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C H E M I S T R Y
Chapter 2
Atoms, Molecules, and Ions
Conservation of Mass and the Law
of Definite Proportions
Law of Conservation of Mass: Mass is neither
created nor destroyed in chemical reactions.
Conservation of Mass and the Law
of Definite Proportions
3.25 g + 3.32 g = 6.57 g
Hg(NO3)2(aq) + 2KI(aq)
HgI2(s) + 2KNO3(aq)
4.55 g + 2.02 g = 6.57 g
Conservation of Mass and the Law
of Definite Proportions
Law of Definite Proportions: Different samples of a pure chemical
substance always contain the same proportion of elements by mass.
By mass, water is:
88.8 % oxygen
11.2 % hydrogen
The Law of Multiple Proportions and
Dalton’s Atomic Theory
Law of Multiple Proportions: Elements can combine in different ways to
form different substances, whose mass ratios are small whole-number
multiples of each other.
nitrogen monoxide (NO):
nitrogen dioxide (NO2):
7 grams nitrogen per 8 grams oxygen
7 grams nitrogen per 16 grams oxygen
Insert Figure 2.2 p37
The Law of Multiple Proportions
and Dalton’s Atomic Theory
•
Elements are made up of tiny particles called atoms.
•
Each element is characterized by the mass of its atoms. Atoms of
the same element have the same mass, but atoms of different
elements have different masses.
•
The chemical combination of elements to make different chemical
compounds occurs when atoms join in small whole-number ratios.
•
Chemical reactions only rearrange how atoms are combined in
chemical compounds; the atoms themselves don’t change.
Atomic Structure: Electrons
Cathode-Ray Tubes: J. J. Thomson (1856-1940) proposed that cathode rays
must consist of tiny negatively charged particles. We now call them
electrons.
Atomic structure: Electrons
 The strength of deflecting magnetic electric field
 The size of the negative charge of electron
 The mass of the electron
Atomic Structure: Electrons
Millikan’s oil drop experiment
Atomic Structure: Protons and
Neutrons
Rutherford’s Gold Foil Experiment
In Rutherford’s gold foil experiment (1871-1937),positively
charged particles
•
were aimed at atoms of gold.
•
mostly went straight through the atoms.
•
were deflected only occasionally.
Conclusion:
There must be a small, dense, positively charged
nucleus in the atom that deflects positive particles that
come close.
Rutherford proposed that the atom must consist mainly of
empty space with the mass concentrated in a tiny
central core—the nucleus
Atomic Structure: Protons and
Neutrons
The mass of the atom is primarily
in the nucleus.
The charge of the proton is opposite in
sign but equal to that of the electron.
Atomic Numbers
Atomic Number (Z): Number of protons in an atom’s nucleus.
Equivalent to the number of electrons around an atom’s nucleus
Mass Number (A): The sum of the number of protons and the number of
neutrons in an atom’s nucleus
Isotope: Atoms with identical atomic numbers but different mass
numbers
Atomic Numbers
Atomic Numbers
carbon-12
or C-12
mass number
12
6
C
6 protons
6 electrons
6 neutrons
C
6 protons
6 electrons
7 neutrons
atomic number
carbon-13 or C-13
mass number
13
6
atomic number
Isotopic symbols
Examples
1. How many protons, electrons and neutrons are
present in an atom of
2. Write isotopic symbols in both forms for aluminum
isotope with 14 neutrons
3. An atom has 82 electrons and 126 neutrons. What
is its mass number and what is the element?
Atomic Masses and the Mole
Atomic Mass: weighed according to the natural abundance of each
isotope. Sometimes called average atomic mass.
The mass of 1 atom of carbon-12 is defined to be 12 amu.
Atomic Masses and the Mole
Why is the atomic mass of the element carbon 12.01 amu?
carbon-12:
98.89 % natural abundance
12 amu
carbon-13:
1.11 % natural abundance
13.0034 amu
mass of carbon = (12 amu)(0.9889) + (13.0034 amu)(0.0111)
= 11.87 amu + 0.144 amu
= 12.01 amu
Atomic Masses and the Mole
 Atomic mass unit (amu) is the mass in grams of a
single atom
 1 amu = 1.6605 x 10-24g
 E.g 1 H atom = 1.01 amu
Calculating Atomic Mass
The calculation for atomic mass requires the
• percent(%) abundance of each isotope.
• atomic mass of each isotope of that element.
• sum of the weighted averages.
mass of isotope(1)x (% abundance) + mass of isotope(2) x (% abundance) +
100
100
Calculating Atomic Mass for Copper
Copper has two naturally occurring isotopes: Cu-63 with
mass 62.9396 amu and a natural abundance of 69.17% and
Cu-65 with mass 64.9278 amu and a natural abundance of
30.83%. Calculate the atomic mass of copper
• Use atomic mass and percent of each isotope to calculate
the contribution of each isotope to the weighted average.
Atomic mass Cu-63 x % abundance =
Atomic mass Cu-65 x % abundance =
• Sum is atomic mass of Cu is
22
Example
 Bromine has two naturally occurring isotopes (Br-79
and Br-81) and has an atomic mass of 79.904 amu.
The mass of Br-81 is 80.9163 amu, and its natural
abundance is 49.31%. Calculate the mass and natural
abundance of Br-79
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
Copyright © 2009 by Pearson Education, Inc.
2
Atomic Masses and the Mole
Avogadro’s Number (NA): One mole of any substance contains 6.022 x 1023
formula units.
1 mole of anything = 6.02 x1023
Molar Mass: The mass in grams of
one mole of any element. It is
numerically equivalent to its atomic
mass.
E.g 1 H atom = 1.01 amu
1 mol H = 1.01 g
A Mole of Atoms
A mole is a collection that contains
• the same number of particles as there are carbon atoms in
12.0 g of carbon.
• 6.02 x 1023 atoms of an element (Avogadro’s number).
3
Samples of 1 Mole Quantities
1 mole of C atoms
= 6.02 x 1023 C atoms
1 mole of Al atoms
= 6.02 x 1023 Al atoms
1 mole of S atoms
= 6.02 x 1023 S atoms
5
Avogadro’s Number and the Mole
Avogadro’s Number (NA): One mole of any substance
contains 6.022 x 1023 formula units.
One mole of any substance is equivalent to its molar mass.
H
1 mole = 1.08 g
6.022 x 1023 molecules = 1.08 g
C:
1 mole = 12.01 g
6.022 x 1023 molecules = 12.01 g
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter
3/28
Molar Mass from Periodic Table
Molar mass
is the atomic
mass
expressed in
grams.
1 mole of Ag
= 107.9 g
1 mole of C
= 12.01 g
1 mole of S
= 32.07 g
Copyright © 2009 by Pearson Education, Inc.
29
Learning Check
Give the molar mass for each
A. 1 mole of K atoms
=
________
B. 1 mole of Sn atoms
=
________
30
Avogadro’s Number
Avogadro’s number, 6.02 x 1023, can be written as an
equality and two conversion factors.
Equality:
1 mole = 6.02 x 1023 particles
Conversion Factors:
6.02 x 1023 particles and
1 mole
1 mole
6.02 x 1023 particles
6
Examples
1. Calculate the number of copper atoms in 2.45 mol of
copper
2. The number of moles of S in 1.8 x 1024 atoms of S
3. The number of atoms in 2.0 g of Al atoms is
10