SCH4U – Grade 11 Review
Grade 11 Review for SCH4U
1. Significant Digits
Zero as a Significant Digit:
Zero is significant ONLY if it is between two significant figures, OR if it is to the right of a significant
figure AND to the right of the decimal.
Egs. 0.0018
2 sig. fig.
1.8 x 10-3
18 000
2 sig. fig.
1.8 x 104
18.00
4 sig. fig.
1.800 x 101
Multiplication and Division:
The answer to the calculation will have the same number of significant digits as the measurement
with the fewest significant digits.
Eg.
0.01208
0.0236
= 0.51186... but 0.0236 has only 3 sig. fig.
therefore
= 0.512 or 5.12 x 10-1
Addition and Subtraction:
The answer to the calculation will have the same number of decimal places as the measurement
with the fewest number of significant figures.
Eg.
0.12 + 1.6 + 10.976
= 12.696
but 0.12 and 1.6 each have 2 sig. fig.
and 1.6 has only 1 decimal place
therefore
= 12.7 or 1.27 x 101
2. Calculating Average Atomic Masses from Isotopic Abundances
Example:
Chlorine is a mixture of 2 isotopes
▪ 75.77% of the atoms are 35Cl (34.9689μ)
▪ 24.23% of the atoms are 37Cl (36.9659μ)
What is the average atomic mass of chlorine?
Solution:
0.7577 x 34.9689μ = 26.50μ
0.2423 x 36.9659μ = 8.957μ
Total mass of average atom = 26.50μ + 8.957μ = 35.46μ
3. Atomic Structure
𝐴
𝑍𝑋
A = mass #
= # of protons + # of neutrons
Z = atomic #
= # protons
Isotopes = same # of protons, different # of neutrons
Ions
= same # of protons and neutrons, different # of electrons
Page 1 of 9
SCH4U – Grade 11 Review
4. Bonding
Atoms will bond if their new configuration provides equal or greater stability.
This is often due to the acquiring of a “full” valence electron shell
Octet = 8 valence electrons
Duplet = 2 valence electrons
Between:
Ionic Bonding
Covalent / Molecular Bonding
Metal and Non-metal
Two or more non-metals
Metal loses e-  cation
Mechanism:
-
Non-metal gains e  anion
Valence electrons are shared
(up to three shared pairs)
Electrons are transferred from the metal to the non-metal
lithium + chlorine  lithium chloride
Li
+

Cl
LiCl
bromine + chlorine  bromine chloride
Br
+
Cl

BrCl
Br
+
Cl

Br Cl
Example:
Li
Force of
Attraction:
+
Cl

[ Li ]+ + [ Cl ]-
Electrostatic interaction
(attraction between positively
and negatively charged ions)
Physically shared eletron pair(s)
Crystal Lattice
Individual molecules.
Structures vary greatly depending
on how individual molecules
interact with each other.
Depends mostly on:
polarity, size and 3D shape
Conceptual
Diagram:
Structure:
Page 2 of 9
SCH4U – Grade 11 Review
5. Electronegativity, Polarity and Shape
The ability of an individual atom, when bonded, to attract bonding electrons to itself is known as its
electronegativity. To determine the overall polarity of a molecule you must consider polarity of
bonds and 3D shape. S.N.A.P. (Symmetrical = Non-polar ; Asymmetrical = Polar)
Electronegativity
6. Periodic Trends
Page 3 of 9
SCH4U – Grade 11 Review
6. Nomenclature

Recall common polyatomic Ions:
Name
ammonium
hydroxide
nitrate
chlorate
Formula
NH4+
OHNO3ClO3-
bromate
iodate
carbonate
sulfate
phosphate
permanganate
acetate
chromate
dichromate
cyanide
BrO3IO3CO32SO42PO43MnO4CH3COOCrO42Cr2O72CN-
hydrogen carbonate
hydrogen sulfate
hydrogen phosphate
dihydrogen phosphate
HCO3HSO4HPO42H2PO4-
related ions
nitrite
hypochlorite
chlorite
perchlorate
NO2ClO ClO2ClO4-
phosphite
PO33-
cyanate
thiocyanate
OCNSCN-
hydrogen sulfite
HSO3hydrogen phosphite HPO32dihydrogen phosphite H2PO3-
IUPAC Naming Method
 Molecular Compounds (2 non-metals) - use prefixes
1
2
3
4
5
6
mono
di
tri
tetra
penta
hexa
7
hepta
8
octa
9
nona
10
deca
*note: omit mono for first element in compound name
Eg.

CCl4 - carbon tetrachloride
P2Cl5 - diphosphorus pentachloride
Ionic Compounds (metal and non-metal) *note use stock system for multivalent metals
Binary:
Name first element + second element replace ending with “ide”
Polyatomic: Name first element + name polyatomic
Eg.
NaCl - sodium chloride
CuCl - copper(I) chloride
CuCl2 - copper(II) chloride
Mg3(PO4)2 - magnesium phosphate
ZnClO3 - zinc (I) chlorate
Zn(ClO3)2 – zinc (II) chlorate
Page 4 of 9
SCH4U – Grade 11 Review
7. Types of Chemical Reactions
Combustion:
Synthesis:
A + B  AB
Decomposition:
AB  A + B
Single Displacement:
A + BC  AC + B
Describes a chemical reaction in which substances react with oxygen
producing heat and light produces the most common oxides of the elements
making up the substance that is burned
Occur when 2 or more substance combine to produce a more complex
substance
Reverse of an synthesis reaction. A compound is broken down (decomposed)
to give 2 or more
simpler substances.
A reaction in which one element replaces another in a compound.
Activity Series can be used to predict whether or not the reaction will occur.
Each metal will displace any metal ion that appears below it in the series
Double Displacement: Always involves two ionic compounds the cation of one compound changes
AB + CD  AD + CB
places with the cation of the second compound
Solubility Table can be used to predict whether or not a precipitate will form
if product is very soluble, it will stay in solution
if product is slightly soluble, solid is produced  called precipitate
Page 5 of 9
SCH4U – Grade 11 Review
8. Measuring Moles
×𝑚
mass
𝑚 = 𝑛 ×𝑀
÷𝑀
× 6.02×10 23
moles
particles
𝑚
𝑛=
𝑀
÷ 6.02×10 23
𝑁 = 𝑛 × 𝑁𝐴
Molar Mass (M): Sum of the atomic masses of the atoms in the formula.
Eg.
Find the molar mass of calcium phosphate, Ca3(PO4)2
MCa3(PO4)2
=
Convert between mass (m) and moles (n): Use the molar mass (M) as a conversion factor.
Eg.
How many moles are there in 0.58 g of Ca3(PO4)2?
nCa3 (PO4)2
Eg.
=
What is the mass of 2.67 moles of Ca3(PO4)2?
mCa3(PO4)2
=
9. Empirical and Molecular Formulas
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”
Example:
A compound contains 69.58% Ba, 6.090% C, and 24.32% O. What is its
empirical formula?
Element Mass (g)
Ba
C
O
Moles (mol)
Ratio
Whole-number ratio
Therefore the empirical formula is __________________
Page 6 of 9
SCH4U – Grade 11 Review
10. Percentage Composition
% by mass of element =
mass of element
mass of whole sample
x 100%
Example: What is the percentage by mass of hydrogen in H2O?
%H=
11. Stoichiometry and Limiting Reagents


Start with a balanced equation
Always use coefficient ratio with moles (Moles Good! Grams Bad!)
Example:
How many moles of sodium phosphate can be made from 0.240 mol NaOH
according to the following equation?
3 NaOH(aq) + H3PO4 (aq) ↓ Na3PO4 (aq) + 3 H2O
nNa3 PO4(aq) =
12. Percentage Yield
percentage yield =
actual yield x 100%
theoretical yield
13. Ionic Reactions that Produce Precipitates
Example:
Balanced Equation
HNO3 (aq) + KOH (aq) ↓ KNO3 (aq) + H2O (l)
Ionic Equation
Net Ionic Equation
Page 7 of 9
SCH4U – Grade 11 Review
14. Molar Concentration
𝐶=
Example:
𝑛
𝑚𝑜𝑙𝑒𝑠
= 𝑣𝑜𝑙𝑢𝑚𝑒
𝑉
How many moles of each ion are in 2 L of aqueous 0.20 mol/L Al2(SO4)3?
Al2(SO4)3 ↓ 2 Al3+ + 3 SO420.20 mol/L
CAl3+ =
CSO 2- =
nAl3+ =
nSO
4
4
2-
=
15. Dilution of a Solution
C1V1 = C2V2
Example:
What volume of 0.200 M K2Cr2O7 is required to prepare 100.0 mL of 0.0400 M
K2Cr2O7?
C1 = 0.0400 M
V1 = 100.0 mL = 0.1000 L
C2 = 0.200 M
V2 = ?
Page 8 of 9
SCH4U – Grade 11 Review
16. Combined Gas Law
P 1 V1
T1
Example:
=
P2 V2
T2
A sample of argon is trapped in a gas bulb at a pressure of 760 kPa when the
volume is 100 mL and the temperature is 35°C. What must its temperature be if its
pressure becomes 720 kPa and its volume 200 mL?
P1 = 760 kPa
P2 = 720 kPa
T2 =
P 2 V 2 T1
P 1 T2
V1 = 100 mL = 0.1 L
V2 = 200 mL = 0.2 L
T1 = 308 K (35 + 273)
T2 = ?
=
17. Standard Molar Volume
*Remember:
Example:
- ONE mole of any gas at STP (0°C, 101.3 kPa) occupies 22.4 L
- NA = 6.022 x 10-23 atoms or molecules per mole
At STP, how many molecules of water vapour are in 22.4L?
18. Ideal Gas Law
PV = nRT,
Example:
where R = 0.0821 L atm or
mol K
8.314 L kPa
mol K
A sample of oxygen at 24.0°C and 0.980 atm was found to have a volume of
455 mL. How many grams of O2 were in the sample?
Page 9 of 9