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Chapter 3 Objectives
• Understand 4 general reaction types
• Be able to write and balance chemical equations
• Understand the concepts of formula weight and the mole
as a counting number for particles (atoms or molecules)
• Be able to use balanced chemical equations to covert
between particles/mole/mass of one reactant or product
to particles/mol/mass of another
• Understand the concepts of limiting reactant, theoretical
yield and percent yiels and be able use balanced equations
and given information to calculate each.
Words to Sentences
Δ
• Formulas are our words and we have to get them right:
o Correct chemical symbols
o Correct order
o Correct subscripts
• Equations are our sentences: get the ‘grammar’ right
o Must be balanced (same # and kind of atoms on each side)
o Must include states of reactants and products
o May show other conditions for the reaction (Δ = heat, hn = light)
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Chemical Equations
2 Na(s) + 2 HCl(aq) → 2 NaCl(aq) + H2(g)
• Short-hand way of describing a reaction.
• Provides information about the reaction.
• Formulas of reactants and products.
• Relative numbers of reactant and product molecules
• Allow us to determine masses of reactants and products
needed for complete reaction and the resulting masses of
the products
Chemical Reactions:
where the Action is
Chemical reactions involve rearrangement and exchange of
atoms to produce new molecules.
1. Synthesis (aka formation or combination)
o A+BgAB
2. Decomposition (aka falling apart)
o ABgA+B
3. Single Displacement (or single replacement)
o AB + C g AC + B
4. Double Displacement (or double replacement)
o AB + CD g AD + BC
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Decomposition Reactions
AB g A+B
• Compounds break into
either their constituent
elements or to simpler
compounds
• 2H20 g 2H2 + O2
• Decomposition of
Calcium Carbonate
ammonium dichromate
decomposition
• CaCO3 g CaO + CO2
http://www.youtube.c
om/watch?v=kP_kIX
PjuWI
Single Displacement Reactions:
A + BC g AC + B
Mg + 2HCl —› H2 + Mg2Cl
Cu (s)+ 2AgNO3 (aq)—› Cu(NO3)2(aq) + 2Ag (s)
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Double Replacement Reactions:
AB + CD g AD + BC
2KI(aq) + Pb(NO3)2 (aq) g Pb I2 (s)+ K NO3 (aq)
Remember this?
Conservation of Mass
• Matter cannot be created or destroyed.
o The total mass of the reactants = total mass of the products
• In a chemical reaction, all the atoms present at the
beginning are still present at the end.
o Same number and kind of atoms on each side
o Same number of electrons and charges on each side
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Anatomy of a Reaction
• Methane gas burns (reacts with oxygen gas) to produce carbon
dioxide gas and gaseous water – a combustion reaction.
• The equation shows the minimum whole numbers of reactants
needed to make the products.
• Same number and kind of atoms on each side, even though they
are re-arranged into new molecules.
• The mass is also the same on each side.
Writing Balanced Chemical
Equations
1. Write a skeletal equation for the reactants and products
2. Balance the charges in each formula: Use charges for each ion
to adjust subscripts to make formulas neutral
3. Balance equation using coefficients to give the same number
and kind of each atom on each side
a) Count the number of atoms of each element on each side
of the equation (polyatomic ions may often be counted as
if they are one “element”).
b) Pick an element to balance.
 If an element is found in only one compound on both
sides, balance it first. Do metals before non-metals.
 Leave free elements until last. Balance free elements by
adjusting the coefficient where it is a free element.
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Relationship Between
Moles and Mass
molar mass
• 1 mole of anything contains 6 x 1023 pieces
Lorenzo Romano Amedeo Carlo
Avogadro di Quaregna e di Cerreto,
Count of Quaregna and Cerreto
• The mass of one mole of atoms/a compound = molar mass.
• The molar mass of an element, in grams, is numerically equal
to the element’s atomic mass, in amu.
• The lighter the atom, the less a mole weighs.
Which weighs more: a mole of gold (Au) or a mole of lead (Pb)?
• The lighter the atom, the more atoms there are in 1 g.
Which has more atoms: 50 g of sodium (Na) or 50 g of sulfur (S)?
From Molecules to Moles
• A balanced equation is the “recipe” for a chemical reaction.
• The equation

+
3 H2(g) + N2(g)  2 NH3(g)
tells us that 3 molecules of H2 react with exactly 1 molecule of
N2 to make exactly 2 molecules of NH3
• Since we can count molecules by moles, it also tells us that 3
mol of H2 react with exactly 1 mol of N2 to make exactly 2 mol
of NH3
2.02 g
2.02 g
2.02 g
3 mol of H2
6.06 g H2
+
28.00 g

1 mol of N2
+ 28.00 g N2 g
17.03 g
17.03 g
2 mol of NH3
34.06 g NH3
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Molar Mass of Compounds:
The Cheeseburger Approach
100 g
1 hamburger patty
2 buns
2 x 25 = 50 g
3 pickles
3 x 15 = 45 g
1 cheese
1 x 35 = 35 g
230 g
1 Cheeseburger
Molar mass
Introducing the
1 hamburger patty (Hp)
The
Atomic Cheeseburger 2 buns (Bu2)
equation:
Hp + Ch + Bu2 + 3P g ChHpBu2P3
AMU
100
2 x 25 = 50
3 pickles (P)
3 x 15 = 45
1 cheese (C)
1 x 35 = 35
1 Cheeseburger
CHpBu2P3
230
AMU
Formula mass
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Formula Mass of Compounds
• The formula mass of molecules can be calculated from atomic
weights.
Formula mass for 1 molecule of H2O
H
2 x 1.01
= 2.02 amu
O
1 x 16.00
= 16.00 amu
18.02 amu
• 1 mole of H2O (6.023 x 1023 molecules)
contains 2 moles H and 1 mole of O.
Molar mass for 1 mole H2O
H
O
2 x 1.01 g
1 x 16.00 g
= 2.02 g
= 16.00 g
18.02 g
Mole Relationships in
Chemical Formulas
• Since we count atoms and molecules in mole units,
we can find the number of moles of a constituent
element if we know the number of moles of the
compound.
Hp + Ch + Bu2 + 3P g ChHpBu2P3
Moles of compound
Moles of constituents
1 mol NaCl
1 mol H2O
1 mol CaCO3
1 mol Na, 1 mol Cl
2 mol H, 1 mol O
1 mol Ca, 1 mol C, 3 mol O
1 mol C6H12O6
6 mol C, 12 mol H, 6 mol O
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The Mole as a Conversion Factor
• 1 mole of any element has 6.022 x 1023 atoms:
6.022  1023 atoms
1 mole
1 mole
6.022 1023 atoms
• 1 mole of any element has the same mass in grams as the
atomic mass of that element:
Atom
Potassium, K
Silicon, Si
Neon, Ne
grams/mole
39.10 g/mol
28.09 g/mol
20.18 g/mol
moles/gram
1 mole/39.10 g
1 mole/28.09 g
1 mole/20.18 g
atoms/g
6.022 x 1023/39.10 g
6.022 x 1023/28.09 g
6.022 x 1023/20.18 g
Stoichiometry
1. The study of the numerical relationship between
2.
3.
chemical quantities in a chemical reaction is called
stoichiometry.
Using stoichiometry principles, we can determine the
proportions (by weight or number of molecules) in
which elements or compounds react with one another.
That word? Comes from the Greek words στοιχεῖον
(stoikheion, meaning element) and μέτρον (metron,
meaning measure)
Stoichiometry: Number-Mole-Mass Relationships
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Stoichiometry
The rules for determining stoichiometric relationships are
based on:
• The laws of conservation of mass and energy
• The law of combining weights or volumes.
• The tools used are
• chemical formulas
• chemical equations
H2O
2H2(g) + O2(g) g H2O (l)
• atomic weights
• molar mass / formula weights H2O = 18.02 amu = 18.02 g/mol
• Cheer up – you’ve already been doing it when you balance
equations!
Mass g Moles g Mass Conversion
• We know there is a relationship between the mass and number
of moles of a compound: 1 mole = Molar Mass in grams.
• The molar masses and the balanced chemical equation allow us
to convert from the amount of any chemical in the reaction to
the amount of any other.
mole ratio
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How Many Grams of Glucose Can Be
made from 58.5 g of CO 2 in
Photosynthesis?
• Photosynthesis (opposite of combustion):
6 CO2(g) + 6 H2O(g)  C6H12O6(s) + 6 O2(g)
• The equation for the reaction gives the mole relationship
between amount of C6H12O6 and CO2, but we need to
know the mass relationship, so the solution map will be:
58.5 g CO2
mol C6H12O6
mol CO2
g C6H12O6
from balanced equation
mol CO2
g CO2
mol C6H12O6
mol CO2
g C6H12O6
mol C6H12O6
More Cheeseburgers, Please!
Limiting Reactant and Yield
What is the maximum number of cheeseburgers
you can make? 3 – because we only have 3
hamburger patties, everything else is in excess
230 g
What is the theoretical yield in grams?
3 cheeseburgers x 230 g/cheeseburger = 690 g of cheeseburger
We weigh the cheeseburgers we make and they weigh 460 g.
What is our % yield?
% Yield = Experimental yield x 100 = 460 g x 100 = 67%
Theoretical yield
690 g
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More Cheeseburgers, Please
• The number of hamburger patties limited the amount of
cheeseburgers we can make. In chemical reactions we call
this the limiting reactant. (Also known as limiting
reagent.)
• The maximum number of product we can make depends
on this reactant. In chemical reactions, we call this
maximum the theoretical yield.
• We use the balanced equation to determine the ratios of
the reactants and products to calculate theoretical yields
Limiting Reactant
12 total O2
+
2 C3H8
g
? CO2 + ? H2O
propane
• Which reactant is in excess?
o
o
o
o
How many CO2 can be made from 2 C3H8 ?
How many H2O can be made from 2 C3H8?
How many O would be needed?
Do we have enough/excess?
6
8
20-O atoms
yes
O2 is in excess; the limiting reactant is C3H8
10 O2 + 2 C3H8
g 6 CO2 + 8 H2O
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From Atoms /Molecules to Moles
• On the last slide we counted atoms and molecules
• Usually we are working with moles and grams
10 O2 + 2 C3H8 g 6 CO2 + 8 H2O reduces to
5 O2 + C3H8 g 3 CO2 + 4 H2O
• The equation tells us that 5 mol of O2 react with 1 mol of C3H8
to give 3 mol CO2 and 4 mol H2O
• We can use these mole relationships to determine the limiting
reactant:
o If we have 1.5 mole O2 and 0.50 mol C3H8 , which is the limiting
reactant?
o 1.5 mol O2 x 1 mol C3H8 = 0.3 mol C3H8 < 0.5 mol C3H8
5 mole O2
o O2 is the limiting reactant
From Moles to Mass
5 O2 + C3H8 g 3 CO2 + 4 H2O
• We can go from mol to mass (g) by using the molar mass
of each reactant and product:
o O2 = 32.0 g
C3H8 = 44.1 g
CO2 = 44.0 g
H2O = 18.0 g
• So: 160. g of O2 react with 44.1 g of C3H8 to give 132 g CO2
and 72.0 g H2O
• If we have 160 g of O2 but only 40.1 g C3H8 , the C3H8 is now
the limiting reactant.
Note that the total mass of reactants = total mass of products 204 g
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What Is the Limiting Reactant and Theoretical Yield When
0.552 Mol of Al React with 0.887 Mol of Cl2?
2 Al(s) + 3 Cl2(g) → 2 AlCl3
Given:
Find:
Solution Map:
0.552 mol Al, 0.887 mol Cl2
mol AlCl3
mol Al
mol AlCl3
2 mol AlCl 3
2 mol Al
Relationships:
mol Cl2
Pick the
smaller
amount
mol AlCl3
2 mol AlCl 3
3 mol Cl2
Limiting
reactant and
theoretical
yield
3 mol Cl2  2 AlCl3; 2 mol Al  2 mol AlCl3
2 mol AlCl 3
Solution: Limiting
0.552 mol Al 
2 mol Al
Reactant
 0.552 mol AlCl 3
0.877 mol Cl2 
2 mol AlCl 3
3 mol Cl2
 0.5847 mol AlCl 3
Theoretical
Yield
Percent Yield
• The measured amount of product made in a chemical
reaction is called the experimental yield.
• We can determine the percent yield of a reaction or
process by comparing the amount we actually got to the
maximum or theoretical yield:
Experimental Yield
100%  Percent Yield
Theoretical Yield
• The maximum or theoretical yield is calculated based on
the limiting reactant.
• Because of both controllable and uncontrollable factors,
the experimental yield of product is always be less than the
theoretical yield (unless something has gone wrong….)
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For the Coming Week
• Read Chapters 2 and Chapter 3
• Work on text problems for both chapters
Mastering Chemistry Fundamentals /Chap 1 04/14/10 12:00 am
• Nomeclature Worksheet in Lab tonight
• PreLab due on Thursday for 7 Up Lab
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