Unit 6: The Mole
Mr. Blake
23
6.02 X 10
Review: Atomic Masses
Elements occur in nature as
mixtures of isotopes.

 Carbon = 98.89% 12C (6 p+, 6 n)
1.11% 13C (6p+, 7n)
<0.01% 14C (6p+, 8n)
 Carbon’s atomic mass = 12.01 amu
Atomic Weights are relative
C: # and mass
H: # and mass
Mass Ratio
1 C = 12.0 amu
1 H = 1.0 amu
12 to 1
2 C = 24.0 amu
2 H = 2.0 amu
12 to 1
3 C = 36.0 amu
3 H = 3.0 amu
12 to 1
4 C = 48.0 amu
4 H = 4.0 amu
12 to 1
5 C = 60.0 amu
5 H = 5.0 amu
12 to 1
10 C = 120.0 amu
10 H = 10.0 amu
12 to 1
What is the mole?
Not this kind of mole!
A. What is the Mole?
• A counting number (like a dozen).
• 1 mol = 602,000,000,000,000,000,000,000
of anything.
A
large amount!!!!
Similar Words
• Pair: 1 pair of shoelaces = 2
shoelaces.
• Dozen: 1 dozen oranges = 12
oranges.
• Gross: 1 gross of spider rings = 144
rings.
• Ream: 1 ream of paper = 500
sheets of paper.
• Mole: 1 mole of Na atoms = 6.02 X
1023 Na atoms.
The Mole
• So 1 mole of any element has 6.02 X
1023 particles.
• This is a really big number – It’s so big
because atoms are very small!
• Defined as the number of atoms in 12.0
grams of C-12. This is the standard!
• 12.0 g of C-12 has 6.022 X 1023 atoms.
The Mole
• This number is named in honor of
Avogadro
Amedeo ______________
(1776 –
1856), who studied quantities of
gases and discovered that no matter
what the gas was, there were the
same number of molecules present
• 6.02 x 1023 is also called
_______________________.
Avogadro’s number (NA)
I didn’t discover it. Its
just named after me!
Amadeo Avogadro
Just How Big is a Mole?
• Enough soft drink cans to cover
the surface of the earth to a
depth of over 200 miles.
• If you had Avogadro's number of
unpopped popcorn kernels, and
spread them across the United
States of America, the country
would be covered in popcorn to a
depth of over 9 miles.
• If we were able to count atoms at
the rate of 10 million per second,
it would take about 2 billion years
to count the atoms in one mole.
Everybody Knows Avogadro’s Number!
But Where Did it Come From?
• It was NOT just
picked! It was
measured.
• One of the better
methods of measuring
this number was the
Millikan Oil Drop
Experiment.
• Since then we have
found even better ways
of measuring using xray technology.
The Mole
• 1 dozen cookies = 12 cookies
• 1 mole of cookies = 6.02 X 1023 cookies
• 1 dozen cars = 12 cars
• 1 mole of cars = 6.02 X 1023 cars
• 1 dozen Al atoms = 12 Al atoms
• 1 mole of Al atoms = 6.02 X 1023 atoms
- Note that the NUMBER is always the
same, but the MASS is very different!
• Mole is abbreviated mol
A Mole of Particles
Contains 6.02 x 1023 particles
1 mole C
= 6.02 x 1023 C atoms
1 mole H2O = 6.02 x 1023 H2O molecules
1 mole
NaCl
= 6.02 x 1023 NaCl “molecules”
(technically, ionics are compounds not molecules
so they are called formula units)
6.02 x 1023 Na+ ions and
6.02 x 1023 Cl– ions
Learning Check
Suppose we invented a new collection unit
called a widget. One widget contains 8
objects.
1. How many paper clips in 1 widget?
a) 1
b) 4
c) 8
2. How many oranges in 2.0 widgets?
a) 4
b) 8
c) 16
3. How many widgets contain 40 gummy
bears?
a) 5
b) 10
c) 20
Avogadro’s Number as
Conversion Factor
6.02 x 1023 particles
1 mole
or
1 mole
6.02 x 1023 particles
- Note that a particle could be an atom OR a
molecule!
Learning Check
1. Number of atoms in 0.500 mole of Al
a) 500 Al atoms
b) 6.02 x 1023 Al atoms
c) 3.01 x 1023 Al atoms
2.Number
atoms
a) 1.0
b) 3.0
c) 1.1
of moles of S in 1.8 x 1024 S
mole S atoms
mole S atoms
x 1048 mole S atoms
B. Molar Mass
• Mass of 1 mole of an element or compound.
• Equal to the numerical value of the average
atomic mass (get from periodic table)
• Atomic mass tells the...
– atomic mass units per atom (amu)
– grams per mole (g/mol)
• Round to 2 decimal places (hundredths)
Molar Masses of Elements
(Found in Periodic Table)
1 mole of C atoms
=
12.01 g
1 mole of Mg atoms
=
24.31 g
1 mole of Cu atoms
=
63.55 g
Carried to the hundredths place!!
B. Molar Mass Examples
• carbon
12.01 g/mol
• aluminum
26.98 g/mol
• zinc
65.39 g/mol
Calculating Molecular Mass
Calculate the molecular mass of magnesium
carbonate, MgCO3.
24.31 g + 12.01 g + 3(16.00 g) =
84.32 g/mol
Molar Mass of Molecules and
Compounds
A substance’s molecular mass (molecular weight) is
the mass in grams of one mole of the compound.
1 mole Ca x 40.08 g/mol = 40.08 g/mol
+2 moles Cl x 35.45 g/mol = 70.90 g/mol
Therefore,the mass of the entire compound is…
1 mole of CaCl2
= 110.98 g/mol
B. Molar Mass Examples
• water
– H2O
– 2(1.01) + 16.00 = 18.02 g/mol
• sodium chloride
– NaCl
– 22.99 + 35.45 = 58.44 g/mol
B. Molar Mass Examples
• sodium bicarbonate
– NaHCO3
– 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01
g/mol
• sucrose
– C12H22O11
– 12(12.01) + 22(1.01) + 11(16.00) = 342.34
g/mol
C. Molar Conversions
molar
mass
MASS
6.02  1023
MOLES
IN
GRAMS
NUMBER
OF
PARTICLES
(g/mol)
(particles/mol)
Calculations with Molar Mass
molar mass
Grams
Moles
Converting moles to grams
Ex. #1 How many grams of lithium are in 3.50
moles of lithium?
3.50 mol Li
6.94 g Li
1 mol Li
=
24.3
g Li
Converting Moles and Grams
Ex. #2 Aluminum is often used for the
structure of light-weight bicycle frames. How
many grams of Al are in 3.00 moles of Al?
Goal: 3.00 moles Al
? g Al
1. Molar mass of Al 1 mole Al = 26.98 g Al
2. Conversion factors for Al
26.98 g Al
1 mol Al
3. Setup
3.00 moles Al
Answer
or
x
1 mol Al
26.98 g Al
26.98 g Al
1 mole Al
= 81.0 g Al
Molar Conversion Examples
3. How many moles of carbon are
in 26 g of carbon?
26 g C 1 mol C
12.01 g C
= 2.2 mol C
4) How many molecules are in 2.50
moles of C12H22O11?
6.02  1023
2.50 mol molecules
= 1.51  1024
1 mol
molecules
C12H22O11
Calculations
molar mass
Grams
Avogadro’s number
Moles
particles
Everything must go through
Moles!!!
Ex. #1 Find the mass of 2.1  1024
molecules of NaHCO3.
2.1  1024
molecules
1 mol
84.01 g
6.02  1023 1 mol
molecules
= 290 g NaHCO3
Ex. #2 How many atoms of Cu
are present in 35.4 g of Cu?
35.4 g Cu
1 mol Cu
63.5 g Cu
6.02 X 1023 atoms Cu
1 mol Cu
= 3.4 X 1023 atoms Cu
V. Percent Composition: The percent by
______
in a compound.
mass of each _________
element
Mass of element
x 100  % compositio n
Mass of compound
What is the % composition of Ba(CN)2 ?
1(137.33 g Ba) + 2(12.01 g C) + 2(14.01 g N) = 189.37 g Ba(CN)2
% Ba 
137.33 g Ba
x 100  72.52 % Ba
189.37 g Ba(CN)2
24.02 g C
%C
x 100  12.68 % C
189.37 g Ba(CN)2
28.02 g N
%N
x 100  14.80 % N
189.37 g Ba(CN)2
What is the % composition of Iron III sulfate,
___________?
Fe2(SO4)3
2(55.85 g Fe) + 3(32.07 g S) + 12 (16.00 g O) = 399.91 g Fe2(SO4)3
2(55.85 g Fe)
% Fe 
x 100 
399.91 g Fe2(SO4)3
27.93 % Fe
3(32.07 g S)
%S
x 100 
399.91 g Fe2(SO4)3
24.06 % S
12(16.00 g O)
%O
x 100 
399.91 g Fe2(SO4)3
48.01 % O
Percent Composition (Hydrates):
1. Ba(CN)2 ∙ 2 H2O is what % water by mass?
1(Ba) + 2(C) + 2(N) + 2(H2O) = 225.41 g Ba(CN)2 • 2 H2O
2(18.02 g H2O)
% H2O 
x 100 
225.41 g Ba(CN)2  2 H2O
15.99 % H2O
Formulas
•Empirical formula: the lowest whole number
ratio of atoms in a compound.
•Molecular formula: the true number of atoms
of each element in the formula of a compound.
 molecular formula = (empirical formula)n
[n = integer]
 molecular formula = C6H6 = (CH)6
 empirical formula = CH
Formulas (continued)
Formulas for ionic compounds are
ALWAYS empirical (lowest whole
number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas (continued)
Formulas for molecular compounds might be
empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Empirical Formula
Determination
1. Base calculation on 100 grams of
compound.
2. Determine moles of each element in
100 grams of compound.
3. Divide each value of moles by the
smallest of the values.
4. Multiply each number by an integer to
obtain all whole numbers.
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical formula
of adipic acid?
(49.32g C) (1 mol C ) =
4.107 molC
(12.01 gC )
(6.851gH)(1 molH ) =
6.78 molH
(1.01 gH )
(43.84g O)( 1molO ) =
2.74 molO
(16.00 gO )
Empirical Formula Determination
(part 2)
Divide each value of moles by the smallest of the
values.
4.107
molC
Carbon:
= 1.50
2.74 molO
6.78 molH
= 2.47
Hydrogen:
2.74 molO
2.74 molO
= 1.00
Oxygen:
2.74 molO
Empirical Formula Determination
(part 3)
Multiply each number by an integer to obtain all
whole numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Empirical formula:
C3H5O2
Oxygen: 1.00
x 2
2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic
acid is 146 g/mol. What is the molecular
formula of adipic acid?
1. Find the empirical mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic
acid is 146 g/mol. What is the
molecular formula of adipic acid?
2. Divide the molecular mass by the
mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
3. Multiply the empirical formula by this
number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
=2
73
(C3H5O2) x 2
=
C6H10O4
Hydrates
ionic
H2O trapped in
A. An _______
compound that has _______
the crystal lattice structure.
Hydrated salt
1. ________________:
A compound with water (i.e. BaI2
•2 H2O).
2. __________________:
A compound without water (i.e.
Anhydrous salt
BaI2).
B. Determination of a hydrate:
H 2O
1. Determine the mass of ________.
2. Determine the mass of the _________________.
anhydrous salt
3. Find the __________
of water and anhydrous salt.
moles
4. Find the mol ratio.
X=
mol H2O
mol anhydrous salt
* Whole number
C. Example:
1. A barium iodide hydrate is heated in a crucible. Determine
the formula of the hydrate given the following data:
Before heating:
10.407 g BaI2 • X H2O (Hydrated Salt)
_ 9.520 g BaI
After heating:
(Anhydrous Salt)
____________ 2
Step 1: Mass of water?
0.887 g H2O
Step 2: Mass of Anhydrous salt? 9.520 g BaI
2
Step 3: Calculate the mol of the water & the anhydrous salt.
0.887 g H2O 1 mol H2O
= 0.0492 mol H2O
18.02 g H2O
9.520 g BaI2 1 mol BaI2
= 0.0243 mol BaI2
391.13 g BaI2
Step 4: Calculate the number of waters.
mol H 2O
0.0492 mol H 2O
=
x
= 2
mol anhydrous salt =
0.0243 mol BaI2
Step 5: BaI2• 2 H2O
2. Data: Before heating: 5.20 g ZnSO3 • X H2O (Hydrated
_ 2.60 g H2O
salt)
____________
After heating:
2.60 g ZnSO3
Step 1: Mass of water?
(Anhydrous salt)
Step 2: Mass of Anhydrous salt? 2.60 g ZnSO3
Step 3: Calculate the mol of the water & the anhydrous salt.
2.60 g H2O
1 mol H2O
= 0.144 mol H2O
18.02 g H2O
2.60 g ZnSO3 1 mol ZnSO3
145.43 g ZnSO3 = 0.0179 mol ZnSO3
Step 4: Calculate the number of waters.
mol H2O
0.144 mol H2O
x

=8
mol anhydrous salt 0.0179 mol ZnSO 3
Step 5: ZnSO3 • 8 H2O
IQ #2
Using your knowledge of mole
calculations and unit conversions,
determine how many atoms there are in
1 gallon of gasoline. Assume that the
molecular formula for gasoline is C6H14
and that the density of gasoline is
approximately 0.85 grams/mL.
Using a conversion factor of 3785 mL per gallon,
we can determine that the mass of gasoline in one
gallon is 3785 mL x 0.8500 g/mL = 3217 grams.
Because the molar mass of C6H14 is 86 g/mole,
there are 3217 / 86 moles of gasoline molecules,
or 37.4 moles of molecules present.
Multiplying 37.4 x 20 (the number of atoms per
mole of gasoline), there are 748 moles of atoms.
Finally, multiplying 748 moles of atoms by 6.02 x
1023 atoms/mole, we can find that there are
4.50 x 1026 atoms present in the sample.
CHEMICAL REACTIONS
Reactants: Zn + I2
Product: Zn I2
Chemical Equations
• Their Job: Depict the kind of
reactants and products and their
relative amounts in a reaction.
4 Al (s) + 3 O2 (g) ---> 2 Al2O3 (s)
• The numbers in the front are called
coefficients
stoichiometric ____________.
• The letters (s), (g), and (l) are the
physical states of compounds.
Introduction
– Chemical reactions occur when bonds
between the outermost parts of atoms
are formed or broken.
– Chemical reactions involve changes in
matter, the making of new materials
with new properties, and energy
changes.
– Symbols represent elements, formulas
describe compounds, chemical equations
describe a chemical reaction.
Parts of a Reaction Equation
– Chemical equations show the conversion
of reactants (the molecules shown on the
left of the arrow) into products (the
molecules shown on the right of the
arrow).
• A “+” sign separates molecules on the
same side
• The arrow is read as “yields”
• Example
C + O2  CO2
• This reads “carbon plus oxygen react
to yield carbon dioxide”
• The charcoal used in a grill is basically carbon. The
carbon reacts with oxygen to yield carbon dioxide.
The chemical equation for this reaction, C + O2 
CO2, contains the same information as the English
sentence but has quantitative meaning as well.
Chemical Equations
 Because of the principle of
the conservation of matter,
an equation must be
balanced.
 It must have the same
number of atoms of the
same kind on both sides.
Lavoisier, 1788
Symbols Used in Equations
(s)
• Solid ___
• Liquid (l)
(g)
• Gas ___
• Aqueous solution (aq)
H2SO4
• Catalyst
• Escaping gas ()
• Change of temperature ()
Balancing Equations
– When balancing a chemical reaction you
may add coefficients in front of the
compounds to balance the reaction, but
you may
not
change the subscripts.
– Changing the subscripts changes the
compound.
– Subscripts are determined by the
valence electrons (charges for ionic or
sharing for covalent)
Subscripts vs. Coefficients
• The subscripts
tell you how
many atoms of
a particular
element are in
a compound.
The coefficient
tells you about
the quantity,
or number, of
molecules of
the compound.
Chemical Equations
4 Al(s) + 3 O2(g)
---> 2 Al2O3(s)
This equation means:
4 Al atoms + 3 O2 molecules
---produces--->
2 molecules of Al2O3
AND/OR
4 moles of Al + 3 moles of O2
---produces--->
2 moles of Al2O3
Steps to Balancing Equations
There are four basic steps to balancing a chemical
equation.
1.Write the correct formula for the reactants and
the products. DO NOT TRY TO BALANCE IT
YET! DO NOT CHANGE THE FORMULAS!
2.Find the number of atoms for each element on
the left side. Compare those against the
number of the atoms of the same element on the
right side.
3.Determine where to place coefficients in front
of formulas so that the left side has the same
number of atoms as the right side for EACH
element in order to balance the equation.
4.Check your answer to see if:
– The numbers of atoms on both sides of the
equation are now balanced.
– The coefficients are in the lowest possible
whole number ratios. (reduced)
Some Suggestions to Help You
• Take one element at a time, working left to
right except for H and O. Save H for next
to last, and O until last.
• If everything balances except for O, and
there is no way to balance O with a whole
number, double all the coefficients and
try again. (Because O is diatomic as an
element)
• (Shortcut) Polyatomic ions that appear on
both sides of the equation should be
balanced as independent units
Balancing Equations
2 H2(g) + ___ O2(g) ---> ___
2 H2O(l)
___
What Happened to the
Other Oxygen Atom?????
This equation is not
balanced!
Now, follow your rules for
balancing equations and
see if you can do it!
IQ 2
Rewrite these word equations as balanced
chemical equations
1) Hydrogen + sulfur  hydrogen sulfide
2) Iron III chloride + calcium hydroxide 
iron III hydroxide + calcium chloride
3) Heating copper (II) sulfide in the presence
of oxygen gas produces pure copper and
sulfur dioxide gas.
Balancing
Equations
___
2 Al(s) + ___
3 Br2(l) ---> ___ Al2Br6(s)
Balancing
Equations
Since we can’t have a fraction
Take
the
number
you
need
and
for
a coefficient,
we must
Need
11
Oxygen
atoms
but
5 11/2).
____C
+ (i.e.
_____
O2(g) ---->
put
it
over
3H
8(g) 2
multiply
it come
out byinmultiplying
they only
pairs! How all
of
2.2O(g)
_____CO
(g) + _____
4by H
canthe
we3coefficients
fix 2this?
11
2
=11
____B
H
(g)
+
_____
O2(g) ---->
1
?
4 10
2
10
2 B2O3(g) + _____
4
___
H2O(g)
5
Learning Check!
Sodium phosphate + iron (III) oxide 
sodium oxide + iron (III) phosphate
Na3PO4 + _____ Fe2O3 ---->
____ Na2O
+ _____ FePO4
Algebraic Method
Consider: Ca3(PO4)2 + H2SO4
CaSO4 + H3PO4
1. Assign letters to unknown coefficients:
A Ca3(PO4)2 + B H2SO4
C CaSO4 + D H3PO4
2. Make a grid indicating the appearance of element (or
ion) in each species of the equation. Use a whole
number and the coefficient “letter” to indicate each
appearance.
Ca 3a + 0 = c + 0
PO4 2a + 0 = 0 + d
H
0 + 2b = 0 + 3d
SO4 0 + b = c + 0
3. Reduce each equation:
Ca
PO4
H
SO4
3a = c
2a = d
2b = 3d
b = c
4. Assume a = 1; solve for coefficients:
A= 1
C= 3
D = 2
B = 3
5. Write equation with coefficients:
Ca3(PO4)2 + 3 H2SO4
3 CaSO4 +2 H3PO4
Example:
A
B
C
Na2CO3 + C + Sb2S3
Na: 2a = 2e
C: a + b = f
O: 3a = 2f
Sb: 2c = d
S: 3c = e
a = 1
e = 1
f = 3/2
c = 1/3
d = 2/3
b = 1/2
6Na2CO3 + 3 C + 2 Sb2S3
D
E
F
Sb + Na2S + CO2
x 6
A = 6
E = 6
F = 9
C = 2
D =4
B = 3
4 Sb + 6 Na2S + 9 CO2
Topics for Unit #5
•
•
•
•
The Mole & Quantifying Substances (ch. 10)
Avagadro’s Number (ch. 10)
Chemical Equations & Balancing (ch. 11)
Empircal vs. Molecular Formulas (ch. 11)
Introductory Questions #1
1) How do you measure matter if there’s a lot of it
there? Name the three ways suggested in the
text (pg. 287)
2) What is the mass of 90 average-sized apples if 1
dozen of apples has a mass of 2.0 kg?
3) How many representative parts do you find in
one mole?
4) Magnesium is a light metal used in the
manufacture of aircraft, automobile wheels, tools,
and garden furniture. How many moles of
magnesium is 1.25 x1023 atoms of magnesium?
(see practice problem 10.2 on pg 291)
IQ#1-answers
1. Count the items, determine the mass, & Volume
2. We know: 1 dozen apples = 2.0 Kg
# of apples = 90
12 apples = 1 dozen
**Dimensional analysis:
90 apples x 1 doz. apples x 2.0 kg
12 apples
1 doz apples
= 15 kg
3. 1 mol = 6.02 x 1023 parts
IQ#1-answers cont’d
4. Count the items, determine the mass, &
Volume
**We know:
# of atoms present = 1.25 x1023 Mg atoms
1 mol Mg = 6.02 x1023 atoms Mg
Conversion: # atoms → # moles
Use Dimensional analysis:
1.25 x1023 atoms Mg x
1 mol Mg
= 0.208 mol Mg
23
6.02 x10 atoms Mg
Answers to Homework
Practice Problems #1-5
1) 5.0 kg
2) 672 seeds
3) 4.65 mol Si
4) 0.360 mol Br2
5) 2.75 x1024 atoms
6) 7.72 mol NO2
(pg. 289)
(pg. 289)
(pg. 291)
(pg. 291)
(pg. 292)
(pg. 292)
Introductory Questions #2
1. Using your periodic table determine the
molecular mass of ammonium nitrate
and Potassium permanganate.
2. Which compound has higher molar mass
ammonium nitrate or Potassium
permanganate? Show your work.
3. If I had 38.5 g of potassium
permanganate and 58.6g of ammonium
nitrate which do I have more of? Convert
each into the number ofmoles present to
compare.
Homework
• Do Quest. #7-13 from “The Mole Worksheet I”
• Practice problems:
pg. 296 #7 & 8
pg. 298 #16 & 17
pg. 299 #18 & 19
Read/Review Percent Composition Pgs 305-307
*Do Practice problems #34 & 35 on pg. 307
Introductory Questions #3
1. Which of the following substances has
more molecules/compounds present? Be
sure to show all calculations and work.
242.38g Heptanitrogen pentoxide
85.56g Aluminum arsenate
154.28g hydrogen perioxide
2. How much would 1000 atoms of
magnesium weigh in grams?
3. Determine the % composition of barium
perchlorate.
Assignments to Check/Review
•
•
•
•
•
% Composition Worksheet answers
Chemistry WS II: % Comp & Empirical Formulas
Practice Problems from text: #36 & 37 (pg.310)
Hydrate Formula----Worksheet Today’s Handout
Lect/Disc: Chemical Reactions & equation writing
**Homework:
Hydrate Formula WS
Chemical Equation Worksheet
Signs of a Chemical Reaction
• Evolution of heat and light
• Formation of a gas
• Formation of a precipitate
• Color change
B.Law of Conservation of Mass
• Mass is neither created nor
destroyed in a chemical reaction
• Total mass stays the same
• Atoms can only rearrange
4 H
36 g
2 O
4g
32 g
4 H
2 O
C. Chemical Equations
A+B  C+D
REACTANTS
PRODUCTS
Chemical Symbols
Writing Equations
2H2(g) + O2(g)  2H2O(g)
• Identify the substances involved.
• Use symbols to show:
– How many? - coefficient
– Of what? - chemical formula
– In what state? - physical state
• Remember the diatomic elements.
D. Writing Equations
Two atoms of aluminum react with three
units of aqueous copper(II) chloride to
produce three atoms of copper and two units
of aqueous aluminum chloride.
• How many?
• Of what?
• In what state?
2 Al (s) +3 CuCl2 (aq)  3 Cu(s)+ 2AlCl3 (aq)
E. Describing Equations
• Describing Coefficients:
– individual atom = “atom”
– covalent substance = “molecule”
– ionic substance = “unit”
3CO2  3 molecules of carbon dioxide
2Mg
 2 atoms of magnesium
4MgO  4 units of magnesium oxide
E. Describing Equations
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
• How many?
• Of what?
• In what state?
One atom of solid zinc reacts with
two molecules of aqueous
hydrochloric acid
one unit
to produce
of aqueous zinc chloride and one
molecule of hydrogen gas.
Unit 5 HW Packet
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Book Notes (15 total) 10-1 practice problems #1-4*
Page 315 #50-56, 58-59
Moles, Molecules, and Grams WS
Percent Composition WS
p. 315 # 63-68
Percent Comp., Emp. Formula WS
Writing Complete Equations WS
Balancing Chemical Equations WS
Balancing Equations: Algebraic Technique WS
Unit 5 Sample Test