1. MATTER AND MEASUREMENT
Matter: is anything which has mass and which
occupies space.
Mass and Weight: Mass, a constant property, is a
measure of the quantity of matter in a body. Weight
of an object depends on the gravitational pull.
Volume: The space which matter occupies is called
its volume.
Density: Density is the mass per unit volume. Mathematically
we would write:
Density = mass/volume g/cm3
The Phases of Matter
Solid:
Liquid:
Gas:
Phase Transitions
Physical and Chemical
Changes
***Physical changes
*Ice is H2O, liquid water is H2O, and water
steam is H2O.
***Chemical Changes
*When sugar is heated
*Burning of wood produces ashes and hot
gases
*Corrosion of iron
Energy
Energy is the ability to perform work
Exothermic Process: A chemical or physical
change which produces heat is
Endothermic Process: A process which requires
heat energy
Identification of a Substance
Physical properties are color, odor, density, melting
point or boiling point. The most common chemical
property is the ability of certain substances to burn
(react with Oxygen).
The Classification Of Matter
Elemental state
Compounds
Mixtures:
Homogeneous Mixture:
Heterogeneous Mixture:
9. Measurement
The International System of Units, which
is generally abbreviated to SI base units.
THE MOLE CONCEPT &
STOICHIOMETRY
1 mole of substance = 6.02 × 1023 atoms (Avogadro's number)
Examples:
1 mole of carbon atoms = 6.02 × 1023 carbon atoms.
2 mole of sulfur atoms = 2 × Avogadro's number = 2 × 6.02 × 1023
Problem 1: Calculate the number of atoms present in a 0.2 moles of sulfur.
Problem 2: Calculate the number of moles in 3.01 × 1020 molecules of
sulfur dioxide (SO2).
Molar mass of carbon atom = mass of 1 mole of carbon atoms
= mass of Avogadro's number of carbon atoms
= mass of 6.02 × 1023 carbon atoms
 CS2: Molecular mass = 12 u + 2(32.0 u) =
76.0 u; Molar mass = 76.0 g.mol-1.
 NaCl: Molecular mass = 23 u + 35.5 u =
58.5 u; Molar mass = 58.5 g.mol-1.
Mass of Substance
(grams)
÷ by molar mass
Amount
(moles) × by 6.02 1023
Number of Atoms
Percentage Composition
 Example 1: Each carbon monoxide is made up of
one carbon and one oxygen atom, as we know
one mol of this substance consists of 6.02 × 1023
molecules.
 Thus, 1 mol of CO is made up of 1 mol of C atoms
and 1 mol of oxygen atoms.
 Molecular mass of CO = (12.0 u) + (16.0 u) = 28 u
 Molar mass of CO = 28 g.mol-1.
 Percent of carbon (by mass) = (12.0 ÷ 28) × 100%
= 42.9 %.
 Percent of oxygen (by mass) = (16.0 ÷ 28) × 100%
= 57.1 %.
 Example 2: Acetic acid has the formula CH3COOH or
C2H4O2
 Molecular mass of C2H4O2 = 2(12.0 u) + 4(1.01 u) +
2(16.0 u) = 60.0 u;
 Molar mass = 60.0 g.mol-1.
 thus, Molecular mass of C2H4O2 = 2(12.0 u) +
22(1.01 u) + 2(16.0 u)=
 24.0 u + 4.04 u + 32.0 u = 60.0 u , Molar
mass = 60 g.mol-1
 Percent of carbon = (24.0 ÷ 60) × 100% = 40.0 %.
 Percent of hydrogen = (4.04 ÷ 60) × 100% = 6.73 %.
 Percent of oxygen = (32.0 ÷ 60) × 100% = 53.3 %.
Empirical Formula
 Ex.1: A certain substance is found to contain
59.9% oxygen and 40.1% sulfur by mass.
 Solution
 Amount of oxygen = 59.9 ÷ 16.0 = 3.74 mol O;
 Amount of sulfur = 40.1 ÷ 32.1 = 1.25 mol S
 Divide the ratio by the smallest number in the
ratio: S : O = 1.25 : 3.74 = 1.25/1,25 : 3.74/1.25 =
1.0 : 2.99
 Thus the empirical formula is SO3.
Molecular Formula
 Butyl butyrate has a molecular mass of 144 u. Analysis of this
compound shows that it contains 66.7 % carbon, 11.1 % hydrogen, and
22.2 % oxygen. Determine the molecular formula of this compound.
 Solution
 Amount of carbon = 66.7 ÷ 12.0 = 5.56 mol C;
 Amount of hydrogen = 11.1 ÷ 1.01 = 11.00 mol H.
 Amount of oxygen = 22.2 ÷ 16.0 = 1.39 mol O.
 Dividing through by 1.39, we get the empirical formula:
 C : H : O = 5.56 : 11.00 : 1.39 = 5.56/1.39 : 11.00/1.39 : 1.39/1.39 = 4 :
8:1
 The empirical formula is C4H8O,
 Empirical formula mass = 4(12.0 u) + 8(1.01 u) + 16.0 u = 72.1 u
 Molecular formula = 2 × Empirical formula = 2 × C4H8O = C8H16O2.
Avogadro’s Hypothesis
 Equal volumes of all gases, measured at the same temp and pressure,
contain the same number of molecules and thus contain also equal amounts
of substance.
 Example:
 We find that the mass of 1 L O2, measured at 0 °C and atmospheric
pressure, is 1.43 g whereas the mass of 1 L Cl2, under the same conditions
is 3.17 g.The calculation proved that we have the same number of moles in
each case. How?

Mass of O2
1.43
 Amount of O2 = ------------------- = -------- = 4.47 x 10-2 moles

Molar mass of O2
32.0

Mass of Cl2
3.17
 Amount of Cl2 = ------------------- = -------- = 4.47 x 10-2 moles

Molar mass of Cl2
70.9
 In each case the number of molecules = (X moles) (Avogadro’s number)

= 4.47 x 10-2 x 6.02 x 10 23

= 2.69 x 1022 molecules
Example:
Suppose we have 1 L H2 under certain conditions of
T and P has a mass of 8.99 x 10-2 g , if an equal
volume of an unknown has a mass of 1.96 g under
the same conditions ?
What is the molar mass of this unknown gas ?
 Solution:
NET IONIC EQUATIONS
 The molecular equation that represents this
process is:
 AgNO3 (aq) +
NaCl(aq) → AgCl(s) +
NaNO3 (aq)
 We would write
 Ag+(aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) →
AgCl(s) + Na+ (aq) + NO3- (aq)
 The net ionic equation is:
 Ag+(aq) + Cl- (aq) → AgCl(s)
.
 The molecular equation
is:
2AgNO3 (aq) + BaCl2 (aq) → 2AgCl(s) + Ba(NO3)2
(aq)
writing all the ionic compounds that are
present in the solution as ions, we get the
total ionic equation:
2Ag+(aq) + 2NO3- (aq) + Ba2+(aq) + 2Cl-(aq)
→ 2AgCl(s) + Ba2+(aq) + 2NO3- (aq)
Elimination of the spectator ions, Ba2+ and
NO3-, simplifies this equation to:
2Ag+(aq) + 2Cl-(aq) → 2AgCl(s)
The net ionic equation:
Ag+(aq) + Cl-(aq) → AgCl(s)
.
 Example: Write (a) the balanced molecular
equation, (b) the total ionic equation, and (c) the
net ionic equation for the reaction that occurs
when hydrochloric acid is added to a solution of
sodium hydrogen carbonate. The products are
carbon dioxide gas, water, and sodium chloride;
the latter remains dissolved in solution.
 (a) The molecular equation is
HCl (aq) + NaHCO3 (aq) → NaCl(aq) + CO2 (g)+H2O(ℓ)
 (b) The total ionic equation is
H+(aq) + Cl- (aq) + Na+(aq) + HCO3- (aq) → Na+(aq)
+ Cl-(aq) + CO2 (g) H2O(ℓ)
 (c ) The net ionic equation is
H+(aq) + HCO3+ (aq) → H2O(ℓ) + CO2 (g)
3. Chemical Formulas and
Nomenclature
1. Oxidation Number
Rule (1): The oxidation number of an atom in the elemental state is
zero.
A single neutral atom, such as helium (He), neon (Ne), aluminum (Al) and
iron (Fe), the oxidation number of the atom is zero. Similarly, in the
diatomic molecules of elements such as chlorine (Cl2) oxygen (O2)
Rule (2): The oxidation number of a monatomic ion is equal to its
charge.
 For example, sodium fluoride as an ionic compound contains sodium
ions (Na+) and fluoride ions (F-). Thus, the oxidation number sodium is
+1 and the oxidation number of fluorine is -1.
Rule (3): The sum of the oxidation numbers of all the atoms in a
neutral compound is zero.
 Example: In the ionic compound sodium fluoride (NaF), the sum of the
oxidation numbers is zero: (+1) + (-1) = 0.
Example: What is the oxidation number of each of the elements
present in Al2O3?
(2 × Oxidation number of Al) + (3 × Oxidation number of O)
= [2 × (+3) + [3 × (-2)] = (+6) + (-6) = 0
Example: The same result is obtained in magnesium chloride
(MgCl2): (+2) + 2(-1) = 0.
Rule (4): The oxidation number of hydrogen is +1 when it is
combined with a nonmetal as in CH4, NH3 and H2O. While when it is
combined with a metal as in metal hydrides, such as sodium hydride
(NaH) and calcium hydride (CaH2), the oxidation number is -1.
Rule (5): The oxidation number of oxygen is always -2 except in:
peroxides, such as hydrogen peroxides (H2O2) and sodium
peroxides (Na2O2), the oxidation number is -1.
the compound oxygen difluoride (OF2), the oxidation number is +2.
elemental oxygen (O2)
The oxidation numbers of [O] is -1 in H2O2 and +2 in OF2
Rule (6): In a binary covalent compound the positive oxidation
number is assigned to the less electronegative element, while the
negative oxidation number is given to the more electronegative
element.
Remember that fluorine is the most electronegative element with
oxygen.
F > O > Cl > Br > I
Example: In the compound carbon tetrachloride (CCl4), the chlorine atom
is more electronegative than carbon, so chlorine must have a negative
oxidation number (-1). Therefore, the oxidation number of carbon is +4.
(4 × Oxidation number of Cl) + (Oxidation number of C) = 0
4(-1) + (Oxidation number of C) = 0, thus: (Oxidation number of C) = +4
Problem: Determine the oxidation number of iodine in ICl3.
Chlorine is more electronegative than iodine, and will have its normal
oxidation number of -1. Thus, the oxidation number of iodine (I) is +3.
Rule (7): The sum of the oxidation numbers of all the atoms in a
polyatomic ion is equal to the net charge on the ion.
Let us to use the carbonate ion (CO32-) to illustrate this point.
(Oxidation number of C) + (3 × Oxidation number of O) = -2,
(Oxidation number of C) + 3(-2) = -2
(Oxidation number of C) -6 = -2,
thus: Oxidation number of C = -2 + 6 = +4
Problem: Arrange the following compounds in order of
increasing oxidation state for the carbon atom: (a) CO
(b) CO2
(c) H2CO
(d) CH3OH
(e) CH4
Problem: Determine the oxidation numbers for each atom in
the following compounds.
N2O3 , NO , P2O4 , NO2 , NH4+1, NO3-1 , OF2 , H2O2
, Na2O2
2.
The
Formulas
of
Compounds
Binary ionic compounds
In ZnBr2, the Zn2+ ions has an oxidation number +2 while, Br- ions has an
oxidation number of -1. The formula will be ZnBr2.
Example: Predict the simplest chemical formula for the binary ionic
compounds formed by the following pair of elements: Aluminum and sulfur
Al3+ ion is +3, sulfur forms the S2- ion which has an oxidation number of -2.
The formula will be Al2S3
Example: Predict the simplest formula of the compound formed by calcium
(Ca2+) and phosphate (PO43-) ions.
Ca3(PO4)2
Example: Predict the simplest formula of the compound formed by
ammonium ions and sulfide ions.
(NH4)2S
Ionic compounds containing
polyatomic ions
Name
Formula
Acetate
CH3CO2-
Ammonium
NH4+
Carbonate
CO32-
Chlorate
ClO3-
Chromate
CrO42-
Dichromate
Cr2O72-
5. GASES
Measurements of gases:
1- Volume of gas: 1L = 103 cm3 = 10-3 m3
2- Amount of gas: n = m / M
3- Temperature of gas: Tk = t °C + 273.15
4- Pressure of gas:
1 atm = 101325 pa = 101.325 Kpa = 760
mmHg = 760 torr
 Example:
 A ballon with a volume of 2.36 x 104 m3 contains 4.68 x
106 g of helium at 18 °C and 120.0 kPa. Express the
volume of the balloon in liters, the amount in moles,
the temperature in K, and the pressure in both
atmospheres and millimeters of mercury.
 Solution:
 The volume of He in liters = 2.36 x 104 m3 x (10)3 = 2.36 x 107 L
(1m3 = 100 cm x 100 cm x 100 cm =106 cm3 = 106 ml = 103 L)
 The amount of He in moles = mass (m) / molar mass (M)
= 4.68 x 106 g / 4.003 g = 1.17 x 106 mol He
 Tk = t °C + 273 = 18 + 273 = 291 k
 1 atm = 101.325 kPa
The pressure (atm) = 120.0 kPa / 101.325 kPa = 1.185 atm
 The pressure (mm Hg) = 1.185 atm x 760 mm Hg = 900.3 mm Hg
The ideal gas law:
Volume is directly proportional to amount
PV = nRT
n= m/M
PV = mRT/M
d= m/V
P = mRT/VM
P = dRT/M
Or
PM = dRT
 Example:
The ozone friendly composed now used as a
refrigerant in car air conditioners has the molecular
formula C2F4H2. If 2.50 g of this compound is
introduced into an evacuated 500 ml container at
10 °C, what pressure in atmospheres is developed
?
Solution:
PV =nRT
 V = 500.0 ml = 0.5 L
 n= m/M
= 2.5 g / 102.04 g C2F4H2 = 0.0245 mol
T = 10 + 273 = 283 K
&
P=?
 Thus,
P=nRT/V

= 0.0245 x 0.0821 x 283 / 0.5 = 1.14 atm
Example: Taking the molar mass of dry air
to be 29.0 g / mol, calculate the density of
air at 27 °C and 1 atm.
Solution:
PV =nRT
,
n=m/M
P V = m R T / M ……………………. (1)
d=m/V
PV =dVRT/M
P =dRT/M
PM =dRT
d = PM / R T = 1.00 x 29.0 / 0.821 x (27 + 273) =
1.18 g / L
Gas Mixtures
 For a mixture of two gases A and B:
Ptotal = n total R T / V
= (nA + nB) RT/ V
= nA RT/V + nB RT/ V
 The terms nA RT/ V and nB RT/ V are,
according to the ideal gas law, referred to as
partial pressures, PA and PB
 Ptotal = PA + PB
(Dalton's law)
 Example: A student prepares a sample of hydrogen
gas by electrolyzing water at 25 °C. He collects 152 ml
of H2 at a total pressure of 758 Hg. If the vapor
pressure of H2O is 23.76 mm Hg at 25 °C. Calculate:
 The partial pressure of H2
 The number of moles of H2 collected
 Solution:
 PH2 = Ptotal - PH2O

= 758 – 23.76 = 734 mm Hg = 734 / 760 = 0.9657atm
 PH2 V = nH2 RT
 nH2 = PH2 V / RT
= (734 / 760 atm) (152 / 1000 L) / (0.0821 L.atm.mol-1.
K-1) (25+273)
= 0.006 mol H2
 As pointed out earlier,
 For mixture containing gas A and gas B:
 PA = nA R T / V
& Ptotal = ntotal R T / V
 Dividing PA by Ptotal gives:
 PA / Ptotal = nA / ntotal
 The fraction nA / ntotal is referred to as the
mole fraction of A in the mixture.
 Using XA to represent the mole fraction of A
(i.e., XA = nA / ntotal)
 PA = XA Ptotal
5. RADIOACTIVITY &
RADIATION UNITS
 Matter and energy:
 Matter is characterized by its quantity, called the mass, and
is composed of the smallest unit, the atom.
 Energy is the capacity to do work and can exist in several
forms:
 Kinetic energy (which is due to the motion of matter).
 Potential energy (which is due to the position and
configuration of matter).
 Thermal energy (which is due to the motion of atoms or
molecules in matter.
 Electrical energy (which is due to the flow of electrons
across an electric potential).
 Chemical energy (which is due to chemical reaction).
Einstein's mass-energy relationship:
E = mc2
Radiation:
Particulate radiations: Examples of these
radiations are energetic electrons, protons, neutrons, α
particles, and so forth. They have mass and charge, except
neutrons.
Electromagnetic radiations:
These radiations
are a form of energy in motion that does not have mass and
charge and can propagate as either waves or discrete
packets of energy, called the photons or quanta. These
radiations travel with the velocity of light. Various examples
of electromagnetic radiations include radio waves, visible
light, heat waves, γ-radiations, and so forth,
The energy of electromagnetic radiations is given by:
E = hν = hc / λ
where (h) is the Planck constant given as 6.625 x
10 -27erg.s/cycle, (ν) is the frequency in hertz, defined as 1
cycle per second, (λ) is the wavelength in centimeters, and
(c) is the velocity of light in vacuum, which is equal to nearly
3 x 10 10 cm/s.
Using 1 eV = 1.602 x 10 -12 erg, the previous equation
becomes
E (eV) = 1.24 x 10 -4 / λ (cm)
Structure of the nucleus
 The symbolic representation of an element, X is
given by AZXN
 Example: 2311Na12
 The ratio of the number of neutrons to the number
of protons (N/Z) is an approximate indicator of the
stability of a nucleus. The N/Z ratio is 1 in low-Z
elements such as 12 6C , 14 7N , and 16 8O , but it
increases with increasing atomic number of
elements.
 For example, N/Z = 1.40 for 127 53I and 1.54 for
208 Pb .
82
Nuclear binding energy
The stability of the nucleus is explained by the
existence of a strong binding force called the
nuclear force, which overcomes the repulsive
force of the protons.
The mass of a nucleus is always less than the
combined masses of the nucleons in the nucleus.
The difference in mass is termed the mass
defect.
mass defect= Real mass –Calculated mass
Nuclear nomenclature
Isotopes: nuclides having the same atomic number Z but
different mass number. Examples 116C, 126C, and 136C.
Isotones: nuclides having the same number of neutrons N but
different numbers of protons. Examples : 13455Cs, 13354Xe,
and 13253I, each having 79 neutrons.
Isobars: 82Y, 82Sr,
mass number 82.
82Rb,
and
82Kr
are all isobars having the
Isomers: different energy states and spins. 99Tc and 99mTc
These excited states are called the isomeric states.
Radioactive Decay
 Radionuclides can decay by one or more of the
five modes:
1- Alpha (α) decay
2- Beta (β-) decay
3- Positron (β+) decay
4- Electron capture (EC)
5- Isomeric transition (IT)
1- Alpha (α) Decay:
The α-decay occurs mostly in heavy nuclides
such as uranium, radon, plutonium, and so
forth, decays by α particle emission. After αdecay, the atomic number of the nucleus is
reduced by 2 and the mass number by 4.
222 Rn
86

218 Po
84
+
α
2- Beta (β-) Decay
n 
p +
β-
+
ν
3- Positron (β+) Decay
p  n + β+ + ν
4- Electron Capture
p + e-  n +
ν
5- Isomeric transition: nuclear transition from an
upper excited state to a lower excited state is by
the emission of electromagnetic radiation, called
the gamma (γ)-radiation.
Units of Radioactivity
1 curie (Ci) = disintegration rate of 1 g of radium. (=3.7 x 1010
disintegrations per second (dps).
1 curie (Ci) = 3.7 x 1010 disintegrations per second (dps)
= 2.22 x 1012 disintegrations per minute (dpm)
1 millicurie (mCi) = 3.7 x 107 dps = 2.22 x 109 dpm
1 microcurie (μCi) = 3.7 x 104 dps = 2.22 x 106 dpm
The system international (SI) unit for radioactivity is Becquerel
(Bq), which is defined as 1 dps.
1 Becquerel (Bq) = 1dps = 2.7 x 10-11 Ci
Half-life:
λ =0.693 / t½
λ is the decay constant of radionuclide