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WELCOME TO SKO16
CHEMISTRYCHEMISTRYCHEMISTRY
08/16/1108/16/11 mattermatter 22
CHEMISTRY SK016
1.0 Matter 7
2.0 Atomic Structure 7
3.0 Periodic Table 4
4.0 Chemical Bonding 2
5.0 State of Matter 7
6.0 Chemical Equilibrium 5
7.0 Ionic Equilibria 12
Total 54
Chapter Topics Hours
08/16/1108/16/11 mattermatter 33
CHEMISTRY SK026
8.0 Thermochemistry 4
9.0 Electrochemistry 6
10.0 Reaction Kinetics 7
11.0 Intro To Organic Chemistry 4
12.0 Hydrocarbons 8
13.0 Aromatic Compounds 3
14.0 Haloalkanes (Alkyl halides) 4
15.0 Hydroxy compounds 3
Chapter Topic Hours
08/16/1108/16/11 mattermatter 44
CHEMISTRY SK026
16.0 Carbonyl 4
17.0 Carboxylic acids & Derivatives 4
18.0 Amines 5
19.0 Amino acids and Proteins 2
20.0 Polymers 1
Chapter Topic Hour
08/16/1108/16/11 mattermatter 55
ASSESSMENT
1. COURSEWORK (20%) Continuous evaluation (tutorial/test/quiz) - 10% Practical work - 10%
2. MID-SEMESTER EXAMINATION - 10%
3. FINAL EXAMINATION (70%) Paper 1 (30 multiple choice questions) - 30% Paper 2 (Part A-structured)
(Part B-long structured) -100%
08/16/1108/16/11 mattermatter 66
REFERENCE BOOKS
CHEMISTRY ,9th Ed. – Raymond Chang, McGraw-Hill
CHEMISTRY –The Molecular Nature of Matter and Change, 3rd Ed.– Martin Silberberg, McGraw Hill
CHEMISTRY – The Central Science, 9th Ed. Theodore L.Brown, H.Eugene LeMay,Jr, Bruce E Bursten, Pearson Education
GENERAL CHEMISTRY – Principle & Structure, 6th Ed. James E Brady, John Wiley and Sons.
08/16/1108/16/11 mattermatter 77
GENERAL CHEMISTRY – Principle and Modern Applications, 8th Ed. Ralph H. Petrucci, William S. Harwood, Prentice-Hall
ORGANIC CHEMISTRY, 7th Ed – T.W.Graham
Solomon,Craig B.Fryhle, John Wiley and Sons
ORGANIC CHEMISTRY, 4th Ed – L.G. Wade, Jr, Prentice Hall
ORGANIC CHEMISTRY, 6th Ed – John McMurry, Thompson – Brooks/Cole
08/16/1108/16/11 mattermatter 88
Chapter 1 : MATTER
1.1 Atoms and Molecules
1.2 Mole Concept
1.3 Stoichiometry
08/16/1108/16/11 mattermatter 99
1.1Atoms and Molecules
08/16/1108/16/11 mattermatter 1010
Learning Outcome At the end of this topic, students should be able to:
(a) Describe proton, electron and neutron in terms of the relative mass and relative charge.
(b) Define proton number, Z, nucleon number, A and isotope.
(c) Write isotope notation.
08/16/1108/16/11 mattermatter 1111
Introduction
Matter
Anything that occupies space and has mass.
e.g: air, water, animals, trees, atoms, etc
Matter may consists of atoms, molecules or ions.
Classifying Matter
08/16/1108/16/11 mattermatter 1212
A substance is a form of matter that has a definite or constant composition and distinct properties.
Example: H2O, NH
3, O
2
A mixture is a combination of two or more substances in which the substances retain their identity.
Example : air, milk, cement
08/16/1108/16/11 mattermatter 1313
An element is a substance that cannot be separated into simpler substances by chemical means. Example : Na, K, Al,Fe
A compound is a substance composed of atoms of two or more elements chemically united in fixed proportion. Example : CO
2, H
2O, CuO
08/16/1108/16/11 mattermatter 1515
Three States of Matter
SOLID LIQUID GAS
08/16/1108/16/11 mattermatter 1616
a) AtomsAn atom is the smallest unit of a chemical
element/compound.In an atom, there are 3 subatomic particles:
- Proton (p)
- Neutron (n)
- Electron (e)
1.1 Atoms and Molecules
08/16/1108/16/11 mattermatter 1717
Modern Model of the Atom
Electrons move around the region of the atom.
All neutral atoms can be identified by the number of protons and neutrons they contain.
Proton number (Z) is the number of protons in the nucleus of the atom of an element (which is equal to the number of electrons). Protons number is also known as atomic number.
Nucleon number (A) is the total number of protons and neutrons present in the nucleus of the atom of an element. Also known as mass number.
08/16/1108/16/11 mattermatter 1919
Subatomic Particles
Particle Mass
(gram)
Charge
(Coulomb)
Charge
(units)
Electron (e) 9.1 x 10-28 -1.6 x 10-19 -1
Proton (p) 1.67 x 10-24 +1.6 x 10-19 +1
Neutron (n) 1.67 x 10-24 0 0
08/16/1108/16/11 mattermatter 2020
Isotope
Isotopes are two or more atoms of the same element that have the same proton number in their nucleus but different nucleon number.
Examples:
H11 (D) H21 (T) H3
1
U23592
U23892
08/16/1108/16/11 mattermatter 2222
Isotope Notation
X = Element symbol
Z = Proton number of X (p)
A = Nucleon number of X
= p + n
An atom can be represented by an isotope notation ( atomic symbol )
08/16/1108/16/11 mattermatter 2323
Total charge on the ion
Proton number of mercury,
Z = 80
Nucleon number of mercury, A = 202
The number of neutrons= A – Z= 202 – 80= 122
In a neutral atom:
number of protons equals number of
electrons
In a positive ion:
number of protons is more than number of
electrons
In a negative ion:
number of protons is less than number of
electrons
08/16/1108/16/11 mattermatter 2525
Exercise 1
Symbol Number of : Charge
Proton Neutron Electron
Give the number of protons, neutrons, electrons and charge in each of the following species:
Hg20080
Cu6329
2178O
35927Co
08/16/1108/16/11 mattermatter 2626
Exercise 2
Species Number of : Notation for nuclideProton Neutron Electron
A 2 2 2
B 1 2 0
C 1 1 1
D 7 7 10
Write the appropriate notation for each of the following nuclide :
08/16/1108/16/11 mattermatter 2727
A molecule consists of a small number of atoms joined together by bonds.
b) Molecules
08/16/1108/16/11 mattermatter 2828
A diatomic molecule Contains only two atoms
Ex : H2, N2, O2, Br2, HCl, CO
A polyatomic moleculeContains more than two atoms
Ex : O3, H2O, NH3, CH4
08/16/1108/16/11 mattermatter 2929
Learning Outcomes
At the end of this topic, student should be able
to :
(a) Define relative atomic mass, Ar and
relative molecular mass, Mr based on
the C-12 scale.
(b) Calculate the average atomic mass of an
element given the relative abundance of
isotopes or a mass spectrum.
08/16/1108/16/11 mattermatter 3030
Relative Mass
i. Relative Atomic Mass, Ar
A mass of one atom of an element compared to 1/12 mass of one atom of 12C with the mass 12.000 amu
Cof atom one of Mass X 121
element of atom one of Mass Amass,atomic lativeRe
12r
08/16/1108/16/11 3131
Mass of an atom is often expressed in atomic mass unit, amu (or u).
Atomic mass unit, amu is defined to be one twelfth of the mass of 12C atom
Mass of a 12C atom is given a value of exactly 12 amu
1 u = 1.660538710-24 g
The relative isotopic mass is the mass of an atom, scaled with 12C.
08/16/1108/16/11 mattermatter 3232
Example 1
Determine the relative atomic mass of an element Y if the ratio of the atomic mass of Y to carbon-12 atom is 0.45
ANSWER:
08/16/1108/16/11 mattermatter 3333
ii) Relative Molecular Mass, Mr
A mass of one molecule of a compound compared to 1/12 mass of one atom of 12C with the mass 12.000amu
C of atom one of Mass x 121
molecule one of Mass
Mr mass, molecular Relative
12
The relative molecular mass of a compound is the summation of the relative atomic masses of all atoms in a molecular formula.
08/16/1108/16/11 mattermatter 3535
Example 2
Calculate the relative molecular mass of C5H5N,
Ar C = 12.01
Ar H = 1.01
Ar N = 14.01
08/16/1108/16/11 3636
MASS SPECTROMETER
An atom is very light and its mass cannot be measured directly
A mass spectrometer is an instrument used to measure the precise masses and relative quantity of atoms and molecules
08/16/1108/16/11 mattermatter 3737
08/16/1108/16/11 3838
Mass Spectrum of Monoatomic Elements
Modern mass spectrum converts the abundance into percent abundance
08/16/1108/16/11 mattermatter 3939
Mass Spectrum of Magnesium
The mass spectrum of Mg shows that Mg consists of 3 isotopes: 24Mg, 25Mg and 26Mg.
The height of each line is proportional to the abundance of each isotope.
24Mg is the most abundant of the 3 isotopes
Re
lati
ve
ab
un
da
nce
63
8.1 9.1
24 25 26 m/e (amu)
08/16/1108/16/11 mattermatter 4040
Learning Outcomes
At the end of this topic, student should be able At the end of this topic, student should be able
to :
(a) Calculate the average atomic mass of an element given the relative abundances of isotopes or a mass spectrum.
08/16/1108/16/11 mattermatter 4141
How to calculate the relative atomic mass, A
r from mass spectrum?
abundance
)abundancemass (isotopic mass atomic Average
Ar is calculated using data from the mass spectrum.
The average of atomic masses of the entire element’s isotope as found in a particular environment is the relative atomic mass, A
r of
the atom.
08/16/1108/16/11 4242
Example 1:
Calculate the relative atomic mass of neon from the mass spectrum.
Solution:
Average atomic =
mass of Ne
=
= 20.2 u
Relative atomic mass Ne = 20.2
e% abundanc
isotopicabundance mass) (%
)2.93.05.90()u 222.9()u 213.0()u 205.90(
08/16/1108/16/11 4444
Example 2:
Copper occurs naturally as mixture of 69.09% of 63Cu and 30.91% of 65Cu. The isotopic masses of 63Cu and 65Cu are 62.93 u and 64.93 u respectively. Calculate the relative atomic mass of copper.
Solution:
Average atomic =
mass of Cu
=
= 63.55 u
Relative atomic mass Cu = 63.55
e% abundanc
isotopicabundance mass) (%
)9 1.3 00 9.6 9()u 9 3.6 49 1.3 0()u 9 3.6 20 9.6 9(
08/16/1108/16/11 4646
Example 3:
Naturally occurring iridium, Ir is composed of two isotopes, 191Ir and 193Ir in the ratio of 5:8. The relative isotopic mass of 191Ir and 193Ir are 191.021 u and 193.025 u respectively. Calculate the relative atomic mass of Iridium
08/16/1108/16/11 4747
Solution:
Average atomic =
mass of Ir
=
= 192.254 u
Relative atomic mass Ir = 192.254
abundance
massisotopic abundance
)85()u 025.1938()u 021.1915(
Mass Spectrum of Molecular Elements
A sample of chlorine which contains 2 isotopes with nucleon number 35 and 37 is analyzed in a mass spectrometer. How many peaks would be expected in the mass spectrum of chlorine?
MASS SPECTROMETER
+ _ _
Cl2 + e Cl2+ + 2e
Cl2
Cl2 + e 2Cl+ + 2e
35Cl-35Cl35Cl-37Cl37Cl-37Cl
35Cl-35Cl+35Cl-37Cl+37Cl-37Cl+
35Cl+37Cl+
Mass Spectrum of Diatomic Elements
Exercise:
How many peaks would be expected in a mass spectrum of X2 which consists of 3 isotopes?
MATTER
1.2 Mole Concept
Learning Outcome
At the end of this topic, students should be able to:
a) Define mole in terms of mass of carbon-12 and Avogadro’s
constant, NA
Avogadro’s Number, NA
Atoms and molecules are so small – impossible to count
A unit called mole (abbreviated mol) is devised to count chemical substances by weighing them
A mole is the amount of matter that contains as many objects as the number of atoms in exactly 12.00 g of carbon-12 isotope
The number of atoms in 12 g of 12C is called Avogadro’s number, NA = 6.02 x 1023
Example:
1 mol of Cu contains Cu atoms
1 mol of O2 contains O2 molecules
O atoms
1 mol of NH3 contains NH3 molecules
N atoms
H atoms
6.02 6.02 10102323
33 6.02 6.02 10102323
22 6.02 6.02 10102323
6.02 6.02 10102323
6.02 6.02 10102323
6.02 6.02 10102323
1 mol of CuCl2 contains Cu2+ ions
Cl- ions
6.02 1023
2 6.02 1023
Mole and MassExample:
Relative atomic mass for carbon, C = 12.01
Mass of 1 C atom = 12.01 amu
Mass of 1 mol C atoms = 12.01 g
Mass of 1 mol C atoms consists of 6.02 x 1023 C atoms
= 12.01 g
Mass of 1 C atom =
= 1.995 x 10-23 g
2310 x 6.02
g 01.12
12.01 amu = 1.995 x 10-23 g
1 amu =
= 1.66 x 10-23 g
am u 12.01
g 10 x .9951 23
Example:
From the periodic table, Ar of nitrogen, N is
The mass of 1 N atom =
The mass of 1 mol of N atoms =
The molar mass of N atom =
The molar mass of nitrogen gas =
The nucleon number of N =
14.0114.01
14.01 amu14.01 amu
14.01 g14.01 g
14.01 g mol14.01 g mol11
28.02 g mol28.02 g mol11
1414
Mr of CH4 is
The mass of 1 CH4 molecule =
The mass of 1 mol of CH4 molecules =
The molar mass of CH4 molecule =
16.016.055
16.05 amu16.05 amu
16.05 g16.05 g
16.05 g 16.05 g molmol11
08/16/1108/16/11 MATTERMATTER 6161
Learning Outcome
At the end of this topic, students should be
able to:(a) Interconvert between moles, mass,
number of particles, molar volume of gas at STP and room temperature.
(b) Define the terms empirical & molecular formulae
(c) Determine empirical and molecular formulae from mass composition or from mass composition or combustion data.combustion data.
Example 1:
Calculate the number of moles of molecules for 3.011 x 1023 molecules of oxygen gas.
Solution:
6.02 x 1023 molecules of O2
3.011 x 1023 molecules of O2
= 0.5000 mol of O2 molecules
1 mol of O1 mol of O22 molecules molecules
molecules 10 6. 02mol 1 molecules
23 2 31 00 11.3
Example 2:
Calculate the number of moles of atoms for 1.204 x 1023 molecules of nitrogen gas.
Solution:
6.02 x 1023 molecules of N2
2 mol of N atoms
1.204 x 1023 molecules of N2
= 0.4000 mol of N atoms
1 mol of N1 mol of N22 molecules molecules
molecules mol 2 molecules
2 3
2 3
1 00 2.61 02 0 4.1
Example 3:
Calculate the mass of 0.25 mol of chlorine gas.
Solution:
1 mol Cl2
0.25 mol Cl2
18 g
or
mass = mol x molar mass
= 0.25 mol x (2 x 35.45 g mol-1)
= 18 g
2 2 35.45 g 35.45 g
m o l 1m o l 0 . 2 5 g 3 5 . 4 5 2
Example 4:
Calculate the mass of 7.528 x 1023 molecules of methane, CH4
Solution:
6.02 x 1023 CH4 molecules (12.01 + 4(1.01)) g
7.528 x 1023 CH4 molecules
= 20.06 g
23
23
1002.6107. 528 g 05.16
Molar Volume of Gases Avogadro (1811) stated that equal volumes of
gases at the same temperature and pressure contain equal number of molecules
Molar volume is a volume occupied by 1 mol of gas
At standard temperature and pressure (STP), the molar volume of an ideal gas is 22.4 L mol1
Standard Temperature and Pressure
273.15 K 1 atm 760 mmHg
0 C 101325 N m-2
101325 Pa
08/16/1108/16/11 MATTERMATTER 6767
Standard Molar Volume
At room conditions (1 atm, 25 C), the molar volume of a gas = 24 L mol-1
08/16/1108/16/11 MATTERMATTER 6969
Example 1:
Calculate the volume occupied by 1.60 Calculate the volume occupied by 1.60 mol of Clmol of Cl22 gas at STP. gas at STP.
Solution:
At STP,At STP,
1 mol Cl1 mol Cl22 occupiesoccupies
1.60 mol Cl1.60 mol Cl22 occupies occupies
= 35.8 = 35.8 LL
22.4 L
m o l 1L 4.2 2m o l 6 0.1
08/16/1108/16/11 MATTERMATTER 7070
Example 2:
Calculate the volume occupied by 19.61 Calculate the volume occupied by 19.61 g of Ng of N22 at STP at STP
Solution:
1 mol of N1 mol of N22 occupiesoccupies 22.4 L22.4 L
of Nof N22 occupies occupies
= 15.7 L= 15.7 L
mol 1
L 22. 4mol )01.14(2
61.19
1mol g 2(14. 01)g 61.19
08/16/1108/16/11 MATTERMATTER 7171
Example 3:0.50 mol methane, CH0.50 mol methane, CH44 gas is kept in a cylinder gas is kept in a cylinder at STP. Calculate:at STP. Calculate:
(a)(a) The mass of the gasThe mass of the gas
(b)(b) The volume of the cylinderThe volume of the cylinder
(c)(c) The number of hydrogen atoms in the The number of hydrogen atoms in the cylindercylinderSolution:
(a)(a) Mass of 1 mol CHMass of 1 mol CH44 ==
Mass of 0.50 mol CHMass of 0.50 mol CH44 ==
= 8.0 g= 8.0 g
16.05 g
m o l 1m o l 0 . 5 0g 0 5.1 6
08/16/1108/16/11 MATTERMATTER 7272
(b)(b) At STP;At STP; 1 mol CH1 mol CH44 gas gas occupiesoccupies
0.50 mol CH0.50 mol CH44 gas gas occupies occupies
= 11 L= 11 L
(c)(c)1 mol of CH1 mol of CH44 molecules molecules 4 mol of H atoms4 mol of H atoms
0.50 mol of CH0.50 mol of CH44 molecules molecules 2 mol of H atoms2 mol of H atoms
1 mol of H atoms1 mol of H atoms
2 mol of H atoms2 mol of H atoms
1.2 x 101.2 x 102424 atoms atoms
22.4 L
6.02 x 1023 atoms
2 x 6.02 x 1023 atoms
m o l 1m o l 0 . 5 0 L 4.2 2
08/16/1108/16/11 MATTERMATTER 7373
Exercise A sample of CO2 has a volume of 56 cm3 at
STP. Calculate:
a) The number of moles of gas molecules (0.0025 mol)
a) The number of CO2 molecules (1.506 x 1021 molecules)
a) The number of oxygen atoms in the sample (3.011x1021atoms)
Notes: 1 dm3 = 1000 cm3
1 dm3 = 1 L
08/16/1108/16/11 MATTERMATTER 7474
Empirical And Molecular Formulae
- Empirical formula => => chemical formula that shows the simplest ratio of all elements in a molecule.
- Molecular formula => => formula that show the actual number of atoms of each element in a molecule.
08/16/1108/16/11 MATTERMATTER 7575
- The relationship between empirical formula and molecular formula is :
Molecular formula = n ( empirical formula )
08/16/1108/16/11 MATTERMATTER 7676
Example:
A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56. Determine the empirical formula and molecular formula of the compound.
== = = == = =
Mass
Num ber of m oles
S im plest ra tio
85.7
21.984
14.1584
1
7.1357
14.3
01.12
7.85
01.1
3.14
C H
Solution:
Em pirica l form ula = Em pirica l form ula = CH2
1 4 . 0 3
5 6 n
4
3 . 9 9
84
2
HC form ula M olecular
)n(C H form ula M olecular
08/16/1108/16/11 MATTERMATTER 7979
Exercise: A combustion of 0.202 g of an organic sample that contains carbon, hydrogen and oxygen produce 0.361g carbon dioxide and 0.147 g water. If the relative molecular mass of the sample is 148, what is the molecular formula of the sample?
Answer : C6H12O4
08/16/1108/16/11 MATTERMATTER 8080
At the end of this topic, students should be At the end of this topic, students should be able to:(a) (a) Define and perform calculation for each for each of of the the following concentration following concentration measurements :measurements : i) molarity (M) ii) molality(m) iii) mole fraction, X iv) percentage by mass, % w/w v) percentage by volume, %v/v
Learning Outcome
08/16/1108/16/11 MATTERMATTER 8181
Concentration of Solutions
A solution is a homogeneous mixture of two or more substances:
solvent + solute(s)
e.g: sugar + water –
solution
sugar – solute
water – solvent
08/16/1108/16/11 MATTERMATTER 8282
08/16/1108/16/11 MATTERMATTER 8383
Concentration of a solution can be expressed in various ways :
a) molarityb) molalityc) mole fractiond) percentage by mass e) percentage by volume
08/16/1108/16/11 MATTERMATTER 8484
a) Molarity
Molarity is the number of moles of solute in 1 litre of solution
Units of molarity:mol L-1
mol dm-3
M
( L ) s o l u t i o n o f v o l u m e
( m o l ) s o l u t e o f m o l e sM m o l a r i t y ,
Example 1: Determine the molarity of a solution
containing 29.22 g of sodium chloride, NaCl in a 2.00 L solution.
Solution:
solution
NaC lNaC l V
n M
L 00.2
m ol )45.3599.22(
22.29
= 0.250 mol L-1
Example 2:How many grams of calcium chloride, CaCl2 should be used to prepare 250.00 mL solution with a concentration of 0.500 M
Solution:
solut ionCaClCaCl x VM n22
= 0.500 m ol L 1 250.00 10 3 L
massmolar x n CaClof mass2CaCl2
= (0.500 250.00 10 3) m ol (40.08 + 2(35.45)) g m ol 1
= 13.9 g
08/16/1108/16/11 MATTERMATTER 8989
b) Molality
Molality is the number of moles of solute dissolved in 1 kg of solvent
Units of molality: mol kg-1
molal m
( k g ) s o l v e n t o f m a s s
( m o l ) s o l u t e o f m o l e s m m o l a l i t y ,
Example:What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?
Solution:
1-CaClmol g 2(35. 45) 08.40
g 0.32 n
2
kg 10271
mol 98.1100.32
CaClof Molal ity 32
1kg mol 1. 06
Exercise:Calculate the molality of a solution prepared by dissolving 24.52 g of sulphuric acid in 200.00 mL of distilled water. (Density of water = 1 g mL-1)
Ans = 1.250 mol kg-1
08/16/1108/16/11 MATTERMATTER 9393
c) Mole Fraction (X)
Mole fraction is the ratio of number of moles of one component to the total number of moles of all component present.
For a solution containing A, B and C:
T
A
CBA
AA
nn
n n nn
Xof A, f ract ion Mol
08/16/1108/16/11 MATTERMATTER 9494
Mol fraction is always smaller than 1
The total mol fraction in a mixture (solution) is equal to one.
XA + XB + XC + X….. = 1
Mole fraction has Mole fraction has no unit (dimensionless) since it is a ratio of two similar quantities.
Example:A sample of ethanol, C2H5OH contains 200.0 g of ethanol and 150.0 g of water. Calculate the mole fraction of(a) ethanol(b) waterin the solution.
Solution:
nethanol = 1mol g 16. 00)5(1. 01) (2(12. 01)g 0.200
nwater = 1mol g 16. 00)(2(1. 01)
g 0.150
Xethanol =
mol 02.18
0.150mol
07.450.200
mol 07.45
0.200
= 0. 3477
Xwater = 1 0. 3477= 0. 6523
08/16/1108/16/11 MATTERMATTER 9898
d) Percentage by Mass (%w/w)Percentage by mass is defined as
the percentage of the mass of solute per mass of solution.
Note: Mass of solution = mass of solute + mass of solvent
1 0 0xs o l u t i o no f m a s s
s o l u t eo f m a s sw
w%
Example:A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.362 g of water. What is the percent by mass of KCl in the solution?
Solution:
%1 0 0g 5 4 . 3 6 2g 0 . 8 9 2
g 8 9 2.0 m a s s %
= 1.61%
Exercise:A solution is made by dissolving 4.2 g of sodium chloride, NaCl in 100.00 mL of water. Calculate the mass percent of sodium chloride in the solution.
Answer = 4.0%
08/16/1108/16/11 MATTERMATTER 101101
e) Percentage by Volume (%V / V)
Percentage by volume is defined as the percentage of volume of solute in milliliter per volume of solution in milliliter.
Note:
s o l u t i o no f v o l u m e s o l u t i o no f m a s s
s o l u t i o no f D e n s i t y
1 0 0 x s o l u t i o n o f v o l u m e
s o l u t e o f v o l u m e v
v % 1 0 0 x s o l u t i o n o f v o l u m e
s o l u t e o f v o l u m e v
v %
Example 1: 25 mL of benzene is mixed with 125
mL of acetone. Calculate the volume percent of benzene solution.
Solution:
1 0 0 % m L 1 2 5 m L 2 5
m L 2 5 v o l u m e%
= 17%
Example 2:A sample of 250.00 mL ethanol is labeled as 35.5% (v/v) ethanol. How many milliliters of ethanol does the solution contain?
Solution:
100%mL 00.250%5.35
Vethanol
%100VV
ethanolof volume%solut ion
ethanol
= 88.8 mL
Example 3:A 6.25 m of sodium hydroxide, NaOH solution has has a density of 1.33 g mL-1 at 20 ºC. Calculate the concentration NaOH in:(a) molarity(b) mole fraction(c) percent by mass
Solution:
(a) M = so lu tion
NaO H
Vn
6.25 m of N aO H there is 6.25 m ol of NaO H in 1 kg of water
for a solution consists of 6.25 m ol of NaO H and 1 kg of water;
V solution = so lu tion
so lu tionm ass
m ass solution = m assN aO H + m asswater
m assN aO H = nN aO H m olar m ass of NaO H = 6.25 m ol (22.99 + 16.00 + 1.01) g m ol 1
= 250 g
m ass solution = 250 g + 1000 g = 1250 g
V solution = 1
m L g 1.33
g 1250
M N aO H =
L 10 33.1
1250m ol 25.6
3
= 6.65 mol L 1
(b) X N aO H = waterNaO H
NaO H
nnn
1 kg of water contains 6.25 m ol of N aO H
n water = w aterof m ass m olar
m ass water
= 1
m ol g 16.00) (2(1.01)
g 1000
X N aO H =
m ol
02.181000
m ol 25.6
m ol 25.6
= 0.101
(c) % (w /w) of N aO H = waterN aO H
N aO H
m assm assm ass
100%
= g 1000 g 250
g 250
100%
= 20.0%
Exercise:An 8.00%(w/w) aqueous solution of ammonia has a density of 0.9651 g mL-1. Calculate the
(a) molality
(b) molarity
(c) mole fraction of the NH3 solution
Answer: a) 5.10 mol kg-1
b) 4.53 mol L-1
c) 0.0842
MATTER
1.3 Stoichiometry
Learning Outcome
At the end of the lesson, students should be able to:
a) Determine the oxidation number of an element in a chemical formula.
b) Write and balance :i) Chemical equation by inspection methodii) redox equation by ion-electron method
08/16/1108/16/11 MATTERMATTER 114114
Balancing Chemical Equation
A chemical equation shows a chemical reaction using symbols for the reactants and products.
The formulae of the reactants are written on the left side of the equation while the products are on the right.
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Example:
x A + y B z C + w D
Reactants Products
08/16/1108/16/11 MATTERMATTER 116116
A chemical equation must have an equal number of atoms of each element on each side of the arrow
The number x, y, z and w, showing the relative number of molecules reacting, are called the stoichiometric coefficients.
A balanced equation should contain the smallest possible whole-number coefficients
The methods to balance an equation: a) Inspection method b) Ion-electron method
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Inspection Method
1. Write down the unbalanced equation. Write the correct formulae for the reactants and products.
1. Balance the metallic atom, followed by non-metallic atoms.
1. Balance the hydrogen and oxygen atoms.
1. Check to ensure that the total number of atoms of each element is the same on both sides of equation.
08/16/1108/16/11 MATTERMATTER 118118
Example: Balance the chemical equation by applying the inspection method.
NH3 + CuO → Cu + N2 + H2O
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Exercise
Balance the chemical equation below by applying inspection method.
1. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O
2. C6H6 + O2 → CO2 + H2O
3. N2H4 + H2O2 → HNO3 + H2O
4. ClO2 + H2O → HClO3 + HCl
08/16/1108/16/11 MATTERMATTER 120120
Redox Reaction
Mainly for redox (reduction-oxidation) reaction
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Oxidation is defined as a process of electron loss.
The substance undergoes oxidation loses one or more electrons.
increase in oxidation number
act as an reducing agent (electron donor)
Half equation representing oxidation: Mg Mg2+ 2e Fe2+ Fe3+ + e 2Cl- Cl2 + 2e
Reduction is defined as a process of electron gain.
The substance undergoes reduction gains one or more electrons.
decrease in oxidation number
act as an oxidizing agent (electron acceptor)
Half equation representing reduction: Br2 + 2e → Br-
Sn4+ + 2e → Sn2+
Al3+ + 3e → Al
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Oxidation numbers of any atoms can be determined by applying the following rules:
1. For monoatomic ions, oxidation number = the charge on the ione.g: ion oxidation number
Na+ +1Cl- -1Al3+ +3S2- -2
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2. For free elements, e.g: Na, Fe, O2, Br2, P4, S8 oxidation number on each atom = 0
1. For most cases, oxidation number for O = -2 H = +1
Halogens = -1
08/16/1108/16/11 MATTERMATTER 125125
Exception:
1. H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1
1. Halogen bonded to oxygen (e.g: Cl2O7) oxidation number for halogen = +ve
1. In a neutral compound (e.g: H2O, KMnO4) the total of oxidation number of every atoms that made up the molecule = 0
1. In a polyatomic ion (e.g: MnO4-, NO3
-) the total oxidation number of every atoms that made up the molecule = net charge on the ion
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Exercise
1. Assign the oxidation number of Mn in the following chemical compounds.i. MnO2 ii. MnO4
-
1. Assign the oxidation number of Cl in the following chemical compounds.i. KClO3 ii. Cl2O7
2-
1. Assign the oxidation number of following:i. Cr in K2Cr2O7
ii. U in UO22+
iii. C in C2O42-
08/16/1108/16/11 MATTERMATTER 127127
Balancing Redox Reaction
Redox reaction may occur in acidic and basic solutions.
Follow the steps systematically so that equations become easier to balance.
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Balancing Redox Reaction In Acidic Solution
Fe2+ + MnO4- → Fe3+ + Mn2+
1. Separate the equation into two half-reactions: reduction reaction and oxidation reaction
i. Fe2+ → Fe3+ ii. MnO4
- → Mn2+
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1. Balance atoms other than O and H in each half-reaction separately
i. Fe2+ → Fe3+
ii. MnO4- → Mn2+
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130130
3. Add H2O to balance the O atoms
Add H+ to balance the H atoms
i. Fe2+ → Fe3+ ii. MnO4
- + → Mn2+ +
4. Add electrons to balance the charges
i. Fe2+ → Fe3+ + ii. MnO4
- + 8H+ + → Mn2+ + 4H2O
4H2O8H+
1 e
5 e
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3. Multiply each half-reaction by an integer, so that number of electron lost in one half-reaction equals the number gained in the other.
i. 5 x (Fe2+ → Fe3+ + 1e)5Fe2+ → 5Fe3+ + 5e
ii. MnO4- + 8H+ + 5e → Mn2+ +
4H2O
1. Add the two half-reactions and simplify where possible by canceling the species appearing on both sides of the equation.
i. 5Fe2+ → 5Fe3+ + 5eii. MnO4
- + 8H+ + 5e → Mn2+ + 4H2O
___________________________________5Fe2+ + MnO4
- + 8H+ → 5Fe3+ + Mn2+ + 4H2O ___________________________________
5. Check the equat ion to make sure that there are the same number of atoms of each kind and the same total charge on both sides.
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
T otal charge reactant= 5(+2) + (-1) + 8(+1)= + 10 - 1 + 8= +17
T otal charge product= 5(+3) + (+2) + 4(0)= + 15 + (+2)= +17
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
T otal charge reactant= 5(+2) + (-1) + 8(+1)= + 10 - 1 + 8= +17
T otal charge product= 5(+3) + (+2) + 4(0)= + 15 + (+2)= +17
08/16/1108/16/11 MATTERMATTER 134134
Exercise: In Acidic Solution
C2O42- + MnO4
- + H+ → CO2 + Mn2+ + H2O
Solution :
08/16/1108/16/11 MATTERMATTER 135135
Balancing Redox Reaction In Basic Solution
1. Firstly balance the equation as in acidic solution.
1. Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O.
1. The number of OH- added is equal to the number of H+ in the equation.
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Example: In Basic Solution
Cr(OH)3 + IO3- + OH- → CrO3
2- + I- + H2O
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Exercise:
1. H2O2 + MnO4- + H+ O2 + Mn2+ + H2O
(acidic medium)
2. Zn + SO42- + H2O Zn2+ + SO2 + 4OH-
(basic medium)
3. MnO4- + C2O4
2- + H+ Mn2+ + CO2 + H2O
(acidic medium)
4. Cl2 ClO3- + Cl- (basic medium)
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Stoichiometry
Stoichiometry is the quantitative study of reactants and products in a chemical reaction.
A chemical equation can be interpreted in terms of molecules, moles, mass or even volume.
C3H8 + 5O2 3CO2 + 4H2O
1 molecule of C3H8 reacts with 5 molecules of O2 to produce 3 molecules of CO2 and 4 molecules of H2O
6.02 x 1023 molecules of C3H8 reacts with 5(6.02 x 1023) molecules of O2 to produce 3(6.02 x 1023) molecules of CO2 and 4(6.02 x 1023) molecules of H2O
C3H8 + 5O2 3CO2 + 4H2O
1 mol of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O
44.09 g of C3H8 reacts with 160.00 g of O2 to produce 132.03 g of CO2 and 72.06 g of H2O
5 moles of C3H8 reacts with 25 moles of O2 to produce 15 moles of CO2 and 20 moles of H2O
At room condition, 25 ºC and 1 atm pressure;22.4 dm3 of C3H8 reacts with 5(22.4 dm3) of O2 to produce 3(22.4 dm3) of CO2
Example 1:How many grams of water are produced in the oxidation of 0.125 mol of glucose?
C6H12O6(s) + O2(g) CO2(g) + H2O(l)
Solution:From the balanced equation;1 mol C6H12O6 produce 6 mol H2O
0.125 mol C6H12O6 produce H2O
mass of H2O = (0.125 x 6) mol
x (2.02 + 16.00) g mol-1
= 13.5 g
m o l 1m o l 6m o l 1 2 5.0
Example 2:Ethene, C2H4 burns in excess oxygen to form carbon dioxide gas and water vapour.(a) Write a balance equation of the
reaction(b) If 20.0 dm3 of carbon dioxide gas is
produced in the reaction at STP, how many grams of ethene are used?
Solution:(a) C2H4 + O2 CO2 + H2O
(b) 22.4 dm3 is the volume of 1 mol CO2
20.0 dm3 is the volume of CO2
2 mol CO2 produced by 1 mol C2H4
mol CO2 produced by C2H4
3
3
dm 4.22
m ol 1dm 0.20
4.220.20
m ol 2
m ol 1m ol 4.220.20
massethane1-mol g 4(1. 01)] [2(12. 01) x mol
24.220.20
= 12.5 g
Learning Outcome
At the end of this topic, students should be able to:
a) Define the limiting reactant and percentage yield
b) Perfome stoichiometric calculations using mole concept including
limiting reactant and percentage yield.
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Limiting Reactant/Reagent
Limiting reactant is the reactant that is completely consumed in a reaction and limits the amount of product formed
Excess reactant is the reactant present in quantity greater than necessary to react with the quantity of limiting reactant
Example:3H2 + N2 2NH3
If 6 moles of hydrogen is mixed with 6 moles of nitrogen, how many moles of ammonia will be produced?
Solution:3 mol H2 reacts with 1 mol N2
6 mol H2 reacts with
m o l 3
m o l 1 m o l 6
= 2 m ol N 2
N2 is the excess reactant
H2 is the limiting reactant
limits the amount of products formed
3 mol H2 produce 2 mol NH3
6 mol H2 produce m o l 3
m o l 2 m o l 6
= 4 mol NH 3
or1 mol N2 react with 3 mol H2
6 mol N2 react with mol NH3
m o l 1m o l 3 m o l 6
= 18 mol H2
H2 is not enough
H2 l imits the amount of products formed
lim iting reactant
3 mol H2 produce 2 mol NH3
6 mol N2 produce mol NH3
= 4 mol NH3
m o l 3m o l 2 m o l 6
08/16/1108/16/11 MATTERMATTER 154154
Exercise:
Consider the reaction:
2 Al(s) + 3Cl2(g) 2 AlCl3(s)
A mixture of 2.75 moles of Al and 5.00 moles of Cl2 are allowed to react.
(a) What is the limiting reactant?
(b) How many moles of AlCl3 are formed?
(c) How many moles of the reactant remain at the end of the reaction?
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PERCENTAGE YIELD
The amount of product predicted by a balanced equation is the theoretical yield
The theoretical yield is never obtain because: 1. The reaction may undergo side reaction 2. Many reaction are reversible 3. There may be impurities in the
reactants
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4. The product formed may react further to form other product
5. It may be difficult to recover all of the product from the reaction medium
The amount product actually obtained in a reaction is the actual yield
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Percentage yield is the percent of the actual yield of a product to its theoretical yield
100 x yield
yield yield %
ltheoretica
actual
Example 1:
Benzene, C6H6 and bromine undergo reaction as follows:
C6H6 + Br2 C6H5Br + HBr
In an experiment, 15.0 g of benzene are mixed with excess bromine
(a) Calculate the mass of bromobenzene, C6H5Br that would be produced in the reaction.
(b) What is the percent yield if only 28.5 g of bromobenzene obtain from the experiment?
