Transcript CHAPTER 2
CHAPTER 7
Chemical Calculations 7.1-7.7
1
The Mole
A number of atoms, ions, or molecules
that is large enough to see and handle.
A mole = number of things
Avogadro’s number = 6.022 x 1023
2
Just like a dozen = 12 things
One mole = 6.022 x 1023 things
Symbol for Avogadro’s number is NA.
The Mole
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The Mole
How do we know when we have a mole?
Molar mass - mass in grams numerically equal
to the atomic weight of the element in grams.
H has an atomic weight of 1.00794 g
1.00794 g of H atoms = 6.022 x 1023 H atoms
Mg has an atomic weight of 24.3050 g
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count it out
weigh it out
24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms
The Mole
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The Mole
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Example 2-1: Calculate the mass of a
single Mg atom in grams to 3 significant
figures.
? g Mg
The Mole
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Example 2-1: Calculate the mass of a
single Mg atom in grams to 3 significant
figures.
? g Mg 1 Mg atom
The Mole
Example 2-1: Calculate the mass of a
single Mg atom in grams to 3 significant
figures.
1 mol Mg atoms
? g Mg 1 Mg atom
23
6.022 10 Mg atoms
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The Mole
Example 2-1: Calculate the mass of a
single Mg atom, in grams, to 3 significant
figures.
1 mol Mg atoms
? g Mg 1 Mg atom
23
6.022
10
Mg atoms
24.30gMg
4.04 10 23 g Mg
1 mol Mg atoms
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The Mole
Example 2-2: Calculate the number of atoms in
one-millionth of a gram of Mg to 3 significant
figures.
? Mg atoms
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The Mole
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Example 2-2: Calculate the number of
atoms in one-millionth of a gram of Mg to
3 significant figures.
1 mol Mg
6
? Mg atoms 1.00 10 g Mg
24.30 g Mg
The Mole
Example 2-2: Calculate the number of
atoms in one-millionth of a gram of Mg to
3 significant figures.
? Mg atoms 1.00 10
6
6.022 1023 Mg atoms
1 mol Mg atoms
12
1 mol Mg
g Mg
24.30 g Mg
The Mole
Example 2-2: Calculate the number of
atoms in one-millionth of a gram of Mg to
3 significant figures.
1 mol Mg
? Mg atoms 1.00 10 g Mg
24.30 g Mg
6
6.022 1023 Mg atoms
2.48 1016 Mg atoms
1 mol Mg atoms
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The Mole
Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
? Mg atoms
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The Mole
Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
? Mg atoms 1.67 mol Mg
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The Mole
Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
6.022 10 23 Mg atoms
? Mg atoms 1.67 mol Mg
1 mol Mg
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The Mole
Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
6.022 10 23 Mg atoms
? Mg atoms 1.67 mol Mg
1 mol Mg
1.00 10 24 Mg atoms
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The Mole
Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
6.022 1023 Mg atoms
? Mg atoms 1.67 mol Mg
1 mol Mg
1.00 1024 Mg atoms
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The Mole
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Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
You do it!
The Mole
Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
? mol Mg 73.4 g Mg
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The Mole
Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
1 mol Mg atoms
? mol Mg 73.4 g Mg
24.30 g Mg
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The Mole
Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
1 mol Mg atoms
? mol Mg 73.4 g Mg
24.30 g Mg
3.02 mol Mg
IT IS IMPERATIVE THAT YOU KNOW
HOW TO DO THESE PROBLEMS
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Formula Weights, Molecular
Weights, and Moles
How do we calculate the molar mass of a
compound?
add atomic weights of each atom
The molar mass of propane, C3H8, is:
3 C 3 12.01 amu 36.03 amu
8 H 8 1.01 amu 8.08 amu
Molar mass
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44.11 amu
Formula Weights, Molecular
Weights, and Moles
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The molar mass of calcium nitrate,
Ca(NO3)2 , is:
You do it!
Formula Weights, Molecular
Weights, and Moles
1 Ca 1 40.08 amu 40.08 amu
2 N 2 14.01 amu 28.02 amu
6 O 6 16.00 amu 96.00 amu
Molar mass
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164.10 amu
Formula Weights, Molecular
Weights, and Moles
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One Mole of
Cl2 or 70.90g
C3H8 You do it!
Contains
6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
Formula Weights, Molecular
Weights, and Moles
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One Mole of
Cl2 or 70.90g
C3H8 or 44.11 g
Contains
6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
6.022 x 1023 C3H8 molecules
3 (6.022 x 1023 ) C atoms
8 (6.022 x 1023 ) H atoms
Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3H8 molecules 74.6 g C3H8
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Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3 H 8 molecules 74.6 g C3 H 8
1 mole C3 H 8
44.11 g C3 H 8
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Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3H8 molecules 74.6 g C3H8
1 mole C3H8 6.022 1023 C3H8 molecules
44.11 g C3H8
44.11 g C3H8
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Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3 H 8 molecules 74.6 g C3 H 8
1 mole C3 H 8 6.022 10 23 C3 H 8 molecules
44.11 g C3 H 8
44.11 g C3H 8
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1.02 10 molecules
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Formula Weights, Molecular
Weights, and Moles
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Example 2-8. Calculate the number of O
atoms in 26.5 g of Li2CO3.
You do it!
Formula Weights, Molecular
Weights, and Moles
Example 2-8. Calculate the number of O
atoms in 26.5 g of Li2CO3.
1 mol Li 2 CO3
? O atoms 26.5 g Li 2 CO3
73.8 g Li 2 CO3
6.022 10 23 form. units Li 2 CO3
3 O atoms
1 mol Li 2 CO3
1 formula unit Li 2 CO3
6.49 10 23 O atoms
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Percent Composition
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% composition = mass of an individual
element in a compound divided by the
total mass of the compound x 100%
Determine the percent composition of C
in C3H8.
mass C
%C
100%
mass C3 H 8
3 12.01 g
100%
44.11 g
81.68%
Percent Composition and
Formulas of Compounds
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What is the percent composition of H in C3H8?
You do it!
Percent Composition
What is the percent composition of H in C3H8?
mass H
%H
100%
mass C3 H 8
8 H
100%
C3H 8
8 1.01 g
100% 18.32%
44.11 g
or
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18.32% 100% 81.68%
Percent Composition
Example 2-10: Calculate the percent
composition of Fe2(SO4)3 to 3 significant
figures.
You do it!
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Percent Composition
Example 2-10: Calculate the percent
composition of Fe2(SO4)3 to 3 sig. fig.
2 Fe
2 55.8 g
% Fe
100%
100% 27.9% Fe
Fe 2 (SO 4 ) 3
399.9 g
3 S
3 32.1 g
%S
100%
100% 24.1% S
Fe 2 (SO 4 ) 3
399.9 g
12 O
12 16.0 g
% O
100%
100% 48.0% O
Fe 2 (SO 4 ) 3
399.9 g
Total
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100%
Percent Composition
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Empirical and Molecular Formulas
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Empirical Formula - smallest whole-number ratio
of atoms present in a compound
Molecular Formula - actual numbers of atoms of
each element present in a molecule of the
compound
We determine the empirical and molecular
formulas of a compound from the percent
composition of the compound.
Empirical And Molecular
Formulas
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Empirical Formulas
Example 2-11: A compound contains 24.74% K,
34.76% Mn, and 40.50% O by mass. What is its
empirical formula?
Make the simplifying assumption that we have
100.0 g of compound.
In 100.0 g of compound there are:
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24.74 g of K
34.76 g of Mn
40.50 g of O
Empirical Formulas
1 mol K
? mol K 24.74g K
0.6327mol K
39.10g K
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Empirical Formulas
1 mol K
? mol K 24.74g K
0.6327mol K
39.10g K
1 mol Mn
? mol Mn 34.76g Mn
0.6327mol Mn
54.94g Mn
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Empirical Formulas
1 mol K
? mol K 24.74g K
0.6327mol K
39.10g K
1 mol Mn
? mol Mn 34.76g Mn
0.6327mol Mn
54.94g Mn
1mol O
? mol O 40.50g O
2.531mol O
16.00g O
obtain smallest whole number ratio
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Empirical Formulas
1 mol K
? mol K 24.74g K
0.6327mol K
39.10g K
1 mol Mn
? mol Mn 34.76g Mn
0.6327mol Mn
54.94g Mn
1mol O
? mol O 40.50g O
2.531mol O
16.00g O
obtain smallest whole number ratio
for K
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0.6327
1K
0.6327
for Mn
0.6327
1 Mn
0.6327
Empirical Formulas
? mol K
24.74 g K
1 mol K
0.6327 mol K
39.10 g K
1 mol Mn
? mol Mn 34.76 g Mn
0.6327 mol Mn
54.94 g Mn
1mol O
? mol O 40.50 g O
2.531 mol O
16.00 g O
obtain smallest whole number ratio
0.6327
for K
1K
0.6327
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0.6327
for Mn
1 Mn
0.6327
2.531
for O
4O
0.6327
thus the chemical formula is KMnO 4
Empirical Formulas
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Example 2-12: A sample of a compound
contains 6.541g of Co and 2.368g of O.
What is empirical formula for this
compound?
You do it!
Empirical Formulas
Example 2-12: A sample of a compound
contains 6.541g of Co and 2.368g of O.
What is empirical formula for this
compound?
1 mol Co
? mol Co 6.541 g Co
0.1110 mol Co
58.93 gCo
1mol O
? mol O 2.368 g O
0.1480 mol O
16.00 g O
find smallest whole number ratio
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Empirical Formulas
Example 2-12: A sample of a compound
contains 6.541g of Co and 2.368g of O.
What is empirical formula for this
compound?
0.1110
0.1480
for Co
1 Co for O
1.333O
0.1110
0.1110
multipy both by 3 to turn fraction to whole number
1 Co 3 3 Co 1.333 O 3 4 O
Thus the compound' s formula is :
Co3O 4
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Molecular Formulas
Example 2-13: A compound is found to
contain 85.63% C and 14.37% H by mass. In
another experiment its molar mass is found to
be 56.1 g/mol. What is its molecular formula?
short cut method
1 mol contains 56.1 g
85.63% is C and 14.37% is H
56.1 g 0.8563 48.0 g of C
56.1 g 0.1437 8.10 g of H
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Determination of Molecular
Formulas
convert masses to moles
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1 mol C
48.0 g of C
4 mol C
12.0 g C
1 mol H
8.10 g of H
8 mol H
1.01 g H
Thus the formula is :
C4 H8
Calculations Based on Chemical
Equations
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Calculations Based on Chemical
Equations
Fe2O3 + 3 CO 2 Fe + 3 CO2
Example 3-1: How many CO molecules
are required to react with 25 formula
units of Fe2O3?
3 CO molecules
? CO molecules= 25 formulaunits Fe2O3
1 Fe2O3 formulaunit
75 moleculesof CO
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Calculations Based on Chemical
Equations
Example 3-2: How many iron atoms can
be produced by the reaction of 2.50 x 105
formula units of iron (III) oxide with excess
carbon monoxide?
? Fe atoms= 2.5010 formulaunits Fe2O3
5
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Calculations Based on Chemical
Equations
Example 3-2: How many iron atoms can
be produced by the reaction of 2.50 x 105
formula units of iron (III) oxide with excess
carbon monoxide?
? Fe atoms= 2.5010 formulaunits Fe2O3
5
2 Fe atoms
1 formulaunits Fe2O3
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Calculations Based on Chemical
Equations
Example 3-2: How many iron atoms can
be produced by the reaction of 2.50 x 105
formula units of iron (III) oxide with excess
carbon monoxide?
? Fe atoms= 2.5010 formulaunits Fe2O3
5
2 Fe atoms
5
5.0010 Fe atoms
1 formulaunits Fe2O3
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Calculations Based on Chemical
Equations
Example 3-3: What mass of CO is
required to react with 146 g of iron (III)
oxide?
1 molFe2O3
? g CO = 146g Fe2O3
159.7 g Fe2O3
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Calculations Based on Chemical
Equations
Example 3-3: What mass of CO is
required to react with 146 g of iron (III)
oxide?
1 molFe2O3
3 molCO
? g CO = 146g Fe2O3
159.7 g Fe2O3 1 molFe2O3
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Calculations Based on Chemical
Equations
Example 3-3: What mass of CO is
required to react with 146 g of iron (III)
oxide?
1 mol Fe2O3
3 molCO
? g CO = 146g Fe2O3
159.7 g Fe2O3 1 mol Fe2O3
28.0 g CO
76.8 g CO
1 molCO
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Calculations Based on Chemical
Equations
Example 3-4: What mass of carbon dioxide can
be produced by the reaction of 0.540 mole of
iron (III) oxide with carbon monoxide?
3 molCO 2
? g CO 2 0.540 molFe2O3
1 mol Fe2O3
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Calculations Based on Chemical
Equations
Example 3-4: What mass of carbon dioxide can
be produced by the reaction of 0.540 mole of
iron (III) oxide with carbon monoxide?
3 molCO 2
44.0 g CO 2
? g CO 2 0.540 mol Fe2O3
1 mol Fe2O3 1 molCO 2
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Calculations Based on Chemical
Equations
Example 3-4: What mass of carbon
dioxide can be produced by the reaction of
0.540 mole of iron (III) oxide with carbon
monoxide?
3 mol CO 2
44.0 g CO 2
? g CO 2 0.540 mol Fe 2 O 3
1 mol Fe 2 O 3 1 mol CO 2
= 71.3 g CO 2
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Calculations Based on Chemical
Equations
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Example 3-5: What mass of iron (III) oxide
reacted with carbon monoxide if the
carbon dioxide produced by the reaction
had a mass of 8.65 grams?
You do it!
Calculations Based on Chemical
Equations
Example 3-5: What mass of iron (III) oxide
reacted with carbon monoxide if the
carbon dioxide produced by the reaction
had a mass of 8.65 grams?
1 molCO2 1mol Fe2O3
? g Fe2O3 8.65g CO2
44.0g CO2 3 molCO2
159.7 g Fe2O3
10.5 g Fe2O3
1 mol Fe2O3
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Calculations Based on Chemical
Equations
66
Example 3-6: How many pounds of carbon
monoxide would react with 125 pounds of
iron (III) oxide?
You do it!
Calculations Based on Chemical
Equations
454 g Fe2 O3
? lb CO = 125lb Fe2 O3
1 lb Fe 2 O 3
1 mol Fe 2 O3
3 molCO
159.7 g Fe 2 O3 1 mol Fe2 O 3
28 g CO
1 lb CO
65.7 lb CO
1 mol CO 454 g CO
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YOU MUST BE PROFICIENT WITH THESE
TYPES OF PROBLEMS!!!
Limiting Reactant Concept
Kitchen example of limiting reactant concept.
1 packet of muffin mix + 2 eggs + 1 cup of milk
12 muffins
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How many muffins can we make with the
following amounts of mix, eggs, and milk?
Limiting Reactant Concept
69
Mix Packets Eggs
Milk
1
1 dozen
1 gallon
limiting reactant is the muffin mix
2
1 dozen
1 gallon
3
1 dozen
1 gallon
4
1 dozen
1 gallon
5
1 dozen
1 gallon
6
1 dozen
1 gallon
7
1 dozen
1 gallon
limiting reactant is the dozen eggs
Limiting Reactant Concept
Example 3-7: Suppose a box contains 87 bolts,
110 washers, and 99 nuts. How many sets,
each consisting of one bolt, two washers, and
one nut, can you construct from the contents of
one box?
1 bolt 87 sets
55sets
110 washers 1 set
2 washers
99 sets
99 nuts1 set
1 nut
87 bolts1 set
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the maximumnumber we can makeis 55 sets
determinedby thesmallest number
Limiting Reactant Concept
Example 3-8: What is the maximum mass of
sulfur dioxide that can be produced by the
reaction of 95.6 g of carbon disulfide with 110. g
of oxygen?
CS2 3 O2 CO2 2 SO 2
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Limiting Reactant Concept
Example 3-8: What is the maximum mass of
sulfur dioxide that can be produced by the
reaction of 95.6 g of carbon disulfide with 110. g
of oxygen?
CS2 3 O2 CO2 2 SO 2
1 mol
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3 mol
1 mol
2 mol
Limiting Reactant Concept
Example 3-8: What is the maximum mass of
sulfur dioxide that can be produced by the
reaction of 95.6 g of carbon disulfide with 110. g
of oxygen?
CS2 3 O 2 CO 2 2 SO 2
1 mol 3 mol
1 mol 2 mol
76.2 g 3(32.0 g) 44.0 g 2(64.1 g)
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Limiting Reactant Concept
Example 3-8: What is the maximum mass of
sulfur dioxide that can be produced by the
reaction of 95.6 g of carbon disulfide with 110. g
of oxygen?
Determine which mass makes the most product
CS2 3 O2 CO2 2 SO 2
1 molCS2
? molSO 2 95.6 g CS2
76.2g
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Limiting Reactant Concept
CS2 3 O 2 CO 2 2 SO 2
1 molCS2
? molSO 2 95.6 g CS2
76.2g
2 molSO 2
64.1 g SO 2
161g SO 2
1 molCS2
1 molSO 2
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Limiting Reactant Concept
CS2 3 O 2 CO 2 2 SO 2
1 molCS2 2 molSO 2 64.1 g SO 2
? molSO 2 95.6 g CS2
161g SO 2
76.2g
1 molCS2 1 molSO 2
1 molO 2 2 molSO 2 64.1 g SO 2
? molSO 2 110 g O 2
147 g SO 2
32.0g O 2 3 molO 2 1 molSO 2
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Which is limiting reactant?
Limiting reactant is O2.
What is maximum mass of sulfur dioxide?
Maximum mass is 147 g.
Percent Yields from Reactions
Theoretical yield is calculated by assuming
that the reaction goes to completion.
Actual yield is the amount of a specified pure
product made in a given reaction.
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Determined from the limiting reactant calculation.
In the laboratory, this is the amount of product that
is formed in your beaker, after it is purified and
dried.
Percent yield indicates how much of the
product is obtained from a reaction.
actual yield
% yield =
100%
theoretical yield
Percent Yields from Reactions
Example 3-9: A 10.0 g sample of ethanol,
C2H5OH, was boiled with excess acetic acid,
CH3COOH, to produce 14.8 g of ethyl acetate,
CH3COOC2H5. What is the percent yield?
CH 3COOH + C2 H 5OH CH 3COOC2 H 5 H 2O
1. Calculate the theoretical yield
78
Percent Yields from Reactions
CH 3COOH + C 2 H 5OH CH 3COOC2 H 5 H 2 O
1. Calculate the theoretical yield
88.0g CH 3COOC2 H 5
? g CH 3COOC2 H 5 = 10.0g C 2 H 5OH
46.0 g C 2 H 5OH
19.1 g CH 3COOC2 H 5
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Percent Yields from Reactions
CH 3COOH + C 2 H 5OH CH 3COOC2 H 5 H 2 O
1. Calculat e t he t heoret ical yield
88.0g CH 3COOC2 H 5
? g CH 3COOC2 H 5 = 10.0g C 2 H 5OH
46.0 g C 2 H 5OH
19.1 g CH 3COOC2 H 5
2. Calculat e t hepercentyield.
80
Percent Yields from Reactions
CH 3COOH + C 2 H 5OH CH 3COOC2 H 5 H 2 O
1. Calculate the theoretical yield
88.0g CH 3COOC2 H 5
? g CH 3COOC2 H 5 = 10.0g C 2 H 5OH
46.0 g C 2 H 5OH
19.1 g CH 3COOC2 H 5
2. Calculate thepercentyield.
14.8g CH 3COOC2 H 5
% yield =
100% 77.5%
19.1g CH 3COOC2 H 5
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