Atoms, Molecules and Ions Chapter 2. Dalton’s Atomic Theory (1808) 1. Elements are composed of extremely small particles called atoms. All atoms of a

Atoms, Molecules and Ions

Chapter 2

Dalton’s Atomic Theory (1808)1. Elements are composed of extremely small

particles called atoms. All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of one element are different from the atoms of all other elements.

2. Compounds are composed of atoms of more than one element. The relative number of atoms of each element in a given compound is always the same.

3. Chemical reactions only involve the rearrangement of atoms. Atoms are not created or destroyed in chemical reactions. 2.1

2

2.1

8 X2Y16 X 8 Y+

2.1

J.J. Thomson, measured mass/charge of e-

(1906 Nobel Prize in Physics)2.2

Cathode Ray Tube

2.2

Laws Conservation of Mass Law of Definite Proportion- compounds

have a constant composition by mass. They react in specific ratios by mass. Multiple Proportions- When two elements

form more than one compound, The different masses of one element that combine with the same mass of the other element are in the ration of small whole numbers.

What?!

Water has 8 g of oxygen per 1g of hydrogen.

Hydrogen peroxide has 16 g of oxygen per 1g of hydrogen.

Comparing the grams of O per 1 g of H in each compound:

16/8 = 2/1 Small whole number ratios

Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 9.10

1 Analyze List the knowns and the unknown.

Apply the law of multiple proportions to the two compounds. For each compound, find the grams of carbon that combine with 1.00 g of oxygen. Then find the ratio of the masses of carbon in the two compounds. Confirm that the ratio is the lowest whole-number ratio.

KNOWNS

Compound A = 2.41 g C and 3.22 g OCompound B = 6.71 g C and 17.9 g O

UNKNOWN Mass ratio of C per g O in two compounds = ?


Sample Problem 9.10

Solve Solve for the unknown.

First, calculate grams of carbon per gram of oxygen in compound A.

2

2.41 g C

3.22 g O

0.748 g C

1.00 g O=


Sample Problem 9.10


Then, calculate grams of carbon per gram of oxygen in compound B.

2

6.71 g C

17.9 g O

0.375 g C

1.00 g O=


Sample Problem 9.10


Calculate the mass ratio to compare the two compounds.

2

To calculate the mass ratio, compare the masses of one element per gram of the other element in each compound.

0.748 g C

0.375 g O1.99

1= ≈ 2

1


Sample Problem 9.10

Evaluate Does this result make sense?

The ratio is a low whole-number ratio, as expected. For a given mass of oxygen, compound A contains twice the mass of carbon as compound B.

3

Two different compounds are formed by the elements carbon and oxygen. The first compound contains 42.9% by mass carbon and 57.1% by mass oxygen. The second compound contains 27.3% by mass carbon and 72.7% by mass oxygen. Show that the data are consistent with the

Law of Multiple Proportions.

• The Law of Multiple Proportions is the third postulate of Dalton's atomic theory. It states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole

numbers.

• Therefore, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole-number ratio. In 100 g of the first compound (100 is chosen to make calculations easier) there are 57.1 g O and 42.9 g C.

• The mass of O per gram C is:57.1 g O / 42.9 g

1.33 g O per g C • In the 100 g of the second compound, there are

72.7 g O and 27.3 g C. The mass of oxygen per gram of carbon is:72.7 g O / 27.3 g C = 2.66 g O per g C

• Dividing the mass O per g C of the second (larger value) compound:2.66 / 1.33 = 2

Gay-Lussac- under the same conditions of temperature and pressure, compounds always react in whole number ratios by volume.

Avagadro- interpreted that to mean at the same temperature and pressure, equal

volumes of gas contain the same number of particles

(called Avagadro’s hypothesis)

A Helpful Observation

e- charge = -1.60 x 10-19 C

Thomson’s charge/mass of e- = -1.76 x 108 C/g

e- mass = 9.10 x 10-28 g

Measured mass of e-

(1923 Nobel Prize in Physics)

2.2

(Uranium compound)2.2

0-1e

42He

2.2

The modern view of the atom was developed by Ernest Rutherford (1871-1937).

1. atoms positive charge is concentrated in the nucleus2. proton (p) has opposite (+) charge of electron (-)3. mass of p is 1840 x mass of e- (1.67 x 10-24 g)

a particle velocity ~ 1.4 x 107 m/s(~5% speed of light)

(1908 Nobel Prize in Chemistry)

2.2

atomic radius ~ 100 pm = 1 x 10-10 m

nuclear radius ~ 5 x 10-3 pm = 5 x 10-15 m

Rutherford’s Model of the Atom

2.2

“If the atom is the Houston Astrodome, then the nucleus is a marble on the 50-yard line.”

Chadwick’s Experiment (1932)

H atoms - 1 p; He atoms - 2 p

mass He/mass H should = 2

measured mass He/mass H = 4

a + 9Be 1n + 12C + energy

neutron (n) is neutral (charge = 0)

n mass ~ p mass = 1.67 x 10-24 g2.2

There must be something else in the nucleus

mass p = mass n = 1840 x mass e-

2.2

Atomic number (Z) = number of protons in nucleus

Mass number (A) = number of protons + number of neutrons

= atomic number (Z) + number of neutrons

Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei

XAZ

H11 H (D)2

1 H (T)31

U23592 U238

92

Mass Number

Atomic NumberElement Symbol

2.3

2.3

6 protons, 8 (14 - 6) neutrons, 6 electrons

6 protons, 5 (11 - 6) neutrons, 6 electrons

Do You Understand Isotopes?

2.3

How many protons, neutrons, and electrons are in C14

6 ?

How many protons, neutrons, and electrons are in C11

6 ?

Average atomic mass from isotopes


Atomic MassThe atomic mass of an element is a weighted average mass of the atoms in a naturally occurring sample of the element.• A weighted average mass reflects both

the mass and the relative abundance of the isotopes as they occur in nature.


Atomic Mass• To calculate the atomic mass of an element,

multiply the mass of each isotope by its natural abundance, expressed as a decimal, and then add the products.


Atomic MassCarbon has two stable isotopes: carbon-12, which has a natural abundance of 98.89 percent, and carbon-13, which has a natural abundance of 1.11 percent.• The mass of carbon-12 is 12.000 amu; the mass

of carbon-13 is 13.003 amu.

• The atomic mass of carbon is calculated as follows:

Atomic mass of carbon = (12.000 amu x 0.9889) + 13.003 amu x 0.0111)

= (11.867 amu) + (0.144 amu)

= 12.011 amu


•Calculating Atomic Mass

•Element X has two naturally occurring isotopes. The isotope with a mass of 10.012 amu (10X) has a relative abundance of 19.91 percent. The isotope with a mass of 11.009 amu (11X) has a relative abundance of 80.09 percent. Calculate the atomic mass of element X.

Sample Problem 4.5


Sample Problem 4.5

The mass each isotope contributes to the element’s atomic mass can be calculated by multiplying the isotope’s mass by its relative abundance. The atomic mass of the element is the sum of these products.

Analyze List the knowns and the unknown.1

atomic mass of X = ?KNOWNS UNKNOWN• Isotope 10X:

mass = 10.012 amurelative abundance = 19.91% = 0.1991

• Isotope 11X:mass = 11.009 amurelative abundance = 80.09% = 0.8009


•Use the atomic mass and the decimal form of the percent abundance to find the mass contributed by each isotope.

• for 10X: 10.012 amu x 0.1991 = 1.993 amu• for 11X: 11.009 amu x 0.8009 = 8.817 amu

Sample Problem 4.5

Calculate Solve for the unknowns.2


•Add the atomic mass contributions for all the isotopes.

Sample Problem 4.5

Calculate Solve for the unknowns.2

For element X, atomic mass = 1.953 amu + 8.817 amu= 10.810 amu


Sample Problem 4.5

Evaluate Does the result make sense?3

The calculated value is closer to the mass of the more abundant isotope, as would be expected.

Period

Group

Alkali M

etal

Noble G

as

Halogen

Alkali E

arth Metal

2.4

A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical bonds

H2 H2O NH3 CH4

A diatomic molecule contains only two atoms

H2, N2, O2, Br2, HCl, CO

A polyatomic molecule contains more than two atoms

O3, H2O, NH3, CH4

2.5

ELEMENTS THAT EXIST AS DIATOMIC MOLECULES

Remember:

BrINClHOFThese elements only exist as

PAIRS. Note that when they

combine to make compounds, they

are no longer elements so they are no longer in

pairs!P: 1 or 4 S: 1 or 8

An ion is an atom, or group of atoms, that has a net positive or negative charge.

cation – ion with a positive chargeIf a neutral atom loses one or more electronsit becomes a cation.

anion – ion with a negative chargeIf a neutral atom gains one or more electronsit becomes an anion.

Na 11 protons11 electrons Na+ 11 protons

10 electrons

Cl 17 protons17 electrons Cl-

17 protons18 electrons

2.5

Forming Cations & Anions

A CATION forms when an atom loses one or more electrons.

An ANION forms when an atom gains one or more electrons

Mg --> Mg2+ + 2 e- F + e- --> F-

A monatomic ion contains only one atom

A polyatomic ion contains more than one atom

2.5

Na+, Cl-, Ca2+, O2-, Al3+, N3-

OH-, CN-, NH4+, NO3

-

13 protons, 10 (13 – 3) electrons

34 protons, 36 (34 + 2) electrons

Do You Understand Ions?

2.5

How many protons and electrons are in ?Al2713

3+

How many protons and electrons are in ?Se7834

2-

2.5

2.6

A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance

An empirical formula shows the simplest whole-number ratio of the atoms in a substance

H2OH2O

molecular empirical

C6H12O6 CH2O

O3 O

N2H4 NH2

2.6

ionic compounds consist of a combination of cation(s) and an anion(s)• the formula is always the same as the empirical formula

• the sum of the charges on the cation(s) and anion(s) in each formula unit must equal zero

The ionic compound NaCl

2.6

Formula of Ionic Compounds

Al2O3

2.6

2 x +3 = +6 3 x -2 = -6

Al3+ O2-

CaBr2

1 x +2 = +2 2 x -1 = -2

Ca2+ Br-

Na2CO3

1 x +2 = +2 1 x -2 = -2

Na+ CO32-

2.6

2.7

Examples of Older Names of Cations formed from Transition Metals

From Zumdahl

Chemical Nomenclature• Ionic Compounds

– often a metal + nonmetal– anion (nonmetal), add “ide” to element name

BaCl2 barium chloride

K2O potassium oxide

Mg(OH)2 magnesium hydroxide

KNO3 potassium nitrate

2.7

• Transition metal ionic compounds– indicate charge on metal with Roman numerals

FeCl2 2 Cl- -2 so Fe is +2 iron(II) chloride

FeCl3 3 Cl- -3 so Fe is +3 iron(III) chloride

Cr2S3 3 S-2 -6 so Cr is +3 (6/2) chromium(III) sulfide

2.7

• Molecular compounds• nonmetals or nonmetals + metalloids• common names

• H2O, NH3, CH4, C60

• element further left in periodic table is 1st

• element closest to bottom of group is 1st

• if more than one compound can be formed from the same elements, use prefixes to indicate number of each kind of atom

• last element ends in ide

2.7

HI hydrogen iodide

NF3 nitrogen trifluoride

SO2 sulfur dioxide

N2Cl4 dinitrogen tetrachloride

NO2 nitrogen dioxide

N2O dinitrogen monoxide

Molecular Compounds

2.7

TOXIC!

Laughing Gas

2.7

An acid can be defined as a substance that yields hydrogen ions (H+) when dissolved in water.

HCl• Pure substance, hydrogen chloride• Dissolved in water (H+ Cl-), hydrochloric acid

An oxoacid is an acid that contains hydrogen, oxygen, and another element.

HNO3 nitric acid

H2CO3 carbonic acid

H2SO4 sulfuric acid

2.7HNO3

2.7

2.7

2.7

A base can be defined as a substance that yields hydroxide ions (OH-) when dissolved in water.

NaOH sodium hydroxide

KOH potassium hydroxide

Ba(OH)2 barium hydroxide

2.7

2.7

Mixed Practice

1. Dinitrogen monoxide

2. Potassium sulfide

3. Copper (II) nitrate

4. Dichlorine heptoxide

5. Chromium (III) sulfate

6. Ferric sulfite

7. Calcium oxide

8. Barium carbonate

9. Iodine monochloride

1. N2O

2. K2S

3. Cu(NO3)2

4. Cl2O7

5. Cr2(SO4)3

6. Fe2(SO3)3

7. CaO

8. BaCO3

9. ICl

Mixed Practice

1. BaI2

2. P4S3

3. Ca(OH)2

4. FeCO3

5. Na2Cr2O7

6. I2O5

7. Cu(ClO4)2

8. CS2

9. B2Cl4

1. Barium iodide

2. Tetraphosphorus trisulfide

3. Calcium hydroxide

4. Iron (II) carbonate

5. Sodium dichromate

6. Diiodine pentoxide

7. Cupric perchlorate

8. Carbon disulfide

9. Diboron tetrachloride

65

Organic chemistry is the branch of chemistry that deals with carbon compounds

C

H

H

H OH C

H

H

H NH2 C

H

H

H C OH

O

methanol methylamine acetic acid

Functional Groups

66

Documents

Atoms, Molecules and Ions Chapter 2. Dalton’s Atomic Theory (1808) 1. Elements are composed of extremely small particles called atoms. All atoms of a