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1
Chapter 6
Chemical Quantities
6.1
Atomic Mass and Formula Mass
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
2
Atomic Mass
Atomic mass is the
• Mass of a single atom in
atomic mass units (amu).
• Mass of an atom
compared to a 12C atom.
• Number below the
symbol of an element.
3
Periodic Table and Atomic Mass
Ag has
atomic
mass =
107.9 amu
C has atomic mass
= 12.01 amu
S has
atomic mass
= 32.07 amu
4
Atomic Mass Factors
The atomic mass
• Can be written as an equality.
Example: 1 P atom = 30.97 amu
• Can be written as two conversion factors.
Example: 1 P atom and 30.97 amu
30.97 amu 1 P atom
5
Uses of Atomic Mass Factors
The atomic mass factors are used to convert
• A specific number of atoms to mass (amu).
• An amount in amu to number of atoms.
6
Using Atomic Mass Factors
1. What is the mass in amu of 75 P atoms?
75 P atoms x 30.97 amu = 2323 amu (2.323 x103 amu)
1 P atom
2. How many Cu atoms have a mass of 4.500 x 105
amu?
4.500 x 105 amu x 1 Cu atom = 7081 Cu atoms
63.55 amu
7
Learning Check
What is the mass in amu of
75 silver atoms?
1) 107.9 amu
2) 8093 amu
3) 1.439 amu
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8
Solution
What is the mass in amu of 75 silver atoms?
2) 8093 amu
75 Ag atoms x 107.9 amu = 8093 amu
1 Ag atom
9
Learning Check
How many gold atoms have
a mass of 1.85 x 105 amu?
1) 939 Au atoms
2) 3.64 x 107 Au atoms
3) 106 Au atoms
Copyright © 2008 by Pearson Education, Inc.
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10
Solution
How many gold atoms have a mass of
1.85 x 105 amu?
1) 939 Au atoms
1.85 x 105 amu x 1 Au atom = 939 Au atoms
197.0 amu
11
Formula Mass
The formula mass is
• The mass in amu of a compound.
• The sum of the atomic masses of the elements
in a formula.
Copyright © 2008 by Pearson Education, Inc.
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12
Calculating Formula Mass
To calculate the formula mass of Na2SO4
• Multiply the atomic mass of each element by its subscript, then total the masses of the atoms.
2 Na x 22.99 amu = 45.98 amu
1 Na
1 S x 32.07 amu = 32.07 amu 142.05 amu
1 S
4 O x 16.00 amu = 64.00 amu
1 O
13
Learning Check
Using the periodic table, calculate the formula
mass of aluminum sulfide Al2S3.
14
Solution
Using the periodic table, calculate the formula
mass of aluminum sulfide Al2S3.
2 Al x 26.98 amu = 53.96 amu
1 Al
3 S x 32.07 amu = 96.21 amu
1 S 150.17 amu
15
Chapter 6
Chemical Quantities
6.2
The Mole
Copyright © 2008 by Pearson Education, Inc.
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16
Collection Terms
A collection term states a specific number of items.
• 1 dozen donuts = 12 donuts
• 1 ream of paper = 500 sheets
• 1 case = 24 cans
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17
A mole (mol) is a collection that contains
• The same number of particles as there are
carbon atoms in 12.01 g of carbon.
• 6.022 x 1023 atoms of an element (Avogadro’s
number).
1 mol C = 6.022 x 1023 C atoms
1 mol Na = 6.022 x 1023 Na atoms
1 mol Au = 6.022 x 1023 Au atoms
A Mole of Atoms
18
A mole
• Of a covalent compound has Avogadro’s number
of molecules.
1 mol CO2 = 6.022 x 1023 CO2 molecules
1 mol H2O = 6.022 x 1023 H2O molecules
• Of an ionic compound contains Avogadro’s
number of formula units.
1 mol NaCl = 6.022 x 1023 NaCl formula units
1 mol K2SO4 = 6.022 x 1023 K2SO4 formula units
A Mole of A Compound
19
Samples of One Mole
Quantities
Table 6.1
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20
Avogadro’s number 6.022 x 1023 can be written as
an equality and two conversion factors.
As an equality:
1 mol = 6.022 x 1023 particles
As conversion Factors:
6.022 x 1023 particles and 1 mol
1 mol 6.022 x 1023 particles
Avogadro’s Number
21
Using Avogadro’s Number
Avogadro’s number
• Converts moles of a substance to the number of particles.
How many Cu atoms are in 0.50 mol
Cu?
0.50 mol Cu x 6.022 x 1023 Cu atoms
1 mol Cu
= 3.0 x 1023 Cu atoms Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
22
Using Avogadro’s Number
Avogadro’s number
• Converts particles of a substance to moles.
How many moles of CO2 are
2.50 x 1024 CO2 molecules?
2.50 x 1024 CO2 x 1 mol CO2
6.022 x 1023 CO2
= 4.15 mol CO2
23
1. The number of atoms in 2.0 mol Al is
A. 2.0 Al atoms
B. 3.0 x 1023 Al atoms
C. 1.2 x 1024 Al atoms
2. The number of moles of S in 1.8 x 1024 atoms S is
A. 1.0 mol S atoms
B. 3.0 mol S atoms
C. 1.1 x 1048 mol S atoms
Learning Check
24
C. 1.2 x 1024 Al atoms
2.0 mol Al x 6.022 x 1023 Al atoms
1 mol Al
B. 3.0 mol S atoms
1.8 x 1024 S atoms x 1 mol S
6.022 x 1023 S atoms
Solution
25
Subscripts and Moles
The subscripts in a formula state
• The relationship of atoms in the formula.
• The moles of each element in 1 mol of
compound.
Glucose
C6H12O6
In 1 molecule: 6 atoms C 12 atoms H 6 atoms O
In 1 mol: 6 mol C 12 mol H 6 mol O
26
Subscripts State Atoms and
Moles
1 mol C9H8O4 = 9 mol C 8 mol H 4 mol O
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27
Factors from Subscripts
Subscripts used for conversion factors
• Relate moles of each element in 1 mol compound.
• For aspirin C9H8O4 can be written as:
9 mol C 8 mol H 4 mol O
1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4 and
1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4
9 mol C 8 mol H 4 mol O
28
Learning Check
A. How many moles O are in 0.150 mol aspirin
C9H8O4?
B. How many O atoms are in 0.150 mol aspirin
C9H8O4?
29
Solution
A. How many mol O are in 0.150 mol aspirin C9H8O4?
0.150 mol C9H8O4 x 4 mol O = 0.600 mol O
1 mol C9H8O4
subscript factor
B. How many O atoms are in 0.150 mol aspirin C9H8O4?
0.150 mol C9H8O4 x 4 mol O x 6.022 x 1023 O atoms
1 mol C9H8O4 1 mol O subscript Avogadro’s
factor Number = 3.61 x 1023 O atoms
30
Learning Check
How many O atoms are in 0.150 mol aspirin C9H8O4?
31
Solution
How many O atoms are in 0.150 mol aspirin C9H8O4?
0.150 mol C9H8O4 x 4 mol O x 6.022 x 1023 O atoms
1 mol C9H8O4 1 mol O
subscript Avogadro’s
factor number
= 3.61 x 1023 O atoms
32
Chapter 6
Chemical Quantities
6.3
Molar Mass
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33
Molar Mass
The molar mass
• Is the mass of one
mol of an element or
compound.
• Is the atomic mass
expressed in grams.
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34
Molar Mass from the Periodic
Table
Molar mass
• Is the
atomic
mass
expressed
in grams.
Copyright © 2008 by Pearson Education, Inc.
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35
Give the molar mass for:
A. 1 mol K atoms = ________
B. 1 mol Sn atoms = ________
Learning Check
36
Give the molar mass for:
A. 1 mol K atoms = 39.10 g
B. 1 mol Sn atoms = 118.7 g
Solution
37
Molar Mass of a Compound
The molar mass of a compound is the sum of the
molar masses of the elements in the formula.
Example: Calculate the molar mass of CaCl2.
Element Number
of Moles
Atomic Mass Total Mass
Ca 1 40.08 g/mol 40.08 g
Cl 2 35.45 g/mol 70.90 g
CaCl2 110.98 g
38
Molar Mass of K3PO4
Calculate the molar mass of K3PO4.
Element Number
of Moles
Atomic Mass Total Mass
in K3PO4
K 3 39.10 g/mol 117.3 g
P 1 30.97 g/mol 30.97 g
O 4 16.00 g/mol 64.00 g
K3PO4 212.3 g
39
Some One-Mol Quantities
32.07 g 55.85 g 58.44 g 294.20 g 342.30 g
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40
What is the molar mass of each of the following?
A. K2O
B. Al(OH)3
Learning Check
41
A. K2O 94.20 g/mol
2 mol K (39.10 g/mol) + 1 mol O (16.00 g/mol)
78.20 g + 16.00 g
= 94.20 g
B. Al(OH)3 78.00 g/mol
1 mol Al (26.98 g/mol) + 3 mol O (16.00 g/mol)
+ 3 mol H (1.008 g/mol)
26.98 g + 48.00 g + 3.024 g
= 78.00 g
Solution
42
Prozac, C17H18F3NO, is an antidepressant that
inhibits the uptake of serotonin by the brain. What
is the molar mass of Prozac?
1) 40.06 g/mol
2) 262.0 g/mol
3) 309.4 g/mol
Learning Check
43
Prozac, C17H18F3NO, is an antidepressant that
inhibits the uptake of serotonin by the brain. What is
the molar mass of Prozac?
3) 309.3 g/mol
17C (12.01) + 18H (1.008) + 3F (19.00)
+ 1N (14.01) + 1 O (16.00) =
204.2 +18.14 + 57.00 + 14.01 + 16.00
= 309.4 g/mol
Solution
44
Chapter 6
Chemical Quantities
6.4
Calculations Using
Molar Mass
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45
Molar mass conversion factors
• Are written from molar mass.
• Relate grams and moles of an element or
compound.
Example: Write molar mass factors for methane
CH4 used in gas cook tops and gas heaters.
Molar mass:
1 mol CH4 = 16.04 g
Conversion factors:
16.04 g CH4 and 1 mol CH4
1 mol CH4 16.04 g CH4
Molar Mass Factors
46
Acetic acid C2H4O2 gives the sour taste to
vinegar. Write two molar mass conversion
factors for acetic acid.
Learning Check
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47
Acetic acid C2H4O2 gives the sour taste to vinegar.
Write two molar mass factors for acetic acid.
Calculate molar mass:
24.02 + 4.032 = 32.00 = 60.05 g/mol
1 mol of acetic acid = 60.05 g acetic acid
Molar mass factors
1 mol acetic acid and 60.05 g acetic acid
60.05 g acetic acid 1 mol acetic acid
Solution
48
Molar mass factors are used to convert between the
grams of a substance and the number of moles.
Calculations Using Molar Mass
Grams
Molar mass factor Moles
49
Aluminum is used to build lightweight bicycle
frames. How many grams of Al are 3.00 mol Al?
Molar mass equality: 1 mol Al = 26.98 g Al
Setup with molar mass as a factor:
3.00 mol Al x 26.98 g Al = 80.9 g Al 1 mol Al molar mass factor
for Al
Moles to Grams
50
Learning Check
Allyl sulfide C6H10S is a
compound that has the
odor of garlic. How many
moles of C6H10S are in
225 g C6H10S?
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51
Calculate the molar mass of C6H10S.
(6 x 12.01) + (10 x 1.008) + (1 x 32.07)
= 114.21 g/mol
Set up the calculation using a mole factor.
225 g C6H10S x 1 mol C6H10S
114.21 g C6H10S
molar mass factor(inverted)
= 1.97 mol C6H10S
Solution
52
Grams, Moles, and Particles
A molar mass factor and Avogadro’s number
convert
• Grams to particles
molar mass Avogadro’s
number
(g mol particles)
• Particles to grams
Avogadro’s molar mass
number
(particles mol g)
53
Learning Check
How many H2O molecules are in 24.0 g H2O?
1) 4.52 x 1023
2) 1.44 x 1025
3) 8.02 x 1023
54
Solution
How many H2O molecules are in 24.0 g H2O?
3) 8.02 x 1023
24.0 g H2O x 1 mol H2O x 6.022 x 1023 H2O molecules
18.02 g H2O 1 mol H2O
= 8.02 x 1023 H2O molecules
55
Learning Check
If the odor of C6H10S can be detected from 2 x
10-13 g in one liter of air, how many molecules
of C6H10S are present?
56
Solution
If the odor of C6H10S can be detected from
2 x 10-13 g in one liter of air, how many molecules
of C6H10S are present?
2 x 10-13 g x 1 mol x 6.022 x 1023 molecules
114.21 g 1 mol
= 1 x 109 molecules C6H10S
57
Chapter 6 Chemical Quantities
6.5
Percent
Composition
and
Empirical
Formulas
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58
Percent Composition
Percent composition
• Is the percent by mass of each element in a
formula.
Example: Calculate the percent composition of CO2.
CO2 = 1 C(12.01g) + 2 O(16.00 g) = 44.01 g/mol)
12.01 g C x 100 = 27.29 % C
44.01 g CO2
32.00 g O x 100 = 72.71 % O
44.01 g CO2 100.00 %
59
What is the percent composition of lactic acid,
C3H6O3, a compound that appears in the blood
after vigorous activity?
Learning Check
60
STEP 1
3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol
36.03 g C + 6.048 g H + 48.00 g O
STEP 2
%C = 36.03 g C x 100 = 40.00% C
90.08 g
%H = 6.048 g H x 100 = 6.714% H
90.08 g
%O = 48.00 g O x 100 = 53.29% O
90.08 g
Solution
61
Learning Check
The chemical isoamyl
acetate C7H14O2 gives
the odor of pears. What
is the percent carbon in
isoamyl acetate?
1) 7.102 %C
2) 35.51 %C
3) 64.58 %C
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62
3) 64.58 %C
Molar mass C7H14O2 = 7C(12.01) + 14H(1.008)
+ 2O(16.00) = 130.18 g/mol
Total C = 7C(12.01) = g
% C = total g C x 100 total g % C = 84.07 g C x 100 = 64.58 % C 130.18 g
Solution
63
The empirical formula
• Is the simplest whole number ratio of the atoms.
• Is calculated by dividing the subscripts in the
actual (molecular) formula by a whole number
to give the lowest ratio.
C5H10O5 5 = C1H2O1 = CH2O
actual (molecular) empirical formula
formula
Empirical Formulas
64
Some Molecular and Empirical
Formulas • The molecular formula is the same or a multiple of
the empirical.
Table 6.3
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65
A. What is the empirical formula for C4H8?
1) C2H4 2) CH2 3) CH
B. What is the empirical formula for C8H14?
1) C4H7 2) C6H12 3) C8H14
C. Which is a possible molecular formula for CH2O?
1) C4H4O4 2) C2H4O2 3) C3H6O3
Learning Check
66
A. What is the empirical formula for C4H8?
2) CH2 C4H8 4
B. What is the empirical formula for C8H14?
1) C4H7 C8H14 2
C. Which is a possible molecular formula for CH2O?
2) C2H4O2 3) C3H6O3
Solution
67
A compound has an empirical formula SN. If
there are 4 atoms of N in one molecule, what is
the molecular formula? Explain.
1) SN
2) SN4
3) S4N4
Learning Check
68
A compound has an empirical formula SN. If
there are 4 atoms of N in one molecule, what
is the molecular formula? Explain.
3) S4N4
In this molecular formula 4 atoms of N and 4
atoms of S and N are related 1:1. Thus, it has
an empirical formula of SN.
Solution
69
A compound contains 7.31 g Ni and 20.0 g Br.
Calculate its empirical (simplest) formula.
Learning Check
70
Convert 7.31 g Ni and 20.0 g Br to moles.
7.31 g Ni x 1 mol Ni = 0.125 mol Ni
58.69 g Ni
20.0 g Br x 1 mol Br = 0.250 mol Br
79.90 g Br
Divide by smallest:
0.125 mol Ni = 1 Ni 0.250 mol Br = 2 Br
0.125 0.125
Write ratio as subscripts: NiBr2
Solution
71
Converting Decimals to Whole
Numbers
When the number of moles for an element is a
decimal, all the moles are multiplied by a small
integer to obtain whole number.
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
Table 6.4
72
Aspirin is 60.0% C, 4.5 % H and 35.5 % O.
Calculate its empirical (simplest) formula.
Learning Check
73
STEP 1. Calculate the moles of each element in
100 g.
100 g aspirin contains 60.0% C or 60.0 g C,
4.5% H or 4.5 g H, and 35.5% O or 35.5 g O.
60.0 g C x 1 mol C = 5.00 mol C
12.01 g C
4.5 g H x 1 mol H = 4.5 mol H
1.008 g H
35.5 g O x 1mol O = 2.22 mol O
16.00 g O
Solution
74
Solution (continued)
STEP 2. Divide by the smallest number of mol.
5.00 mol C = 2.25 mol C (decimal)
2.22
4.5 mol H = 2.0 mol H
2.22
2.22 mol O = 1.00 mole O
2.22
75
Solution (continued)
3. Use the lowest whole number ratio as subscripts
When the moles are not whole numbers, multiply
by a factor to give whole numbers, in this case x 4.
C: 2.25 mol C x 4 = 9 mol C
H: 2.0 mol H x 4 = 8 mol H
O: 1.00 mol O x 4 = 4 mol O
Using these whole numbers as subscripts the
simplest formula is
C9H8O4
76
Chapter 6 Chemical
Quantities
6.6
Molecular Formulas
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77
A molecular formula
• Is a multiple (or equal) of its empirical formula.
• Has a molar mass that is the empirical formula
mass multiplied by a whole number.
molar mass = a whole number
empirical mass
• Is obtained by multiplying the empirical formula by
a whole number.
Relating Molecular and
Empirical Formulas
78
Diagram of Molecular and
Empirical Formulas
A small integer links
• A molecular formula and its empirical
formula.
• A molar mass and its empirical formula
mass.
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79
Determine the molecular formula of compound that
has a molar mass of 78.11 g and an empirical
formula of CH.
STEP 1. Empirical formula mass of CH = 13.02 g
STEP 2. Divide the molar mass by the empirical
mass.
78.11 g = 5.999 ~ 6
13.02 g
STEP 3. Multiply each subscript in C1H1 by 6.
molecular formula = C1x 6 H1 x 6 = C6H6
Finding the Molecular Formula
80
Some Compounds with
Empirical Formula CH2O
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Table 6.5
81
A compound has a molar mass of 176.1g and an
empirical formula of C3H4O3. What is the
molecular formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9
Learning Check
82
A compound has a formula mass of 176.1 and an
empirical formula of C3H4O3. What is the
molecular formula?
2) C6H8O6
C3H4O3 = 88.06 g/EF
176.1 g (molar mass) = 2.00
88.06 g (empirical mass)
Molecular formula = 2 x empirical formula
C3 x 2H4 x 2O3 x 2 = C6H8O6
Solution
83
A compound contains C 24.27%, H 4.07%, and Cl
71.65%. The molar mass is about 99 g. What are
the empirical and molecular formulas?
STEP 1. Calculate the empirical formula.
Write the mass percents as the grams in a
100.00-g sample of the compound.
C 24.27 g H 4.07 g Cl 71.65 g
Molecular Formula
84
Finding the Molecular Formula
(Continued)
Calculate the number of moles of each element.
24.27 g C x 1 mol C = 2.021 mol C
12.01 g C
4.07 g H x 1 mol H = 4.04 mol H
1.008 g H
71.65 g Cl x 1 mol Cl = 2.021 mol Cl
35.45 g Cl
85
Finding the Molecular Formula
(Continued) Divide by the smallest number of moles:
2.021 mol C = 1 mol C
2.021
4.04 mol H = 2 mol H
2.021
2.02 mol Cl = 1 mol Cl
2.021
Empirical formula = C1H2Cl1 = CH2Cl
Calculate empirical mass (EM) CH2Cl = 49.48 g
86
Finding the Molecular Formula
(Continued)
STEP 2. Divide molar mass by empirical mass.
Molar mass = 99 g = 2
Empirical mass 49.48 g
STEP 3. Multiply the empirical formula by the small
integer to determine the molecular formula.
2 x (CH2Cl)
C1 x2 H2 x 2 Cl1x 2 = C2H4Cl2
87
A compound is 27.4% S, 12.0% N and 60.6
% Cl. If the compound has a molar mass of
351 g, what is the molecular formula?
Learning Check
88
In 100. g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl.
27.4 g S x 1 mol S = 0.854 mol S
32.07 g S
12.0 g N x 1 mol N = 0.857 mol N
14.01 g N
60.6 g Cl x 1mol Cl = 1.71 mol Cl
35.45 g Cl
Solution
89
Divide by the smallest number of moles
0.854 mol S /0.854 = 1.00 mol S
0.857 mol N/0.854 = 1.00 mol N
1.71 mol Cl/0.854 = 2.00 mol Cl
empirical formula = SNCl2 = 116.98 g
Molar Mass/ Empirical mass
351 g = 3
116.98 g
molecular formula = (SNCl2)3 = S3N3Cl6
Solution (continued)
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