Transcript Modern Chemistry Chapter 3 Atoms: The Building Blocks of Matter

Modern Chemistry Chapter 3
Atoms: The Building Blocks of Matter

law of conservation of mass- mass is
neither created nor destroyed during
ordinary chemical reactions or physical
changes
e.g.
20 g A + 20 g B  40 g AB

law of definite proportions- a
chemical compound contains the same
elements in exactly the same proportions
by mass regardless of the size of the
sample or the source of the compound

e.g. If 10 grams of A combine with 20 grams of
B to form compound AB, how many grams of B
will be necessary to combine with 20 grams of A
to form AB? Answer 40: grams of B.

law of multiple proportions- if two or
more different compounds are composed
of the same two elements, then the ratio
of the masses of the second element
combined with a certain mass of the first
element is always a ratio of small whole
numbers

e.g. H2O & H2O2
or
NO2 & N2O5
Dalton’s Atomic Theory
In 1808, an English
school teacher named
John Dalton proposed
an explanation for the
law of conservation of
mass, the law of
definite proportions,
and the law of
multiple proportions.
Dalton’s Atomic Theory:
1- All matter is composed of extremely
small particles called atoms.
2- Atoms of a given element are identical
in size, mass and other properties.
3- Atoms cannot be subdivided, created or
destroyed.
4- Atoms of different elements combine in
simple whole-number ratios to form
chemical compounds.
5- In chemical reactions, atoms are
combined, separated, or rearranged.
Modern Atomic Theory

Dalton’s theory was a good one, but it has
since been modified.
– Atoms are divisible into even smaller
particles.
– Atoms of a given element can have
different masses.
– Atoms can be destroyed.
 Do Section Review #3 on page 71.
Section review #3 page 71
IF each compound contains 1.0 g oxygen and the
three samples contain:
compound A: K = 1.22 g / 1.22 = 1.0
compound B: K = 2.44 g / 1.22 = 2.0
compound C: K = 4.89 g / 1.22 = 4.0
1:2:4 ratios of potassium  multiple
proportions
Chapter 3 Section 2
The Structure of the Atom

atom- the
smallest particle of
an element that
retains the
chemical properties
of that element

atomic nucleus- the small, densely packed,
positively charged central portion of the atom
that contains nearly all of its mass but nearly
none of its volume
– neutron- The neutral particle of the nucleus
of an atom.
– proton- The positively charged particle of
the nucleus of an atom.

electron cloud- The large area surrounding
the nucleus of an atom in which the electrons
are located.
– electron- the negatively charged particles of
an atom
The Discovery of Electrons

In the late 1800’s, electric current was passed
through cathode ray tubes. It was discovered
that the cathode ray was attracted to the
positive pole of a magnet and repelled by the
negative pole.
– This led to the discovery of electrons.
– In 1909, Robert Millikan measured the negative
charge of the electron.
– From this, it was found that the mass of an
electron is 9.109 x 10-31 kg.
– The mass of an electron is 1/1837th the mass of
the simplest hydrogen atom.
– The negative charges of the electrons equal the
positive charges of an atom (protons).
The Discovery of the Atomic Nucleus

In 1911, Ernest Rutherford, Hans Geiger,
& Ernest Marsden bombarded a thin foil of
gold with positively charged particles
called alpha particles. They were
surprised to find that a few of the particles
(1 in 8000) were reflected from the foil
straight back toward their source. They
reasoned that this must mean that there
was a small positively charged portion of
each gold atom.
Composition of the Nucleus

protons- are positively charged particles
in the nucleus (their + charge = - charge
of an electron)
– mass = 1.673 x 10-27 kg or 1836/1837 the
mass of a protium atom

neutrons- neutral particles of the nucleus
– mass = 1.675 x 10-27 kg = mass of 1 electron
+ 1 proton
Composition of the Nucleus

How can the numerous positively charged
protons exist packed into the nucleus
without flying apart due to their like
charges repelling one another?

nuclear forces- are short range forces
(proton-proton, proton-neutron, &
neutron-neutron) that hold the nuclear
particles together.
The Size of Atoms

Atomic radius ranges between 40 &
270 pm (10-12 m)
Discuss how small a picometer would be.

Nuclear radius = 0.001 pm.
The size of the nucleus to the entire atom
would be about the same as if you placed a
dime at the center of the football stadium.
Section 3.3
Counting Atoms

atomic number- is equal to the number
of protons in the nucleus of each atom of
an element

mass number- is the number of protons
plus neutrons in a single atom of an
element.
Counting Atoms

isotopes- are atoms of the same element that
have different masses due to different numbers of
neutrons
 nuclide- is a general term for a specific isotope of
an element
– We designate isotopes using one of two different
designations.
– Hydrogen has three isotopes; protium (m# = 1),
deuterium (m# = 2), and tritium (m# = 3).
 hyphen notation
 nuclear symbol notation
H-1
1
1
H
H-2
2
1
H
H-3
3
1
H
Isotopes

Do practice problems
#1, 2, & 3 on page
80.
Practice problems page 80
1-
bromine-80
2-
carbon-13 
35 protons
35 electrons
80-35 = 45 neutrons
13
C
6
3-
15 electrons, so 15 protons & element is
phosphorus. 15 + 15  phosphorus-30
Identifying Isotopes
element
symbol
atomic #
calcium
uranium
92
uranium
13
I-127
mass #
p+ n0 e-
40
____
235
____
238
____
27
____
____
Relative Atomic Masses

atomic mass unit (amu)- exactly 1/12th
of the mass of a carbon-12 atom

average atomic mass- is the weighted
average of the atomic masses of the
naturally occurring isotopes of an element
– see table 4 on page 82
Calculating Average Atomic Mass
We will be calculating the “average mass” of the science
textbooks in the classroom. This is the approximate method
used to determine average atomic mass of the isotopes of an
element.
1- Using a bathroom scale, find the weight of the physics
textbooks and count their number. What is the average
weight of each book?
2- Find the weight of the chemistry textbooks and count their
number. What is the average weight of each book?
3- Add the weights of the textbooks and add the numbers of
books.
4- Divide the total weight by the total number of books to find
the average weight. How does this compare to the
average weight of each book?
Relating Mass to Numbers of Atoms

mole (mol)- the

The concept is similar
to that of a dozen.
amount of a
substance that
contains the same
number of particles as
there are in 12 grams
of carbon-12

Avogadro’s
number- is equal to
the number of
particles in one mole
of a substance and is
equal to 6.022 x 1023
Molar Mass

molar mass- the
mass of one mole of a
pure substance
– We can find molar
mass by using the
average atomic
mass found on the
periodic table and
changing the units
from amu to grams.
Mole-molar mass conversions
#mole x molar mass = mass (#grams)
mass (#grams) ÷ molar mass = #moles
Mole Hill
#moles
(mol)
÷ molar mass
(g/mol)
x molar mass
(g/mol)
mass in grams
(g)
mass in grams
(g)
Gram to Mole Conversions

We can use conversion factors to convert
between grams and moles.
2.00 mol He x 4.00 g He = 8.00 g He
1 mol He
8.00 g He x 1 mol He = 2.00 mol He
4.00 g He
Do practice problems #1-4 and #1-3 on page 85
Mole to Gram Conversions pg 85
1)
2.25 mol Fe x 55.85 g/mol = 126 g Fe
2)
0.375 mol K x 39.10 g/mol = 14.7 g K
3)
.0135 mol Na x 22.99 g/mol = 0.310 g Na
4)
16.3 mol Ni x 58.69 g/mol = 957 g Ni
Gram to Mole Conversions pg 85
1)
5.00 g Ca ÷ 40.08 g/mol = 0.125 mol Ca
2)
3.60 x 10-5 g Au ÷ 196.97 g/mol = 1.83 x 10-7 mol Au
3)
0.535 g Zn ÷ 65.39 g/mol = 8.18 x 10-3 mol Zn
Conversions Using Avogadro’s Number
3.01 x 1023 Ag atoms x
1 mole Ag atoms
6.022 x 1023 Ag atoms
= 0.500 moles of Ag
1.20 x 108 atoms Cu x
1 mol Cu atoms x 63.55 g
6.022 x 1023 Cu atoms
= Cu = 1.27 x 10-14 g Cu
Do section review problems #2-6 on page 87.
Section review page 87
2abcd-
sodium-23
calcium-40
copper-64
silver-108
3ab-
silicon-28
iron-56
4-
potassium  39.10 amu
5a-
2.00 mol x 14.0 g/mol = 28.0 g N
6a-
12.15 g / 24.3 g/mol = 0.50 mol Mg




11 protons, 11 electrons, 12 neutrons
20 protons, 20 electrons, 20 neutrons
29 protons, 29 electrons, 35 neutrons
47 protons, 47 electrons, 61 neutrons
&
39.10 g/mol
Section Review page 87
#7-
2.06 mol Cu
222 g Ag
Which has the larger mass?
2.06 mol Cu x 63.5 g/mol = 130.8 g Cu
222 g Ag > 130.8 g Cu
Which beaker has the larger number of atoms?
222 g Ag ÷ 107.9 g/mol = 2.06 mol Ag
SINCE both beakers have the same number of
moles, they have equal numbers of atoms.
Chapter 3- Practice problems
1-
U-235
92 electrons & 92 protons
235-92 = 143 neutrons
2-
U-238
3-
C-14
92 electrons & 92 protons
238-92 = 146 neutrons
6 electrons & 6 protons
14-6 = 8 neutrons
4-
5-
I-127
K-41
53 electrons & 53 protons
127-53 = 74 neutrons
19 electrons & 19 protons; 41 – 19 = 22 neutrons
#6) 4.25 mol Na
x 22.99 g/mol = 97.7 g Na
#7) 0.0013 mol Au
x 196.97 g/mol = 0.26 g Au
#8) 111.5 mol Ca
x 40.08 g/mol = 4469 g Ca
#9) 2.5 mol C
x 12.01 g/mol = 30 g C
#10) 0.025 mol Ag
x 107.87 g/mol = 2.7 g Ag
#11) 100.3 g Ca
÷ 40.07 g/mol = 2.503 mol Ca
#12) 72.0 g O
÷ 16.00 g/mol = 4.50 mol O
#13) 0.06 g C
÷ 12.01 g/mol = 0.005 mol C
#14) 5.4 g Au
÷ 196.97 g/mol = 0.027 mol Au
#15) 3.449 x 1011 g He
÷ 4.00 g/mol = 8.62 x 1010 mol He
Chapter 3 Review

Do problems #2, 6-11, 1719, 21-24, & 28 on pages
89 & 90 of the textbook.

Do the Math Tutor
problems #1 & 2 on page
92.

Do the Standardized Test
Prep on page 93.
Chapter 3 vocabulary
law of conservation of mass
law of multiple proportionsatomic nucleus********
electronneutronsatomic numberisotopesaverage atomic massAvogadro’s number-
law of definite proportionsatomneutronelectron cloudprotonsnuclear forcesmass number-.
atomic mass unit
mole
molar mass-
Chemistry Chapter 3 Test
30 multiple choice Questions:
 definitions & uses of the Laws of Conservation of Mass, Definite
Proportions, & Multiple Proportions
 Dalton’s Atomic Theory: its 5 points & modifications
 the cathode ray experiment & the discovery of electrons
 Rutherford’s experiment & the discovery of the atomic nucleus
 definitions of proton, neutron, electron, atomic nucleus (& its
characteristics), nuclear forces, atomic number, mass number,
isotopes, average atomic mass, mole, molar mass, &
Avogadro’s number
 determine the number of protons, electrons, & neutrons of an
element from its atomic and mass numbers
 mass to mole & mole to mass calculations
Honors Chemistry Chapter 3 Test
50 Multiple Choice questions
 definitions & implications of the Laws of Conservation of Mass,
Definite Proportions, Multiple Proportions, Dalton’s Atomic
Theory (and its modifications)
 implications of the cathode ray experiment & the discovery of
the electron
 Rutherford’s experiment & the discovery of the atomic nucleus
 description of an atom, atomic nucleus, & electron cloud
 definitions & implications of atom, proton, neutron, electron,
nuclear forces, isotopes, average atomic mass, atomic number,
mass number, Avogadro’s number, mole, molar mass
 describe the isotopes of hydrogen (protium, deuterium, tritium)
 determine numbers of protons, electrons & neutrons from
atomic & mass numbers
 mass to mole & mole to mass conversions