dr. s. m. condren atoms, molecules & ions chapter 2

Dr. S. M. Condren

Atoms, Molecules

& Ions

Chapter 2

Dr. S. M. Condren

Quantum Corral

http://www.almaden.ibm.com/vis/stm/corral.html

Dr. S. M. Condren

Scanning Tunneling Microscope

Dr. S. M. Condren

Scanning Tunneling Microscope

Dr. S. M. Condren

Scanning Tunneling Microscope

Dr. S. M. Condren

http://www.cbu.edu/~mcondren/SeeAtoms.htm

Dr. S. M. Condren

http

://m

rsec

.wis

c.ed

u/

http://mrsec.wisc.edu/

Developed in collaboration with theInstitute for Chemical Education and the

Magnetic Microscopy CenterUniversity of Minnesota

http://www.physics.umn.edu/groups/mmc/

Dr. S. M. Condren

Pull Probe StripProbe

Sample

Pull Probe Strip

http://w

ww

.nsf.g

ov/m

ps/dm

r/mrse

c.htm

http://www.nsf.gov/mps/dmr/mrsec.htm

Dr. S. M. Condren

(a) (b)

North South

(c)

Which best represents the poles?

Dr. S. M. Condren

Atoms & MoleculesAtoms

• can exist alone or enter into chemical combination

• the smallest indivisible particle of an element

Molecules

• a combination of atoms that has its own characteristic set of properties

Dr. S. M. Condren

Law of Constant Composition

A chemical compound always contains the same elements in the same proportions by mass.

Dr. S. M. Condren

Law of Multiple Proportions

• the same elements can be combined to form different compounds by combining the elements in different proportions

Dr. S. M. Condren

Dalton’s Atomic Theory

Postulates

• proposed in 1803

• know at least 2 for first exam

Dr. S. M. Condren

Dalton’s Atomic Theory

Postulate 1

• An element is composed of tiny particles called atoms.

• All atoms of a given element show the same chemical properties.

Dr. S. M. Condren

Dalton’s Atomic Theory

Postulate 2• Atoms of different elements have different

properties.

Dr. S. M. Condren

Dalton’s Atomic Theory

Postulate 3

• Compounds are formed when atoms of two or more elements combine.

• In a given compound, the relative number of atoms of each kind are definite and constant.

Dr. S. M. Condren

Dalton’s Atomic Theory

Postulate 4

• In an ordinary chemical reaction, no atom of any element disappears or is changed into an atom of another element.

• Chemical reactions involve changing the way in which the atoms are joined together.

Dr. S. M. Condren

Radioactivity

Dr. S. M. Condren

Radioactivity

• Alpha – helium-4 nucleus

• Beta – high energy electron

• Gamma – energy resulting from transitions from one nuclear energy level to another

Dr. S. M. Condren

Alpha Radiation

• composed of 2 protons and 2 neutrons

• thus, helium-4 nucleus

• +2 charge

• mass of 4 amu

• creates element with atomic number 2 lower

• Ra226 Rn222 + He4()

Dr. S. M. Condren

Beta Radiation

• composed of a high energy electron which was ejected from the nucleus

• “neutron” converted to “proton”

• very little mass

• -1 charge

• creates element with atomic number 1 higher

• U239 Np239 + -1

Dr. S. M. Condren

Gamma Radiation

• nucleus has energy levels

• energy released from nucleus as the nucleus changes from higher to lower energy levels

• no mass

• no charge

• Ni60* Ni60 +

Dr. S. M. Condren

Cathode Ray Tube

Dr. S. M. Condren

Thompson’s Charge/Mass Ratio

Dr. S. M. Condren

Millikin’s Oil Drop

Dr. S. M. Condren

Rutherford’s Gold Foil

Dr. S. M. Condren

Rutherford’s Model of the Atom

Dr. S. M. Condren

Rutherford’s Model of the Atom

• atom is composed mainly of vacant space

• all the positive charge and most of the mass is in a small area called the nucleus

• electrons are in the electron cloud surrounding the nucleus

Dr. S. M. Condren

Structure of the Atom Composed of:

• protons

• neutrons

• electrons

Dr. S. M. Condren

Structure of the Atom

Composed of:• protons• neutrons• electrons

• protons– found in nucleus

– relative charge of +1

– relative mass of 1.0073 amu

Dr. S. M. Condren

Structure of the Atom

Composed of:• protons• neutrons• electrons

• neutrons– found in nucleus

– neutral charge

– relative mass of 1.0087 amu

Dr. S. M. Condren

Structure of the Atom

Composed of:• protons• neutrons• electrons

• electrons– found in electron cloud– relative charge of -1– relative mass of 0.00055 amu

Dr. S. M. Condren

Size of Nucleus

If the nucleus were1” in diameter,

the atom would be 1.5 miles in diameter.

Dr. S. M. Condren

Ions

• charged single atom

• charged cluster of atoms

Dr. S. M. Condren

Ions

• cations– positive ions

• anions– negative ions

• ionic compounds– combination of cations and anions– zero net charge

Dr. S. M. Condren

Atomic number, Z

• the number of protons in the nucleus

• the number of electrons in a neutral atom

• the integer on the periodic table for each element

Dr. S. M. Condren

Isotopes

• atoms of the same element which differ in the number of neutrons in the nucleus

• designated by mass number

Dr. S. M. Condren

Mass Number, A

• integer representing the approximate mass of an atom

• equal to the sum of the number of protons and neutrons in the nucleus

Dr. S. M. Condren

Masses of Atoms

Carbon-12 Scale

Dr. S. M. Condren

Isotopes of Hydrogen H-1, 1H, protium

• 1 proton and no neutrons in nucleus

• only isotope of any element containing no neutrons in the nucleus

• most common isotope of hydrogen

Dr. S. M. Condren

Isotopes of Hydrogen H-2 or D, 2H, deuterium

• 1 proton and 1 neutron in nucleus

Dr. S. M. Condren

Isotopes of Hydrogen H-3 or T, 3H, tritium

• 1 proton and 2 neutrons in nucleus

Dr. S. M. Condren

Isotopes of Oxygen

O-16

• 8 protons, 8 neutrons, & 8 electrons

O-17

• 8 protons, 9 neutrons, & 8 electrons

O-18

• 8 protons, 10 neutrons, & 8 electrons

Dr. S. M. Condren

The radioactive isotope 14C has how many neutrons?

6, 8, other

Dr. S. M. Condren

The identity of an element is determined by the number of which particle?

protons, neutrons, electrons

Dr. S. M. Condren

Mass Spectrometer

Dr. S. M. Condren

Mass Spectra of Neon

Dr. S. M. Condren

Measurement of Atomic Masses

Mass Spectrometer

a simulation is available at

http://www.colby.edu/chemistry/

OChem/DEMOS/MassSpec.html

Dr. S. M. Condren

Atomic Masses andIsotopic Abundances

natural atomic masses =

sum[(atomic mass of isotope)

*(fractional isotopic abundance)]

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?

let x = fraction Cl-35 y = fraction Cl-37

x + y = 1 y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

Thus:34.96885*x + 36.96590*y = 35.453

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35 y = fraction Cl-37x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.45334.96885*x + 36.96590*(1-x) = 35.453(34.96885 - 36.96590)x + 36.96590 =

35.453

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

- 1.99705x = - 1.5129

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

- 1.99705x = - 1.5129

1.99705x = 1.5129

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

- 1.99705x = - 1.5129

1.99705x = 1.5129

x = 0.7553 <=> 75.53% Cl-35

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

- 1.99705x = - 1.5129

1.99705x = 1.5129 x = 0.7553 <=> 75.53% Cl-35

y = 1 - x

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

- 1.99705x = - 1.5129

1.99705x = 1.5129 x = 0.7553 <=> 75.53% Cl-35

y = 1 - x = 1.0000 - 0.7553

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

- 1.99705x = - 1.5129

1.99705x = 1.5129 x = 0.7553 <=> 75.53% Cl-35

y = 1 - x = 1.0000 - 0.7553 = 0.2447

Dr. S. M. Condren

Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35 y = fraction Cl-37x + y = 1 <=> y = 1 - x(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.45334.96885*x + 36.96590*y = 35.45334.96885*x + 36.96590*(1-x) = 35.453(34.96885 - 36.96590)x + 36.96590 = 35.453(34.96885 - 36.96590)x = (35.453 - 36.96590)- 1.99705x = - 1.5129

1.99705x = 1.5129 x = 0.7553 <=> 75.53% Cl-35

y = 1 - x = 1.0000 - 0.7553 = 0.2447 24.47% Cl-37

Dr. S. M. Condren

Development of Periodic Table

Newlands - English

1864 - Law of Octaves - every 8th element has similar

properties

Dr. S. M. Condren

Development of Periodic Table

Dmitri Mendeleev - Russian

1869 - Periodic Law - allowed him to predict properties of

unknown elements

Dr. S. M. Condren

Mendeleev’s Periodic Table

the elements are arranged according to increasing atomic weights

Dr. S. M. Condren

Missing elements: 44, 68, 72, & 100 amu

Mendeleev’s Periodic Table

Dr. S. M. Condren

Properties of Ekasilicon

Dr. S. M. Condren

Modern Periodic TableMoseley, Henry Gwyn Jeffreys

1887–1915, English physicist.

Studied the relations among bright-line spectra of different elements.

Derived the ATOMIC NUMBERS from the frequencies of vibration of X-rays emitted by each element.

Moseley concluded that the atomic number is equal to the charge on the nucleus.

This work explained discrepancies in Mendeleev’s Periodic Law.

Dr. S. M. Condren

Modern Periodic Table

the elements are arranged according to increasing atomic numbers

Dr. S. M. Condren

I A II A III B IV B V B VI B VII B VIII B I B II B III A IV A V A VI A VII A VIII A1 1 2

1 H H He1.008 1.008 4.0026

3 4 5 6 7 8 9 10

2 Li Be B C N O F Ne6.939 9.0122 10.811 12.011 14.007 15.999 18.998 20.183

11 12 13 14 15 16 17 18

3 Na Mg Al Si P S Cl Ar22.99 24.312 26.982 28.086 30.974 32.064 35.453 39.948

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr39.102 40.08 44.956 47.89 50.942 51.996 54.938 55.847 58.932 58.71 63.54 65.37 69.72 72.59 74.922 78.96 79.909 83.8

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe85.468 87.62 88.906 91.224 92.906 95.94 * 98 101.07 102.91 106.42 107.9 112.41 114.82 118.71 121.75 127.61 126.9 131.29

55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

6 Cs Ba **La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.2 192.22 195.08 196.97 200.29 204.38 207.2 208.98 * 209 * 210 * 222

87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 118

7 Fr Ra ***Ac Rf Ha Sg Ns Hs Mt Uun Uuu Uub Uut Uuq Uup Uuh Uuo* 223 226.03 227.03 * 261 * 262 * 263 * 262 * 265 * 268 * 269 * 272 * 277 *284 *285 *288 *292 *294

Based on symbols used by ACS S.M.Condren 2007

58 59 60 61 62 63 64 65 66 67 68 69 70 71

* Designates that **Lanthanum Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Luall isotopes are Series 140.12 140.91 144.24 * 145 150.36 151.96 157.25 158.93 162.51 164.93 167.26 168.93 173.04 174.97

radioactive 90 91 92 93 94 95 96 97 98 99 100 101 102 103

*** Actinium Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Series 232.04 231.04 238.03 237.05 * 244 * 243 * 247 * 247 * 251 * 252 * 257 * 258 * 259 * 260

Periodic Table of theElements

Periodic Table of the ElementsPeriodic Table of the Elements

Dr. S. M. Condren

Organization of Periodic Table

• period - horizontal row

• group - vertical column

Dr. S. M. Condren

Family Names

Group IA alkali metals

Group IIA alkaline earth metals

Group VIIA halogens

Group VIIIAnoble gases

transition metals

inner transition metals

• lanthanum series rare earths

• actinium series trans-uranium series

Dr. S. M. Condren

Types of Elements

metals

nonmetals

metalloids - semimetals

Dr. S. M. Condren

Elements, Compounds, and Formulas

Elements

• can exist as single atoms or molecules

Compounds

• combination of two or more elements

• molecular formulas for molecular compounds

• empirical formulas for ionic compounds

Dr. S. M. Condren

Organic CompoundsOrganic Chemistry

• branch of chemistry in which carbon compounds and their reactions are studied.

• the chemistry of carbon-hydrogen compounds

Dr. S. M. Condren

Inorganic Compounds Inorganic Chemistry

• field of chemistry in which are studied the chemical reactions and properties of all the chemical elements and their compounds, with the exception of the hydrocarbons (compounds composed of carbon and hydrogen) and their derivatives.

Dr. S. M. Condren

Molecular and Structural Formulas

Dr. S. M. Condren

Bulk Substances

• mainly ionic compounds– empirical formulas– structural formulas

Dr. S. M. Condren

Models of Sodium Chloride

NaCl “table salt”

Dr. S. M. Condren

How many atoms are in the formula Al2(SO4)3?

3, 5, 17

Dr. S. M. Condren

Naming Binary Molecular Compounds

• For compounds composed of two non-metallic elements, the more metallic element is listed first.

• To designate the multiplicity of an element, Greek prefixes are used:mono => 1; di => 2; tri => 3; tetra => 4; penta => 5; hexa => 6; hepta => 7; octa => 8

Dr. S. M. Condren

Common CompoundsH2O

water

NH3

ammonia

N2Onitrous oxide

COcarbon

monoxide

CS2

carbon disulfide

SO3

sulfur trioxide

CCl4

carbon tetrachloride

PCl5

phosphorus pentachloride

SF6

sulfur hexafluoride

Dr. S. M. Condren

Alkanes - CnH2n+2

• methane - CH4

• ethane - C2H6

• propane - C3H8

• butanes - C4H10

• pentanes - C5H12

• hexanes - C6H14

• heptanes - C7H16

• octanes - C8H18

• nonanes - C9H20

• decanes - C10H22

Dr. S. M. Condren

Burning of Propane Gas

Dr. S. M. Condren

Butanes

Dr. S. M. Condren

Ionic Bonding

Characteristics of compounds with ionic bonding:

• non-volatile, thus high melting points

• solids do not conduct electricity, but melts (liquid state) do

• many, but not all, are water soluble

Dr. S. M. Condren

Ion Formation

Dr. S. M. Condren

ValanceCharge on Ions

• compounds have electrical neutrality

• metals form positive monatomic ions

• non-metals form negative monatomic ions

Dr. S. M. Condren

Valence of Metal Ions

Monatomic Ions

Group IA => +1

Group IIA => +2

Maximum positive valence

equals

Group A #

Dr. S. M. Condren

Valence of Non-Metal Ions

Monatomic Ions

Group VIA => -2

Group VIIA => -1

Maximum negative valence

equals

(8 - Group A #)

Dr. S. M. Condren

Charges of Some Important Ions

Dr. S. M. Condren

Polyatomic Ions

• more than one atom joined together

• have negative charge except for NH4+ and

its relatives

• negative charges range from -1 to -4

Dr. S. M. Condren

Polyatomic Ions

ammonium NH4+

perchlorate ClO41-

cyanide CN1-

hydroxide OH1-

nitrate NO31-

sulfate SO42-

carbonate CO32-

phosphate PO43-

Dr. S. M. Condren

Names of Ionic Compounds

1. Name the metal first.

If the metal has more than one oxidation state, the oxidation state is specified by Roman numerals in parentheses.

2. Then name the non-metal,

changing the ending of the non-metal to

-ide.

Dr. S. M. Condren

Nomenclature

NaCl

sodium chloride

Fe2O3

iron(III) oxide

N2O4

dinitrogen tetroxide

KI

potassium iodide

Mg3N2

magnesium nitride

SO3

sulfur trioxide

Dr. S. M. Condren

NomenclatureNH4NO3

ammonium nitrate

KClO4

potassium perchlorate

CaCO3

calcium carbonate

NaOH

sodium hydroxide

Dr. S. M. Condren

Nomenclature Drill

Available for PCs:– on your disk to use at home or in the

dorm– in the Chemistry Resource Center– off the web under Chapter 2, Links

http://www.cbu.edu/~mcondren/c115lkbk.html

Dr. S. M. Condren

How many moles of ions are there per mole of Al2(SO4)3?

2, 3, 5

Dr. S. M. Condren

Chemical Equation

• reactants

• products

• coefficients

reactants -----> products

Dr. S. M. Condren

Writing and BalancingChemical Equations

• Write a word equation.

• Convert word equation into formula equation.

• Balance the formula equation by the use of prefixes (coefficients) to balance the number of each type of atom on the reactant and product sides of the equation.

Dr. S. M. Condren

Example

Hydrogen gas reacts with oxygen gas to produce water.

Step 1.

hydrogen + oxygen -----> water

Step 2.

H2 + O2 -----> H2O

Step 3.

2 H2 + O2 -----> 2 H2O

Dr. S. M. Condren

Example

Iron(III) oxide reacts with carbon monoxide to produce the iron oxide (Fe3O4) and carbon dioxide.

iron(III) oxide + carbon monoxide -----> Fe3O4 + carbon dioxide

Fe2O3 + CO -----> Fe3O4 + CO2

3 Fe2O3 + CO -----> 2 Fe3O4 + CO2