1
CHEMICAL REACTIONS
Example: Hydrogen + Oxygen  Water
H2 + O2  H2O

+
+
- Note there is not enough hydrogen to react with oxygen
- It is necessary to “balance” equation.
reactants
products

+
2 H2 + O2  2 H2O
(balanced equation)
CONSERVATION OF MASS
During a chemical reaction, matter is neither created nor destroyed.
- i. e. the number of atoms of each element remains constant
BALANCING CHEMICAL EQUATIONS
- ensures # of reactant atoms = # of product atoms
C2H4 +
O2 
CO2 +
Cu +
S8 
Cu2S
Fe2O3 +
C
Fe +
Sc2O3 +
H2O 
Sc(OH)3
CH3NH2 +
O2 
H2O
CO2
CO2 +
H2O +
N2
- Sometimes it is more convenient to balance groups of atoms (polyatomic
ions) than individual atoms.
AgNO3 +
CaCl2  Ca(NO3)2 +
AgCl
- Balance NO3- ions rather than N and O atoms.
Ba(ClO4)2 +
Na2SO4 
- Balance ClO4- and SO42- ions.
BaSO4 +
NaClO4
2
GENERAL TYPES OF REACTIONS
1) Combination
A+BC
- two or more substances react two form a single substance
a) formation reactions
elements  compound
4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s)
8 C (s) + 9 H2 (g)  C8H18 (l)
b) hydration reactions
- addition of water
- metal oxide + H2O  metal hydroxide
CaO (s) + H2O (l)  Ca(OH)2 (s)
CaO (lime) when heated emits a very bright white light (limelight).
Ca(OH)2 (hydrated lime) is added to plaster to make it harder.
- nonmetal oxide + H2O  acid (oxidation # are unchanged.)
P2O5 (s) + 3 H2O (l)  2 H3PO4 (aq)
P2O5 is a very effective dehydration agent.
2) Combustion
- combination with O2 (as in burning fuel)
- products are CO2, H2O and N2
2 C8H18 (l) + 25 O2 (g)  16 CO2 (g) + 18 H2O (g)
2,2,4-trimethylpentane (isooctane) is the major component of gasoline.
4 (CH3)3N (l) + 21 O2 (g)  12 CO2 (g) + 18 H2O (g) + 4 N2 (g)
Trimethylamine is responsible for the smell of rotten fish.
3) Decomposition
AB+C
- one substance decomposing into two or more substances
a) decomposition of carbonates
- metal carbonate  metal oxide + CO2
CaCO3 (s)  CaO (s) + CO2 (g)
Calcium carbonate is the major component of seashells and egg shells.
b) decomposition of chlorates and perchlorates
- metal chlorate  metal chloride + O2
2 NaClO3 (s)  2 NaCl (s) + 3 O2 (g)
- metal perchlorate  metal chloride + O2
Fe(ClO4)3 (s)  FeCl3 (s) + 6 O2 (g)
Sodium chlorate is used in the manufacture of bleach (sodium hypochlorite, NaClO).
3
c) decomposition of sulfites
- metal sulfite  metal oxide + SO2
Na2SO3 (s)  Na2O (s) + SO2 (g)
Sodium sulfite is added to wine to stop the fermentation process and to help preserve the wine.
d) many others
4) Single replacement reactions
- when a metal ion in a compound is replaced with another metal ion
SnSO4 (aq) + Ni (s)  NiSO4 (aq) + Sn (s)
Tin bonds with iron very well and helps prevent the corrosion of iron. “Tin cans” are
steel(iron) with thin covering of tin.
3 AgNO3 (aq) + Cr (s)  Cr(NO3)3 (aq) + 3 Ag (s)
Chromium is added to iron to make stainless steel.
5) Halogen replacement reactions
- A halogen with a higher electron affinity will replace a halogen with a
lower electron affinity within an ionic compound.
2 NaI (s) + F2 (g)  2 NaF (s) + I2 (s)
6) Metathesis (double replacement) reactions
- cations and anions in two ionic compounds switch places
Na2SO4 (aq) + Ba(C2H3O2)2 (aq)  BaSO4 (s) + 2 NaC2H3O2 (aq)
Barium sulfate is ingested by X-ray patients to improve the visibility of GI tract
to X-ray radiation.
Pb(ClO3)2 (aq) + 2 KCl (aq)  PbCl2 (s) + 2 KClO3 (aq)
Lead(II) chloride combined with lead(II) hydroxide is used as a white paint pigment.
4
FORMULA AND MOLECULAR MASS
Formula Mass (Weight) – sum of atomic masses in chemical formula
Molecular Mass (Weight) – same as formula mass for molecular compounds
- technically not defined for ionic compounds
Clarification of definitions
- often the words mass and weight are used interchangeably even though they
are technically not the same thing.
- often formula mass and molecular mass are used interchangeably even
though they are technically not the same things
Calculating Formula Mass
Example: C3H6 (cyclopropane)
3 (12.011 amu) + 6 (1.00794 amu) = 42.806 amu
Example: Ca(OH)2 (calcium hydroxide)
40.08 amu + 2 (15.9994 amu) + 2 (1.00794 amu) = 74.09 amu
THE MOLE
***A mole is 6.022 x 1023 items.***
1 mole = 6.022 x 1023 items
Analogy
1 dozen = 12 items
18 eggs = 1.5 dz.
- to convert from eggs from dozen, we need to multiply by conversion factor
18 eggs  18 eggs 
1 dz
 15
. dz
12 eggs
Example: How many eggs in 3.2 dozen?
3.2 dz  3.2 dz 
12 eggs
 38 eggs
1 dz
Analogy #2
1 gross = 144 items
Example: How many gross is 68 pencils?
68 pencils  68 pencils 
1 gross
 0.47 grosspencils
144 pencils
5
1 mole = 6.022 x 1023 items
- 6.022 x 1023 is called Avogadro’s number and is abbreviated NA.
6.022 x 1023 molecules = 1 molemolecule
12.044 x 1023 molecules = 2 molemolecule
Example: How many moles of atoms is 7.43 x 1021 atoms?
7.43 x1021 atoms  7.43 x1021 atoms 
1mole
 0.0123mol atoms
6.022 x1023 atoms
Example: How many moles of ions is 2.5 x 1025 ions?
2.5 x 1025 ions  2.5 x 1025 ions 
1 mole
 42 mol ions
6.022 x 1023 ions
Example: How many molecules are in 8.333 mol of molecules?
8.333molmolecules  8.333molmolecules 
6.022 x1023 molecules
 5.018 x1024 molecules
1mole
MOLAR MASS
Definition of Molar Mass
Mass of 1 mole 12C is 12.000000 g BY
DEFINITION
Recall 1 atom of 12C is 12.000000 amu
*This is not a coincidence*
Definition of amu and a mole are made to ensure “coincidence”
1 amu = 1.66 x 10-27 kg = 1/NA grams
1 molC-12 = 12.000000 g  2 molC-12 = 24.000000 g
47.36 mol C 12  47.36 mol C 12 
12.000000 g C 12
 568.3 g C 12
mol C 12
ATOMIC MOLAR MASS
C: 1 atom = 12 amu 
1 mol = 12 g
24
Mg: 1 atom = 24 amu 
1 mol = 24 g
56
Fe: 1 atom = 56 amu 
1 mol = 56 g
12

M(12C) = 12 g/mol

M(24Mg) = 24 g/mol

M(56Fe) = 56 g/mol
6
AVERAGE ATOMIC MOLAR MASS
- Elements often have two or more naturally occurring isotopes
- Average atomic molar mass is the average of all atomic molar masses of the
naturally occurring isotopes according to each isotope’s relative abundance
- Except for units, identical to average atomic mass
- Note: mass of isotopes not integers because of nuclear forces
Example: Neon
nuclide
abundance
20
Ne
90.48%
21
Ne
0.27%
22
Ne
9.25%
mass(g/mol)
19.992
20.994
21.991
M(Ne) = 0.9048 (19.992 g/mol) + 0.0027 (20.994 g/mol) + 0.0925 (21.991 g/mol)
= 20.18 g/mol
- Note: example is the same as example for average atomic mass
Example: How many moles of atoms are in 96.3 grams of carbon?
96.3 g C  96.3 g C 
1 mol
 8.02 mol C
12.011g
Example: How much mass does 0.0840 moles of uranium have?
0.0840 mol U  0.0840 mol U 
238.03 g
 20.0 g U
mol
FORMULA MOLAR MASS
- add molar masses for all atoms within a chemical formula
Example: ethene, C2H4
M(C2H4) = 2 x 12.011 g/mol + 4 x 1.00794 g/mol = 28.054 g/mol
Ethene (ethylene) is used to ripen fresh fruit. It is also used to make polyethylene, which is used
to make milk jugs.
Example: How many moles are in 538 g of Ba(NO3)2
First calculate formula weight.
2x
6x
137.33 g/mol
14.0067 g/mol
15.9994 g/mol
261.35 g/mol
538 g  538 g 
Barium nitrate is used to color
fireworks green.
1mol
 2.06 mol
26135
. g
7
SCHEME: Converting mass to moles to number
M
Molar mass
Mass (g)
NA
Avogadro’s number
Moles (mol)
Number
(atoms or
molecules)
Note: Molar mass and Avogadro’s number are conversion factors
Example: How many atoms are in 20.21 g of He?
1 mol 6.022 x1023 atoms
20.21g 

 3.040 x1024 atoms
4.003 g
mol
Example: How many F atoms are in 13.11 g of CaF2?
13.11g 
1mol 6.022 x1023 molecules 2atoms F


 2.022 x1023 atoms F
78.08g
mol
moleculeCaF2
EMPIRICAL FORMULA FROM MASS PERCENT ANALYSIS
Given: Percent Mass Composition
Find: Empirical Formula
Strategy:
1) Assume 100 g of matter.
2) Multiply 100 g by mass percent to find amount of each element.
3) Convert mass of each element to moles using molar mass.
4) Find whole number ratios by dividing each number of moles by lowest
number of moles.
8
Example: Find the empirical formula for a compound with the following mass
percentages:
79.73 %
Cl
18.01 %
C
2.26 %
H
2
Cl: 100.0 g x 0.7973 = 79.73 g
C: 100.0 g x 0.1801 = 18.01 g
H: 100.0 g x 0.0226 = 2.26 g
3
H:
2.26 g 
C:
18.01g 
1mol
 1.500 molC
12.01g
Cl:
79.73g 
1mol
 2.249 molCl
35.45g
4
1 mol
 2.242 mol H
1008
.
g
3molH
mol H 2.242 molH

 1.495  1.5 
mol C 1.500 molC
2 molC
3molCl
mol Cl 2.249 molCl

 1.499  1.5 
mol C 1.500 molC
2 molC
 Empirical Formula is C2H3Cl3
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COMBUSTION ANALYSIS
A mass of hydrocarbon is burnt with oxygen to produce CO2, H2O and N2. From
the mass of CO2, H2O and N2 produced and mass of oxygen used, find the
empirical formula for the compound.
Ultimately, we want to find molar ratios that we’ll use to find the empirical
formula.
1. Convert mass of CO2 to moles of carbon atoms.
- calculation yields moles of carbon in hydrocarbon sample.
CO2 (g) + KOH (s)  KHCO3 (s)
2. Convert mass of H2O to moles of hydrogen atoms.
- because there are two moles of hydrogen to one mole of water, multiply
result by 2 to yield moles of hydrogen in hydrocarbon sample.
H2O (g) + CaCl2 (s)  CaCl2 • 2H2O (s)
3. Convert mass of N2 to moles of nitrogen atoms.
- because there are two moles of nitrogen atoms in one mole of nitrogen
molecules, multiply result by 2 to yield moles of nitrogen in hydrocarbon
sample.
4. Calculate moles of oxygen in products.
- moles of oxygen = 2 × moles of CO2 + moles of H2O
5. Calculate moles of oxygen in hydrocarbon
- moles of hydrocarbon oxygen
= moles of oxygen atoms in products – moles of oxygen atoms used
6. Divide molar amounts by smallest of the values to find molar ratios in terms
of whole numbers.
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Example: Find the empirical formula for a sample of hydrocarbon that produces
7.61 g of CO2, 4.15 g of H2O and uses 7.392 g of O2 during combustion.
Moles of carbon
1molCO2
7.61g 

1molC
 0.173molC
1molCO2

2 molH
 0.461molH
1molH2O
44.0g CO2
Moles of hydrogen
4.15g 
1molH2O
18.0g H2O
Moles of oxygen in products
molO products   7.61g 
1molCO2
44.0g CO2

1molH2O 1molO
2 molO
 4.15g 

1molCO2
18.0g H2O 1molH2O
 0.346 molO  0.231molO  0.577 molO
Moles of oxygen in hydrocarbon
molO hydrocarbon   molO products   molO  0.577 molO  7.39g 
 0.577 molO  0.462 molO  0.115molO
Molar ratios
0.173molC
3molC
 1.50 
0.115 molO
2 molO
0.461molH
4 molH
 4.01 
0.115 molO
1molO
Empirical formula is C3H8O2
1molO2
32.0g O2

2 molO
1molO2
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THEORETICAL STOICHIOMETRY
- coefficients of balanced equations relate moles of reactants to moles of products
Example: N2 (g) + 3 H2 (g)  2 NH3 (g)

+
- 1 mole of N2 is “stoichiometrically equivalent” to 2 moles of NH3.
- in other words, for every 1 mole of N2 reacted, 2 moles of NH3 are
produced.
- 1 mol N 2  2 mol NH 3
- equivalence is only true for specific chemical reaction
- equivalence can be considered a conversion factor
1 mol N 2  2 mol NH 3 
1 mol N 2
2 mol NH 3
or
2 mol NH 3
1 mol N 2
- other equivalences are
- 1 mol N 2  3 mol H 2
- 3 mol H 2  2 mol NH 3
The Haber process is essential in the production of fertilizer, which in turn, is essential for
sustenance of Earth’s 7,000,000,000 people.
Example: a) What are all of the stoichiometric equivalences for the reaction?
2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (g)?
1 mol C 2 H 2  2 mol CO 2
5 mol O 2  4 mol CO 2
2 mol CO 2  1 mol H 2 O
2 mol C 2 H 2  5 mol O 2
1 mol C 2 H 2  1 mol H 2 O
5 mol O 2  2 mol H 2 O
b) How many moles of carbon dioxide are formed when 5 moles of
acetylene (C2H2) is combusted?
5 mol C2 H 2 
2 mol CO2
1 mol C2 H 2
 10 mol CO2
c) How many moles of oxygen are needed to fully burn 29.8 moles of
acetylene (C2H2)?
29.8 mol C2 H 2 
5 mol O2
2 mol C2 H 2
 74.5 mol O2
Acetylene is a welder’s fuel.
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PRACTICAL STOICHIOMETRY
- can’t measure moles directly in the “real” world.
- must measure amount of substance with grams.
***- cannot compare substances stoichiometrically by mass, must convert to
moles.***
Mass of
reactant
(g)
SCHEME:
Mass of
product
(g)
M
Molar
mass
M
Molar
mass
Balanced
equation
Moles of
reactant
(mol)
Moles of
product
(mol)
Example: For the reaction,
NH3 (g) + HCl (g)  NH4Cl (s),
a) how much NH3 is needed to react with 92.3 g of HCl?
1) First convert grams of reactant to moles of reactant
92.3 g HCl 
1 mol HCl
  2.53 mol HCl
36.5 g HCl
2) Compare moles of one reactant to other reactant.
2.53 mol HCl 
1 mol NH 3
1 mol HCl
 2.53 mol NH 3
3) Convert moles of other reactant to grams.
2.53 mol NH 3 
17.0 g NH 3
mol NH 3
  43.0 g NH 3
The reaction of ammonia with hydrogen chloride gas is used to create a “smokescreen”.
b) How much ammonium chloride is produced when 92.3 g of HCl is
fully reacted?
- Note with dimensional analysis, we can do problems all on one line.
92.3g HCl 
. g NH 4Cl
1mol HCl 1mol NH 4Cl 535


 135g NH 4Cl
36.5g HCl 1mol HCl mol NH 4Cl
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Example: For the reaction 4BaCO3 (s) + Y2(CO3)3 (s) + 6 CuCO3 (s)
→ 2 YBa2Cu3O7 (s) + 13 CO2 (g) + 3 O2 (g)
a) calculate how many grams of CuCO3 is needed to fully react with 0.104 g
of BaCO3,
0.104 g BaCO3 
1molBaCO3
197.336 g BaCO3

6 molCuCO3 123.555g CuCO3

 0.0976 g CuCO3
4 mol BaCO3
molCuCO3
b) calculate how many grams of YBa2Cu3O7 is formed from 0.104 g of BaCO3
fully reacting.
0.104g BaCO3 
1molBaCO3
197.336g BaCO3

2 molYBCO 666.19g YBCO

 0.176g YBCO
4 molBaCO3
molYBCO
Yttrium barium copper oxide (YBCO) is a superconducting ceramic. It is superconducting below
a temperature of 95 K.
14
Example: Epsom salts are used in foot baths to soften skin and relieve itching.
Epsom salt is a hydrated crystal of magnesium sulfate. An intense
heat source such as a Bunsen burner flame can drive off the water
from the crystal leaving the anhydrous salt. Given the data below,
write the correct chemical formula for Epsom salt.
Mass of beaker
Mass of beaker and Epsom salt
Mass of beaker and dried salt
67.920 g
79.012 g
73.348 g
Find the mass of the Epsom salt
79.012 g – 67.920 g = 11.092 g
Find the mass of the dried salt
73.348 g – 67.920 g = 5.428 g
Find the mass of water that has left
11.092 g – 5.428 g = 5.664 g
Find the moles of water
5.664g H2O 
1molH2O
18.015g H2O
 0.3144 mol H2O
Find the moles of MgSO4
5.428g MgSO4 
1molMgSO4
120.37 g MgSO4
 0.04509 mol MgSO4
Find the number of water molecules per MgSO4 unit
0.3144 molH2O
0.04509 molMgSO4
 6.973  7 
The correct chemical formula is MgSO4  7 H2O
7 molH2O
1molMgSO4
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LIMITING REAGENTS
- Often starting materials are not available in proper stoichiometric
proportions.
- Given “unbalanced” amounts of reactants, we would like to know how
much product could be produced.
Analogy: Bicycle Factory
The equation to make a bicycle is
2 wheels + 1 frame + 1 handlebar  1 bicycle
If the parts inventory is as follows:
240 wheels
150 frames
135 handlebars,
we ask ourselves
- What “reactant” limits production?
- How much product can be produced?
Limiting reactant: wheels
Production: 120 bicycles
***In limiting reagent problems, we need to compare moles to moles***
- need to convert all masses to moles
- To find limiting reactant, calculate number of moles of product formed from
each number of moles of reactant
- Limiting reactant will yield lowest number of moles produced. (The lowest
number of moles produced is the actual number of moles of product produced.)
Example: For the reaction 2 SO2 (g) + O2 (g) + 2 H2O (l)  2 H2SO4 (aq),
if 5.6 mol of SO2, 4.8 mol of O2 and 6.0 mol of H2O are reacted together,
a) how many moles of H2SO4 are produced?
For the SO2: 5.6 mol SO2 
For the O2:
4.8 mol O2 
1mol H 2SO4
1mol SO2
2 mol H 2SO4
For the H2O: 6.0 mol H2O 
1mol O2
1mol H2SO4
1mol H2O
 5.6 mol H 2SO4
 9.6 mol H 2SO4
 6.0 mol H2SO4
SO2 is limiting reactant and therefore 5.6 moles of H2SO4 is produced.
16
b) how much O2 is remaining after reaction is complete?
- to answer the question, we need to know how much O2 was used.
- then O2 remaining is the amount of O2 reacted subtracted from the starting
amount of O2.
O2 used: 5.6 mol SO2 
1mol O2
2 molSO2
 2.8 mol O2
O2 remaining = O2 starting – O2 used = 4.8 mol – 2.8 mol = 2.0 mol
Sulfur dioxide is a pollutant from burning coal that is a contributor to acid rain. Sulfur dioxide is
removed from air with calcium oxide.SO2 (g) + CaO (s)  CaSO3 (s)
Example: For the reaction Zn (s) + CuCl2 (aq)  ZnCl2 (aq) + Cu (s)
a) How much copper metal is produced from the reaction of 2.00 g of Zn and
2.00 g of CuCl2?
1 mol Zn 1 mol Cu

 0.0306 mol Cu
65.38 g Zn 1 mol Zn
1 mol CuCl 2
1 mol Cu


 0.0149 mol Cu
134.56 g CuCl 2 1mol CuCl 2
2.00 g Zn 
2.00 g CuCl 2
Thus CuCl2 is the limiting reactant and the amount of copper produced is
0.0149 mol Cu 
63546
.
g Cu
 0.945 g Cu
1 mol Cu
b) How much reactant was left over?
Since CuCl2 is the limiting reactant, all of it was consumed in the reaction.
Thus some Zn is left over.
Moles of Zn used
0.0149 mol Cu 
1 mol Zn
 0.0149 mol Zn
1 mol Cu
Total original moles – moles used = moles left over
0.0306 molZn – 0.0149 molZn = 0.0157 molZn
0.0157 molZn 
65.38g Zn
 1.03g Zn
1mol Zn
17
REACTION YIELDS
- an actual chemical process is rarely perfect
- actual yield is less than theoretical “perfect” yield
- we have been calculating theoretical yields
- we often want to compare actual yield to theoretical yield
Example: For the reaction Cr2O3 (s) + 2 Al (s)  2 Cr (s) + Al2O3 (s)
18.7 g of Chromium (III) oxide reacts to form 10.8 g of chromium
metal. What the percent yield of this process?
% yield 
actual yield
 100%
theoretical yield
Theoretical yield of chromium metal is
18.7 g Cr2O3 
1mol Cr2 O3
152 g Cr2 O3

2 mol Cr 51996
.
g Cr

 12.8 g Cr
1mol Cr2O3 1mol Cr
Thus the percent yield is
% yield 
10.8 g Cr
 100%  84.4%
12.8 g Cr