2
The Mole
A number of atoms, ions, or molecules that is large enough to see and handle.
A mole = number of things Just like a dozen = 12 things One mole = 6.022 x 1023 things
Avogadro’s number = 6.022 x 1023 Symbol for Avogadro’s number is NA.
4
The Mole
How do we know when we have a mole? count it out weigh it out
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 24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms
6
The Mole
Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.
Mgg ?
7
The Mole
Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.
atom Mg1 Mgg ?
8
The Mole
Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.
atoms Mg 106.022
atoms Mg mol 1atom Mg 1Mg g ?
23
9
The Mole
Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures.
Mg g 104.04atoms Mg mol 1
Mgg24.30
atoms Mg 106.022
atoms Mg mol 1atom Mg 1Mg g ?
23
23
10
The Mole
Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.
atoms Mg?
11
The Mole
Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.
Mgg 24.30
Mg mol 1 Mgg 101.00atoms Mg? 6
12
The Mole
Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.
atoms Mgmol 1
atoms Mg106.022
Mgg 24.30
Mg mol 1 Mgg 101.00atoms Mg?
23
6
13
The Mole
Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.
atoms Mg102.48atoms Mgmol 1
atoms Mg106.022
Mgg 24.30
Mg mol 1 Mgg 101.00atoms Mg?
1623
6
16
The Mole
Example 2-3. How many atoms are contained in 1.67 moles of Mg?
Mg mol 1
atoms Mg 106.022 Mg mol 1.67atoms Mg ?
23
17
The Mole
Example 2-3. How many atoms are contained in 1.67 moles of Mg?
atoms Mg 101.00
Mg mol 1
atoms Mg 106.022 Mg mol 1.67atoms Mg ?
24
23
18
The Mole
Example 2-3. How many atoms are contained in 1.67 moles of Mg?
atoms Mg101.00
Mgmol 1
atoms Mg106.022Mg mol 1.67atoms Mg?
24
23
19
The Mole
Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?
You do it!You do it!
21
The Mole
Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?
Mg g 24.30
atoms Mg mol 1 Mg g 4.73Mg mol ?
22
The Mole
Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?
Mg mol 02.3
Mg g 24.30
atoms Mg mol 1 Mg g 4.73Mg mol ?
IT IS IMPERATIVE THAT YOU KNOWHOW TO DO THESE PROBLEMS
23
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:
amu 44.11 mass Molar
amu 8.08 amu 1.01 8H 8
amu 36.03amu 12.01 3C 3
24
Formula Weights, Molecular Weights, and Moles
The molar mass of calcium nitrate, Ca(NO3)2 , is:
You do it!You do it!
25
Formula Weights, Molecular Weights, and Moles
amu 164.10 massMolar
amu 96.00 amu 16.006 O6
amu 28.02 amu 14.012 N2
amu 40.08 amu 40.081 Ca1
26
Formula Weights, Molecular Weights, and Moles
One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms C3H8 You do it!You do it!
27
Formula Weights, Molecular Weights, and Moles
One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms C3H8 or 44.11 g 6.022 x 1023 C3H8 molecules 3 (6.022 x 1023 ) C atoms 8 (6.022 x 1023 ) H atoms
28
Formula Weights, Molecular Weights, and Moles
Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.
8383 HC g 74.6molecules HC ?
29
Formula Weights, Molecular Weights, and Moles
Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.
83
83
8383
HC g 44.11
HC mole 1
HC g 74.6molecules HC ?
30
Formula Weights, Molecular Weights, and Moles
Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.
83
8323
83
83
8383
HC g 44.11
molecules HC 106.022
HC g 44.11
HC mole 1
HC g 74.6molecules HC ?
31
Formula Weights, Molecular Weights, and Moles
Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.
molecules 10 02.1
HC g 44.11
molecules HC 106.022
HC g 44.11
HC mole 1
HC g 74.6molecules HC ?
24
83
8323
83
83
8383
32
Formula Weights, Molecular Weights, and Moles
Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.
You do it!You do it!
33
Formula Weights, Molecular Weights, and Moles
Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.
atoms O 106.49
COLi unit formula 1
atoms O 3
COLi mol 1
COLi unitsform.106.022
COLi g 73.8
COLi mol 1COLi g 26.5atoms O ?
23
3232
3223
32
3232
34
Percent Composition % 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.
81.68%
100%g 44.11
g 12.013
100%HC mass
C mass C %
83
35
Percent Composition and Formulas of Compounds
What is the percent composition of H in C3H8?
You do it!You do it!
36
Percent Composition
What is the percent composition of H in C3H8?
81.68%100%18.32%
or
%18.32100%g 44.11
g 1.018
100%HC
H8
100%HC mass
H massH %
83
83
37
Percent Composition
Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures.
You do it!
38
Percent Composition Example 2-10: Calculate the percent
composition of Fe2(SO4)3 to 3 sig. fig.
100% Total
O 48.0% 100%g 399.9
g 16.012 100%
)(SOFe
O12 O %
S 24.1% 100%g 399.9
g 32.13 100%
)(SOFe
S3 S %
Fe 27.9% 100%g 399.9
g 55.82 100%
)(SOFe
Fe2Fe %
342
342
342
40
Empirical and Molecular Formulas
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.
42
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: 24.74 g of K 34.76 g of Mn 40.50 g of O
44
Empirical Formulas
Mn mol 0.6327Mn g 54.94
Mn mol 1Mn g 34.76 Mn mol ?
K mol 0.6327K g 39.10
K mol 1K g 24.74 K mol ?
45
Empirical Formulas
rationumber wholesmallest obtain
O mol 2.531O g 16.00
O mol1O g 40.50 O mol ?
Mn mol 0.6327Mn g 54.94
Mn mol 1Mn g 34.76 Mn mol ?
K mol 0.6327K g 39.10
K mol 1K g 24.74 K mol ?
46
Empirical Formulas
Mn 10.6327
0.6327Mnfor K 1
0.6327
0.6327Kfor
rationumber wholesmallest obtain
O mol 2.531O g 16.00
O mol1O g 40.50 O mol ?
Mn mol 0.6327Mn g 54.94
Mn mol 1Mn g 34.76 Mn mol ?
K mol 0.6327K g 39.10
K mol 1K g 24.74 K mol ?
47
Empirical Formulas
4KMnO is formula chemical thethus
O 40.6327
2.531Ofor
Mn 10.6327
0.6327Mnfor K 1
0.6327
0.6327Kfor
rationumber wholesmallest obtain
O mol 2.531O g 16.00
O mol1O g 40.50 O mol ?
Mn mol 0.6327Mn g 54.94
Mn mol 1Mn g 34.76 Mn mol ?
K mol 0.6327K g 39.10
K mol 1K g 24.74 K mol ?
48
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?
You do it!You do it!
49
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?
ratio number wholesmallest find
O mol 0.1480O g 16.00
O mol1O g 2.368O mol ?
Co mol 0.1110Cog 58.93
Co mol 1Co g 6.541Co mol ?
50
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?
43OCo
:is formula scompound' theThus
O 43O 1.333 Co 33Co 1
number wholetofraction turn to 3 by both multipy
O1.3330.1110
0.1480O for Co 1
0.1110
0.1110Co for
51
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
H of g 8.100.1437g 56.1
C of g 48.00.8563g 56.1
H is 14.37% and C is 85.63%
g 56.1 contains mol 1
52
Determination of Molecular Formulas
84HC
:is formula theThus
H mol 8H g 1.01
H mol 1H of g 8.10
C mol 4C g 12.0
C mol 1C of g 48.0
moles tomassesconvert
54
Calculations Based on Chemical Equations
Example 3-1: How many CO molecules are required to react with 25 formula units of Fe2O3?
CO of molecules 75
unit formula OFe 1
molecules CO 3OFe units formula 25 = molecules CO ?
3232
Fe O + 3 CO 2 Fe + 3 CO2 3 2
55
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?
325 OFe units formula 102.50=atoms Fe ?
56
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?
OFe units formula 1
atoms Fe 2
OFe units formula 102.50=atoms Fe ?
32
325
57
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?
atoms Fe 105.00 OFe units formula 1
atoms Fe 2
OFe units formula 102.50=atoms Fe ?
5
32
325
58
Calculations Based on Chemical Equations
Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?
32
3232 OFe g 7.159
OFe mol 1OFe g 146 = CO g ?
59
Calculations Based on Chemical Equations
Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?
3232
3232 OFe mol 1
CO mol 3
OFe g 7.159
OFe mol 1OFe g 146 = CO g ?
60
Calculations Based on Chemical Equations
Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?
CO g 8.76CO mol 1
CO g 28.0
OFe mol 1
CO mol 3
OFe g 7.159
OFe mol 1OFe g 146 = CO g ?
3232
3232
61
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?
OFe mol 1
CO mol 3OFe mol 540.0CO g ?
32
2322
62
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?
2
2
32
2322 CO mol 1
CO g 0.44
OFe mol 1
CO mol 3OFe mol 540.0CO g ?
63
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?
? g CO mol Fe O3 mol CO
1 mol Fe O
g CO
mol CO
= 71.3 g CO
2 2 32
2 3
2
2
2
0 54044 0
1.
.
64
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?
You do it!You do it!
65
Calculations Based on Chemical Equations
3232
32
2
32
2
2232
O Feg 5.10O Femol 1
O Feg 7.159
CO mol 3
O Femol1
CO g 44.0
molCO 1CO g 8.65O Feg ?
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?
66
Calculations Based on Chemical Equations
Example 3-6: How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide?
You do it!You do it!
67
Calculations Based on Chemical Equations
CO lb 7.65CO g 454
CO lb 1
CO mol 1
CO g 28
OFe mol 1
CO mol 3
OFe g 7.159
OFe mol 1
OFe lb 1
OFe g 454OFe lb 125 = CO lb ?
3232
32
32
3232
YOU MUST BE PROFICIENT WITH THESE
TYPES OF PROBLEMS!!!
68
Limiting Reactant Concept
Kitchen example of limiting reactant concept.1 packet of muffin mix + 2 eggs + 1 cup of milk
12 muffins How many muffins can we make with the
following amounts of mix, eggs, and milk?
69
Limiting Reactant Concept
Mix Packets Eggs Milk1 1 dozen 1 gallon
limiting reactant is the muffin mix2 1 dozen 1 gallon3 1 dozen 1 gallon4 1 dozen 1 gallon5 1 dozen 1 gallon6 1 dozen 1 gallon7 1 dozen 1 gallon
limiting reactant is the dozen eggs
70
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?
numbersmallest by the determined
sets 55 is make can number we maximum the
sets 99nut 1set 1nuts 99
sets 55 washers2set 1 washers110
sets 87bolt 1set 1bolts 87
71
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?
2222 SO 2 CO O 3 CS
72
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?
mol 2 mol 1 mol 3 mol 1
SO 2 CO O 3 CS 2222
73
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?
CS O CO 2 SO
1 mol 3 mol 1 mol 2 mol
76.2 g 3(32.0 g) 44.0 g 2(64.1 g)
2 2 2 2 3
74
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
g 76.2
CS mol 1CS g 6.95 SO mol ?
SO 2 CO O 3 CS
222
2222
75
Limiting Reactant Concept
22
2
2
2
222
2222
SO g 161SO mol 1
SO g 1.64
CS mol 1
SO mol 2
g 76.2
CS mol 1CS g 6.95 SO mol ?
SO 2 CO O 3 CS
76
Limiting Reactant Concept
22
2
2
2
2
222
22
2
2
2222
2222
SO g 147SO mol 1
SO g 1.64
O mol 3
SO mol 2
O g 32.0
O mol 1O g 110SO mol ?
SO g 161SO mol 1
SO g 1.64
CS mol 1
SO mol 2
g 76.2
CS mol 1CS g 6.95 SO mol ?
SO 2 CO O 3 CS
Which is limiting reactant? Limiting reactant is O2. What is maximum mass of sulfur dioxide? Maximum mass is 147 g.
77
Percent Yields from Reactions Theoretical yield is calculated by assuming that
the reaction goes to completion. Determined from the limiting reactant calculation.
Actual yield is the amount of a specified pure product made in a given reaction. 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.
% yield = actual yield
theoretical yield100%
78
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?
yieldal theoretic theCalculate 1.
O HHCOOCCH OH HC +COOH CH 2523523
79
Percent Yields from Reactions
523
52
52352523
2523523
HCOOCCH g 1.19
OHHC g 0.46
HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?
yield al theoretic theCalculate 1.
OH HCOOCCH OHHC + COOHCH
80
Percent Yields from Reactions
yield.percent theCalculate .2
HCOOCCH g 1.19
OHHC g 0.46
HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?
yield al theoretic theCalculate 1.
OH HCOOCCH OHHC + COOHCH
523
52
52352523
2523523