Structure of the Atom

1
Structure of the Atom Size 0.1 to 0.5
nanometers (1 nm 1 x 10-9 m) Nucleus
Electron clouds
5 x 10-15 m
Nucleus- consists of protons and neutrons
2 x 10-10 m
2
Iron atoms on a copper crystal
3
Iron atoms on a copper crystal
Carbon monoxide molecules on platinum crystal
4
Iron atoms on a copper crystal
Carbon monoxide molecules on platinum crystal
Xenon atoms on nickel
5
Atomic numbernumber of protons in an atom.
Symbol Z Atomic mass mass of an atom in a.m.u.
(number of protons neutrons in an atom).
Symbol A
6
Atomic numbernumber of protons in an atom.
Symbol Z Atomic mass mass of an atom in a.m.u.
(number of protons neutrons in an atom).
Symbol A Atom e- p n Z A (a.m.u.)
H 1 1 0 1 1.008 Li 3 3 4 3 7.016
C 6 6 6 6 12.000 U 92 92 146 92 238.05
7
Conventional Notation for Chemical Symbols
Mass number
Number and sign of charge
35
-
Cl
17
2
8
Conventional Notation for Chemical Symbols
Mass number
Number and sign of charge
35
-
Cl
17
2
Atomic number Number of atoms in entity
9
Conventional Notation for Chemical Symbols
Mass number
Number and sign of charge
35
-
Cl
17
2
Atomic number Number of atoms in entity
4
2-
He
OH-
CO
Examples
2
3
10
Ions atoms or molecules which have gained or
lost one or more electrons Isotopes atoms of
the same element which have different
masses, i.e. different numbers of neutrons
11
Ions atoms or molecules which have gained or
lost one or more electrons Isotopes atoms of
the same element which have different
masses, i.e. different numbers of neutrons
Examples of ions
H H- NO3-
12
Ions atoms or molecules which have gained or
lost one or more electrons Isotopes atoms of
the same element which have different
masses, i.e. different numbers of neutrons
Examples of ions
H H- NO3-
1
Examples of isotopes
H
Hydrogen
1
2
H
Deuterium
1
3
H
Tritium
1
13
Elements
Compounds
Identical
atoms
Atoms
Molecules
14
Elements
Compounds
Identical
atoms
Atoms
Molecules
e-
n
e-
n
Ions
Isotopes
15
(Chemical) Atomic Mass The average of the
isotope masses of an element, weighted to reflect
their relative natural abundances.
16
(Chemical) Atomic Mass The average of the
isotope masses of an element, weighted to reflect
their relative natural abundances.
Example Chlorine has two naturally occurring
isotopes, 35Cl (A 34.97 a.m.u.) and 37Cl (A
36.96 a.m.u.). The respective natural abundances
of these isotopes are 75.5 and 24.5.
(34.97 x 75.5) (36.96 x 24.5)
100
Atomic mass of Cl

35.45 a.m.u.
17
(Chemical) Atomic Mass The average of the
isotope masses of an element, weighted to
reflect their relative natural abundances.
Example Chlorine has two naturally occurring
isotopes, 35Cl (A 34.97 a.m.u.) and 37Cl (A
36.96 a.m.u.). The respective natural abundances
of these isotopes are 75.5 and 24.5.
(34.97 x 75.5) (36.96 x 24.5)
100
Atomic mass of Cl

35.45 a.m.u.
Molecular Mass The sum of the atomic masses of
all of the constituent atoms in a molecule.
18
(Chemical) Atomic Mass The average of the
isotope masses of an element, weighted to reflect
their relative natural abundances.
Example Chlorine has two naturally occurring
isotopes, 35Cl (A 34.97 a.m.u.) and 37Cl (A
36.96 a.m.u.). The respective natural abundances
of these isotopes are 75.5 and 24.5.
(34.97 x 75.5) (36.96 x 24.5)
100
Atomic mass of Cl

35.45 a.m.u.
Molecular Mass The sum of the atomic masses of
all of the constituent atoms in a molecule.
Example Molecular mass of H2? (A.M. of H
1.008 a.m.u.)
M.M. of H2 2 x 1.008 2.016 a.m.u.
19
Example 2 M.M. of carbon monoxide, CO? (A.M. of
C 12.01 a.m.u., A.M. of O 15.99 a.m.u)
M.M. of CO 12.01 15.99 29.01 a.m.u.
20
Example 2 M.M. of carbon monoxide, CO? (A.M. of
C 12.01 a.m.u., A.M. of O 15.99 a.m.u)
M.M. of CO 12.01 15.99 29.01 a.m.u.
Example 3 M.M. of lead nitrate, Pb(NO3)2? (Pb
207.2, N 14.01, O 15.99)
M.M. of Pb(NO3)2 207.2 (14.01 x 2) (15.99
x 6) 331.16 a.m.u.
21
Chemical Accounting 1 The Mole Relating Number
of Chemical Entities to their Mass
22
Consider the atomic mass table Element Atomic
mass (of one atom) H 1.008 a.m.u. He 4.003
a.m.u. Li 6.941 a.m.u. Be 9.012 a.m.u. C
12.000 a.m.u. ...etc. ...etc.
23
Consider the atomic mass table Element Atomic
mass (of one atom) H 1.008 a.m.u. He 4.003
a.m.u. Li 6.941 a.m.u. Be 9.012 a.m.u. C
12.000 a.m.u. ...etc. ...etc.
Gives Relative mass of one atom of a given
element w.r.t. to mass of one atom of one
element (carbon)
24
Practical work requires a table which gives
(a) relative mass of a given large number of
atoms of each element w.r.t. mass of the same
number of C atoms (b) a value of 12 g for this
given large number of C atoms.
25
Practical work requires a table which gives
(a) relative mass of a given large number of
atoms of each element w.r.t. mass of the same
number of C atoms (b) a value of 12 g for this
given large number of C atoms. Thus this
practical table should look like...............
Element Mass (of n atoms) H 1.008
g He 4.003 g Li 6.941 g Be 9.012 g C
12.000 g ...etc. ...etc. where n is the
given large number of atoms.
26
What value of n will allow us to scale up from
the a.m.u. scale (single atoms) to the gram
scale (large numbers of atoms)? n 6.022 x
1023 entities Avogadros Number (N) 1
mole (mol)
Now have a system for relating large numbers of
atoms to their total mass, and for comparing the
relative masses of any given large number of
atoms of different elements
27
What value of n will allow us to scale up from
the a.m.u. scale (single atoms) to the gram
scale (large numbers of atoms)? n 6.022 x
1023 entities Avogadros Number (N) 1
mole (mol)
Now have a system for relating large numbers of
atoms to their total mass, and for comparing
the relative masses of any given large number
of atoms of different elements
Element Atomic Mass Scale Mole Mass Scale No.
of Mass No. of Mass Atoms (a.m.u.) Atoms
(g) H 1 1.008 1 mole 1.008 He
1 4.003 1 mole 4.003 Li 1 6.941 1
mole 6.941 Be 1 9.012 1 mole 9.012
...etc.. ...etc.. ...etc..
28
What value of n will allow us to scale up from
the a.m.u. scale (single atoms) to the gram
scale (large numbers of atoms)? n 6.022 x
1023 entities Avogadros Number (N) 1
mole (mol)
Now have a system for relating large numbers of
atoms to their total mass, and for comparing
the relative masses of any given large number
of atoms of different elements
Element Atomic Mass Scale Mole Mass Scale No.
of Mass No. of Mass Atoms (a.m.u.) Atoms
(g) H 1 1.008 1 mole 1.008 He
1 4.003 1 mole 4.003 Li 1 6.941 1
mole 6.941 Be 1 9.012 1 mole 9.012
...etc.. ...etc.. ...etc..
6.022 x 1023 atoms
29
This system can also be used to determine the
masses of large numbers of other chemical
entities (e.g. molecules, ions, etc.). Example
1 Mass of 0.5 mole of plutonium (Pu) atoms?
30
This system can also be used to determine the
masses of large numbers of other chemical
entities (e.g. molecules, ions, etc.). Example
1 Mass of 0.5 mole of plutonium (Pu) atoms?
From table of atomic masses Pu 244. This
means that 1 atom of Pu has a mass of 244 a.m.u.
and that 1 mole of Pu has a mass of 244 g. Since
1 mole Pu 244 g, then 0.5 mole Pu has a mass
of 244 x 0.5 g. Answer 122 g
31
Example 2 Mass of 1 mole of silicon carbide
(SiC)?
32
Example 2 Mass of 1 mole of silicon carbide
(SiC)? A molecule of SiC consists of one atom of
silicon and one atom of carbon bonded together.
Thus 1 mole SiC consists of 1 mole Si 1 mole
C. From atomic mass table, Si 28.086 and C
12.011. So the mass of 1 mole SiC 28.086
12.011 g
33
Example 2 Mass of 1 mole of silicon carbide
(SiC)? A molecule of SiC consists of one atom of
silicon and one atom of carbon bonded together.
Thus 1 mole SiC consists of 1 mole Si 1 mole
C. From atomic mass table, Si 28.086 and C
12.011. So the mass of 1 mole SiC 28.086
12.011 g
Answer 40.097 g
34
Example 3 Mass of 2 moles of SO42- ions? (O
15.999 and S 32.06) 1 SO42- ion contains
1 S and 4 Os, thus 2 mol SO42- contains 2 mol
sulphur and 4 x 2 mol oxygen Mass of 2 mol S 2
x 32.06 g 64.12 g Mass of 4 x 2 mol O 4 x 2
x 15.99 g 127.92 g
Mass of 2 moles of sulphate ions 192.04 g
35
Molarity Molarity moles of solute litres
of solution
36
Molarity Molarity moles of solute litres
of solution Example What is the molarity of a
solution made by dissolving 2.355 g of sulfuric
acid (H2SO4) in water and diluting to a final
volume of 50.0 ml?
37
Molarity Molarity moles of solute litres
of solution Example What is the molarity of a
solution made by dissolving 2.355 g of sulfuric
acid (H2SO4) in water and diluting to a final
volume of 50.0 ml? Molecular weight of H2SO4
2.0 x 1 32.1 16.0 x 4 98.1
a.m.u. Molar mass of H2SO4 98.1 g/mol
38
Molarity Molarity moles of solute litres
of solution Example What is the molarity of a
solution made by dissolving 2.355 g of sulfuric
acid (H2SO4) in water and diluting to a final
volume of 50.0 ml? Molecular weight of H2SO4
2.0 x 1 32.1 16.0 x 4 98.1
a.m.u. Molar mass of H2SO4 98.1 g/mol 2.355 x
1/98.1 0.024 Units g x mol g-1 mol gt
0.024 mol H2SO4 Molarity moles of solute/
Litres of solution 0.024 mol/ 0.05 L
0.48 M
39
Example What is the concentration of water in 1
litre of water?
40
Example What is the concentration of water in 1
litre of water? Molecular weight of H2O 2.0 x
1 16.0 18.0 a.m.u.
41
Example What is the concentration of water in 1
litre of water? Molecular weight of H2O 2.0 x
1 16.0 18.0 a.m.u. Density of water 1.0
g/cm3 or 1.0 kg/L Mass of H2O in 1 L 1
kg molarity moles of solute/ litres of
solution 1000/18.0 g L -1 g-1 mol 55.5
mol L-1 55.5 M
42
Example The molecular mass of sugar
(C12H22O11) How many molecules of sugar are there
in 1 spoon (0.025 kg) and what is the
concentration of sugar in 1 cup of coffee (100
ml)?
43
Example The molecular mass of sugar
(C12H22O11) How many molecules of sugar are there
in 1 spoon (0.025 kg) and what is the
concentration of sugar in 1 cup of coffee (100
ml)? Molecular mass of sugar 12 x 12.00 22 x
1.01 11 x 16.00 342.22 a.m.u.
44
Example The molecular mass of sugar
(C12H22O11) How many molecules of sugar are there
in 1 spoon (0.025 kg) and what is the
concentration of sugar in 1 cup of coffee (100
ml)? Molecular mass of sugar 12 x 12.00 22 x
1.01 11 x 16.00 342.22 a.m.u. Mass of 1 mole
of sugar 342.22 g mol -1
45
Example The molecular mass of sugar
(C12H22O11) How many molecules of sugar are there
in 1 spoon (0.025 kg) and what is the
concentration of sugar in 1 cup of coffee (100
ml)? Molecular mass of sugar 12 x 12.00 22 x
1.01 11 x 16.00 342.22 a.m.u. Mass of 1 mole
of sugar 342.22 g mol -1 of moles of sugar
0.025 x 1000/342.22 0.073 mole
46
Example The molecular mass of sugar
(C12H22O11) How many molecules of sugar are there
in 1 spoon (0.025 kg) and what is the
concentration of sugar in 1 cup of coffee (100
ml)? Molecular mass of sugar 12 x 12.00 22 x
1.01 11 x 16.00 342.22 a.m.u. Mass of 1 mole
of sugar 342.22 g of moles of sugar 0.025
x 1000/342.22 0.073 mole of molecules of
sugar 0.073 x 6.022 x 1023
mole molecule mole-1
47
Example The molecular mass of sugar
(C12H22O11) How many molecules of sugar are there
in 1 spoon (0.025 kg) and what is the
concentration of sugar in 1 cup of coffee (100
ml)? Molecular mass of sugar 12 x 12.00 22 x
1.01 11 x 16.00 342.22 a.m.u. Mass of 1 mole
of sugar 342.22 g of moles of sugar 0.025
x 1000/342.22 0.073 mole of molecules of
sugar 0.073 x 6.022 x 1023
mole molecule mole-1 4.4 x 1022
molecules
48
2000 Exam How many moles of copper atoms are
there in a copper coin of mass 2.5 g given that
the atomic mass of copper is 63.55 a.m.u. and
that the coin contains 55 copper by mass?  

49
2000 Exam How many moles of copper atoms are
there in a copper coin of mass 2.5 g given that
the atomic mass of copper is 63.55 a.m.u. and
that the coin contains 55 copper by mass?   (a)
0.04 (b) 0.02 (c) 0.01 (d) 0.06 (e) 0.03

50
2000 Exam How many moles of copper atoms are
there in a copper coin of mass 2.5 g given that
the atomic mass of copper is 63.55 a.m.u. and
that the coin contains 55 copper by mass?   (a)
0.04 (b) 0.02 (c) 0.01 (d) 0.06 (e) 0.03 Mass
of copper 0.55 x 2.5 g 1.38 g

51
2000 Exam How many moles of copper atoms are
there in a copper coin of mass 2.5 g given that
the atomic mass of copper is 63.55 a.m.u. and
that the coin contains 55 copper by mass?   (a)
0.04 (b) 0.02 (c) 0.01 (d) 0.06 (e) 0.03 Mass
of copper 0.55 x 2.5 g 1.38 g of moles of
copper 1.38/ 63.55 (g/ g mol-1)

52
2000 Exam How many moles of copper atoms are
there in a copper coin of mass 2.5 g given that
the atomic mass of copper is 63.55 a.m.u. and
that the coin contains 55 copper by mass?   (a)
0.04 (b) 0.02 (c) 0.01 (d) 0.06 (e) 0.03 Mass
of copper 0.55 x 2.5 g 1.38 g of moles of
copper 1.38/ 63.55 (g/ g mol-1)
0.02 moles

53
2000 Exam How many moles of copper atoms are
there in a copper coin of mass 2.5 g given that
the atomic mass of copper is 63.55 a.m.u. and
that the coin contains 55 copper by mass?   (a)
0.04 (b) 0.02 (c) 0.01 (d) 0.06 (e) 0.03 Mass
of copper 0.55 x 2.5 g 1.38 g of moles of
copper 1.38/ 63.55 (g/ g mol-1)
0.02 moles

54

55
http//www.ul.ie/ces/Resource.html B3053 CH47
01 2C Mon 16.00 CH4701 2B Fri
10.00 CH4701 2A Wed 12.00 CH4711 2A
Wed 10.00 CH4721 2A Th 10.00
56
Example What mass of NaOH and HCl are require to
make 29.22 g of NaCl? NaOH HCl gt NaCl
H2O (1) Atomic masses Na 22.99 a.m.u. , Cl
35.45 a.m.u. , O 15.999 a.m.u. and H 1.008 a.m.u.
57
Example What mass of NaOH and HCl are require to
make 29.22 g of NaCl? NaOH HCl gt NaCl
H2O (1) Atomic masses Na 22.99 a.m.u. , Cl
35.45 a.m.u. , O 15.999 a.m.u. and H 1.008 a.m.u.
1 molecule NaOH 1 molecule HCl gt
1 molecule NaCl 1 molecule H2O 1
mole NaOH 1 mole HCl gt1 mole NaCl 1 mole
H2O
58
Example What mass of NaOH and HCl are require to
make 29.22 g of NaCl? NaOH HCl gt NaCl
H2O (1) Atomic masses Na 22.99 a.m.u. , Cl
35.45 a.m.u. , O 15.999 a.m.u. and H 1.008 a.m.u.
1 molecule NaOH 1 molecule HCl gt
1 molecule NaCl 1 molecule H2O 1
mole NaOH 1 mole HCl gt1 mole NaCl 1 mole
H2O x mole NaOH x mole HCl gtx mole NaCl x
mole H2O where x is the number of moles
present in 29.22 g NaCl
59
Example What mass of NaOH and HCl are require to
make 29.22 g of NaCl? NaOH HCl gt NaCl
H2O (1) Atomic masses Na 22.99 a.m.u. , Cl
35.45 a.m.u. , O 15.999 a.m.u. and H 1.008 a.m.u.
1 molecule NaOH 1 molecule HCl gt
1 molecule NaCl 1 molecule H2O 1
mole NaOH 1 mole HCl gt1 mole NaCl 1 mole
H2O x mole NaOH x mole HCl gtx mole NaCl x
mole H2O where x is the number of moles
present in 29.22 g NaCl Molecular mass of NaCl
22.99 35.45 58.44 a.m.u. 1 mole of NaCl has
a mass of 58.44 g
60
X 1 x 29.22/58.44 0.5 mole
61
X 1 x 29.22/58.44 0.5 mole Calculate the
mass of 0.5 mole of NaOH and of HCl
62
X 1 x 29.22/58.44 0.5 mole Calculate the
mass of 0.5 mole of NaOH and of HCl Mass of 0.5
mole of NaOH 0.5 x (22.99 15.999 1.008)
20.00g Mass of 0.5 mole HCl 0.5 x (1.008
35.45) 18.23 g
63
X 1 x 29.22/58.44 0.5 mole Calculate the
mass of 0.5 mole of NaOH and of HCl Mass of 0.5
mole of NaOH 0.5 x (22.99 15.999 1.008)
20.00g Mass of 0.5 mole HCl 0.5 x (1.008
35.45) 18.23 g Answer 20.00 g NaCl and
18.23 g HCl.
64

• Q. 8 2001 Exam
• The alcohol content of a beer is 5 by weight.
  How many
• moles of ethanol (C2H5OH) are there in 500 g of
  beer?
•  
• 5.4 (b) 0.54 (c) 0.44 (d) 5
• none of the above

65

• Q. 8 2001 Exam
• The alcohol content of a beer is 5 by weight.
  How many
• moles of ethanol (C2H5OH) are there in 500 g of
  beer?
•  
• 5.4 (b) 0.54 (c) 0.44 (d) 5
• none of the above
• Mass of alcohol 0.05 x 500 g 25 g
• C2H5OH mass of one molecule
• (2 x 12) (5 x 1) 16 1 46 amu

66

• Q. 8 2001 Exam
• The alcohol content of a beer is 5 by weight.
  How many
• moles of ethanol (C2H5OH) are there in 500 g of
  beer?
•  
• 5.4 (b) 0.54 (c) 0.44 (d) 5
• none of the above
• Mass of alcohol 0.05 x 500 g 25 g
• C2H5OH mass of one molecule
• (2 x 12) (5 x 1) 16 1 46 amu
• Mass of one mole of ethanol 46 g mol-1
• of moles 25/46 g/ g mol-1 0.54 moles

67

• Q. 8 2001 Exam
• The alcohol content of a beer is 5 by weight.
  How many
• moles of ethanol (C2H5OH) are there in 500 g of
  beer?
•  
• 5.4 (b) 0.54 (c) 0.44 (d) 5
• none of the above
• Mass of alcohol 0.05 x 500 g 25 g
• C2H5OH mass of one molecule
• (2 x 12) (5 x 1) 16 1 46 amu
• Mass of one mole of ethanol 46 g mol-1
• of moles 25/46 g/ g mol-1 0.54 moles

68
Early Chemical Concepts and Their Present Day
Uses
69

• Basic idea of the atom
• Democritus (400 B.C.)
• Alchemists (Middle Ages),
• Lavoisier (1743 - 1794).

70

• Basic idea of the atom
• Democritus (400 B.C.)
• Alchemists (Middle Ages),
• Lavoisier (1743 - 1794).
• Viewed the notion of atoms from a philosophical
  point of
• view, i.e. did not try to prove their existence.

71

• Basic idea of the atom
• Democritus (400 B.C.)
• Alchemists (Middle Ages),
• Lavoisier (1743 - 1794).
• Viewed the notion of atoms from a philosophical
  point of
• view, i.e. did not try to prove their existence.
• Robert Boyle (1627 - 91)
• Disproved Aristotelian approach (based on a
  priore theories).

72

• Basic idea of the atom
• Democritus (400 B.C.)
• Alchemists (Middle Ages)
• Lavoisier (1743 - 1794).
• Viewed the notion of atoms from a philosophical
  point of
• view, i.e. did not try to prove their existence.
• Robert Boyle (1627 - 91)
• Disproved Aristotelian approach (based on a
  priore theories).
• Introduced the Scientific Method as the
  underlying
• philosophy of science.

73
(No Transcript)
74
Observation
The Scientific Method
75
Observation Hypothesis/Theory
The Scientific Method
76
Observation Hypothesis/Theory Prediction
The Scientific Method
77
Observation Hypothesis/Theory
Prediction Experiment
The Scientific Method
78
Observation Hypothesis/Theory
Prediction Experiment Do results agree with
theory? yes Scientific Law
The Scientific Method
Modify theory
no
79
John Dalton (1767 - 1844) Used the Scientific
Method to develop a basic atomic theory
80
John Dalton (1767 - 1844) Used the Scientific
Method to develop a basic atomic theory 1.
Matter is made up of atoms which are indivisible
and indestructible.
81
John Dalton (1767 - 1844) Used the Scientific
Method to develop a basic atomic theory 1.
Matter is made up of atoms which are indivisible
and indestructible. 2. Atoms of a given element
are identical both in mass and in chemical
proportions.
82
John Dalton (1767 - 1844) Used the Scientific
Method to develop a basic atomic theory 1.
Matter is made up of atoms which are indivisible
and indestructible. 2. Atoms of a given element
are identical both in mass and in chemical
proportions. 3. Atoms of different elements have
different masses and different chemical
properties.
83
John Dalton (1767 - 1844) Used the Scientific
Method to develop a basic atomic theory 1.
Matter is made up of atoms which are indivisible
and indestructible. 2. Atoms of a given element
are identical both in mass and in chemical
proportions. 3. Atoms of different elements have
different masses and different chemical
properties. 4. Atoms of different elements
combine in simple whole numbers to form
compounds.
84
John Dalton (1767 - 1844) Used the Scientific
Method to develop a basic atomic theory 1.
Matter is made up of atoms which are indivisible
and indestructible. 2. Atoms of a given
element are identical both in mass and in
chemical proportions. 3. Atoms of different
elements have different masses and different
chemical properties. 4. Atoms of different
elements combine in simple whole numbers to form
compounds. 5. When a compound is decomposed,
the recovered atoms are unchanged and can form
the same or new compounds.
85
Gay-Lussac (1778 - 1850) Gases react in simple
whole number units of volume, and where the
products are gases they also have simple whole
number units of volume.
86
Gay-Lussac (1778 - 1850) Gases react in simple
whole number units of volume, and where the
products are gases they also have simple whole
number units of volume. Thus 2 volumes H2
1 volume O2 2 volumes H2O
87
Gay-Lussac (1778 - 1850) Gases react in simple
whole number units of volume, and where the
products are gases they also have simple whole
number units of volume. Thus 2 volumes H2
1 volume O2 2 volumes H2O 3 volumes H2 1
volume N2 2 volumes NH3
88
Gay-Lussac (1778 - 1850) Gases react in simple
whole number units of volume, and where the
products are gases they also have simple whole
number units of volume. Thus 2 volumes H2
1 volume O2 2 volumes H2O 3 volumes H2 1
volume N2 2 volumes NH3 Note this relationship
is only true if pressure and temperature are
constant throughout.
89
Avogadro (1776 - 1856) Avogadros Hypothesis
(1811) "Equal volumes of gases under the same
conditions of temperature and pressure contain
the same number of particles (i.e. atoms or
molecules)."
90
Avogadro (1776 - 1856) Avogadros Hypothesis
(1811) "Equal volumes of gases under the same
conditions of temperature and pressure contain
the same number of particles (i.e. atoms or
molecules)." Volume of 1 mole of any gaseous
pure substance at STP (273.15K, 101.3 kPa)
22.41 dm3 Gas Molar Volume Thus for any gas at
STP 1 mole ( atomic or molecular mass in grams)
occupies 22.41 dm3.
91
Avogadro (1776 - 1856) Avogadros Hypothesis
(1811) "Equal volumes of gases under the same
conditions of temperature and pressure contain
the same number of particles (i.e. atoms or
molecules)." Volume of 1 mole of any gaseous
pure substance at STP (273.15K, 101.3 kPa)
22.41 dm3 Gas Molar Volume Thus for any gas at
STP 1 mole ( atomic or molecular mass in grams)
occupies 22.41 dm3. Examples 1 mol of H2 (M.M.
2.02) has a mass of 2.02 g and at STP occupies
22.41 dm3.
92
Avogadro (1776 - 1856) Avogadros Hypothesis
(1811) "Equal volumes of gases under the same
conditions of temperature and pressure contain
the same number of particles (i.e. atoms or
molecules)." Volume of 1 mole of any gaseous
pure substance at STP (273.15K, 101.3 kPa)
22.41 dm3 Gas Molar Volume Thus for any gas at
STP 1 mole ( atomic or molecular mass in grams)
occupies 22.41 dm3. Examples 1 mol of H2 (M.M.
2.02) has a mass of 2.02 g and at STP occupies
22.41dm3. 0.5 mol of CO (M.M. 28.01) has a mass
of 14.005 g and at STP occupies 11.205 dm3.
93
Avogadro's Hypothesis implies gas density is
proportional to molecular mass. Allowed
the early chemists to deduce molecular mass from
gas density measurements.
94
Avogadro's Hypothesis implies gas density is
proportional to molecular mass. Allowed
the early chemists to deduce molecular mass from
gas density measurements. Cannizzaro (1858)
Put all of the foregoing hypotheses together
into a method for determining accurate atomic
and molecular masses, but only for gaseous
compounds. Postulated that over a large number
of compounds of a particular element, at least
one will have only one atom of the element per
molecule. Mass of the element in 22.4 L of this
compound is the atomic mass.
95

Compound Mass of Mass of contained
gas/ g hydrogen/ g Hydrogen
(H2) 2 2 Methane (CH4) 16 4 Ethane
(C2H6) 30 6 Water (H2O) 18 2 Hydrogen
sulfide (H2S) 34 2 Hydrogen cyanide
(HCN) 27 1 Hydrogen chloride (HCl) 36 1 Ammoni
a (NH3) 17 3 Pyridine (C5H5N) 79 5 Minimum
mass of hydrogen 1 g Atomic mass of hydrogen
1 a.m.u.
96
Dulong and Petit (1819) Found a correlation
between atomic mass and specific heat of solid
elements Atomic mass x specific heat ? 25
JK-1mo1-1 where atomic mass is in grams per mole
and specific heat in JK-1g-1.
97
Dulong and Petit (1819) Found a correlation
between atomic mass and specific heat of solid
elements Atomic mass x specific heat ? 25
JK-1mo1-1 where atomic mass is in grams per mole
and specific heat in JK-1g-1.
Element Experimental values of heat
capacity Al 24.3 J K-1 mol-1 Fe 25.2 Ni 2
6.0 Ag 25.5 Au 25.2 Pb 26.8 For Bi,
heat capacity 0.123 J gt atomic mass of 211
a.m.u. measured atomic mass 209 a.m.u.
98
Determination of Empirical and Molecular
Formulae
99
Determination of Empirical and Molecular
Formulae Empirical formula the relative number
of atoms of each element in a molecule or
formula unit
100
Determination of Empirical and Molecular
Formulae Empirical formula the relative number
of atoms of each element in a molecule or
formula unit Molecular formula the actual
number of atoms of each element in a molecule.
101
Determination of Empirical and Molecular
Formulae Empirical formula the relative number
of atoms of each element in a molecule or
formula unit Molecular formula the actual
number of atoms of each element in a
molecule. For ethane molecular formula C2H6,
empirical formula CH3.
102
Determination of Empirical and Molecular
Formulae Empirical formula the relative number
of atoms of each element in a molecule or
formula unit Molecular formula the actual
number of atoms of each element in a
molecule. For ethane molecular formula C2H6,
empirical formula CH3.
Elemental composition data Empirical formula
Molecular formula Molecular mass
103
Procedure 1. If composition data are given,
assume 100 g of compound present. If direct
weights are given, use these data.
104
Procedure 1. If composition data are given,
assume 100 g of compound present. If direct
weights are given, use these data. 2.
Determine the number of moles of each element.
105
Procedure 1. If composition data are given,
assume 100 g of compound present. If direct
weights are given, use these data. 2.
Determine the number of moles of each element. 3.
Divide each value in point 2 by the smallest of
them.
106
Procedure 1. If composition data are given,
assume 100 g of compound present. If direct
weights are given, use these data. 2.
Determine the number of moles of each element. 3.
Divide each value in point 2 by the smallest of
them. 4. If necessary, multiply all the numbers
by the smallest integer possible to obtain
approximate whole number values for each
element.
107
Procedure 1. If composition data are given,
assume 100 g of compound present. If direct
weights are given, use these data. 2.
Determine the number of moles of each element. 3.
Divide each value in point 2 by the smallest of
them. 4. If necessary, multiply all the numbers
by the smallest integer possible to obtain
approximate whole number values for each
element. 5. Write these whole numbers as
subscripts of the element symbols to give an
empirical formula.
108
Procedure 1. If composition data are given,
assume 100 g of compound present. If direct
weights are given, use these data. 2.
Determine the number of moles of each element. 3.
Divide each value in point 2 by the smallest of
them. 4. If necessary, multiply all the numbers
by the smallest integer possible to obtain
approximate whole number values for each
element. 5. Write these whole numbers as
subscripts of the element symbols to give an
empirical formula. 6. If the molecular weight is
known, determine how many empirical formula
weights are required to obtain the molecular
weight. Use this factor to multiply the number
of atoms of each element in the empirical
formula to give the molecular formula.
109
A sample of vitamin C of mass 8.00 g was analysed
and Found to contain 3.27 g of carbon, 0.366 g
of hydrogen and 4.36 g of oxygen. What is the
molecular formula of vitamin C? The molar mass
of vitamin C is 176.14 g mol-1.
110
A sample of vitamin C of mass 8.00 g was analysed
and Found to contain 3.27 g of carbon, 0.366 g
of hydrogen and 4.36 g of oxygen. What is the
molecular formula of vitamin C? The molar mass
of vitamin C is 176.14 g mol-1. 1. Mass of
carbon 3.27g/ 8.00 g x 100 40.9 Mass of
hydrogen 0.366 g/ 8.00 g x 100 4.58 Mass
of oxygen 4.36 g/ 8.00 g x 100 54.5
111
A sample of vitamin C of mass 8.00 g was analysed
and Found to contain 3.27 g of carbon, 0.366 g
of hydrogen and 4.36 g of oxygen. What is the
molecular formula of vitamin C? The molar mass
of vitamin C is 176.14 g mol-1. 1. Mass of
carbon 3.27g/ 8.00 g x 100 40.9 Mass of
hydrogen 0.366 g/ 8.00 g x 100 4.58 Mass
of oxygen 4.36 g/ 8.00 g x 100 54.5 In 100
g, have 40.9 g C, 4.58 g H and 54.5 g O.
112
A sample of vitamin C of mass 8.00 g was analysed
and Found to contain 3.27 g of carbon, 0.366 g
of hydrogen and 4.36 g of oxygen. What is the
molecular formula of vitamin C? The molar mass
of vitamin C is 176.14 g mol-1. 1. Mass of
carbon 3.27g/ 8.00 g x 100 40.9 Mass of
hydrogen 0.366 g/ 8.00 g x 100 4.58 Mass
of oxygen 4.36 g/ 8.00 g x 100 54.5 In 100
g, have 40.9 g C, 4.58 g H and 54.5 g O. 2. of
moles C 40.9 g /12.01 g mol-1 3.41 mol
of moles H 4.58 g /1.008 g mol-1 4.54 mol
of moles O 54.5 g /16.00 g mol-1 3.41 mol
113
A sample of vitamin C of mass 8.00 g was analysed
and Found to contain 3.27 g of carbon, 0.366 g
of hydrogen and 4.36 g of oxygen. What is the
molecular formula of vitamin C? The molar mass
of vitamin C is 176.14 g mol-1. 1. Mass of
carbon 3.27g/ 8.00 g x 100 40.9 Mass of
hydrogen 0.366 g/ 8.00 g x 100 4.58 Mass
of oxygen 4.36 g/ 8.00 g x 100 54.5 In 100
g, have 40.9 g C, 4.58 g H and 54.5 g O. 2. of
moles C 40.9 g /12.01 g mol-1 3.41 mol
of moles H 4.58 g /1.008 g mol-1 4.54 mol
of moles O 54.5 g /16.00 g mol-1 3.41
mol 3. 3.41 C 4.54 H 3.41 O divide
by 3.41 gt1 C 1.33 H 1.33 O
114
A sample of vitamin C of mass 8.00 g was analysed
and Found to contain 3.27 g of carbon, 0.366 g
of hydrogen and 4.36 g of oxygen. What is the
molecular formula of vitamin C? The molar mass
of vitamin C is 176.14 g mol-1. 1. Mass of
carbon 3.27g/ 8.00 g x 100 40.9 Mass of
hydrogen 0.366 g/ 8.00 g x 100 4.58 Mass
of oxygen 4.36 g/ 8.00 g x 100 54.5 In 100
g, have 40.9 g C, 4.58 g H and 54.5 g O. 2. of
moles C 40.9 g /12.01 g mol-1 3.41 mol
of moles H 4.58 g /1.008 g mol-1 4.54 mol
of moles O 54.5 g /16.00 g mol-1 3.41
mol 3. 3.41 C 4.54 H 3.41 O divide
by 3.41 gt1 C 1.33 H 1.33 O 4. Multiply
by 3 3 C 4 H 3 O
115
5. Empirical formula is C3H4O3. Molar mass
of empirical formula 3 x 12.00 4 x 1.008
3 x 16.00 88.032 g mol 1
116
5. Empirical formula is C3H4O3. Molar mass
of empirical formula 3 x 12.00 4 x 1.008
3 x 16.00 88.032 g mol -1 6. of empirical
formula units per molecule 174.14/ 88.06 g
g-1 mol mol-1 2 Molecular formula is C6H8O6
117
5. Empirical formula is C3H4O3. Molar mass
of empirical formula 3 x 12.00 4 x 1.008
3 x 16.00 88.032 g mol -1 6. of empirical
formula units per molecule 174.14/ 88.06 g
g-1 mol mol-1 2 Molecular formula is C6H8O6
The molar mass of oxalic acid is 90 g mol-1 and
its empirical formula is CHO2. What is its
molecular formula?
118
5. Empirical formula is C3H4O3. Molar mass
of empirical formula 3 x 12.00 4 x 1.008
3 x 16.00 88.032 g mol -1 6. of empirical
formula units per molecule 174.14/ 88.06 g
g-1 mol mol-1 2 Molecular formula is C6H8O6
The molar mass of oxalic acid is 90 g mol-1 and
its empirical formula is CHO2. What is its
molecular formula? 1. 90 g mol -1
119
5. Empirical formula is C3H4O3. Molar mass
of empirical formula 3 x 12.00 4 x 1.008
3 x 16.00 88.032 g mol -1 6. of empirical
formula units per molecule 174.14/ 88.06 g
g-1 mol mol-1 2 Molecular formula is C6H8O6
The molar mass of oxalic acid is 90 g mol-1 and
its empirical formula is CHO2. What is its
molecular formula? 1. 90 g mol -1 5. Empirical
formula is CHO2. Molar mass of empirical
formula 12.01 1.008 2 x 16.00 45.018 g mol
-1
120
5. Empirical formula is C3H4O3. Molar mass
of empirical formula 3 x 12.00 4 x 1.008
3 x 16.00 88.032 g mol -1 6. of empirical
formula units per molecule 174.14/ 88.06 g
g-1 mol mol-1 2 Molecular formula is C6H8O6
The molar mass of oxalic acid is 90 g mol-1 and
its empirical formula is CHO2. What is its
molecular formula? 1. 90 g mol -1 5. Empirical
formula is CHO2. Molar mass of empirical
formula 12.01 1.008 2 x 16.00 45.018 g mol
-1 6. of empirical formula units per molecule
90/ 45.018 g g-1 mol mol-1 2 Molecular
formula is C2H2O4
121
Example A compound contains 5.6 g N, 1.6 g H,
20.6 g Cr and 22.2 g O. Find its empirical
formula. (At. masses N 14.0, H 1.01, Cr
52.0, O 16.0)
122
Example A compound contains 5.6 g N, 1.6 g H,
20.6 g Cr and 22.2 g O. Find its empirical
formula. (At. masses N 14.0, H 1.01, Cr
52.0, O 16.0) Solution 1. Use direct mass.
123
Example A compound contains 5.6 g N, 1.6 g H,
20.6 g Cr and 22.2 g O. Find its empirical
formula. (At. masses N 14.0, H 1.01, Cr
52.0, O 16.0) Solution 1. Use direct
mass. 2 moles of each element N 5.6/14.0
0.4 mol N atoms H 1.6/1.01 1.58 mol H
atoms Cr 20.6/52.0 0.4 mol Cr atoms O
22.2/16.0 1.39 mol O atoms
124
Example A compound contains 5.6 g N, 1.6 g H,
20.6 g Cr and 22.2 g O. Find its empirical
formula. (At. masses N 14.0, H 1.01, Cr
52.0, O 16.0) Solution 1. Use direct
mass. 2 moles of each element N 5.6/14.0
0.4 mol N atoms H 1.6/1.01 1.58 mol H
atoms Cr 20.6/52.0 0.4 mol Cr atoms O
22.2/16.0 1.39 mol O atoms 3. Divide by
smallest (0.4) N 1, H 3.95, Cr 1, O 3.48
125
Example A compound contains 5.6 g N, 1.6 g H,
20.6 g Cr and 22.2 g O. Find its empirical
formula. (At. masses N 14.0, H 1.01, Cr
52.0, O 16.0) Solution 1. Use direct
mass. 2 moles of each element N 5.6/14.0
0.4 mol N atoms H 1.6/1.01 1.58 mol H
atoms Cr 20.6/52.0 0.4 mol Cr atoms O
22.2/16.0 1.39 mol O atoms 3. Divide by
smallest (0.4) N 1, H 3.95, Cr 1, O
3.48 4. Multiply by 2 to obtain approx. whole
nos. N2, H8, Cr2, O7
126
Example A compound contains 5.6 g N, 1.6 g H,
20.6 g Cr and 22.2 g O. Find its empirical
formula. (At. masses N 14.0, H 1.01, Cr
52.0, O 16.0) Solution 1. Use direct
mass. 2 moles of each element N 5.6/14.0
0.4 mol N atoms H 1.6/1.01 1.58 mol H
atoms Cr 20.6/52.0 0.4 mol Cr atoms O
22.2/16.0 1.39 mol O atoms 3. Divide by
smallest (0.4) N 1, H 3.95, Cr 1, O
3.48 4. Multiply by 2 to obtain approx. whole
nos. N2, H8, Cr2, O7 5. Write as subscripts
N2H8Cr2O7