00 L.
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4.5 Types of aqueous Solutions and Solubility
Electrolyte and Nonelectrolyte Solutions
Electrolyte: A substance that produces ions in a water solution and therefore conducts a
current of electricity.
Strong Electrolyte: A substance that dissociates completely in water solution to produce a
lot of ions.
Weak Electrolyte: A substance that doesn’t dissociate completely and only produces a few
ions in water solution.
Nonelectrolyte: A substance that does not produce ions when it dissolves in water and
therefore does not conduct a current of electricity.
Solubility of Ionic Compounds Look at Table 4.1 on page 161 in the text.
Example 4.9 Using the Table 4.1 predict whether each compound is soluble or insoluble.
a.
b.
c.
d.
PbCl2 --insoluble
CuCl2 --soluble
Ca(NO3)2 –soluble
BaSO4 --insoluble
4.6 Precipitation Reactions (This is a type of double displacement; the compounds come
together and switch partners.)
When the compounds switch partners, the negative ion of the first compound goes with the
positive ion of the second compound; the positive ion of the first compound goes with the
negative ion of the second compound.
2 KI(aq) + Pb(NO3)2(aq)  2 KNO3(aq) + PbI2(s)
The two reactants are soluble and the first product is soluble, but the second product is
insoluble. This is a precipitation reaction—two compounds in aqueous solution combine-- at
least one product is insoluble and a precipitate forms. If neither product is insoluble, then
no reaction occurs for this type of reaction.
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Example 4.10 Write the equation for the precipitation reaction that occurs(if any) when
solutions of potassium carbonate and nickel((II) chloride are mixed.
K2CO3(aq) + NiCl2(aq)  2 KCl(aq) + NiCO3(s)
1, When the two compounds come together they switch partners. Write the formulas for the
reactants and the products. Remember charges!!!
2. Use Table 4.1 to decide what is soluble (put aq behind it) and if anything is insoluble (put s
behind it).
3. If we find a product is insoluble, then the reaction goes, because something insoluble
precipitates out.
4. Balance the equation.
Example 4.11 Write the equation for the precipitation reaction that occurs(if any) when
solutions of sodium nitrate and lithium sulfate are mixed.
2 NaNO3(aq) + Li2SO4(aq)  2 LiNO3(aq) + Na2SO4 (aq)
Everything is soluble; therefore no reaction.
4.7 Representation Aqueous Reaction: Molecular, Ionic and Complete Ionic Equations
I omit ionic and complete ionic equations in 1411 and do them in 1412.
4.8 Acid-Base and Gas-Evolution Reactions. (These are both types of double displacement;
the compounds come together and switch partners.)
Acid-base reactions are also called neutralizations.
In gas-evolution reactions one of the products is a gas. Two solutions must be the reactants
for this one.
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For both of these types of reactions, two soluble compounds in solution come together and
trade partners to produce the products.
Acid-Base Reactions (Acid-Base Reactions Always Go)
Acid: Substance that produces H+ ions in aqueous solution is the Arrhenius definition of acid.
Base: Substance that produces OH- in aqueous solution is the Arrhenius definition of base.
Actually a hydrogen ion is a bare proton and will associate with water to form the hydronium
ion: H+ (aq) + H2O(l)  H3O+ Chemists use the hydrogen ion or hydronium ion to mean the
same thing and use them interchangeably.
This is an example of an acid-base reaction:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Notice they exchange partners.
Example 4.13 Write the molecular equation for the reaction between aqueous HI and
aqueous Ba(OH)2.
2 HI (aq) + Ba(OH)2  BaI2(aq) + 2 H2O(l)
Have them change partners; write the formulas correctly based on charges.
Balance the equation.
Gas-Evolution Reactions Remember two solutions must come together for this type.
Example:
H2SO4(aq) + Li2S(aq)  H2S(g) + Li2SO4(aq)
Example 4.15
HNO3(aq) + Na2CO3(aq)  H2CO3(aq) + NaNO3(aq)
Have them switch partners and write the equations correctly based on charges.
However H2CO3(aq) decomposes to water and carbon dioxide. Thus, the equation:
2 HNO3(aq) + Na2CO3(aq)  H2O(l) + CO2(g) + 2 NaNO3(aq)
Finish by balancing the equations with the correct coefficients.
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Redox Reactons
Oxidation: Electrons are lost by a substance.
Reduction: Electrons are gained by a substance.
These always occur together—you can’t have oxidation without reduction occurring at the
same time. Oxidation-reduction also called redox is a transfer of electrons.
LEO the lion says GER. Loss of electrons is oxidation; gain of electrons is reduction.
Oxidation States (Also called oxidation numbers.)
Rules for Assigning Oxidation States (numbers):
Note: Oxidation states are not the same as real charges in ionic substances.
1. The oxidation state of an atom in a free element in its most stable state is 0.
2. The oxidation state of a monoatomic ion is equal to its charge.
3. The sum of the oxidation states of all atoms in:
a. A neutral molecule or formula unit is 0.
b. An ion is equal to the charge of the ion.
4. In their compounds, metals have a positive oxidation states.
a. Group 1A metals always have an oxidation state of +1
b. Group 2A always have an oxidation state of +2.
5. In their compounds, nonmetals are assigned oxidation states according to the Table on
page 177. Look at directions for the Table’s use.
I could summarize:
H in compounds is usually +1
O is usually -2
Halogens in binary compounds when they are the negative part are -1.
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Sulfur in binary compounds when it is the negative part is -2.
Example 4.16 Assign an oxidation state to each atom, in each element, ion, or
compound.
Cl2 0
Rule 1
Na+ +1 Rule 2
KF K is +1 and F is -1
Rule 4a and 5
CO2 C is +4 and O is -2 Rule 5 for O and Rule 3a to calculate C
SO42- O is -2 Rule 5 and S is +6 using Rule 3b to calculate
Identifying Redox Reactions
Is this redox? C + 2 S  CS2
If we see a change in oxidation number as a substance goes from reactant to product, it is
redox.
C at 0  +4 Carbon is oxidized as it loses 4 electrons to form the +4.
S at 0  -2 Sulfur is reduced as it gains 2 electrons to form -2.
Example 4.17 Use oxidation states to identify the element is oxidized and the element that
is reduced in the following redox reaction.
Mg(s) + 2 H2O(l)  Mg(OH)2(aq) + H2(g)
Mg goes from 0 to +2; therefore it lost 2 electrons is oxidized.(Gains in oxidation state.)
H goes from +1 to 0 and therefore gained electrons and is reduced.(Reduced in oxidation
state.)
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Oxidizing agent: The substance that causes the oxidation of another substance and therefore,
it itself, is being reduced.
Reducing agent: The substance that causes the reduction of another substance and therefore,
it itself, is being oxidized.
Example 4.18 Determine whether each reaction is an oxidation-reduction reaction. For each
redox reaction, identify the oxidizing agent and the reducing agent.
(a) 2 Mg(s) + O2(g)  2 MgO(s) Yes It is redox since Mg changes from 0 to +2 and O changes
from 0 to -2. Mg is oxidized therefore it is the reducing agent. O is reduced therefore it is the
oxidizing agent.
(b) 2 HBr(aq) + Ca(OH)2(aq)  2 H2O(l) + CaBr2 (aq) Not redox; there are no changes in
oxidation numbers.
(c) Zn(s) + Fe2+ (aq)  Zn2+(aq) + Fe(s)
Yes it is redox since Zn changes from 0 to +2 and Fe
changes from +2 to 0. Zn is oxidized and is therefore the reducing agent. Iron is reduced and
is therefore the oxidizing agent.
Combustion Reactions (A special kind of redox with an organic substance reacting with
oxygen forming carbon dioxide and water.)
Combustion reactions are characterized by an organic substance reacting with oxygen. This is
a type of oxidation-reduction.
Example: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
Example 4.19 Write a balanced equation for the combustion of liquid methyl alcohol (CH 3OH).
1 CH3OH(l) +1.5 O2(g) 1 CO2(g) +2 H2O(g)
To get rid of fractions, multiply by two.
2 CH3OH(l) +3 O2(g) 2 CO2(g) +4 H2O(g)
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SUMMARY OF REACTION TYPES:
I. Double Displacement : two compounds come together and switch partners.
a. Precipitation Reactions: Two aqueous solutions are mixed and one product is
solid when the reactants switch partners. Must have a solid precipitate form or
it won’t go.
b. Acid-Base Reaction: An acid and a base are mixed and we get water and a salt,
when the acid and base switch partners. Acid-base reactions always go.
c. Gas-Evolution Reaction: Two aqueous solutions are mixed and one product is a
gas. Must have gas formed or it won’t go.
II.
Redox: There is an exchange of electrons and substances change oxidation
numbers.
a. Combustion—special case of redox—organic material combine with oxygen to
produce carbon dioxide and water.
b. Single Displacement—element and a compound and get different element and
different compound.
III.
Combination two or more elements and/or compounds combine to give one type of
product. This may also be redox, but not always.
IV.
Decomposition: one compound decomposes and forms elements and/or
compounds. This may also be redox, but not always.
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