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Chapter 2 NOR AKMALAZURA JANI Chapter 2 NOR AKMALAZURA JANI

Chapter 2 NOR AKMALAZURA JANI - PowerPoint Presentation

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Chapter 2 NOR AKMALAZURA JANI - PPT Presentation

BASIC CHEMISTRY CHM 138 Daltons Atomic Theory 1808 1 Elements are composed of extremely small particles called atoms 2 All atoms of a given element are identical having the same size mass and chemical properties ID: 744635

mass mol formula atoms mol mass atoms formula number element amu chemical molar atomic mole compounds examples reagent empirical

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Slide1

Chapter 2

NOR AKMALAZURA JANI

BASIC CHEMISTRY

CHM 138Slide2

Dalton’s Atomic Theory (1808)

1.

Elements are composed of extremely small particles called

atoms

.

2. 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

.

3.

Compounds

are composed of

atoms of more than one element

. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an integer or a simple fraction.

4. A

chemical reaction

involves only the

separation, combination, or rearrangement of atoms

; it does not result in their creation or destruction.Slide3

8 X

2Y

16 X

8 Y

+

Law of Conservation of Mass

- Matter can be neither created nor destroyedSlide4

Element

- A substance that cannot be separated into simpler substances by chemical means.

AtomThe basic unit of an element that can enter into chemical combination

Proton

- The positively charged particles in the nucleus

Neutron

- Electrically neutral particles having a mass slightly greater than that of protons

Electron

- Negatively charged particles

THE STRUCTURE OF THE ATOMSlide5

THE STRUCTURE OF THE ATOM

electronSlide6

6

mass p

≈ mass n ≈ 1840 x mass e

-Slide7

Atomic number

(Z) = number of protons in nucleus

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

= atomic number (Z) + number of neutrons

X

A

Z

Mass Number

Atomic Number

Element Symbol

Atomic number, Mass number and IsotopesSlide8

Example:

Mass number (A) = 16Atomic number (Z) = 8 (indicating 8 protons in nucleus)

Number of neutrons = 16-8 = 8Number of electrons = 8 (when the element is neutral)

O

16

8

Element Symbol

Mass Number

Atomic NumberSlide9

Isotopes

are atoms of the same element (X) with

different numbers of neutrons in their nuclei or atoms that have the same atomic number but different mass number.

Examples:

H

1

1

H (D)

2

1

H (T)

3

1

U

235

92

U

238

92

1) Hydrogen

2) UraniumSlide10
Slide11

The Modern Periodic Table

Period

Group

Alkali Metal

Noble Gas

Halogen

Alkali Earth MetalSlide12

A

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

H

2

H

2

O

NH

3

CH

4

A

diatomic molecule

contains only two atoms

Examples: H

2

, N

2

, O

2

, Br

2

, HCl, CO

A

polyatomic molecule

contains more than two atoms

Examples: O

3

, H

2

O, NH

3

, CH

4

diatomic elements

MOLECULES AND IONSSlide13

An

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

Cation

– ion with a positive charge

- If a neutral atom loses one or more electrons it becomes a cation.

anion

– ion with a negative charge

- If a neutral atom gains one or more electrons it becomes an anion.

Na

11 protons

11 electrons

Na

+

11 protons

10 electrons

Cl

17 protons

17 electrons

Cl

-

17 protons

18 electronsSlide14

A

monatomic ion contains only one atom

A polyatomic ion contains more than one atom

Examples: Na

+

, Cl

-

, Ca

2+

, O

2-

, Al

3+, N3-

Examples: OH

-

, CN-, NH4

+

, NO

3

-Slide15

Common Ions Shown on the Periodic Table

metals tend to form cations

nonmetals tend to form anionsSlide16

1) How many protons and electrons are in ?

Al

27

13

3+

2) How many protons and electrons are in ?

Se

78

34

2-

Examples:

No. of protons = 13

Charge = 3+ (loss of 3 electrons)

No. of electrons = 13 – 3 = 10

No. of protons = 34

Charge = 2- (accept of 2 electrons)

No. of electrons = 34 + 2 = 36 Slide17

A

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

Allotrope: one of two or more distinct forms of an element. - Example: two allotropic forms of the element carbon which

are diamond and graphite.

An

empirical formula

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

H

2

O

H

2

O

Molecular formula

Empirical formula

C

6

H

12

O

6

CH

2

O

O

3

O

N

2

H

4

NH

2

CHEMICAL FORMULAS

A

structural formula

shows how atoms are bonded to one another in a moleculeSlide18

Formulas and ModelsSlide19
Slide20

ionic compounds consist of a combination of cations and an anions The formula is usually 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 Examples: NaCI (consists of equal numbers of Na

+

and Cl

-

)

Formula of Ionic CompoundsSlide21

The most reactive

metals

(green) and the most reactive nonmetals (blue) combine to form ionic compounds.Slide22

Method of Writing Chemical Formula for Ionic Compounds

1) Aluminium oxide (containing Al

3+

and O

2-

)

Al

3+

O

2-

Charge

3+

2-

Simplest ration of ion combined

2

3

Sum of charges is 2(+3) + 3(-2) = 0

So, 2 cation Al

3+

combined with 3 anion O

2-

to form aluminium oxide

Formula:

Al

2

O

3Slide23

Method of Writing Chemical Formula for Ionic Compounds

2) Ammonium carbonate (containing NH

4

+

and CO

3

2-

)

NH

4

+

CO

3

2-

Charge

1+

2-

Simplest ration of ion combined

2

1

Sum of charges is 2(+1) + 1(-2) = 0

So, 2 cation NH

4

+

combined with 1 anion CO

3

2-

to form ammonium carbonate

Formula:

(NH

4

)

2

CO

3Slide24

1) Elements:Refer to the periodic table

- Examples: i) Na = sodium ii) Si = silicon

CHEMICAL NOMENCLATURESlide25

2) Ionic Compounds

Often a metal (cation) + nonmetal (anion)Binary compounds (compounds formed from two elements) - first element named is the metal cation followed by the nonmetallic anion.

Anion (nonmetal), add “ide” to element name

Examples:

i) BaCl

2

= barium chlor

ide

ii)K

2

O = potassium ox

ide

iii) Mg(OH)2

= Magnesium hydroxideSlide26
Slide27
Slide28

Transition metal ionic compounds - older nomenclature system:

- ending “ous” cation with fewer positive charges - ending “ic” to the cation with more positive charges

- examples: Fe2+ ferrous ion Fe

3+

ferric ion

- indicate charge on metal with Roman numerals

i) FeCl

2

2 Cl

-

-2 so Fe is +2

iron(II) chloride

ii) FeCl

3

3 Cl

-

-3 so Fe is +3

iron(III) chloride

iii) Cr

2

S

3

3 S

-2

-6 so Cr is +3 (6/2)

chromium(III) sulfide

Examples: Slide29
Slide30
Slide31

3) Molecular compounds

- place the name of the first element in the formula first and second element is named by adding “-ide” to the root of element name- Nonmetals or nonmetals + metalloids

- Common names: H2O, NH

3

, CH

4

- Element furthest to the left in a period and closest to the bottom of a group on periodic table is placed first in formula

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

- Last element name ends in

“-

ide”Slide32

Guidelines in naming compounds with prefixes

The prefix ‘mono-’ maybe omitted for the first element.For oxides, the ending ‘a’ in the prefix is sometimes omitted.

- for example: N2O4

maybe called dinitrogen teroxide rather than dinitrogen teraoxide.Slide33

HI

hydrogen iodide

NF

3

nitrogen

tri

fluoride

SO

2

sulfur

di

oxide

N

2

Cl

4

di

nitrogen

tetra

chloride

NO

2

nitrogen

di

oxide

N

2

O

di

nitrogen

mo

noxide

Molecular CompoundsSlide34
Slide35
Slide36

36Slide37

An

acid can be defined as a substance that yields hydrogen ions (H+

) when dissolved in water.

For example: HCl gas and HCl in water

- Pure substance, hydrogen chloride

- Dissolved in water (H

3

O

+

and Cl

), hydrochloric acid

Anions whose names end in “-ide” form acids with a “hydro-” prefix and an “-ic” ending.

4) Acids and basesSlide38

An

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

i) HNO3

nitric acid

ii) H

2

CO

3

carbonic acid

iii) H

3

PO4

phosphoric acidiv) HCIO3 chloric acid

v) H2SO4

sulfuric acidvi) HIO3

iodic acidvii)HBrO3

bromic acid

Examples: Slide39

39

Naming Oxoacids and OxoanionsSlide40

The rules for naming

oxoanions, anions of oxoacids, are as follows:

1. When all the H ions are removed from the “-ic” acid, the anion’s name ends with

“-ate”.

2. When all the H ions are removed from the

“-ous” acid

, the anion’s name ends with

“-ite.”

3. The names of anions in which

one or more but not all the hydrogen ions

have been removed must indicate the number of H ions present.

For example:

H

3PO4

Phosphoric acidH

2PO

4

-

dihydrogen phosphate

HPO

4

2-

hydrogen phosphate

PO

4

3-

phosphateSlide41
Slide42
Slide43

A

base can be defined as a substance that yields hydroxide ions (OH-

) when dissolved in water. Examples:

NaOH

sodium hydroxide

KOH

potassium hydroxide

Ba(OH)

2

barium hydroxideSlide44

Hydrates

are compounds that have a specific number of water molecules attached to them.

Examples:

i) BaCl

2

•2H

2

O

barium chloride

dihydrate

ii) LiCl

•H

2

O

iii) MgSO

4

•7H

2

O

iv) Sr(NO

3

)

2

•4H

2

O

lithium chloride

monohydrate

magnesium sulfate

heptahydrate

strontium nitrate

tetrahydrate

CuSO

4

•5H

2

O

CuSO

4

5) HydratesSlide45
Slide46

By definition:

1 atom 12C “weighs” 12 amu

On this scale:

1

H = 1.008 amu

16

O = 16.00 amu

Atomic mass

is the mass of an atom in atomic mass units (amu)

One atomic mass unit – a mass exactly equal to one-twelfth the mass of one carbon-12 atom.

ATOMIC MASSSlide47

The average atomic mass

is the weighted average of all of the naturally occurring isotopes of the element.

Average atomic mass of natural carbon

= (0.9890)(12.00000 amu)+(0.0110)(13.00335 amu)

= 12.01 amuSlide48

Naturally occurring lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

(7.42 x 6.015) + (92.58 x 7.016)

100

= 6.941 amu

Average atomic mass

of lithium

:

Example: Slide49

Average atomic mass (6.941)Slide50
Slide51

The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of

12C

1 mol =

N

A

= 6.0221367 x 10

23

Avogadro’s number (

N

A

)

Dozen = 12

Pair = 2

The Mole (mol):

A unit to count numbers of particles

AVOGADRO’S NUMBER AND THE MOLAR MASSSlide52

Molar mass

is the mass of 1 mole of in grams

atoms

1 mole

12

C atoms = 6.022 x 10

23

atoms = 12.00 g

1

12

C atom = 12.00 amu

1 mole

12

C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)Slide53

N

A = Avogadro’s number = 6.022 x 1023

atoms Mass of element (m)

No. of moles (n)

No. of atoms/molecules (N)

÷ molar mass (g/mol)

x N

A

÷ N

A

x molar mass (g/mol)Slide54

How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 10

23

atoms K

Example:

No. of moles = 0.551 g

39.10 g/mol

= 0.014 mol

No. of atoms = 0.014 mol x 6.022 x 10

23

atoms/mol

=

8.43 x 10

21

atoms KSlide55
Slide56
Slide57

Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

1S

32.07 amu

2O

+ 2 x 16.00 amu

SO

2

64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO

2

= 64.07 amu

1 mole SO

2

= 64.07 g SO

2

SO

2

MOLECULAR MASSSlide58
Slide59

How many H atoms are in 72.5 g of C

3H8

O ?

1 mol C

3

H

8

O = (3 x 12) + (8 x 1) + 16 = 60 g C

3

H

8

O

1 mol H = 6.022 x 1023 atoms H

= 5.82 x 10

24 atoms H

1 mol C

3

H

8

O molecules = 8 mol H atoms

72.5 g C

3

H

8

O

1 mol C

3

H

8

O

60 g C

3

H

8

O

x

8 mol H atoms

1 mol C

3

H

8

O

x

6.022 x 10

23

H atoms

1 mol H atoms

x

ExampleSlide60
Slide61
Slide62

Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound.

1Na

22.99 amu

1Cl

+ 35.45 amu

NaCl

58.44 amu

For any ionic compound

formula mass (amu) = molar mass (grams)

1 formula unit NaCl = 58.44 amu

1 mole NaCl = 58.44 g NaCl

Na

ClSlide63

What is the formula mass of Ca

3(PO4

)2 ?

1 formula unit of Ca

3

(PO

4

)

2

3 Ca

3 x 40.08

2 P

2 x 30.97

8 O

+ 8 x 16.00

310.18 amu

Example:Slide64

Percent composition

of an element in a compound =

n x molar mass of element

molar mass of compound

x 100%

n

is the number of moles of the element in

1 mole

of the compound

C

2

H

6

O

%C =

2

x (12.01 g)

46.07 g

x 100% = 52.14%

%H =

6

x (1.008 g)

46.07 g

x 100% = 13.13%

%O =

1

x (16.00 g)

46.07 g

x 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%Slide65
Slide66

Determination of empirical formula

Elements

K

Mn

O

Mass (g)

24.75

34.77

40.51

mol

24.75

g

39.10 g/mol

= 0.6330

34.77

g

54.94 g/mol

= 0.6329

40.51

g

16.00 g/mol

= 2.532

Simplest ratio

0.6330

0.6329

≈1

0.6329

0.6329

=1

2.532

0.6329

4

Empirical formula = KMnO

4

Determine the empirical formula of a compound that has the following percent composition by mass:

* Assume we have 100 g of the compound, then each percentage can be converted directly to grams.

K: 24.75%, Mn: 34.77%, O: 40.51%Slide67
Slide68

Determination of empirical formula

Elements

C

H

O

Mass (g)

40.92

4.58

54.50

mol

40.92

g

12.01 g/mol

= 3.407

4.58

g

1.008 g/mol

= 4.54

54.50

g

16.00 g/mol

= 3.406

Simplest ratio

3.407

3.406

≈1 x 3

= 3

4.54

3.406

=1.33 x 3

= 4

3.406

3.406

=

1 x 3

= 3

Empirical formula = C

3

H

4

O

3Slide69
Slide70

Determination of Molecular Formula

Elements

N

O

Mass (g)

1.52

3.47

mol

1.52

g

14.01 g/mol

= 0.108

3.47

g

16.00 g/mol

= 0.217

Simplest ratio

0.108

0.108

≈1

0.217

0.108

2

Empirical formula = NO

2

Determination of empirical formulaSlide71

Determination of molecular formula

1) Empirical molar mass = 14.01 g/mol + 2(16.0g/mol) = 46.01 g/ mol

molar mass compound between 90 g/mol-95 g/mol

2) Determine the ratio between the molar mass and empirical formula

Molar mass = 90 g/mol

≈ 2

Empirical molar mass 46.01 g/mol

Molecular formula = 2(NO

2

)

=

N

2

O

4

Actual molecular molar mass = 2(14.01 g/mol) + 4(16.00) = 92.02 g/molSlide72
Slide73

3 ways of representing the reaction of H

2

with O2 to form H2

O

A process in which one or more substances is changed into one or more new substances is a

chemical reaction

A

chemical equation

uses chemical symbols to show what happens during a chemical reaction

reactants

products

CHEMICAL REACTIONS AND CHEMICAL EQUATIONSSlide74

How to “Read” Chemical Equations

2 Mg + O

2 2 MgO

2 atoms Mg + 1 molecule O

2

makes 2 formula units MgO

2 moles Mg + 1 mole O

2

makes 2 moles MgO

48.6 grams Mg + 32.0 grams O

2

makes 80.6 g MgO

NOT

2 grams Mg + 1 gram O

2

makes 2 g MgOSlide75

Balancing Chemical Equations

Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C

2

H

6

+ O

2

CO

2

+ H

2

O

Change the numbers in front of the formulas (

coefficients

) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

2C

2

H

6

NOT

C

4

H

12Slide76

Start by balancing those elements that appear in only one reactant and one product.

C

2

H

6

+ O

2

CO

2

+ H

2

O

start with C or H but not O

2 carbon

on left

1 carbon

on right

multiply CO

2

by 2

C

2

H

6

+ O

2

2

CO

2

+ H

2

O

6 hydrogen

on left

2 hydrogen

on right

multiply H

2

O by 3

C

2

H

6

+ O

2

2

CO

2

+

3

H

2

O

Balancing Chemical EquationsSlide77

Balance those elements that appear in two or more reactants or products.

2 oxygen

on left

4 oxygen

(2x2)

C

2

H

6

+ O

2

2CO

2

+ 3H

2

O

+ 3 oxygen

(3x1)

multiply O

2

by

7

2

= 7 oxygen

on right

C

2

H

6

+ O

2

2CO

2

+ 3H

2

O

7

2

remove fraction

multiply both sides by 2

2C

2

H

6

+ 7O

2

4CO

2

+ 6H

2

O

Balancing Chemical EquationsSlide78

Check to make sure that you have the same number of each type of atom on both sides of the equation.

2C

2

H

6

+ 7O

2

4CO

2

+ 6H

2

O

Reactants

Products

4 C

12 H

14 O

4 C

12 H

14 O

4 C (2 x 2)

4 C

12 H (2 x 6)

12 H (6 x 2)

14 O (7 x 2)

14 O (4 x 2 + 6)

Balancing Chemical EquationsSlide79
Slide80

AMOUNTS OF REACTANTS AND PRODUCTS

Stoichiometry:

- comparison of coefficients in a balanced equation- The quantitative study of reactants and products in a chemical reactionSlide81

Write balanced chemical equation

Convert quantities of known substances into moles

Use coefficients in balanced equation to calculate the number of moles of the sought quantityConvert moles of sought quantity into desired unitsSlide82

Methanol burns in air according to the equation

2CH

3

OH + 3O

2

2CO

2

+ 4H

2

O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH

3

OH

moles CH

3

OH

moles H

2

O

grams H

2

O

molar mass

CH

3

OH

coefficients

chemical equation

molar mass

H

2

O

Example:

Moles of CH

3

OH = 209 g

32 g/mol

= 6.53 molSlide83

2) From the equation, 2 mol CH

3OH is used to give 4 mol H2O, if we have 6.53 mol CH

3OH, how many mole that H2O will produce?

2 mol CH

3

OH = 4 mol H

2

O

6.53 mol CH

3

OH = ? mol H

2O

= 4 mol H2O x 6.53 mol CH

3OH

2 mol CH3OH

= 13.06 mol H

2

O

3) Mass of H

2

O

= mol x molar mass H

2

O

= 13.06 mol x 18 g/mol

=

235.1 g

2CH

3

OH + 3O

2

2CO

2

+ 4H

2

OSlide84
Slide85
Slide86
Slide87

2NO + O

2

2NO2

NO is the limiting reagent

O

2

is the excess reagent

Reactant used up first in the reaction.

Excess reagents

: the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent

LIMITING REAGENTSlide88

LIMITING REAGENT

In one process, 124 g of Al are reacted with 601 g of Fe

2O

3

2Al + Fe

2

O

3

Al

2

O

3

+ 2Fe

Calculate the mass of Al

2

O

3

formed.

Mole of Al

= 124 g

27.0 g/mol

= 4.59 mol

2) Mole of Fe

2

O

3

= 601 g

160 g/mol

= 3.76 mol

Determination of limiting reagent and excess reagentSlide89

Divide moles of Al and Fe

2O

3 with their stoichiometric coefficientsAl ii) Fe

2

O

3

= 4.59 mol = 2.295 mol = 3.76 mol = 3.76 mol

2 1

The reagent that show the

smallest no. of mole

is a

limiting reagent

, while another reagent is a excess reagent.

So, Al is a limiting reagent

, while Fe2

O3

is a excess reagent

.Slide90

4)

From the equation, 2 mol Al is used to give 1 mol Al2

O3 , if we have 4.59 mol Al, how many mole that Al

2

O

3

will produce?

2 mol Al produce 1 mol Al

2

O

3

4.59 mol Al = 1mol Al2O

3 x 4.59 mol Al

2 mol Al

= 2.295 mol Al2

O

3

5) Mass of Al

2

O

3

= mol x molar mass Al

2

O

3

= 2.295 mol x 102.0 g/mol

=

234 gSlide91

Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.

Actual Yield

is the amount of product actually obtained from a reaction.

% Yield =

Actual Yield

Theoretical Yield

x 100%

REACTION YIELDSlide92
Slide93