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08/16/11matter*CHEMISTRY SK0161.0Matter72.0Atomic Structure73.0Periodic Table44.0Chemical Bonding25.0State of Matter76.0Chemical Equilibrium57.0Ionic Equilibria12Total 54

ChapterTopicsHours

08/16/11matter*CHEMISTRY SK026 8.0Thermochemistry 4 9.0Electrochemistry 610.0Reaction Kinetics 711.0Intro To Organic Chemistry 412.0Hydrocarbons 813.0Aromatic Compounds 314.0Haloalkanes (Alkyl halides) 415.0Hydroxy compounds 3

ChapterTopicHours

08/16/11matter*CHEMISTRY SK026

16.0 Carbonyl 417.0 Carboxylic acids & Derivatives 418.0 Amines 519.0 Amino acids and Proteins 220.0 Polymers 1

ChapterTopicHour

08/16/11matter*ASSESSMENT1. 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%

08/16/11matter*REFERENCE BOOKSCHEMISTRY ,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.

08/16/11matter*

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

08/16/11matter*Chapter 1 : MATTER1.1Atoms and Molecules1.2 Mole Concept1.3Stoichiometry

08/16/11matter*1.1Atoms and Molecules

08/16/11matter*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.

08/16/11matter*IntroductionMatter Anything that occupies space and has mass.e.g: air, water, animals, trees, atoms, etc

Matter may consists of atoms, molecules or ions.

Classifying Matter08/16/11matter*

A substance is a form of matter that has a definite or constant composition and distinct properties.Example: H2O, NH3, O2

A mixture is a combination of two or more substances in which the substances retain their identity. Example : air, milk, cement

08/16/11matter*

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 : CO2, H2O, CuO

08/16/11matter*Three States of MatterSOLIDLIQUIDGAS

08/16/11matter*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

08/16/11matter* 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.

08/16/11matter*Subatomic Particles

ParticleMass(gram)Charge(Coulomb)Charge (units)Electron (e)9.1 x 10-28-1.6 x 10-19-1Proton (p)1.67 x 10-24+1.6 x 10-19+1Neutron (n)1.67 x 10-2400

08/16/11matter* IsotopeIsotopes are two or more atoms of the same element that have the same proton number in their nucleus but different nucleon number.

Examples:

08/16/11matter*Isotope Notation X=Element symbol

Z=Proton number of X (p)

A=Nucleon number of X=p + nAn atom can be represented by an isotope notation ( atomic symbol )

08/16/11matter*

Total charge on the ion

Proton number of mercury, Z = 80

Nucleon number of mercury, A = 202The 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

08/16/11matter*Exercise 1Give the number of protons, neutrons, electrons and charge in each of the following species:

Symbol Number of :ChargeProtonNeutronElectron

08/16/11matter*Exercise 2Write the appropriate notation for each of the following nuclide :

Species Number of :Notation for nuclideProtonNeutronElectronA222B120C111D7710

08/16/11matter*A molecule consists of a small number of atoms joined together by bonds.b) Molecules

08/16/11matter*A diatomic molecule Contains only two atomsEx : H2, N2, O2, Br2, HCl, CO

A polyatomic moleculeContains more than two atomsEx : O3, H2O, NH3, CH4

08/16/11matter*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.

08/16/11matter* Relative MassRelative 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

08/16/11*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 amu1 u = 1.660538710-24 g

The relative isotopic mass is the mass of an atom, scaled with 12C.

08/16/11matter*Example 1Determine the relative atomic mass of an element Y if the ratio of the atomic mass of Y to carbon-12 atom is 0.45ANSWER:

08/16/11matter*ii) Relative Molecular Mass, MrA mass of one molecule of a compound compared to 1/12 mass of one atom of 12C with the mass 12.000amu

The relative molecular mass of a compound is the summation of the relative atomic masses of all atoms in a molecular formula.

08/16/11matter*Example 2Calculate the relative molecular mass of C5H5N,Ar C= 12.01Ar H= 1.01Ar N= 14.01

08/16/11*MASS SPECTROMETERAn atom is very light and its mass cannot be measured directlyA mass spectrometer is an instrument used to measure the precise masses and relative quantity of atoms and molecules

08/16/11matter*

08/16/11*Mass Spectrum of Monoatomic ElementsModern mass spectrum converts the abundance into percent abundance

08/16/11matter*Mass Spectrum of MagnesiumThe 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 isotopesRelative abundance

08/16/11matter*Learning OutcomesAt 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.

08/16/11matter*How to calculate the relative atomic mass, Ar from mass spectrum?Ar is calculated using data from the mass spectrum. The average of atomic masses of the entire elements isotope as found in a particular environment is the relative atomic mass, Ar of the atom.

08/16/11*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

08/16/11*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 uRelative atomic mass Cu = 63.55

08/16/11*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

08/16/11*Solution: Average atomic =mass of Ir

=

= 192.254 u

Relative atomic mass Ir = 192.254

Mass Spectrum of Molecular ElementsA 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+ + 2eCl2Cl2 + e 2Cl+ + 2e35Cl-35Cl35Cl-37Cl37Cl-37Cl35Cl-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?

MATTER1.2 Mole Concept

Learning OutcomeAt the end of this topic, students should be able to:a)Define mole in terms of mass of carbon-12 and Avogadros constant, NA

Avogadros Number, NAAtoms and molecules are so small impossible to countA unit called mole (abbreviated mol) is devised to count chemical substances by weighing themA mole is the amount of matter that contains as many objects as the number of atoms in exactly 12.00 g of carbon-12 isotopeThe number of atoms in 12 g of 12C is called Avogadros number, NA = 6.02 x 1023

Example:

1 mol of Cu contains Cu atoms

1 mol of O2 containsO2 moleculesO atoms

1 mol of NH3 containsNH3 moleculesN atomsH atoms6.02 10233 6.02 10232 6.02 10236.02 10236.02 10236.02 1023

1 mol of CuCl2 contains Cu2+ ions Cl- ions6.02 10232 6.02 1023

Mole and MassExample: Relative atomic mass for carbon, C = 12.01Mass of 1 C atom = 12.01 amuMass 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

12.01 amu = 1.995 x 10-23 g

1 amu =

= 1.66 x 10-23 g

Example:From the periodic table, Ar of nitrogen, N isThe 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 amu14.01 g14.01 g mol128.02 g mol114

Mr of CH4 isThe mass of 1 CH4 molecule =The mass of 1 mol of CH4 molecules =The molar mass of CH4 molecule =16.0516.05 amu16.05 g16.05 g mol1

08/16/11MATTER*Learning OutcomeAt the end of this topic, students should be able to:Interconvert between moles, mass, number of particles, molar volume of gas at STP and room temperature. Define the terms empirical & molecular formulaeDetermine empirical and molecular formulae from mass composition or 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 molecules1 mol of O2 molecules

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 atoms1 mol of N2 molecules

Example 3:Calculate the mass of 0.25 mol of chlorine gas.Solution:1 mol Cl2

0.25 mol Cl2

18 g ormass= mol x molar mass= 0.25 mol x (2 x 35.45 g mol-1)= 18 g2 35.45 g

Example 4:Calculate the mass of 7.528 x 1023 molecules of methane, CH4Solution:6.02 x 1023 CH4 molecules (12.01 + 4(1.01)) g

7.528 x 1023 CH4 molecules

= 20.06 g

Molar Volume of GasesAvogadro (1811) stated that equal volumes of gases at the same temperature and pressure contain equal number of moleculesMolar volume is a volume occupied by 1 mol of gasAt standard temperature and pressure (STP), the molar volume of an ideal gas is 22.4 L mol1Standard Temperature and Pressure 273.15 K1 atm 760 mmHg 0 C101325 N m-2101325 Pa

08/16/11MATTER*Standard Molar Volume

At room conditions (1 atm, 25 C), the molar volume of a gas = 24 L mol-1

08/16/11MATTER*Example 1:Calculate the volume occupied by 1.60 mol of Cl2 gas at STP.Solution:At STP,1 mol Cl2 occupies

1.60 mol Cl2occupies

= 35.8 L22.4 L

08/16/11MATTER*Example 2:Calculate the volume occupied by 19.61 g of N2 at STPSolution:1 mol of N2 occupies22.4 L

of N2 occupies

= 15.7 L

08/16/11MATTER*Example 3:0.50 mol methane, CH4 gas is kept in a cylinder at STP. Calculate:(a)The mass of the gas(b)The volume of the cylinder(c)The number of hydrogen atoms in the cylinderSolution:(a)Mass of 1 mol CH4=Mass of 0.50 mol CH4=

= 8.0 g16.05 g

08/16/11MATTER*(b)At STP;1 mol CH4 gas occupies

0.50 mol CH4 gas occupies

= 11 L

(c)1 mol of CH4 molecules 4 mol of H atoms0.50 mol of CH4 molecules2 mol of H atoms 1 mol of H atoms

2 mol of H atoms

1.2 x 1024 atoms22.4 L6.02 x 1023 atoms2 x 6.02 x 1023 atoms

08/16/11MATTER*Exercise A sample of CO2 has a volume of 56 cm3 at STP. Calculate:

The number of moles of gas molecules (0.0025 mol)

The number of CO2 molecules (1.506 x 1021 molecules)

The number of oxygen atoms in the sample (3.011x1021atoms) Notes:1 dm3= 1000 cm31 dm3= 1 L

08/16/11MATTER*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.

08/16/11MATTER*The relationship between empirical formula and molecular formula is :

Molecular formula = n ( empirical formula )

08/16/11MATTER* 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.

Solution:

08/16/11MATTER* 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

08/16/11MATTER*At the end of this topic, students should be able to:(a) Define and perform calculation for each of the following concentration measurements : i) molarity (M) ii) molality(m) iii) mole fraction, X iv) percentage by mass, % w/w v) percentage by volume, %v/vLearning Outcome

08/16/11MATTER*Concentration of SolutionsA solution is a homogeneous mixture of two or more substances:solvent + solute(s)

e.g:sugar + water solutionsugar solutewater solvent

08/16/11MATTER*

08/16/11MATTER*Concentration of a solution can be expressed in various ways : a)molarityb)molalityc)mole fractiond)percentage by mass e)percentage by volume

08/16/11MATTER*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

Example 1: Determine the molarity of a solution containing 29.22 g of sodium chloride, NaCl in a 2.00 L solution.

Solution:

= 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:

08/16/11MATTER*b) MolalityMolality is the number of moles of solute dissolved in 1 kg of solvent

Units of molality:mol kg-1 molal m

Example:What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?

Solution:

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

08/16/11MATTER*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:

08/16/11MATTER*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 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:

08/16/11MATTER*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

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:

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%

08/16/11MATTER*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.

Example 1: 25 mL of benzene is mixed with 125 mL of acetone. Calculate the volume percent of benzene solution.

Solution:

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:

Example 3:A 6.25 m of sodium hydroxide, NaOH solution has has a density of 1.33 g mL-1 at 20 C. Calculate the concentration NaOH in:(a)molarity(b)mole fraction(c)percent by mass

Solution:

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

MATTER1.3 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

08/16/11MATTER*Balancing Chemical EquationA 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.

08/16/11MATTER*Example:x A+y Bz C+w DReactantsProducts

08/16/11MATTER*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

08/16/11MATTER*Inspection MethodWrite down the unbalanced equation. Write the correct formulae for the reactants and products.

Balance the metallic atom, followed by non-metallic atoms.

Balance the hydrogen and oxygen atoms.

Check to ensure that the total number of atoms of each element is the same on both sides of equation.

08/16/11MATTER*Example: Balance the chemical equation by applying the inspection method.

NH3 + CuO Cu + N2 + H2O

08/16/11MATTER* ExerciseBalance the chemical equation below by applying inspection method. Fe(OH)3 + H2SO4 Fe2(SO4)3 + H2O C6H6 + O2 CO2 + H2O N2H4 + H2O2 HNO3 + H2O ClO2 + H2O HClO3 + HCl

08/16/11MATTER*Redox ReactionMainly for redox (reduction-oxidation) reaction

08/16/11MATTER*Oxidation is defined as a process of electron loss.The substance undergoes oxidation loses one or more electrons.increase in oxidation numberact 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 numberact as an oxidizing agent (electron acceptor)Half equation representing reduction: Br2 + 2e Br- Sn4+ + 2e Sn2+ Al3+ + 3e Al

08/16/11MATTER* Oxidation numbers of any atoms can be determined by applying the following rules:

For monoatomic ions, oxidation number = the charge on the ione.g:ionoxidation numberNa++1Cl--1Al3++3S2--2

08/16/11MATTER*2.For free elements, e.g: Na, Fe, O2, Br2, P4, S8 oxidation number on each atom = 0

For most cases, oxidation number for O= -2 H= +1Halogens= -1

08/16/11MATTER*Exception:

H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1

Halogen bonded to oxygen (e.g: Cl2O7) oxidation number for halogen = +ve

In a neutral compound (e.g: H2O, KMnO4) the total of oxidation number of every atoms that made up the molecule = 0

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

08/16/11MATTER*ExerciseAssign the oxidation number of Mn in the following chemical compounds.i.MnO2ii.MnO4-

Assign the oxidation number of Cl in the following chemical compounds.i.KClO3ii.Cl2O72-

Assign the oxidation number of following:i. Cr in K2Cr2O7ii. U in UO22+iii. C in C2O42-

08/16/11MATTER*Balancing Redox ReactionRedox reaction may occur in acidic and basic solutions.

Follow the steps systematically so that equations become easier to balance.

08/16/11MATTER*Balancing Redox Reaction In Acidic SolutionFe2+ + MnO4- Fe3+ + Mn2+

Separate the equation into two half-reactions: reduction reaction and oxidation reactioni.Fe2+ Fe3+ ii.MnO4- Mn2+

08/16/11MATTER*Balance atoms other than O and H in each half-reaction separatelyi.Fe2+ Fe3+

ii.MnO4- Mn2+

08/16/11MATTER*3.Add H2O to balance the O atomsAdd H+ to balance the H atoms

i.Fe2+ Fe3+ ii.MnO4- + Mn2+ +4.Add electrons to balance the chargesi.Fe2+ Fe3+ + ii.MnO4- + 8H+ + Mn2+ + 4H2O4H2O8H+1 e5 e

08/16/11MATTER*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

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 ___________________________________

08/16/11MATTER*Exercise: In Acidic Solution C2O42- + MnO4- + H+ CO2 + Mn2+ + H2O

Solution :

08/16/11MATTER*Balancing Redox Reaction In Basic SolutionFirstly balance the equation as in acidic solution.

Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O.

The number of OH- added is equal to the number of H+ in the equation.

08/16/11MATTER*Example: In Basic SolutionCr(OH)3 + IO3- + OH- CrO32- + I- + H2O

08/16/11MATTER*Exercise:

1.H2O2 + MnO4- + H+ O2 + Mn2+ + H2O(acidic medium)2.Zn + SO42- + H2O Zn2+ + SO2 + 4OH-(basic medium)3.MnO4- + C2O42- + H+ Mn2+ + CO2 + H2O (acidic medium)4.Cl2 ClO3- + Cl-(basic medium)

08/16/11MATTER*StoichiometryStoichiometry 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 H2O6.02 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 C 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 molC6H12O6 produce H2O

mass of H2O = (0.125 x 6) mol x (2.02 + 16.00) g mol-1= 13.5 g

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

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.

08/16/11MATTER*Limiting Reactant/ReagentLimiting 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 with1 mol N2

6 mol H2 reacts with

N2 is the excess reactantH2 is the limiting reactant limits the amount of products formed

3 mol H2 produce 2 mol NH3

6 mol H2produce

or1 mol N2 react with 3 mol H26 mol N2 react with mol NH3

3 mol H2 produce 2 mol NH3

6 mol N2produce mol NH3= 4 mol NH3

08/16/11MATTER*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?

08/16/11MATTER*PERCENTAGE YIELDThe amount of product predicted by a balanced equation is the theoretical yieldThe 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

08/16/11MATTER*4. The product formed may react further to form other product5. 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

08/16/11MATTER*Percentage yield is the percent of the actual yield of a product to its theoretical yield

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?

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