1
•These slides can only highlight certain parts
of the material we have covered
(not everything is here, but it maybe useful)
• The final will have 50 questions,120 minutes
and is scheduled for
Dec 14
th
(Thursday) 9:30am to 11:30am
• Please remember: an eraser, pencil(s) and a
non-programmable scientific calculator
Optional: coffee, hot chocolate, tea,
bagel+schmear, or pop tarts, or bacon pop-tarts, or
bacon Pringles, or bacon cheese crickets, etc.
The front page of
the final will be similar
to this
2
3
3
Concentration Terms
Parts per hundred (percent or %)
Parts per million (ppm)
Parts per billion (ppb)
Midday ozone (O
3
) levels reach about 0.4 ppm =
ppm = 0.4 ozone molecules
1 x 10
6
molecules and atoms in air
Sulfur dioxide in the air should not exceed 30 ppb =
ppb = 30 sulfur dioxide molecules
1 x 10
9
molecules and atoms in air
Section 1.2
Atmosphere is 21% oxygen = 21 oxygen molecules
% = 21 oxygen molecules
100 molecules and atoms in air
4
4
Another Example – Model 2
Section 1.4
molecules N
2
O
1 breath
USING UNIT ANALYSIS TO CARRY OUT CALCULATIONS
The concentration of dinitrogen monoxide in the atmosphere is 45 parts per billion (ppb).
If there are 2 x 10
22
molecules of air in a breath,
how many molecules of dinitrogen monoxide are there in each breath?
Molecules of N2O in a single breath:
45 N
2
O molecules
1
×
10
9
molecules of air
2×10
22
molecules of air
1
breath
45 N
2
O molecules
1×10
9
molecules of air ×2×10
22
molecules of air
1 breath =
9×10
14
N
2
O molecules
1 breath
Description of Step
in Process
Result
1.
Identify the quantity being calculated paying
attention to the units asked for in the answer
2.
Pull any useful information out of the
problem statement
concentration = 45 ppb
2 x 1022 molecules in one breath
3.
Put this information in the form of a fraction
(a
conversion factor)
4.
Set up a mathematical equation that will
allow you to calculate the desired quantity in the
appropriate units
3
5
5
Example:Express 0.0000306 in scientific notation
Answer:
Writing Numbers in Scientific Notation (d)
(Move decimal point 5places to the right)
3.06 x 10-5
0 . 0 0 0 0306
Appendix 2 (p A-2; back of the book)
6
6
Exercise:Express 4.7350 x 10-6 as a decimal number
(Move decimal point 6 places to the left)
Answer:
4.7350 = 0.0000047350
4.7350 x 10-6
Writing Numbers in Scientific Notation (e)
000000
Appendix 2 (p A-2; back of the book)
4
7
7
Classifying Matter (a)
All Matter
Pure
Substances Mixtures
Can it be separated by
a physical process?
CompoundsElements
YES
NO
Can it be broken
down into
simpler ones by
chemical means?
NO YES
Section 1.6
In other words: Classifying Matter (b)
Section 1.6
Element Compound Mixture
Observable
properties
Cannot be broken
down into simpler
substances
Fixed composition.
Can be broken down
into elements.
Variable composition of
elements, compounds or both
Microscopic
level
Only one kind of
atom
Two or more atoms in
a fixed composition
Variable assortment of
atoms, molecules, or both
Example 1
Example 2
* In formulae, B = black atoms, G = gray atoms
Formula: BG2
Formula: GB3
31
5
9
9
Classify each of these as an element, a compound, or a mixture:
carbon dioxide
nickel
caffeine
water
compound
element
compound
compound
fluorine
table salt
soap
sea water
element
compound
mixture
mixture
Classifying Matter (d)
Section 1.6
Group
Period
The Periodic Table (b)
Section 1.6
Keys
37
37
6
40
molecules →atoms →
electrons/protons
/neutrons
C = carbon
H = hydrogen
O = oxygen
N = nitrogen
Caffeine
One of my favorite things
Section 1.7
Atoms and Molecules (a)
Need to Remember: Components of chemicals
41
41
Name this molecule.
(Hint: It has nothing to
do with Disney)
Two hydrogen atoms
Oxygen atom
Section 1.7
Atoms and Molecules (b)
Models of a water
molecule, H
2
O
Molecule: – fixed number of atoms held together by chemical bonds
in a certain spatial arrangement.
Chemical formula: – symbols that represents the type and number of
each element present in each molecule (e.g. H
2
O).
H
O
H
7
Molecular Compounds and Bonds (c)
Section 1.7
Covalent Compounds:
– Non-metals can also react with other non-metals to
form covalent compounds (example: Br-Br or Br
2
)
– covalent bonds,
– electrons shared between two atoms (e.g. C-H bonds)
Ionic Compounds:
– ionic bonds,
– electrons transferred between two atoms (e.g. Na
+
Cl
–
)
– charges often balance (more on this later)
– Cations / Anions
42
42
Covalent Compounds – covalent bonds (d)
“…formed by two or more atoms (usually nonmetals)
attached by covalent bonds forming molecules
and are called molecular compounds (aka. covalent
compounds or molecules).”
Examples of Chemical Formulas of Covalent Compounds:
NH
3
(ammonia)
H
2
O
Atomic
symbol Subscript indicates
number of atoms
– Each water molecule contains 2 H atoms and 1 Oxygen atom
CO
2
(carbon dioxide)
-Each carbon dioxide molecule is made up of one Carbon
atom, and 2 O atoms
Section 1.7
43
43
8
Ionic Compounds – ionic bonds (e)
“… compounds made up of a metal and a non-metal.”
(Sometimes a metal and a group of elements or polyatomic
ions bound together through an ionic bond)
Ionic Formula
Form of Compound
Table Salt Na
+
Cl
–
NaCl
Drano cleaner N a
+
OH
–
NaOH
Limestone Ca
2+
CO
32-
CaCO
3
Section 1.7
Name
Compound
sodium chloride
sodium hydroxide
calcium carbonate
Examples of Chemical Formulas of Ionic Compounds:
44
44
3.) Lets apply this “prefix” rules to some chemicals:
Naming Binary Compounds
Number
of atoms prefix
1 mono
2 di
3 tri
4 tetra
5 penta
6 hexa
7 hepta
8 octa
Section 1.8
Formula Name
of compound of compound
CO carbon monoxide
NO
2
nitrogen dioxide
CrO
3
chromium trioxide
PbO
4
lead tetroxide
P
2
O
5
diphosphorus pentoxide
PF
6
phosphorus hexafluoride
– –
– –
Table 1.7
49
49
Names and Formulas: The Vocabulary or Chemistry (c)
9
Naming Ionic Compounds (a)
– Ionic compounds are named by stating the
name of the cation first followed by anion.
Naming Ionic Compounds from Chemical Formulas
– Monoatomic cations keep their “metal” name.
–Anions change from “-ine” to “-ide“
{Cation name}{Anion name}
e.g. sodium chloride
Note:
What is the chemical formula for sodium chloride?
NaCl
Section 1.8
50
50
Cation Anion
Na
+
(sodium) Cl
–
(chloride)
K
+
(potassium) I
–
(iodide)
Ca
2+
(calcium) F
–
(fluoride)
Formula
of compound
NaCl
KI
CaF
2
Examples:
– Lets combine what we know about Cations and Anions
Name
of compound
sodium chloride
potassium iodide
calcium fluoride
Naming Ionic Compounds (d)
Summarize naming ionic compounds
– Naming Ionic Compounds from Chemical Formulas
1) What does the “2” in CaF
2
mean?
2) Why is the “2” there?
2 F per Ca atom
CaF
2
is a neutral ionic compound,
the Ca ion is +2, so two “F-” is needed.
Section 1.8
55
55
10
Name the following Ionic Compounds
a) Na
2
O
b) CaF
2
c) Sr
3
N
2
d) MgCl
2
cation: sodium (Na
+
),
anion: oxide (O
2-
) Sodium oxide
cation: calcium (Ca
2+
)
anion: fluoride (F
–
)
cation: strontium (Sr
2+
)
anion: nitride (N
3-
)
cation: Magnesium (Mg
2+
)
anion: chloride (Cl
–
)
Calcium fluoride
Strontium nitride
Magnesium chloride
Activities 2 (g)
Section 1.8
58
58
Write the Chemical Formulas for the following
Ionic Compounds:
a) Potassium chloride ions: K
+
, Cl
–
KCl
b) Magnesium Iodide ions: Mg
2+
, I
–
MgI2
c) Sodium oxide ions: Na
+
, O
2-
Na2O
Practice Naming (i)
(Remember: We’re going backwards.)
Section 1.8
60
60
11
65
65
More practice – lets bring it all together. (n)
Give the name or formula of the following:
a) N
2
O
5
b) Sulfur trioxide
c) Aluminium bromide
d) CO
e) O
2
f) titanium dioxide
Section 1.8
= dinitrogen pentoxide
= SO
3
= AlBr
3
= carbon monoxide
= oxygen
= TiO
2
Chemical Change: The Role of Oxygen in Burning
• Any change where a new
chemical is formed.
• Indications of a chemical reaction:
–Change in temperature
–Production of light/color change
–Change in state
–Change in smell and/or texture
Section 1.9
66
66
What is a chemical reaction?
Tips for Balancing equations:
– if an element is present in just one compound on each side,
balance it first (e.g. Carbon first)
– balance anything that exists as a free element last (e.g. O
2
)
– balance polyatomic ions as a unit (e.g. NO
3–
)
– re-check (Count number of atoms again)
# atoms left (reactants) =# atoms right (products)
C
3
H
8
+ O
2
CO
2
+ H
2
O
3
8
2
Balancing Chemical Equations (a)
x 3 = 3 C
x 4 = 8 H
6 + 1 Oxy
1
2
3
Final balanced equation:
Section 1.10
3 4
= 10 Oxy = 5 O
2
2
5
X 5 = 10
C
H
O+ 4 Oxy
