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CHEM 101 Second Semester 1438/1439 Dr. Abdulaziz Alghamdi

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1 CHEM 101 Second Semester 1438/1439 Dr. Abdulaziz Alghamdi
مكتب : 2أ 146

2 “Raymond Chang, Chemistry, 10th edition, 2010”
Reference Topics Pages: “Raymond Chang, Chemistry, 10th edition, 2010”

3 Text book: Raymond Chang, Chemistry, 10th edition, 2010
CHEM 101 CURRICULUM Text book: Raymond Chang, Chemistry, 10th edition, 2010 Topics Text book pages Number of Lectures Chapter 1: Chemistry: The Study of Change 1.4. Classifications of Matter: substances and mixtures, elements and compounds How to right symbols of Elements (the table and the explanation (P 12) 1.5. The Three States of Matter 1.6. Physical and Chemical properties of Matter: intensive and extensive properties 1.7. Measurement: SI units, mass and weight, volume, density, temperature scales 1.8. Handling Numbers: scientific notation, significant figures, accuracy and precision 1.9. Dimensional Analysis in Solving Problems: conversion factors, a note on problem solving 8 Review and Exercises

4 CHEM 101 CURRICULUM Chapter 2: Atoms, Molecules and Ions
Text book: Raymond Chang, Chemistry, 10th edition, 2010 Topics Text book pages Number of Lectures Chapter 2: Atoms, Molecules and Ions 2.2. The Structure of the Atoms: the electron, the proton and the neutron Only definitions, masses, and charges [Radioactivity is excluded] 2.3. Atomic Number, Mass Number and Isotopes 2.4. The Periodic Table Periods and groups 1 to 18 - Metals and nonmetals - Alkaline, alkaline earth, halogens, and noble gases. 2.5. Molecules and Ions: molecules, ions Diatomic molecules and polyatomic molecules - Homonuclear monatomic molecules, homonuclear multiatomic molecules, and heteronuclear molecules (= Covalent compounds) - Ions (monatomic ions and polyatomic ions) 8 2.7. Naming Compounds: ionic compound, molecular compound, acids and bases, familiar inorganic compound Review and Exercises FIRST MIDTERM EXAM (15 GRADS)

5 Text book: Raymond Chang, Chemistry, 10th edition, 2010
CHEM 101 CURRICULUM Text book: Raymond Chang, Chemistry, 10th edition, 2010 Topics Text book pages Number of Lectures Chapter 3: Mass Relationships in Chemical Reactions 3.1. Atomic Mass: average atomic mass 3.2. Avogadro's Number and the Molar Mass of an Element 3.3. Molecular Mass 6 3.5. Percent Composition of Compounds 3.6. Experimental Determination of Empirical Formulas: determination of molecular formulas 3.7. Chemical Reactions and Chemical Equations: writing chemical equations, balancing chemical equations 3.8. Amounts of reactants and products 3.9. Limiting Reagents 3.10. Reaction Yield 4.4. Combination reactions, decomposition reactions, combustion reactions Only definition and examples [Oxidation numbers are excluded] Review and Exercises

6 CHEM 101 CURRICULUM Chapter 5: Gases SECOND MIDTERM EXAM (15 GRADS)
Text book: Raymond Chang, Chemistry, 10th edition, 2010 Topics Text book pages Number of Lectures Chapter 5: Gases 5.1. Substances That Exist as Gases 5.2. Pressure of a Gas: SI units of pressure, atmospheric pressure. [Manometer is excluded] 5.3. The Gas Laws: the pressure-volume relationship: Boyle's Law, the temperature-volume relationship: Charles's and Gay-Lussac's law, the volume-amount relationship: Avogadro's Law 5.4. The Ideal Gas Equation: density calculation, the molar mass of a gaseous substance 5 5.6. Dalton's law of Partial Pressures Review and Exercises SECOND MIDTERM EXAM (15 GRADS)

7 Text book: Raymond Chang, Chemistry, 10th edition, 2010
CHEM 101 CURRICULUM Text book: Raymond Chang, Chemistry, 10th edition, 2010 Topics Text book pages Number of Lectures Chapter 6: Thermochemistry 6.3. Introduction to Thermodynamics: the first law of thermodynamics, work and heat 6.4. Enthalpy of Chemical Reactions: enthalpy, enthalpy of reactions, thermochemical equations, a comparison of ∆H and ∆E 6.5. Calorimetry: Only specific heat and heat capacity 5 6.6. Standard Enthalpy of Formation and Reaction: the direct method, the indirect method The direct method (use of enthalpies of formation to calculate enthalpies of other reaction). The indirect method (Hess’s law and its use to calculate enthalpies of other reaction) 11.8. Phase change: vapor pressure, and molar heat of vaporization and boiling point. Some types of Endothermic and exothermic physical changes and chemical reactions (Enthalpy of fusion, vaporization 490, 491, 496, 497 Review and Exercises

8 Text book: Raymond Chang, Chemistry, 10th edition, 2010
CHEM 101 CURRICULUM Text book: Raymond Chang, Chemistry, 10th edition, 2010 Topics Text book pages Number of Lectures Chapter 12: Physical Properties of Solutions 12.1. Types of Solutions [Supersaturated solution is excluded] 12.2. A Molecular View of the Solution Process 10 4.5 Concentration of solution 12.3. Concentration Units: types of concentration units, comparison of concentration units Molarity and dilution of solutions, Percent by mass, mole fraction, molarity 12.4. The Effect of Temperature od Solubility: solid solubility and temperature, gas solubility and temperature [Fractional crystallization is excluded] 12.5. The Effect of Pressure on the Solubility of Gases 12.6. Colligative Properties of Nonelectrolyte Solutions: vapor-pressure lowering (Raoult's Law), boiling-point elevation, freezing-point depression, osmotic pressure, using colligative properties to determine molar mass [Fractional distillation is excluded] Review and Exercises TOTAL HOURS 42

9 Grades Practical 30 1st midterm 15 2nd midterm Final exam 40 Total 100
DISTRIBUTION OF THE 100 GRADES OVER SEMESTER: Grades Practical 30 1st midterm 15 2nd midterm Final exam 40 Total 100 FINAL EXAM WILL BE IN ALL TOPICS الإختبار النهائي سيكون في جميع مواضيع المقرر

10 Important Dates First Midterm Exam Second Midterm Exam Final Exam
سيتم إبلاغكم بمواعيد الإختبارات عندما تحدد

11 Chemistry: The Study of Change
Chapter 1

12 Chemistry is the study of matter and the changes it undergoes
Matter is anything that occupies space and has mass. A substance is a form of matter that has a definite composition and distinct properties. liquid nitrogen gold ingots silicon crystals

13 A mixture is a combination of two or more substances in which the substances retain their distinct identities. Homogenous mixture – composition of the mixture is the same throughout. soft drink, milk, solder Heterogeneous mixture – composition is not uniform throughout. cement, iron filings in sand

14 Physical means can be used to separate a mixture into its pure components.
distillation magnet

15 114 elements have been identified
An element is a substance that cannot be separated into simpler substances by chemical means. 114 elements have been identified 82 elements occur naturally on Earth gold, aluminum, lead, oxygen, carbon, sulfur 32 elements have been created by scientists technetium, americium, seaborgium

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17 dry ice – carbon dioxide
A compound is a substance composed of atoms of two or more elements chemically united in fixed proportions. Compounds can only be separated into their pure components (elements) by chemical means. lithium fluoride quartz dry ice – carbon dioxide

18 Classifications of Matter

19 A Comparison: The Three States of Matter

20 The Three States of Matter: Effect of a Hot Poker on a Block of Ice
solid liquid gas

21 Types of Changes A physical change does not alter the composition or identity of a substance. ice melting sugar dissolving in water A chemical change alters the composition or identity of the substance(s) involved. hydrogen burns in air to form water

22 Extensive and Intensive Properties
An extensive property of a material depends upon how much matter is is being considered. mass length volume An intensive property of a material does not depend upon how much matter is is being considered. density temperature color

23 Matter - anything that occupies space and has mass.
mass – measure of the quantity of matter SI unit of mass is the kilogram (kg) 1 kg = 1000 g = 1 x 103 g weight – force that gravity exerts on an object weight = c x mass on earth, c = 1.0 on moon, c ~ 0.1 A 1 kg bar will weigh 1 kg on earth 0.1 kg on moon

24 International System of Units (SI)

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26 Volume – SI derived unit for volume is cubic meter (m3)
1 cm3 = (1 x 10-2 m)3 = 1 x 10-6 m3 1 dm3 = (1 x 10-1 m)3 = 1 x 10-3 m3 1 L = 1000 mL = 1000 cm3 = 1 dm3 1 mL = 1 cm3

27 Density – SI derived unit for density is kg/m3
1 g/cm3 = 1 g/mL = 1000 kg/m3 density = mass volume d = m V A piece of platinum metal with a density of 21.5 g/cm3 has a volume of 4.49 cm3. What is its mass? d = m V m = d x V = 21.5 g/cm3 x 4.49 cm3 = 96.5 g

28 The SI units of density is kg/m3 , but is not common!
The common units of density for solid and liquid substances is g/cm3 or g/mL But for gas is g/L or g/dm3, because it is very low!

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32 A Comparison of Temperature Scales
K = 0C 273 K = 0 0C 373 K = 100 0C 0F = x 0C + 32 9 5 32 0F = 0 0C 212 0F = 100 0C

33 Convert 172.9 0F to degrees Celsius.
0F = x 0C + 32 9 5 0F – 32 = x 0C 9 5 x (0F – 32) = 0C 9 5 0C = x (0F – 32) 9 5 0C = x (172.9 – 32) = 78.3 9 5

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36 Scientific Notation The number of atoms in 12 g of carbon:
602,200,000,000,000,000,000,000 6.022 x 1023 The mass of a single carbon atom in grams: 1.99 x 10-23 N x 10n N is a number between 1 and 10 n is a positive or negative integer

37 Scientific Notation Addition or Subtraction 568.762 0.00000772
move decimal left move decimal right n > 0 n < 0 = x 102 = 7.72 x 10-6 Addition or Subtraction Write each quantity with the same exponent n Combine N1 and N2 The exponent, n, remains the same 4.31 x x 103 = 4.31 x x 104 = 4.70 x 104

38 Scientific Notation Multiplication Division
(4.0 x 10-5) x (7.0 x 103) = (4.0 x 7.0) x (10-5+3) = 28 x 10-2 = 2.8 x 10-1 Multiply N1 and N2 Add exponents n1 and n2 Division 8.5 x 104 ÷ 5.0 x 109 = (8.5 ÷ 5.0) x = 1.7 x 10-5 Divide N1 and N2 Subtract exponents n1 and n2

39 Significant Figures Any digit that is not zero is significant
1.234 kg significant figures Zeros between nonzero digits are significant 606 m significant figures Zeros to the left of the first nonzero digit are not significant 0.08 L significant figure If a number is greater than 1, then all zeros to the right of the decimal point are significant 2.0 mg significant figures If a number is less than 1, then only the zeros that are at the end and in the middle of the number are significant g 3 significant figures

40 How many significant figures are in each of the following measurements?
24 mL 2 significant figures 3001 g 4 significant figures m3 3 significant figures 6.4 x 104 molecules 2 significant figures 560 kg 2 significant figures

41 Significant Figures Addition or Subtraction
The answer cannot have more digits to the right of the decimal point than any of the original numbers. 89.332 1.1 + 90.432 one significant figure after decimal point round off to 90.4 3.70 0.7867 two significant figures after decimal point round off to 0.79

42 Significant Figures Multiplication or Division
The number of significant figures in the result is set by the original number that has the smallest number of significant figures 4.51 x = = 16.5 3 sig figs round to 3 sig figs 6.8 ÷ = = 0.061 2 sig figs round to 2 sig figs

43 Significant Figures Exact Numbers
Numbers from definitions or numbers of objects are considered to have an infinite number of significant figures The average of three measured lengths; 6.64, 6.68 and 6.70? 3 = = 6.67 = 7 Because 3 is an exact number

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46 Accuracy – how close a measurement is to the true value
Precision – how close a set of measurements are to each other accurate & precise precise but not accurate not accurate & not precise

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48 Dimensional Analysis Method of Solving Problems
Determine which unit conversion factor(s) are needed Carry units through calculation If all units cancel except for the desired unit(s), then the problem was solved correctly. given quantity x conversion factor = desired quantity desired unit given unit given unit x = desired unit

49 Dimensional Analysis Method of Solving Problems
How many mL are in 1.63 L? Conversion Unit 1 L = 1000 mL 1L 1000 mL 1.63 L x = 1630 mL 1L 1000 mL 1.63 L x = L2 mL

50 The speed of sound in air is about 343 m/s
The speed of sound in air is about 343 m/s. What is this speed in miles per hour? conversion units meters to miles seconds to hours 1 mi = 1609 m 1 min = 60 s 1 hour = 60 min 343 m s x 1 mi 1609 m 60 s 1 min x 60 min 1 hour x = 767 mi hour

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