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The Components of Matter

Chapter 2

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Chapter 2: The Components of Matter

2.1 Elements, Compounds, and Mixtures: An Atomic Overview

2.2 The Observations That Led to an Atomic View of Matter

2.3 Dalton’s Atomic Theory

2.4 The Observations That Led to the Nuclear Atom Model

2.5 The Atomic Theory Today

2.6 Elements: A First Look at the Periodic Table

2.7 Compounds: Introduction to Bonding

2.8 Compounds: Formulas, Names, and Masses

2.9 Mixtures: Classification and Separation

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Definitions for Components of Matter

Element - the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical orchemical means.

Molecule - a structure that consists of two or more atoms that are chemically bound together and thus behaves as an independent unit.

Figure 2.1Figure 2.1

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Compound - a substance composed of two or more elements which are chemically combined.

Mixture - a group of two or more elements and/or compounds that are physically intermingled.

Definitions for Components of Matter

Figure 2.1

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Figure 2.2 The law of mass conservation:mass remains constant during a chemical reaction.

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The total mass of substances does not change during a chemical reaction.

reactant 1 + reactant 2 product

total mass total mass=

calcium oxide + carbon dioxide calcium carbonate

CaO + CO2 CaCO3

56.08g + 44.00g 100.08g

Law of Mass Conservation:

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No matter the source, a particular compound is composed of the same elements in the same parts (fractions) by mass.

Calcium carbonate Calcium carbonate

Analysis by MassAnalysis by Mass(grams/20.0g)(grams/20.0g)

Mass FractionMass Fraction(parts/1.00 part)(parts/1.00 part)

Percent by MassPercent by Mass(parts/100 parts)(parts/100 parts)

8.0 g calcium8.0 g calcium2.4 g carbon2.4 g carbon9.6 g oxygen 9.6 g oxygen

20.0 g20.0 g

40% calcium40% calcium12% carbon12% carbon48% oxygen 48% oxygen

100% by mass100% by mass

0.40 calcium0.40 calcium0.12 carbon0.12 carbon0.48 oxygen 0.48 oxygen

1.00 part by mass1.00 part by mass

Law of Definite (or Constant) Composition:

Figure 2.3

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Sample Problem 2.1

PROBLEM: What mass of calcium may be obtained from 85.6 g CaCO3?

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If elements A and B react to form two compounds, the different masses of B that combine with a fixed mass of A can be expressedas a ratio of small whole numbers.

Example: Carbon Oxides A & BCarbon Oxide I : 57.1% oxygen and 42.9% carbonCarbon Oxide II : 72.7% oxygen and 27.3% carbon

Assume that you have 100g of each compound. In 100 g of each compound: g O = 57.1 g for oxide I & 72.7 g for oxide II

g C = 42.9 g for oxide I & 27.3 g for oxide II

g Og C =

57.142.9 = 1.33

=g Og C

72.727.3 = 2.66

2.66 g O/g C in II

1.33 g O/g C in I

2

1=

Law of Multiple Proportions:

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DaltonDalton’’s Atomic Theorys Atomic Theory

1. All matter consists of atoms.

2. Atoms of one element cannot be converted into atoms of another element.

3. Atoms of an element are identical in mass and other properties and are different from atoms of any other element.

4. Compounds result from the chemical combination ofa specific ratio of atoms of different elements.

The Postulates

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DaltonDalton’’s Atomic Theorys Atomic Theoryexplains the mass laws

Definite composition

Atoms are combined in compounds in Atoms are combined in compounds in specific ratiosspecific ratiosand each atom has a specific mass.and each atom has a specific mass.

So each element has a fixed fraction of the total mass So each element has a fixed fraction of the total mass in a compound.in a compound.

postulate 3

postulate 4

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DaltonDalton’’s Atomic Theorys Atomic Theoryexplains the mass laws

Multiple proportions

Atoms of an element have the same massAtoms of an element have the same mass

and atoms are indivisible.and atoms are indivisible.

So when different numbers of atoms of elements So when different numbers of atoms of elements combine, they must do so in ratios of small, whole combine, they must do so in ratios of small, whole numbers.numbers.

postulate 3postulate 1

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Figure 2.5

Experiments to determine the properties of cathode rays.

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Experiments to Determine the Properties of Cathode Rays

OBSERVATION

1. Ray bends in magnetic field.2. Ray bends towards positive plate in electric field.

CONCLUSION

consists of charged particles

consists of negative particles3. Ray is identical for any cathode.

particles found in all matter

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Figure 2.6 Millikan’s oil-drop experiment for measuring an electron’s charge.

(1909)(1909)

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Millikan used his findings to also calculate the mass of an electron.

mass of electron =mass

chargeX charge

= (-5.686x10-12 kg/C) X (-1.602x10-19C)

determined by J.J. Thomson and others

= 9.109x10-31kg = 9.109x10-28g

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Figure 2.7 Rutherford’s α-scattering experiment and discovery of the atomic nucleus.

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Figure 2.8 General features of the atom today.•The atom is an electrically neutral, spherical entity composed of a positively charged central nucleus surrounded by one or more negatively charge electrons.•The atomic nucleus consists of protons and neutrons.

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Properties of the Three Key Subatomic Particles

Charge MassRelative

1+

0

1-

Absolute(C)*

+1.60218x10-19

0

-1.60218x10-19

Relative(amu)†

1.00727

1.00866

0.00054858

Absolute(g)

1.67262x10-24

1.67493x10-24

9.10939x10-28

Location in the Atom

Nucleus

Outside Nucleus

Nucleus

Name(Symbol)

Electron (e-)

Neutron (n0)

Proton (p+)

Table 2.2

* The coulomb (C) is the SI unit of charge.

† The atomic mass unit (amu) equals 1.66054x10-24 g.

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Figure 2.9

Atomic Symbols, Isotopes, Numbers

X = Atomic symbol of the element

A = mass number; A = Z + N

Isotope = atoms of an element with the same number of protons, but a different number of neutrons

AZ

Z = atomic number (the number of protons in the nucleus)

N = number of neutrons in the nucleus

X The Symbol of the Atom or Isotope

See Laboratory Tools

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Sample Problem 2.2

PROBLEM: How many protons, neutrons,and electrons are in an atom of 197Au?

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• Mass of an atom measured relative to the mass of an atomic standard: carbon-12 atom.

• 1 atomic mass unit (amu) = 1/12 the mass of a carbon-12 atom = 1 dalton (Da).– So 1H atom has mass 1.008 amu, 4He has mass 4.003 amu.

• Relative masses and isotopic abundance can be measured by a mass spectrometer.

Atomic mass unit.Atomic mass unit.

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Formation of a Positively Charged Neon Particle in a Mass Spectrometer

Figure B2.1

Tools of the Laboratory

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Figure B2.2 The Mass Spectrometer and Its DataThe Mass Spectrometer and Its Data

Tools of the Laboratory

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Sample Problem 2.3

PROBLEM: What is the atomic mass of the element neon?

Sample Problem 2.4

PROBLEM: Naturally occurring chlorine is 75.78 % 35Cl, which has an atomic mass 34.969 amu, and 24.22 % 37Cl, which has atomic mass of 36.966 amu. Calculate the average atomic mass of chlorine.

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Sample Problem 2.5

PROBLEM: Silver(Ag: Z = 47) has 46 known isotopes, but only two occur naturally, 107Ag and 109Ag. Given the following mass spectrometric data, and the fact that silver has an average atomic mass of 107.87 amu, what are the abundances of these two isotopes?

Isotope Mass(amu) Abundance(%)

107Ag

109Ag

106.90509

108.90476

?

?

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The Modern Reassessment of the Atomic Theory

1. All matter is composed of atoms. The atom is the smallest body thatretains the unique identity of the element.

2. Atoms of one element cannot be converted into atoms of another element in a chemical reaction. Elements can only be converted into other elements in nuclear reactions.

3. All atoms of an element have the same number of protons and electrons, which determines the chemical behavior of the element.Isotopes of an element differ in the number of neutrons, and thus in mass number. A sample of the element is treated as though itsatoms have an average mass.

4. Compounds are formed by the chemical combination of two or more elements in specific ratios.

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Figure 2.10 The modern periodic table.The modern periodic table.

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Figure 2.12

The formation of an ionic compound.The formation of an ionic compound.

Transferring electrons from the atoms of one element to those of another results in an ionic compound.

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Figure 2.13 Factors that influence the strength of ionic bonding.

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Figure 2.14 The relationship between ions formed and the nearest noble gas.

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Sample Problem 2.6 Predicting the Ion and Element Forms

PROBLEM: What monatomic ions do the following elements form?

(a) Oxygen (Z = 8) (b) Barium (Z = 56) (c) Bromine (Z = 35)

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Formation of a covalent bond between two H atoms.Figure 2.15

Covalent bonds form when elements share electrons, which usuallyoccurs between nonmetals.

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Figure 2.16 Elements that occur as molecules.

I2

Br2Se8

Cl2S8P4

F2O2N2

H2

(18)(17)(16)(15)(14)(13)(2)(1)

8A7A6A5A4A3A2A1A

diatomic molecules tetratomic molecules octatomic molecules

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A polyatomic ion

Figure 2.17

Elements that are polyatomic.

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Sample Problem 2.7 Covalent or Ionic?

PROBLEM: Which of the following compounds would you expect to be ionic?

(a) N2O (b) Na2O (c) CaCl2

(d) SF4

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Types of Chemical Formulas

An empirical formula indicates the relative number of atoms of each element in the compound. It is the simplest type of formula.

A molecular formula shows the actual number of atoms of each element in a molecule of the compound.

A structural formula shows the number of atoms and the bonds between them, that is, the relative placement and connections of atoms in the molecule.

A chemical formula is comprised of element symbols and numerical subscripts that show the type and number of each atom present in the smallest unit of the substance.

The empirical formula for hydrogen peroxide is HO.

The molecular formula for hydrogen peroxide is H2O2.

The structural formula for hydrogen peroxide is H-O-O-H.

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Figure 2.19Figure 2.19 Some common monatomic ions of the elements.Some common monatomic ions of the elements.

Can you see any patterns?

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Common Monoatomic IonsTable 2.3

H- hydride

Na+ sodium

H+ hydrogen

Li+ lithium fluorideF-

Cs+ cesiumK+ potassium

Ag+ silver

chlorideCl-

bromideBr-

iodideI-

Mg2+ magnesium

Sr2+ strontiumCa2+ calcium

Zn2+ zincBa2+ barium

Cd2+ cadmium

Al3+ aluminum

+1

+2

+3

CationsCharge Formula Name

AnionsCharge Formula Name

-1

-2

-3

oxideO2-

sulfideS2-

nitrideN3-

Common ions are in blue.

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Naming binary ionic compounds

The name of the cation is the same as the name of the metal.

Many metal names end in -ium.

The name of the anion takes the root of the nonmetal name and adds the suffix -ide.

Calcium and bromine form calcium bromide.

The name of the cation is written first, followed by that of the anion.

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Metals With Several Oxidation States

Element

Table 2.4 (partial)

Ion Formula Systematic Name Common Name

Copper Cu+1

Cu+2

copper(I)copper(II)

cuprouscupric

CobaltCo+2

Co+3

cobalt(II)cobalt (III)

ferrousIron

Fe+2 iron(II)Fe+3 iron(III) ferric

ManganeseMn+2 manganese(II)Mn+3 manganese(III)

TinSn+2 tin(II)Sn+4 tin(IV)

stannousstannic

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Sample Problem 2.8 Naming Binary Ionic Compounds

PROBLEM: Name the ionic compound formed from the following pairs of elements:

(a) sodium and phosphorus (b) bromine and zinc

(c) sulfur and germanium

Sample Problem 2.9 Determining Names and Formulas of Ionic Compounds of Elements That Form More Than One Ion

PROBLEM: Give the systematic names for the formulas or the formulas for the names of the following compounds:

(a) lead(IV) chloride (b) WF6

(c) cuprous oxide (d) Fe2S3

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Some Common Polyatomic Ions

Formula

CationsNH4

+

Common Anions

H3O+

Formula

ammonium hydronium

Name Name

CH3COO- acetate

CN- cyanide

OH- hydroxide

ClO3- chlorate

NO2- nitrite

NO3- nitrate

MnO4- permanganate

CO3-2 carbonate

CrO4-2 chromate

Cr2O7-2 dichromate

O2-2 oxide

SO4-2 sulfate

PO4-3 phosphate

Table 2.5 (partial)

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Naming oxoanionsPrefixes Root Suffixes Examples

rootper ate ClO4- perchlorate

ateroot ClO3- chlorate

iteroot ClO2- chlorite

itehypo root ClO- hypochlorite

No.

of O

ato

ms

Figure 2.20

Numerical Prefixes for Hydrates and Binary Covalent Compounds

Number Prefix Number Prefix Number Prefix

1 mono

2 di

3 tri

4 tetra

5 penta

6 hexa

7 hepta

8 octa

9 nona

10 deca

Table 2.6

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Sample Problem 2.10 Determining Names and Formulas of Ionic Compounds Containing Polyatomic Ions

PROBLEM: Give the systematic names or the formula or the formulas for thenames of the following compounds:

(a) Cr(BrO3)3 (b) potassium nitrate (c) MgCO3 6H2O

Sample Problem 2.11 Determining Names and Formulas of Ionic Compounds Containing Polyatomic Ions

PROBLEM: Write the correct formulas for the following combination of cations and anions. Give the correct names of the compounds.

(a) Mg2+, Cl– (b) Na+, SO42– (c) Al3+, SO4

2–

(d) Ti4+, O2–

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Naming Acids

1) Binary acids solutions form when certain gaseous compounds dissolve in water. For example, when gaseous hydrogen chloride(HCl) dissolves in water, it forms a solution called hydrochloric acid. Prefix hydro- + anion nonmetal root + suffix -ic + the word acid - hydrochloric acid

2) Oxoacid names are similar to those of the oxoanions, except for two suffix changes:Anion “-ate” suffix becomes an “-ic” suffix in the acid. Anion “-ite”suffix becomes an “-ous” suffix in the acid.The oxoanion prefixes “hypo-” and “per-” are retained. Thus, BrO4

-

is perbromate, and HBrO4 is perbromic acid; IO2- is iodite, and

HIO2 is iodous acid.

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Sample Problem 2.12 Determining Names of Anions and Acids

PROBLEM: Name the following acids

(a) HCN (b) HBrO (c) HNO3 (d) H2SO3 (e) HClO4

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Naming Binary Molecular Compounds

1) Element name furthest to left of Periodic Table written first (usually – unless compound contains oxygen).

2) When both elements from same group of the PT, lower one named first.

3) Name of second element usually given the ending –ide.4) Use Greek prefixes to indicate atom numbers. c.f. ionic names!

Sample Problem 2.13 Determining Names of Binary Molecular Compounds

PROBLEM: Name the following compounds

(a) Cl2O (b) N2O4 (c) NF3 (d) P4S10

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Molecular Mass (Formula Weight).Molecular Mass (Formula Weight).

Molecular mass = Σ(atomic masses)

Sample Problem 2.14 Calculating the Molecular Mass of a Compound

PROBLEM: Using the data in the periodic table, calculate the molecular (or formula) mass of the following compounds:

(a) Sulfuric acid (b) Ferric nitrate (c) Acetic acid

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Allowed to react chemically therefore cannot be separated by physical means.

Figure 2.21 The distinction between mixtures and compounds.

S

Fe

Physically mixed therefore can be separated by physical means; in this case by a magnet.

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Mixtures

Heterogeneous mixtures : has one or more visible boundaries between the components.

Homogeneous mixtures : has no visible boundaries because the components are mixed as individual atoms, ions, and molecules.

Solutions : A homogeneous mixture is also called a solution. Solutions in water are called aqueous solutions, and are very important in chemistry. Although we normally think of solutions as liquids, they can exist in all three physical states.

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Filtration : Separates components of a mixture based upon differences in particle size. Normally separating a precipitate from a solution, or particles from an air stream.

Crystallization : Separation is based upon differences in solubility of components in a mixture.

Distillation : separation is based upon differences in volatility.

Extraction : Separation is based upon differences in solubility in different solvents (major material).

Chromatography : Separation is based upon differences in solubility in a solvent versus a stationary phase.

Basic Separation Techniques

Tools of the Laboratory

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Basic Separation Techniques

Figure B2.3 Filtration Figure B2.4 Crystallization

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Figure B2.5

Tools of the Laboratory

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Figure B2.6

Tools of the Laboratory

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Procedure for Column Chromatography

Figure B2.7

Tools of the Laboratory

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Figure B2.8 Separation by Gas - Liquid Chromatography

Tools of the Laboratory

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