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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
First exam coming up 1 week from this Thursday
will cover Chapters 1-3 and at least part of Chapter 4.
- read textbook
- attend lecture
- do homework/practice problems
- ask questions in recitation
DO NOT procrastinate!
help session tomorrow night
o Wed. Sept 12, HELD 100, 7-9pm
review session
o Wed. Sept 19, HELD 200, 7-9pm
come with textbook, calculators, and questions!
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
Calculations involving balanced equations. Consider the
following cooking analogy:
chicken
breast +
veggies +
oil +
chicken
stir fry
rice à
we can use the recipe to write a balanced equation:
4 chicken
8
3
4
8
chicken
breasts + cups veggies + tbsp oil + cups rice à servings stir fry
But what if we only have 5 dinner guests and we want to
cut back on the amount of food we make? How much
chicken do we need? How many cups of vegetables?
We set up a ratio that relates the ingredients to one another.
5 servings
x
4 chicken breasts
8 servings
=
2.5 chicken
breasts
5 servings
x
8 cups veggies
8 servings
=
5 cups
veggies
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
We use balanced equations to set up mole ratios to do the
same kind of calculations.
Ex. if we want to make 0.750 moles of AgCl, how much
CaCl2 do we need? How much AgNO3?
CaCl2
+
2AgNO3 à
Ca(NO3)2 + 2AgCl
0.750 mol AgCl x 1 mol CaCl2
2 mol AgCl
=
0.750 mol AgCl x 2 mol AgNO3 =
2 mol AgCl
0.375 mol CaCl2
0.750 mol AgNO3
How much Ca(NO3)2 will we also make?
0.750 mol AgCl x 1 mol Ca(NO3)2 = 0.375 mol AgNO3
2 mol AgCl
This example involves moles. But what about grams?
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
CaCl2
+
2AgNO3 à
Dr. Joy Heising
S 572-580
Ca(NO3)2 + 2AgCl
10.0 g CaCl2 yields how many grams of AgCl?
Strategy: see next slide
Molar mass CaCl2 =
Molar mass AgCl =
40.08 + 2(35.45) = 110.98 g/mol
107.9 + 35.45 = 143.4 g/mol
10.0g CaCl2 x 1 mol CaCl2 x 2 mol AgCl x 143.4g AgCl = 25.8 g AgCl
110.0g CaCl2 1 mol CaCl2 1 mol AgCl
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
Problem solving strategy:
1. what do we want to know? # g AgCl
2. what do we know? #g CaCl2
3. how can we get from #1 to #2?
g CaCl2
g AgCl
? How can you get AgCl from CaCl2?
Use the balanced equation to create a mole ratio
10.0 g
CaCl2
mol
CaCl2
à
mol
AgCl
?g
AgCl
¦ how can I do this kind of problem when the quantities
that are given and/or requested are in grams?
Convert to moles!
Calculate the molar masses of AgCl and CaCl2
Why? to relate grams to moles
10.0 g
CaCl2
à
mol
CaCl2
mol
AgCl
à
?g
AgCl
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
In these calculations, we have been assuming that we have
plenty of the ‘other reactants’
- plenty of AgNO3for CaCl2
but what if I have limited quantities of both reactants?
Which quantity determines the amount of product?
Limiting reactant – a substance that limits the amount of
product(s) that can be formed.
2 oz. Cherry syrup + 1 can coke = 1 cherry coke
8 oz. Cherry syrup
6-pack coke
how many cherry cokes?
8 oz. Syrup
6 cokes x
x
1 cherry coke
=
2 oz. cherry syrup
1 cherry coke
1 coke
=
4 cherry cokes
6 cherry cokes
only have enough to make 4 cherry cokes – syrup is
limiting reactant.
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
What is the maximum mass of sulfur dioxide which can be
obtained from the reaction of 95.6 g of carbon disulfide
with 110. g of oxygen?
Strategy: see next slide
Sulfur dioxide
Carbon disulfide
Oxygen
SO2
CS2
O2
molar mass
molar mass
molar mass
=
=
=
64.07 g/mol
76.14 g/mol
32.00 g/mol
Write equation:
CS2
+
O2
à
SO2
CS2
+
O2
à
SO2
+
CO2
CS2
+
3O2
à
2SO2 +
CO2
à
mol à
carb. disulf.
a. 95.6 g
carbon disulfide
need one more product – CO2
next balance
mol
sulf. diox.
à
?g
sulfur dioxide
95.6 g CS2 x 1 mol CS2 x 2 mol SO2 x 64.07g SO2 = 161 g SO2
76.14 g CS2 1 mol CS2
1 mol SO2
b.
110. g
oxygen
110. g O2
c.
à
mol
oxygen
à
mol
sulf. diox.
à
?g
sulfur dioxide
x 1 mol O2 x 2 mol SO2 x 64.07g SO2 = 147 g SO2
32.00g O2
3 mol O2
1 mol SO2
147g is the maximum mass (there isn’t enough O2 to make 161 g SO2 ).
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
Problem solving strategy:
1. don’t panic.
2. what do we want to know? ‘maximum mass’ sulfur dioxide
3. what do we know? #g carbon disulfide, #g oxygen
4. how can we get from #1 to #2?
a.
95.6 g………………………………………………………? g
carbon disulfide
sulfur dioxide
b.
110. g……………………………………………………….? g
oxygen
sulfur dioxide
c. compare results of a and b and choose the smaller number.
Writing a master plan for steps a and b, then:
a.
95.6 g
à
mol
à
mol
à
carbon disulfide
carb. disulf. sulf. diox.
b.
110. g
oxygen
à
mol
à
mol
à
oxygen
sulf. diox.
?g
sulfur dioxide
?g
sulfur dioxide
? How can I get to sulfur dioxide from carbon disulfide?
From oxygen?
How did I do it before?
Used the balanced equation to create a mole ratio
? How am I going to get a balanced equation?
1. write the formulas for the names
2. figure out on which ‘side’ of the equation each
compound belongs
3. make sure it obeys the law of conservation of
matter (balance it).
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
But what if we had only gotten 100. g of SO 2 with those
amounts of CS2 and O2?
- reaction did not go to completion
- several competing reactions involving CS2
and/or O2 were going on at the same time.
Theoretical yield – the amount of product we should get
when a reactant is completely consumed
according to our chemical equation
Actual yield – the amount of product actually measured
The actual yield is always equal to or lower than the
theoretical yield
Percent yield
actual yield
theoretical yield
x
100
the percent yield of SO 2 is
100. g SO2
146.8 g SO2
x
100
=
68.1 %
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
We have done a lot of calculations between moles and
grams. However, a lot of reactions happen in water.
Solution – a homogeneous mixture of two or more
substances (ex. NaCl in H2O)
Solute – a substance that is dissolved in another substance
(NaCl)
Solvent – the substance in which the other substance is
dissolved (H2O, liquid)
Aqueous solution – when solvent is H2O
For a given quantity of a solution, how can we know:
- amount of solute?
- amount of solvent?
Concentration – the amount of solute per solution
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
Percent by mass:
Percent solute
=
mass of solute x 100
mass of solution
ex. if we have 20.0 g NaOH dissolved in H2O to make 250.
g of solution, what is the percent by mass of NaOH?
20.0 g NaOH =
250 g solution
8.00 % NaOH
(230 g H2O)
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CHEM 101 LECTURE NOTES
Fall 2001
Tuesday, September 11, 2001 Chapter 3
Dr. Joy Heising
S 572-580
TEAM QUIZ #2
5.00 g CH3OH reacts with 8.00 g O2.
What is the limiting reactant?
If only 5.00 g of H2O is the observed yield, what is the
percent yield?
2CH3OH + 3O2
à
2CO2
+
4H2O
molar mass CH3OH = 32.04 g/mol
molar mass H2O
= 18.02 g/mol
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