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
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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) UraniumSlide10Slide11
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 ModelsSlide19Slide20
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 hydroxideSlide26Slide27Slide28
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: Slide29Slide30Slide31
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 CompoundsSlide34Slide35Slide36
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-
phosphateSlide41Slide42Slide43
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) HydratesSlide45Slide46
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)Slide50Slide51
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 KSlide55Slide56Slide57
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 MASSSlide58Slide59
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
ExampleSlide60Slide61Slide62
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%Slide65Slide66
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%Slide67Slide68
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
3Slide69Slide70
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/molSlide72Slide73
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 EquationsSlide79Slide80
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
OSlide84Slide85Slide86Slide87
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 YIELDSlide92Slide93