Transcript Chapter 7 - faculty at Chemeketa

1
The mass of a single atom is too small to
measure on a balance.
mass of hydrogen atom = 1.673 x 10-24 g
2
This is an
infinitesimal
1.673 x 10-24 g
mass
3
The standard to which the masses of all
other atoms are compared to was
originally chosen to be the most
abundant isotope of hydrogen.
1
1
H
4
A mass of exactly 1 atomic mass units
(amu) was assigned to
1
1
H
5
Electric charge and mass of electrons, protons, & neutrons
Particle
symbol Actual charge
(coulombs)
Relative
charge
Actual
mass (g)
Relative
mass
(amu)
Electron
e–
–1.602  10–19
–1
9.11  10–28
0
Proton
p
+1.602  10–19
+1
1.67  10–24
1
Neutron
n
0
1.68  10–24
1
0
6
• Most elements occur as mixtures of
isotopes.
• Isotopes of the same element have
different numbers of neutrons.
• The listed atomic mass of an element is
the average relative mass of the isotopes
of that element compared originally to the
mass of hydrogen: 1amu (now carbon).
7
• Balances in the lab room weigh in
metric ,i.e., grams not in amu’s. So
how do we change the number on the
periodic table from amu’s to grams?
• We need to know how many atoms of
each element we need to obtain the
amu weight in grams.
• This amount (a big pile) would convert
amu’s to grams for each element.
8
1 pile = 1 mole (in Greek)
1 mole = 6.02 x
23
10 objects
9
6.02 x
23
10
is a very
LARGE
number
10
6.02 x
23
10
is
Avogadro’s Number
number
11
Amadeo (Amedeo) Avogadro
12
1 mole of any element contains
6.02 x 1023
particles of that substance.
13
The atomic weight in grams
of any element23
contains 1 mole of atoms.
14
How many seconds in 1 minute?
How many minutes in 1 hour?
How many hours in 1 day?
So how many seconds in 1 day?
60 × 60 × 24 = 86400
15
Suppose you count at a rate of 1 number per second?
How long to count to 6.02 × 1023?
6.02  1023
 6.97  1018 days
86400
How many years is this?
6.97  1018
 1.91  1016 years
365
16
How many years?
19,100,000,000,000,000
How old is the universe?
14 billion years
1.4 × 1010
14,000,000,000
ǃ
17
Examples
18
Species
Quantity
Number of H
atoms
H
1 mole
6.02 x
23
10
19
Species
Quantity
Number of
H2 molecules
H2
1 mole
6.02 x
23
10
20
Species
Quantity
Number of
Na atoms
Na
1 mole
6.02 x
23
10
21
Species
Quantity
Number of
Fe atoms
Fe
1 mole
6.02 x
23
10
22
Species C6H6
Quantity 1 mole
Number of
23
6.02 x 10
C6H6 molecules
23
1 mol of atoms = 6.02 x 1023 atoms
1 mol of molecules = 6.02 x 1023 molecules
1 mol of ions = 6.02 x 1023 ions
24
• The mole weight of an element is its
atomic weight in grams.
• It contains 6.02 x 1023 atoms
(Avogadro’s number) of the element.
25
H
Atomic
Number of
Mole weight
mass
atoms
1.008 amu
1.008 g
6.02 x 1023
Mg
24.31 amu
24.31 g
6.02 x 1023
Na
22.99 amu
22.99 g
6.02 x 1023
Element
26
Problems
27
Convert To
Moles!
28
Weight (mass)
Numbers
Moles

Periodic Table
Avogadro’s Number
6.02 × 1023
29
How many moles of iron does 25.0 g of iron
represent?
Atomic weight iron = 55.85
Conversion sequence: grams Fe → moles Fe
Set up the calculation using a conversion factor
between moles and grams.
 1 mole Fe 
(grams Fe) 

55.85
g
Fe


 1 mole Fe 
(25.0 g Fe) 
0.448 mole Fe

 55.85 g Fe 
30
How many iron atoms are contained in 40.0 grams of
iron?
Atomic weight iron = 55.85
Conversion sequence is:
grams Fe → moles Fe → atoms Fe
 1 mole Fe   6.02 x 1023 atoms Fe 
(40.0 g Fe) 




1 mole Fe
 55.85 g Fe  

4.31 x 1023 atoms Fe
31
What is the mass of 3.01 x 1023 atoms of sodium (Na)?
Mole weight Na = 22.99 g
Conversion sequence:
atoms Na → moles Na → grams Na
1 mole Na

  22.99 g Na  
(3.01 x 10 atoms Na)


23
 6.02 x 10 atoms Na   1 mole Na 
23
11.5 g Na
32
How many oxygen atoms are present in 2.00 moles of
oxygen molecules?
Conversion sequence:
moles O2 → molecules O2 → atoms O
 6.02 x 1023 molecules O2 
(2.00 mole O2 ) 

1 mole O2


 2 atoms O 
 1 molecule O 

2 
= 2.41 x1024 atoms O
33
Mole Weight of
Compounds
34
The mole weight of a compound can be
determined by adding the mole weights
of all of the atoms in its formula.
35
Calculate the mole weight of C2H6O.
2 C = 2(12.01 g) = 24.02 g
6 H = 6(1.01 g) = 6.06 g
1 O = 1(16.00 g) = 16.00 g
46.08 g
36
Calculate the mole weight of lithium perchlorate.
LiClO4
1 Li = 1(6.94 g) = 6.94 g
1 Cl = 1(35.45 g) = 35.45 g
4 O = 4(16.00 g) = 64.00 g
106.39 g
37
Calculate the mole weight of ammonium
phosphate.
(NH4)3PO4
3 N = 3(14.01 g) = 42.03 g
12 H = 12(1.01 g) = 12.12 g
1 P = 1(30.97 g) = 30.97 g
4 O = 4(16.00 g) = 64.00 g
149.12 g
38
Solids that contain water as part of
their crystalline structure are known as
hydrates.
Water in a hydrate is known as water of
hydration or water of crystallization.
39
Formulas of hydrates are written by first writing
the formula for the anhydrous compound and
then adding a dot followed by the number of
water molecules present.
CoCl2 6H2O
40
Calculate the mole weight of NaC2H3O2 · 3 H2O
Sodium acetate trihydrate
1 Na = 1(22.99 g) = 22.99 g
2 C = 2(12.01 g) = 24.02 g
9 H = 9(1.01 g) = 9.09 g
5 O = 5(16.00 g) = 80.00 g
136.10 g
41
Avogadro’s
Number of
Particles
6.02 x 1023
Particles
1 MOLE
Mole Weight
42
Avogadro’s
Number of
Ca atoms
6.02 x 1023
Ca atoms
1 MOLE Ca
40.078 g Ca
43
Avogadro’s
Number of
H2O molecules
6.02 x 1023
H2O
molecules
1 MOLE H2O
18.02 g H2O
44
In dealing with diatomic elements (H2, O2,
N2, F2, Cl2, Br2, and I2), distinguish between
one mole of atoms and one mole of
molecules.
45
Calculate the mole weight of 1 mole of H atoms.
1 H = 1(1.01 g) = 1.01 g
Calculate the mole weight of 1 mole of H2 molecules.
2 H = 2(1.01 g) = 2.02 g
46
Problems
47
How many moles of benzene, C6H6, are present in
390.0 grams of benzene?
The mole weight of C6H6 is 78.12 g.
Conversion sequence: grams C6H6 → moles C6H6
78.12 grams C6 H 6
Use the conversion factor:
1 mole C6 H 6
 1 mole C6 H 6 
= 4.992 moles C6 H 6
(390.0 g C6 H 6 ) 

 78.12 g C6 H 6 
48
How many grams of (NH4)3PO4 are contained in 2.52
moles of (NH4)3PO4?
The mole weight of (NH4)3PO4 is 149.12 g.
Conversion sequence:
moles (NH4)3PO4 → grams (NH4)3PO4
149.12 grams (NH 4 )3PO4
Use the conversion factor:
1 mole (NH 4 )3PO4
 149.12 g (NH 4 )3PO 4 
(2.52 mole (NH 4 )3PO 4 ) 

 1 mole (NH 4 )3PO4 
= 376g (NH 4 )3 PO 4
49
56.04 g of N2 contains how many N2 molecules?
The mole weight of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
Use the conversion factors
1 mole N 2
28.02 g N 2
6.02 x 1023 molecules N 2
1 mole N 2
 1 mole N 2   6.02 x 10 molecules N 2 
(56.04 g N 2 ) 



1 mole N 2

 28.02 g N 2  
23
= 1.20 x 1024 molecules50N 2
56.04 g of N2 contains how many N atoms?
The mole weight of N2 is 28.02 g.
Conversion sequence: g N2 → moles N2 → molecules N2
→ atoms N
Use the conversion factors
1 mole N 2 6.02 x 1023 molecules N 2
2 atoms N
1 molecule N 2
28.02 g N 2
1 mole N 2
 1 mole N 2   6.02 x 1023 molecules N 2 
(56.04 g N 2 ) 


1
mole
N
28.02
g
N


2
2  
 2 atoms N 
 1 molecule N 

2 
= 2.41 x 1024 atoms N
51
Percent Composition
of Compounds
52
Percent composition of a compound is the
mass percent of each element in the compound.
H2O
11.19% H by mass
88.79% O by mass
53
Percent Composition
From Formula
54
If the formula of a compound is known,
a two-step process is needed to calculate
the percent composition.
Step 1 Calculate the mole weight of the
formula.
Step 2 Divide the total mass of each
element in the formula by the
mole weight and multiply by
100.
55
total mass of the element
x 100 = percent of the element
mole weight
56
Calculate the percent composition of hydrosulfuric
acid H2S(aq) .
Step 1 Calculate the mole weight of H2S.
2 H = 2 (1.01 g) = 2.02 g
1 S = 1 (32.07 g) = 32.07 g
34.09 g
57
Calculate the percent composition of hydrosulfuric
acid H2S.
Step 2 Divide the mass of each element
by the mole weight and multiply
by 100.
 2.02 g H 
H: 
 (100) = 5.93%
 34.09 g 
 32.07 g S
S: 
 34.09 g

 (100)  94.07%

S
H
94.07% 5.93%
58
Percent Composition
From Experimental Data
59
Percent composition can be calculated
from experimental data without knowing
the composition of the compound.
Step 1 Calculate the mass of the
compound formed.
Step 2 Divide the mass of each element
by the total mass of the
compound and multiply by 100.
60
A compound containing nitrogen and oxygen is found
to contain 1.52 g of nitrogen and 3.47 g of oxygen.
Determine its percent composition.
Step 1 Calculate the total mass of the compound
1.52 g N
3.47 g O
4.99 g = total mass of product
61
A compound containing nitrogen and oxygen is found
to contain 1.52 g of nitrogen and 3.47 g of oxygen.
Determine its percent composition.
Step 2 Divide the mass of each element
by the total mass of the
compound formed.
 1.52 g N 

 (100) = 30.5%
 4.99 g 
 3.47 g O

 4.99 g

 (100) = 69.5%

O
69.5%
N
30.5%
62
Empirical Formula versus
Molecular Formula
63
• The empirical formula or simplest
formula gives the smallest wholenumber ratio of the atoms present in a
compound.
• The empirical formula gives the
relative number of atoms of each
element present in the compound.
64
• The molecular formula is the true
formula of a compound.
• The molecular formula represents the
total number of atoms of each element
present in one molecule of a compound.
65
Examples
66
Molecular Formula
C2H4
Empirical Formula
CH2
Smallest Whole
Number Ratio
C:H 1:2
67
Molecular Formula
C6H6
Empirical Formula
CH
Smallest Whole
Number Ratio
C:H 1:1
68
Molecular Formula
H2O2
Empirical Formula
HO
Smallest Whole
Number Ratio
H:O 1:1
69
70
Two compounds can have identical
empirical formulas and different molecular
formulas.
71
72
Calculating
Empirical Formulas
73
The analysis of a compound shows that it
contains 27% carbon and 73% oxygen.
Calculate the empirical formula for this
substance.
C___O___
74
Step 1 Assume a definite starting
quantity (usually 100.0 g) of the
compound, if not given, and
express the mass of each
element in grams.
Step 2 Convert the grams of each
element into moles of each
element using each element’s
mole weight.
75
Step 3 Divide the moles of atoms of
each element by the moles of
atoms of the element that had
the smallest value
– If the numbers obtained are whole
numbers, use them as subscripts
and write the empirical formula.
– If the numbers obtained are not
whole numbers, go on to step 4.
76
The analysis of a compound shows that it
contains 27% carbon and 73% oxygen.
Calculate the empirical formula for this
substance.
C___O___
27g
73g
12.0 g/mole 16.0 g/mole
C___O___
2.3
4.6
moles
2.3moles
moles
2.3moles
77
The analysis of a compound shows that it
contains 27% carbon and 73% oxygen.
Calculate the empirical formula for this
substance.
C___O___
1
2
78
Step 4 Multiply the values obtained in
step 3 by the smallest numbers
that will convert them to whole
numbers
Use these whole numbers as the
subscripts in the empirical
formula.
FeO1.5
Fe1 x 2O1.5 x 2
Fe2O3
79
• The results of calculations may differ
from a whole number.
– If they differ ±0.1 round off to the next
nearest whole number.
2.9  3
– Deviations greater than 0.1 unit from a
whole number usually mean that the
calculated ratios have to be multiplied by
a whole number.
80
Problems
81
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73% oxygen
(O). Calculate the empirical formula for this
substance.
Step 1 Express each element in grams. Assume 100
grams of compound.
K = 56.58 g
C = 8.68 g
O = 34.73 g
82
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73% oxygen
(O). Calculate the empirical formula for this
substance.
Step 2 Convert the grams of each element to moles.
 1 mole K atoms 
K:  56.58 g K 
1.447 mole K atoms

 39.10 g K 
 1 mole C atoms  0.723 mole C atoms
mole C atoms
C: 8.68 g C 
 0.723

C has the smallest number
12.01
g
C


of moles
 1 mole O atoms 
O:  34.73 g O 
 2.171 mole O atoms

83
 16.00 g O 
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73% oxygen
(O). Calculate the empirical formula for this
substance.
Step 3 Divide each number of moles by the smallest
value.
1.447 mole
0.723 mole
K=
= 2.00 C:
= 1.00
0.723 mole
0.723 mole
0.723 mole C atoms
2.171 mole
O=
= 3.00
C has the smallest number
0.723 mole
of moles
The simplest ratio of K:C:O is 2:1:3
Empirical formula K2CO3
84
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 1 Express each element in grams. Assume 100
grams of compound.
N = 25.94 g
O = 74.06 g
85
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 2 Convert the grams of each element to moles.
 1 mole N atoms 
N:  25.94 g N 
 1.852 mole N atoms

 14.01 g N 
 1 mole O atoms 
O:  74.06 g O 
 4.629 mole O atoms

 16.00 g O 
86
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 3 Divide each number of moles by the smallest
value.
1.852 mole
4.629 mole
N=
= 1.000 O:
= 2.500
1.852 mole
1.852 mole
This is not a ratio of whole numbers.
87
The percent composition of a compound is 25.94%
nitrogen (N), and 74.06% oxygen (O). Calculate the
empirical formula for this substance.
Step 4 Multiply each of the values by 2.
N: (1.000)2 = 2.000
O: (2.500)2 = 5.000
Empirical formula N2O5
88
Calculating the Molecular Formula
from the Empirical Formula
89
• The molecular formula can be calculated
from the empirical formula if the mole
weight is known.
• The molecular formula will be equal to the
empirical formula or some multiple n of it.
• To determine the molecular formula evaluate
n.
• n is the number of units of the empirical
formula contained in the molecular formula.
mole weight
n=
=
empirical weight
number of empirical
formula units
90
What is the molecular formula of a compound which
has an empirical formula of CH2 and a mole weight of
126.2 g?
Let n = the number of formula units of CH2.
Calculate the mass of each CH2 unit
1 C = 1(12.01 g) = 12.01g
2 H = 2(1.01 g) = 2.02g
14.03g
126.2 g
n
 9 (empirical formula units)
14.03 g
The molecular formula is (CH2)9 = C9H18
91
A compound made of 30.4% N and 69.6% O has a
mole weight of 138 grams/mole. Find the molecular
formula.
Old Way:
1mole
30.4 g 
 2.17 mole of N
14.0 g
1mole
69.6g 
 4.35mole of O
16.0 g
138
 3units
46
N2.17 O4.35  NO2  46g/mole
2.17
2.17
3(NO2 )  N3O6
92
A compound made of 30.4% N and 69.6% O has a
mole weight of 138 grams/mole. Find the molecular
formula.
New Way:
138gcomp'd
30.4 gN
1mole N


 3mole N
1mole comp'd
100 gcomp'd
14.0 g
138gcomp'd
69.6gO
1mole O


 6mole O
1mole comp'd
100 gcomp'd
16.0 g
N3O6
93
94