WHMIS
 workplace hazardous materials information
system
all chemicals are treated with respect
 WHMIS has been developed to provide guidelines
for handling, storage and disposal of
reactive materials
 MSDS is a material safety data sheet
compressed gas
flammable and
combustible
corrosive
poisonous and
infectious material
causing immediate
and serious toxic
effects
poisonous and infectious
causing other toxic effects
oxidizing material
dangerously reactive
material
biohazardous infectious
material
Science 10 Review
A. The Atom
 proton  p+  positive charge
 neutron  n0 
zero charge
 electron e–  negative charge
 electrons are found in a cloud region around
the nucleus
 nucleus contains the protons and neutrons
which make up most of mass of atom
 mass number = # protons + # neutrons
# neutrons = mass number - # protons
 isotope = atoms with the same # of protons
but a different # of neutrons
(different mass numbers)
eg) carbon-12  6 p+, 6 n0
carbon-14  6 p+, 8 n0
B. Periodic Table
arranged in groups (columns) and
periods (rows)
group number = number of outer level
(valence) eperiod number = number of energy levels
occupied by eEx. Na
Group number = 1
Period number = 3
This info is helpful for drawing Energy Level
Diagrams
C. Energy Level Diagrams
atoms are electrically neutral
that the # of p+ = # of e-
which means
maximum number of e-: 3rd level = 8 e2nd level = 8 e1st level = 2 e-
Examples
1 e8 e2 e-
1 e2 e-
11 p+
3 p+
Na
Li
D. Ions
ions are particles or groups of particles that
have a net charge (either positive or
negative)
neutral atoms are unstable if their valence level
is not full
atoms will strive to satisfy the octet rule in order
to become stable…in other words, they strive to have
a full valence level and do so by giving away
or taking e-
metals  give away e- and become positive
ions
eg) Na+, Ca2+, Fe3+
non-metals take e- and become negative
ions
eg) Cl-, P3-, O2-
Examples
1 e8 e2 e11
p+
Na
sodium atom
8 e2 e11 p+
Na+
sodium ion
7 e8 e2 e17
p+
Cl
chlorine atom
8 e8 e2 e17 p+
Cl–
chloride ion
Your Assignment: 1. Draw argon and neon and
compare to Na+ and Cl–
2. pgs 1,2 in workbook
E. Elements
metals exist as single atoms
eg) Li(s), Cu(s), Hg(l)
nonmetals and hydrogen do not exist as single
atoms – flagpole!
H2
N2
O2
F2
P4
S8 Cl2
Br2
I2
Try These:
1. Cu(s) = copper
2. O2(g) = oxygen gas
3. Al(s) = aluminum
4. fluorine gas = F2(g)
5. barium = Ba(s)
6. nitrogen gas = N2(g)
F. Ionic Compounds
 metals + nonmetals or polyatomic ions
 monovalent eg) K+, Be2+ or multivalent metals
eg) Fe3+, Fe2+
charges on the ions are the result of taking or
giving eto go from formula to name: name of first ion,
then brackets for charge if multivalent,
then name for second ion
i.e. first element ( ) second element-ide
eg) AlCl3 = aluminum chloride
Fe2O3 = iron (III) oxide
Try These:
1. Zn3P2 = zinc phosphide
2. NaNO3 = sodium nitrate
3. NiF3 = nickel (III) fluoride
4. MnO2 = manganese (IV) oxide
5. Cr2(SO4)3 = chromium (III) sulphate
to go from name to formula: write the symbol
for each ion, then add subscripts to balance
charges
eg) calcium sulphide = CaS
iron (II) hydroxide = Fe(OH)2
Try These:
1. lithium bromide = LiBr
2. sodium phosphate = Na3PO4
3. magnesium nitride = Mg3N2
4. ammonium sulphate = (NH4)2SO4
5. calcium phosphate =
Ca3(PO4)2
Hydrated Compounds
 ionic compounds containing water in their
structure
water is represented by “xH2O” in the formula
where x is the number of water molecules
prefixes:
1 = mono
6 = hexa
2 = di
7 = hepta
3 = tri
8 = octa
4 = tetra
9 = nona
5 = penta
10 = deca
to go from name to formula: give the ionic name
for the first part of the compound, then name the
“xH2O” part as prefix + “hydrate”
eg) NaF3H2O = sodium fluoride trihydrate
CuSO45H2O = copper (II) sulphate pentahydrate
to go from name to formula: first part is the
same as before …look up the symbol for each
ion then balance the charges using subscripts,
then for the hydrate part…add “xH2O” where
x is the number given in the prefix
eg) iron (III) nitrate nonahydrate = Fe(NO3)39H2O
sodium chlorate tetrahydrate = NaClO34H2O
nickel (II) sulphite heptahydrate = NiSO37H2O
Your Assignment: pgs 3,4 in workbook
G. Molecular Compounds
 nonmetals only
 e- are shared therefore no ions are formed
 no charges involved
 use prefixes in naming
to go from formula to name: name of first
element (including prefix if necessary),
then name for second element with “ide”
ending (including prefix)
i.e. ___first element ___second element -ide
eg)
N2O = dinitrogen monoxide
CO2 = carbon dioxide
P4O10 = tetraphosphorus decaoxide
to go from name to formula: write the symbol
for each element, then use the prefixes to
determine the subscripts
eg)
carbon monoxide = CO
carbon tetrachloride = CCl4
remember the memorizers??????
NH3(g)= ammonia
H2O(l) = water
H2S(g) = hydrogen sulphide
HF, HCl, HBr, HI = no prefixes
CH4(g)= methane
CH3OH(l) = methanol
C2H6(g)= ethane
C2H5OH(l) = ethanol
C6H12O6(s)= glucose
C12H22O11(s)= sucrose
O3(g)= ozone
H2O2(l)= hydrogen peroxide
H. Acids
 always have aqueous (aq) as the state and
always have hydrogen
Rules
1. hydrogen____ide
2. hydrogen ____ate
3. hydrogen ____ite
becomes hydro___ic acid
becomes _______ic acid
becomes ______ous acid
Try These:
1. hydrogen iodide =hydroiodic acid HI(aq)
2. hydrogen phosphate = phosphoric acid H3PO4(aq)
3. hydrogen nitrite = nitrous acid
HNO2(aq)
4. hydrogen sulphite = sulphurous acid
H2SO3(aq)
Your Assignment: pgs 5-7 in workbook
I. States
acids – always (aq)
elements – can be (s), (l) or (g)…see
periodic table
molecular compounds – can be (s), (l), or (g)
ionic compounds - If not in a solution always (s)
- If in a solution either (s) or
(aq)…look up on the solubility
chart
Try These:
1. NaCH3COO(aq )
6. CaCO3( s )
2. BaSO4( s )
7. FeSO4( aq)
3. KOH( aq)
8. (NH4)2S( aq )
4. Pb(NO3)4( aq )
9. Pb(SO4)2( aq )
5. Hg(CH3COO)2(aq )
10. Ca3(PO4)2( s )
J. Chemical Reactions
 endothermic
vs. exothermic
reaction types:
1. hydrocarbon combustion
C?H? + O2(g)  CO2(g)
eg) CH4(g) + 2 O2(g)
2. simple composition
element + element
eg) 2 Mg(s)
+
+
 CO2(g)
H2O(g)
+
2 H2O(g)
 compound
O2(g)
 2 MgO(s)
3. simple decomposition
compound  element + element
eg) 2 H2O(l)  2 H2(g) + O2(g)
4. single replacement
element + compound
 element + compound
eg) Cu(s) + 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
5. double replacement
compound + compound
 compound + compound
eg) Pb(NO3)2(aq) + 2 KI(aq)  2 KNO3(aq) + PbI2(s)
Balancing Reactions
law of conservation of matter says that matter
cannot be created or destroyed, it can only
change forms
we must balance
matter
CH4 (g) +3O2(g) 
chemical equations to conserve
CO2(g) + 2H2O(g)
2 C2H6 (g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
Your Assignment: pg 8, 1st half p. 9
Predicting Reactions
Try the following:
Potassium iodide solution is added to
lead (II) nitrate solution.
2 KI(aq) + Pb(NO3)2(aq)  2 KNO3(aq) + PbI2(s)
NOTE:
-SR and DR reactions always happen in solutions so for
ionic compounds check solubility table
-Composition and decomposition do NOT happen in
solutions so ionic compounds are (s)
Predicting: single replacement
Copper metal is added to a solution of silver nitrate
Cu(s) + 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
Chlorine gas is bubbled through a solution of sodium
phosphide
6 Cl2(g) + 4 Na3P(aq)  P4(s) + 12 NaCl(aq)
Your Assignment:
pg 2nd half p. 9
K. Significant Digits
any digit from 1-9 is significant
trailing zeros are significant eg) 6.3800,
12 000
 “sandwich” zeros are significant eg) 2.04,
1005.002
 leading zeros are not significant eg) 0.0065
 counted objects and constants are not included
in sig digs
/ : multiply or divide then round answer
to the lowest number of sig digs
+/ : add or subtract then round answer to
the lowest number of decimal places
L. The Mole
it is a number= 6.02 x 1023 “items”
1. Molar Mass
sum of the individual atomic masses for
each element in a compound
eg)
CO2 = 44.01 g/mol
Al(OH)3 = 78.01 g/mol
Cu(ClO3)2 = 230.45 g/mol
2. Mole/Mass Calculations
n= m
M
where:
m = nM
n = number of moles in mol
m = mass in g
M = molar mass in g/mol
Example 1
How many moles are in 8.06 g of magnesium
oxide?
m = 8.06 g
M = 40.31 g/mol
n=?
n = m
M
=
8.06 g
40.31 g/mol
= 0.1999503 mol
= 0.200 mol
Example 2
What is the mass of 0.677 mol of potassium
sulphide?
n = 0.677 mol
M = 110.27 g/mol
m=?
m = nM
= (0.677 mol)(110.27 g/mol)
= 74.65…g
= 74.7 g
Your Assignment: p. 10 & 1st half p. 11
Your Review Assignment: finish p. 11 – p. 13