john e. mcmurry robert c. fay c h e m i s t r y sixth edition chapter 2 atoms, molecules, and ions

John E. McMurry • Robert C. Fay

C H E M I S T R YSixth Edition

Chapter 2Atoms, Molecules, and Ions

Law of Conservation of Mass: Mass is neither created nor destroyed in chemical reactions.

HgI2(s) + 2KNO3(aq)Hg(NO3)2(aq) + 2KI(aq)

4.55 g + 2.02 g = 6.57 g

3.25 g + 3.32 g = 6.57 g

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 % oxygen11.2 % hydrogen

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.

Insert Figure 2.2 p37

7 grams nitrogen per 8 grams oxygen

7 grams nitrogen per 16 grams oxygen

nitrogen monoxide (NO):

nitrogen dioxide (NO2):

• 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.

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: ElectronsThe strength of deflecting magnetic electric

field

The size of the negative charge of electron

The mass of the electron

Millikan’s oil drop experiment

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

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 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

carbon-14

C14

6

atomic number

mass number

carbon-12

C12

6

atomic number

mass number

6 protons6 electrons8 neutrons

6 protons6 electrons6 neutrons

The mass of 1 atom of carbon-12 is defined to be 12 amu.

Atomic Mass: The weighted average of the isotopic masses of the element’s naturally occurring isotopes

An atom of zinc has a mass number of 65.A. How many protons are in this zinc atom?

B.How many neutrons are in the zinc atom?

C.What is the mass number of a zinc atom that has

Examples

An atom has 14 protons and 20 neutrons.

A. Its atomic number is

B. Its mass number is

C. The element is

Examples

Example

Write the atomic symbols for atoms with the following subatomic particles:

A. 8 p+, 8 n, 8 e- ___________

B. 47p+, 60 n, 47 e- ___________

carbon-12: 98.89 % natural abundance 12 amu

carbon-13: 1.11 % natural abundance 13.0034 amu

Why is the atomic mass of the element carbon 12.01 amu?

= 12.01 amu

mass of carbon = (12 amu)(0.9889) + (13.0034 amu)(0.0111)

= 11.87 amu + 0.144 amu

Atomic Masses and the MoleAtomic mass unit (amu) is the mass in grams of a

single atomIsotopic mass1 amu = 1.6605 x 10-24g

Atomic Mass of Magnesium

The atomic mass of Mg

• is due to all the Mg isotopes.

• is a weighted average.

• is not a whole number.

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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 (%) + mass of isotope(2) x (%) +

100 100

Calculating Atomic Mass for Cl35Cl has atomic mass 34.97 amu (75.76%) and

37C

has atomic mass 36.97 amu (24.24%).

• Use atomic mass and percent of each isotope to calculate the contribution of each isotope to the weighted average. Atomic mass 35Cl x % abundance =Atomic mass 37Cl x % abundance =

• Sum is atomic mass of Cl is

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Learning CheckGallium is an element found in lasers used in compact disc players. In a sample of gallium, there is 60.10% of 69Ga (atomic mass 68.926) atoms and 39.90% of 71Ga (atomic mass 70.925) atoms.

What is the atomic mass of gallium?Which isotope has more atoms?

26

Avogadro’s Number (NA): One mole of any substance contains 6.022 x 1023 formula units.

Molar Mass: The mass in grams of one mole of any element. It is numerically equivalent to its atomic mass.

Molar Mass from Periodic Table

Molar mass is the atomic mass expressed in grams.

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1 mole of Ag 1 mole of C 1 mole of S = 107.9 g = 12.01 g = 32.07 g

Copyright © 2009 by Pearson Education, Inc.

Learning CheckGive the molar mass for each

A. 1 mole of K atoms = ________

B. 1 mole of Sn atoms = ________

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Collection Terms

A collection term statesa specific number of items.

• 1 dozen donuts

= 12 donuts

• 1 ream of paper

= 500 sheets

• 1 case = 24 cans

2

Copyright © 2009 by Pearson Education, Inc.

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).

1 mole element Number of Atoms

1 mole C = 6.02 x 1023 C atoms

1 mole Na = 6.02 x 1023 Na atoms

1 mole Au = 6.02 x 1023 Au atoms3

Samples of 1 Mole Quantities1 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

Copyright © 2008 Pearson Prentice Hall, Inc.

Chapter 3/33

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.

1 mole = 12.01 gC:

6.022 x 1023 molecules = 12.01 g

1 mole = 1.08 g

6.022 x 1023 molecules = 1.08 g

H

Avogadro’s NumberAvogadro’s number, 6.02 x 1023, can be written

as anequality 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

Learning Check1. The number of atoms in 2.0 g of Al atoms is

2. The number of moles of S in 1.8 x 1024

atoms of S

10


Chapter 2/36

Same compositionVariable composition


Chapter 2/37

Variableproperties

Similarproperties

The same number of protons

Cannot be separated physically

Covalent Bond: Results when two atoms share several (usually two) electrons. Typically a nonmetal bonded to a nonmetal

Covalent Bond: Results when two atoms share several (usually two) electrons. Typically a nonmetal bonded to a nonmetal

Molecule: The unit of matter that results when two or more atoms are joined by covalent bonds.

Insert Figure 2.12 p56

Ionic Bond: A transfer of one or more electrons from one atom to another. A strong electrical attraction between charged particles. Typically a metal bonded to a nonmetal.

Ion: A charged particle

Cation: A positively charged particle. Metals tend to form cations.

Anion: A negatively charged particle. Nonmetals tend to form anions.

Na1+ + Cl1-Na + Cl

11 protons10 electrons




In the formation of sodium chloride, one electron is transferred from the sodium atom to the chlorine atom.

1+ 2+ 3+

Cation Charges for Typical Main-Group Ions

Ionic Compound: A neutral compound in which the total number of positive charges must equal the total number of negative charges.

aluminum sulfide: Al2S3Al3+ S2-

sodium chloride: NaClNa+ Cl-

magnesium oxide: MgOMg2+ O2-

Binary Ionic Compounds

Some transition metals form more than one cation

Use Roman numerals in parentheses to indicate the charge on metals that form more than one kind of cation.

lead(II) fluoride: PbF2Pb2+ F-

iron(III) oxide: Fe2O3Fe3+ O2-

tin(II) chloride: SnCl2Sn2+ Cl-

Binary Ionic Compounds

Naming Ionic Compounds

To name a compound that contains two elements,

• identify the cation and anion.

• name the cation first, followed by the name of the anion with an –ide ending.

48

ExampleWrite the formula and names of the following

compounds:

1) K+ and Br-

2) Ca2+ and O2-

3) Pb4+ and S2-

4) Cu2+ and F-

49

Because nonmetals often combine with one another in different proportions to form different compounds, numerical prefixes are usually included in the names of binary molecular compounds.

Binary Molecular Compounds


Chapter 2/51

N2F4

The second element listed is more anionlike and takes the name of the element with an “ide” modification to the ending.

The first element listed is more cationlike and takes the name of the element.

The prefix is added to the front of each to indicate the number of each atom.

dinitrogen tetrafluoride

Binary Molecular Compounds

dinitrogen tetroxideN2O4

carbon monoxideCO

Whenever the prefix for the first element is “mono,” drop it.

Whenever the prefix ends in “a” or “o” and the element name begins with a vowel, drop the “a” or “o” in the prefix.

carbon dioxideCO2

Learning CheckWrite the name of each covalent compound.

P2O5 _____________________

SiCl4 _____________________

CCl4 _____________________

N2O _____________________

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Polyatomic Ionic Compounds

sodium carbonate: Na2CO3Na+ CO32-

iron(II) hydroxide: Fe(OH)2Fe2+ OH-

magnesium carbonate: MgCO3Mg2+ CO32-

sodium hydroxide: NaOHNa+ OH-

Learning CheckWrite formula and name of the following

compounds:

A. Mg2+ and NO3-

B. Al3+ and BrO3-

C. Fe2+ and SO42-

D. Ba2+ and PO32-

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ExamplesWrite the correct formula for each.

A. aluminum nitrate

B. copper(II) nitrate

C. iron(III) hydrogen sulfite

D. tin(IV) hydroxide

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john e. mcmurry robert c. fay c h e m i s t r y sixth edition chapter 2 atoms, molecules, and ions

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