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Welcome !
• Review : Atomic structure and the periodic table.
• Today: the mol, the law of conservation of mass, stoichiometry
Copyright © Houghton Mifflin Company.All rights reserved.
Presentation of Lecture Outlines, 2–1
Learning Objectives
Today we will:
 Gain an understanding of where major
Chemistry strands spiral throughout the
curriculum.
 Explore the states of matter, the mol, and
the Law of Conservation of Mass.
 Measure and analyze percent composition
of a substance in a compound.
 Explore where an understanding of
stoichiometry can be useful in all content
areas.
2
Definition of Matter
The two properties of matter are:
•Mass – amount of matter
•Volume – amount of space occupied
•Therefore…matter is anything that has
mass and takes up space.
States of Matter
• Solid –
definite shape and volume
• Liquid –
definite volume, and takes the shape
of the container
• Gas –
takes the shape and volume of the
container
• Plasma –
a collection of high energy ions and
electrons.
• Go to lecture on the ideal gas law…..
Copyright © Houghton Mifflin Company.All rights reserved.
Presentation of Lecture Outlines, 2–5
Crushed Can Expt
• When the can is heated, what occurs?
• When the water vaporizes in the can, what
happens to the volume of the can?
• What phase change occurs to the water
vapor in the can when the can is immersed in
the cold water? The water ____________.
• What happens to the pressure of the water
vapor in the can when it condenses?
• Would the same result occur if the can were
immersed into the water with the open side
upward?
Copyright © Houghton Mifflin Company.All rights reserved.
Presentation of Lecture Outlines, 2–6
Now we can introduce
the Mole
A mole is a unit of quantity.
A mole is 6.02 x 1023 things.
6.02 x 1023 is known as Avogadro’s constant (NA)
Number of atoms,
molecules or
fundamental units
÷ 6.02 x 1023
Number of moles (mol)
× 6.02 x 1023
Why the Mole?
Consider one
molecule of water
How many molecules in
2000 mL of water?
6.7 x 1025 molecules
6.7 x 1025 molecules is not a manageable
number. Consider:
We count eggs by the dozen
We measure long periods of time in
centuries.
We measure long distances in our
universe using light years.
The Mole
There are many ways of measuring large
quantities that utilize large units. The mole
is one such unit.
The mole is the SI unit for chemical
quantity used to count the particles in a
sample of pure substance.
One mole = 6.02x1023 particles.
“One mole of anything = 6.02x1023 units of that thing”
The Mole
How many
molecules of water
in 2000 mL?
6.7 x 1025 molecules
Or 111 mol meaning 111 moles of water molecules. This is a much more manageable
number.
Molar Mass
Are you sitting comfortably?
Mass and the Mole: Why 6.02×1023?
1 mole = 6.02×1023 was chosen because this was the number of carbon-12 atoms
that has a mass of 12 grams.
Pure carbon (a mix of isotopes) has a mass of 12.01 g per mole.
We call this value MOLAR MASS
Consider
1 atom of sulfur
AR = 32.07 = 5.326×10-23 grams
Consider
1 mole of sulfur = 6.02×1023 atoms = 32.07 grams
6.02×1023 is a REALLY BIG number
Consider 6.02×1023 sheets of paper stacked.
How many round trips to the Moon would this stack of paper be
equivalent to?
8×1010 (eighty billion ) roundtrips. Create, pick, find a mole
analogy of your own. Show the math.
Language issue
gram atomic mass
gram molecular mass
gram formula mass
Aluminum
MOLAR
MASS
AR = 26.98
gram atomic mass = 26.98 g mol-1
Carbon dioxide (CO2)
MR = 44.00
gram molecular mass = 44.00 g mol-1
Sodium chloride (NaCl) MR = 58.44
gram formula mass = 58.44 g mol-1
We weigh chemical quantities in grams. The molar mass value for a substance allows
us to determine the number of moles from a measured mass.
Consider 5.68 g of MgCl2
Molar mass of MgCl2
= 95.21 g mol-1
mass
number of moles =
molar mass
5.68 g
n
=0.0597 mol
-1
92.51 g mol
Number of ÷ 6.02 x 1023
atoms,
molecules or
fundamental
× 6.02 x 1023
units
× molar mass
MOLES
MASS
÷ molar mass
Molarity
• Used to describe solutions
• M = moles/L
Copyright © Houghton Mifflin Company.All rights reserved.
Presentation of Lecture Outlines, 2–19
Activity – Determine the Molar Mass of Butane
Review safety concerns and protocols !!!!!
1. Identify at least 4 possible sources of experimental error
2. The accepted molar mass of butane is 58.0 g/mole. Determine percentage
error for your data.
3. Use your data, assume a molar mass of 58.0 g/mol, and derive the value of R,
the gas constant.
4. The molar mass of the gas is 58.0 g/mol. Determine the volume of one mole
of butane at STP based on your data. What is the percentage error between
your value and 22.4 L (the accepted value).
Copyright © Houghton Mifflin Company.All rights reserved.
Presentation of Lecture Outlines, 2–20
An Introduction…
• C6H12O6 + O2
H20 + CO2 + Energy
• What is wrong with this equation?
• How did you learn to balance an equation?
21
Chemical Reactions: Equations
• Writing chemical equations
– A chemical equation is the symbolic representation of a chemical reaction
in terms of chemical formulas.
– For example, the burning of sodium and chlorine to produce sodium chloride
is written
2Na  Cl 2  2NaCl
– The reactants are starting substances in a chemical reaction. The arrow
means “yields.” The formulas on the right side of the arrow represent the
products.
Copyright © Houghton Mifflin Company.All rights reserved.
Presentation of Lecture Outlines, 2–22
Chemical Reactions: Equations
• Writing chemical equations
– In many cases, it is useful to indicate the states of the substances in the
equation.
– When you use these labels, the previous equation becomes
2Na(s )  Cl 2 (g )  2NaCl(s )
Balancing chemical reactions video link
Balancing an equation activity…..
Copyright © Houghton Mifflin Company.All rights reserved.
Presentation of Lecture Outlines, 2–23
Law of Conservation of Mass
Activity
24
Law of Conservation of Mass:
• This experiment verifies the Law of Conservation
of Matter: Matter is neither created or destroyed
as a result of chemical changes but may be
changed in form. The balanced equations are as
follows:
• 2NaOH (aq) + CuSO4 (aq) -----> Na2SO4 (aq) +
Cu(OH)2 (s)
• 4NH3 (aq) + CuSO4 (aq) -------> Cu(NH3)4SO4 (s)
• Na2CO3 (aq) + CuSO4 (aq) -------> Na2SO4 (aq) +
CuCO3 (s)
25
Law of Conservation of Mass:
Content Blast
• Matter is neither created nor destroyed in any
process.
• Changes in matter are always accompanied by
changes in energy.
• Energy changes operate under the Law of
Conservation of Energy.
• Atoms are neither created nor destroyed in a
chemical reaction.
• For mass to remain constant during a chemical
reaction, the number of atoms of each element
must be the same before and after a chemical
reaction.
• Equations are balanced using coefficients before
the formulas for reactants and products.
26
Bubble Gum Activity
An Introduction to Stoichiometry
• Complete Task 1 and Procedures 1-3.
– Take a 15 minute break while chewing the
gum.
• Return to your groups and complete the
remainder of the activity.
• NOTE the review hints.
• Track and record content, pedagogical
issues/examples, or diversity issues that
may appear in the classroom.
27
Bubble Gum Activity
Report Out Summary
The manufacturer of one common type of bubble gum indicates on the
package that each piece of bubble gum has a mass of 5 grams. The sugar
content for each piece is 4 grams. Determine the percent by mass of sugar in
this type of gum. How do your results compare to the manufacturer’s stated
content?
Using your data collected calculate the total mass of one piece of gum and the
mass of sugar in one piece of gum.
How many moles of sugar are present in one piece of gum (assume that the
sugar is sucrose)?
How many molecules of sugar are present in one piece of gum if the sugar is
sucrose?
Were any of the questions you or your group had answered by this activity? If
so, which ones?
28
Bubble Gum Activity
Report Out Summary
• What questions did your group have prior
to the activity?
• What challenges did you experience while
completing the activity?
• Were your questions answered? If not,
what adjustments could be made to the
activity to help it better address your
questions?
29
Bubble Gum Activity
Content Blast
• Percent composition is the mass of each
component relative to the total mass of the
substance.
• Molar mass is the mass in grams of one mole of a
substance.
• The mole establishes a relationship between the
atomic mass unit and the gram.
• The mass in grams of 1 mole of a substance is
numerically equal to its atomic mass or formulas
mass in atomic mass units.
• The number of molecules in a mole of any
molecular compound is 6.02 x 1023.
30
Stoichiometry
• Using a balanced chemical equation to
determine how much reactant or product is
consumed or produced in a chemical reaction.
• 2 H2 (g) + O2 (g)  2 H2O (l)
– How many moles of H2O are produced from the
reaction of 2 moles of H2?
– How many moles of O2 are required to produce 4
moles of H2O?
Chemical equation = recipe
Chemical Equations
• 2 H2 (g) + O2 (g)  H2O (l)
4 moles molecules + 2 moles molecules  4 moles molecules
Coefficients in Chemical Equation represents mole ratios
http://www.chemistry.ohio-state.edu/betha/nealChemBal/
Stoichiometry
“Chemical Equation = recipe”
• 2 bread slices + 1 cheese slice  1 sandwich
– Suppose need to make as many cheese sandwiches as
possible for a party and have 20 slices of bread...how
many slices of cheese are needed?
• By inspection of recipe (chemical equation)
• Or use coefficient ratio (mole ratio) from equation:
2 bread slices = 1 cheese slice
•
1cheese _ slice
cheese slices = 20 bread slices x 2bread _ slices
= 10 cheese slices needed
Image source: http://www.fotosearch.com/clip-art/sandwich.html
Stoichiometry
• 4 Fe + 3 O2  2 Fe2O3
• How many moles of Fe2O3 will be produced from
the reaction of 1.50 mol of iron?
2molFe 2O 3
• ….mol Fe2O3 = 1.50 mol Fe x
= 0.75 mol
4molFe
Fe2O3
Mole ratio is coefficient ratio
Stoichiometry – using grams
• Suppose the given is grams, not moles…
• Consider the unbalanced reaction:
Al (s) + Fe2O3 (s)  Al2O3 (s) + Fe (l)
If you react 4.0 g of Al with excess Fe2O3, how many moles of
Fe will be produced??
• First balance the equation:
2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (l)
• Chemical equation speaks to us in moles, not grams! So to
use the mole ratio it gives us (2mol Al: 2 mol Fe),
– first need to convert 4.0 g Al to moles of Al
– Then take moles of Al and “convert” to moles of Fe using mole ratio
from balanced equation (coefficient ratio) as was done on previous
slides.
Stoichiometry – using grams
• Road map:
Stoichiometry – using grams
• Suppose we wanted to solve for grams, not moles
2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (l)
• Question: If you react 4.0 g of Al with excess Fe2O3,
how many grams of Fe will be produced?
• Road map is same as before, except one last step to
convert mol Fe to g Fe.
– Remember that to do any stoichiometry problem, you must
be in the language of moles since the chemical equation
(source of relationship or conversion factor) only speaks to
us in moles, not grams!
Stoichiometry – using grams
• Road map: g → mol → mol → g
Demonstration –
Limiting Reagents
Mg +
2 HCl (aq) 
MgCl2(aq) + H2 (g)
1. 100 mL of 1.0 M HCl would have how many mol HCl present?
2. From the balanced equation, we can do a calculation that will show 0.1 mole HCl
requires exactly how many Mg to completely react?
0.1 mol HCl
x 1 mole Mg x
24.3 g Mg
= 1.2 g Mg
2 mol HCl
1 mole Mg
This is the amount in the Red balloon! Therefore, both the HCl and the
Mg in the 2nd flask are completely consumed! Cool!
3. How much H2 should be produced in the Red balloon?
1.20 g Mg
x 1 mole Mg
24.3 g Mg
x
1 mole H2
1 mole Mg
=
0.050 mol H2
39
Demonstration –
Limiting Reagents
Mg +
2 HCl (aq) 
MgCl2(aq) + H2 (g)
4. How much hydrogen should be produced by the Blue balloon flask?
Since there was 0.60 Mg in the blue balloon flask, the Mg will limit the
amount of hydrogen produced. Therefore, use the amount of Mg to
determine the amount of H2 produced.
0.60 g Mg
x 1 mole Mg
24.3 g Mg
x
1 mole H2
1 mole Mg
=
0.025 mol H2 (Blue balloon)
This is half the amount of gas formed in the red balloon. This explains why the
blue balloon is half the size of the red balloon.
Demonstration –
Limiting Reagents
Mg +
2 HCl (aq) 
MgCl2(aq) + H2 (g)
5. How much HCl would be needed to react with the 0.60 g Mg?
0.60 g Mg
x 1 mole Mg
24.3 g Mg
x
2 mole HCl
1 mole Mg
x ___1 L___ = 50 mL!
1 mole HCl
That means 50 mL of HCl are left over un-reacted in the 1st flask!
6. What about the 3rd flask with the Yellow balloon? You can do similar
calculations to those shown previously.
Content Specific Activities
Report Out Summary
• What questions did your group have prior
to the activity?
• What challenges did you experience while
completing the activity?
• Were your questions answered? If not,
what adjustments could be made to the
activity to help it better address your
questions?
42
Day Two Learning Objectives
Review
Today we:
 Gained an understanding of where major
Chemistry strands spiral throughout the
curriculum.
 Explored the Law of Conservation of Mass.
 Measured and analyzed percent composition of a
substance in a compound.
 Explored stoichiometry
43