1. BASIC CHEMISTRY
CHM 138
Chapter 2
NOR AKMALAZURA JANI

2. 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.

3. Law of Conservation of Mass
- Matter can be neither created nor destroyed
16 X
8 Y +
8 X2Y

4. THE STRUCTURE OF THE ATOM
Element
- A substance that cannot be separated into simpler substances by chemical means.
Atom
The 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

5. THE STRUCTURE OF THE ATOM
electron

6. mass p ≈ mass n ≈ 1840 x mass e-

7. Atomic number, Mass number and Isotopes
Atomic number (Z)
= number of protons in nucleus
Mass number (A)
= number of protons + number of neutrons
= atomic number (Z) + number of neutrons
Mass Number X A Z
Element Symbol
Atomic Number

8. O Example: Mass number (A) = 16
Element Symbol
Atomic Number 8
Mass number (A) = 16
Atomic number (Z) = 8 (indicating 8 protons in nucleus)
Number of neutrons = 16-8
= 8
Number of electrons = 8 (when the element is neutral)

9. 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:
1) Hydrogen H 1
H (D) 2
H (T) 3
2) Uranium U
235 92
238

11. The Modern Periodic Table
Alkali Earth Metal
Noble Gas
Halogen
Alkali Metal
Group
Period

12. MOLECULES AND IONS
A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical forces H2
H2O
NH3
CH4
A diatomic molecule contains only two atoms
diatomic elements
Examples: H2, N2, O2, Br2, HCl, CO
A polyatomic molecule contains more than two atoms
Examples: O3, H2O, NH3, CH4

13. 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. Na
11 protons
11 electrons
Na+
11 protons
10 electrons
anion – ion with a negative charge
- If a neutral atom gains one or more electrons it becomes an anion. Cl
17 protons
18 electrons
17 protons
17 electrons
Cl-

14. A monatomic ion contains only one atom
Examples: Na+, Cl-, Ca2+, O2-, Al3+, N3-
A polyatomic ion contains more than one atom
Examples: OH-, CN-, NH4+, NO3-

15. Common Ions Shown on the Periodic Table
metals tend to form cations
nonmetals tend to form anions

16. Examples: Al Se 1) How many protons and electrons are in ?3+ 27
1) How many protons and electrons are in ? Al 13
No. of protons = 13
Charge = 3+ (loss of 3 electrons)
No. of electrons = 13 – 3 = 10 Se 78 34 2-
2) How many protons and electrons are in ?
No. of protons = 34
Charge = 2- (accept of 2 electrons)
No. of electrons = = 36

17. CHEMICAL FORMULAS
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
Molecular formula
Empirical formula
H2O
H2O
C6H12O6
CH2O O3 O
NH2
N2H4
A structural formula shows how atoms are bonded to one another in a molecule

18. Formulas and Models

20. Formula of Ionic Compounds
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-)

21. The most reactive metals (green) and the most reactive nonmetals (blue) combine to form ionic compounds.

22. Method of Writing Chemical Formula for Ionic Compounds
1) Aluminium oxide (containing Al3+ and O2-)
Al3+
O2- 3+ 2-
Charge
Simplest ration of ion combined 2 3
So, 2 cation Al3+ combined with 3 anion O2- to form aluminium oxide
Sum of charges is 2(+3) + 3(-2) = 0
Formula: Al2O3

23. Method of Writing Chemical Formula for Ionic Compounds
2) Ammonium carbonate (containing NH4+ and CO32-)
NH4+
CO32- 1+ 2-
Charge
Simplest ration of ion combined 2 1
So, 2 cation NH4+ combined with 1 anion CO32- to form ammonium carbonate
Sum of charges is 2(+1) + 1(-2) = 0
Formula: (NH4)2CO3

24. CHEMICAL NOMENCLATURE
1) Elements:
Refer to the periodic table
- Examples:
i) Na = sodium
ii) Si = silicon

25. 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) BaCl2 = barium chloride
ii)K2O = potassium oxide
iii) Mg(OH)2 = Magnesium hydroxide

28. 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
Fe3+ ferric ion
- indicate charge on metal with Roman numerals
Examples:
i) FeCl2
2 Cl- -2 so Fe is +2
iron(II) chloride
ii) FeCl3
3 Cl- -3 so Fe is +3
iron(III) chloride
iii) Cr2S3
3 S-2 -6 so Cr is +3 (6/2)
chromium(III) sulfide

31. 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, NH3, CH4 - 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”

32. 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.

33. Molecular Compounds HI hydrogen iodide NF3 nitrogen trifluoride SO2
sulfur dioxide
N2Cl4
dinitrogen tetrachloride
NO2
nitrogen dioxide
N2O
dinitrogen monoxide

37. 4) Acids and bases
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 (H3O+ and Cl−), hydrochloric acid
Anions whose names end in “-ide” form acids with a “hydro-” prefix and an “-ic” ending.

38. An oxoacid is an acid that contains hydrogen, oxygen, and another element.
Examples:
i) HNO3 nitric acid
ii) H2CO3 carbonic acid
iii) H3PO4 phosphoric acid
iv) HCIO3 chloric acid
v) H2SO4 sulfuric acid
vi) HIO3 iodic acid
vii)HBrO3 bromic acid

39. Naming Oxoacids and Oxoanions

40. 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:
H3PO Phosphoric acid
H2PO4- dihydrogen phosphate
HPO4 2- hydrogen phosphate
PO43- phosphate

43. 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 hydroxide

44. 5) Hydrates
Hydrates are compounds that have a specific number of water molecules attached to them.
Examples:
i) BaCl2•2H2O barium chloride dihydrate
ii) LiCl•H2O
lithium chloride monohydrate
iii) MgSO4•7H2O
magnesium sulfate heptahydrate
iv) Sr(NO3)2 •4H2O
strontium nitrate tetrahydrate
CuSO4•5H2O
CuSO4

46. ATOMIC MASS
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.
By definition:
1 atom 12C “weighs” 12 amu
On this scale:
1H = amu
16O = amu

47. 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)( amu)+(0.0110)( amu)
= amu

48. Average atomic mass of lithium:
Example:
Naturally occurring lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
(7.42 x 6.015) + (92.58 x 7.016)
100
= amu

49. Average atomic mass (6.941)

51. AVOGADRO’S NUMBER AND THE MOLAR MASS Avogadro’s number (NA)
The Mole (mol): A unit to count numbers of particles
Dozen = 12
Pair = 2
The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly grams of 12C
1 mol = NA = x 1023
Avogadro’s number (NA)

52. Molar mass is the mass of 1 mole of in grams atoms
1 mole 12C atoms = x 1023 atoms = g
1 12C atom = amu
1 mole 12C atoms = g 12C
1 mole lithium atoms = g of Li
For any element
atomic mass (amu) = molar mass (grams)

53. No. of atoms/molecules (N)
x NA
÷ molar mass (g/mol)
Mass of element (m)
No. of moles (n)
No. of atoms/molecules (N)
÷ NA
x molar mass (g/mol)
NA = Avogadro’s number = x 1023 atoms

54. No. of atoms = 0.014 mol x 6.022 x 1023 atoms/mol
Example:
How many atoms are in g of potassium (K) ?
1 mol K = g K
1 mol K = x 1023 atoms K
No. of moles = g
39.10 g/mol
= mol
No. of atoms = mol x x 1023 atoms/mol
= 8.43 x 1021 atoms K

57. molecular mass (amu) = molar mass (grams)
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule. 1S
32.07 amu
SO2 2O
+ 2 x amu
SO2
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = amu
1 mole SO2 = g SO2

59. 1 mol C3H8O molecules = 8 mol H atoms
Example
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = x 1023 atoms H
1 mol C3H8O
60 g C3H8O x
8 mol H atoms
1 mol C3H8O x
6.022 x 1023 H atoms
72.5 g C3H8O x
1 mol H atoms
= 5.82 x 1024 atoms H

62. formula mass (amu) = molar mass (grams)
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
NaCl
1Na
22.99 amu
1Cl
amu
NaCl
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = amu
1 mole NaCl = g NaCl

63. What is the formula mass of Ca3(PO4)2 ?
Example:
What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca
3 x 40.08
2 P
2 x 30.97
8 O
+ 8 x 16.00
amu

64. 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 x (12.01 g)
46.07 g
x 100% = 52.14%
C2H6O
%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% % % = 100.0%

66. Determination of empirical formula
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%
Elements K Mn O
Mass (g)
24.75
34.77
40.51
mol
24.75 g
39.10 g/mol =
34.77 g
54.94 g/mol =
40.51 g
16.00 g/mol
= 2.532
Simplest ratio
0.6330
0.6329 ≈1 =1
2.532 ≈4
Empirical formula = KMnO4

68. 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
=1.33 x 3
= 4
=1 x 3
Empirical formula = C3H4O3

70. 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 ≈1
0.217 ≈2
Determination of empirical formula
Empirical formula = NO2

71. Determination of molecular formula
1) Empirical molar mass
= g/mol + 2(16.0g/mol) = 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 g/mol
Molecular formula = 2(NO2)
= N2O4
Actual molecular molar mass = 2(14.01 g/mol) + 4(16.00)
= g/mol

73. CHEMICAL REACTIONS AND CHEMICAL EQUATIONS
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
3 ways of representing the reaction of H2 with O2 to form H2O

74. How to “Read” Chemical Equations
2 Mg + O MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO

75. 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
C2H6 + O2
CO2 + H2O
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.
2C2H6
NOT
C4H12

76. Balancing Chemical Equations start with C or H but not O
Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
2 carbon
on left
multiply CO2 by 2
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2
2CO2 + 3H2O

77. Balancing Chemical Equations
Balance those elements that appear in two or more reactants or products.
multiply O2 by 7 2
C2H6 + O2
2CO2 + 3H2O
2 oxygen
on left
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
= 7 oxygen
on right
C2H O2
2CO2 + 3H2O 7 2
remove fraction
multiply both sides by 2
2C2H6 + 7O2
4CO2 + 6H2O

78. Balancing Chemical Equations
Check to make sure that you have the same number of each type of atom on both sides of the equation.
2C2H6 + 7O2
4CO2 + 6H2O
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)
Reactants
Products
4 C
12 H
14 O

80. AMOUNTS OF REACTANTS AND PRODUCTS
Stoichiometry:
- comparison of coefficients in a balanced equation
- The quantitative study of reactants and products in a chemical reaction

81. 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 quantity
Convert moles of sought quantity into desired units

82. Example:
Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 4H2O
If 209 g of methanol are used up in the combustion, what mass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
Moles of CH3OH = 209 g
32 g/mol
= 6.53 mol

83. 2CH3OH + 3O CO2 + 4H2O
2) From the equation, 2 mol CH3OH is used to give 4 mol H2O, if we have 6.53 mol CH3OH, how many mole that H2O will produce?
2 mol CH3OH = 4 mol H2O
6.53 mol CH3OH = ? mol H2O
= 4 mol H2O x mol CH3OH
2 mol CH3OH
= mol H2O
3) Mass of H2O
= mol x molar mass H2O
= mol x 18 g/mol
= g

87. LIMITING REAGENT Reactant used up first in the reaction. 2NO + O2 2NO2
NO is the limiting reagent
O2 is the excess reagent
Excess reagents: the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent

88. LIMITING REAGENT
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
Determination of limiting reagent and excess reagent
Mole of Al
= g
27.0 g/mol
= 4.59 mol
2) Mole of Fe2O3
= g
160 g/mol
= 3.76 mol

89. Divide moles of Al and Fe2O3 with their stoichiometric coefficients
Al ii) Fe2O3
= 4.59 mol = mol = 3.76 mol = 3.76 mol
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 Fe2O3 is a excess reagent.

90. 4) From the equation, 2 mol Al is used to give 1 mol Al2O3 , if we have 4.59 mol Al, how many mole that Al2O3 will produce?
2 mol Al produce 1 mol Al2O3
4.59 mol Al = 1mol Al2O3 x 4.59 mol Al
2 mol Al
= mol Al2O3
5) Mass of Al2O3
= mol x molar mass Al2O3
= mol x g/mol
= 234 g

91. REACTION YIELD
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%