E.D. Feehan Chemistry 20 Course Outline

Text: Davis, R. E., Metclafe, H. C., Williams, J. E., & Castka, J. F. (2002). Modern chemistry. Toronto, Ontario: Holt, Rinehart, and Winston.

Instructor: Mr. J. Nicholson, room 206 jnicholson@scs.sk.caAvailable: 8:05-8:30am

12:15-1:00pm - except department meetings 3:10-4:30pm - depending on coaching

Course Overview Chapter in TextUnit 1: Introduction (sig figs, conversions, matter and change) Chapters 1 + 2Unit 2: Naming and Organic Chemistry Chapters 7 + 20Unit 3: The Mole Chapters 3, 7 + 13Unit 4: Chemical Equations and Reactions Chapters 8 + 17Unit 5: Stoichiometry Chapter 9Unit 6: The Atom (history, quantum theory and electron configuration) Chapters 3, 4, + 5Unit 7: Chemical Bonding (VSEPR theory, intermolecular forces) Chapter 6

EvaluationLaboratory 15 %Unit Exams and Quizzes 45 %Project 10 %Final Exam 30 % Total 100 %

Chemistry 20 is an introductory course into chemistry. This course is intensive and often extraassistance is required. The text is your most valuable resource. It also lists numerous web siteswith supplementary information and explanations. I am also available for tutorials, howeverbooking this time is essential. Although no mark is assigned for most assignments, failure tocomplete them will likely lead to a low mark. Therefore daily revision of topics is stronglyrecommended. Many students enter this course who have excelled in science classes with littleeffort and expect to continue this in chemistry 20. They are usually disappointed with theirresulting mark. Your success depends ultimately on your effort.

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Unit 1: Introduction Chapters 1 + 2Outline-SI units, derived units -sig figs and scientific notation-metric conversion and conversion factors -functions and graphing-lab setup -components of matter-senses and technology -phases-accuracy vs. precision -chemical vs. physical change-% error -elements and the periodic table, metals, nonmetals

Metric Units: Le Système International d'Unités (SI)SI Units Symbol Unit Unit Abbreviation p.34Length l metre mMass m kilogram kgTime t second sTemperature T degrees Celsius °CAmount of substance n mole molElectric current I ampere ALuminous intensity Iv candela cd(Derived units)Area A square meter m2 (l x w)Volume V cubic metre m3 (l x w x h)Density D grams per cubic cm g/cm3 (mass ÷ vol)Molar mass M or mm grams per mole g/mol (mass ÷ mol)Concentration c moles per litre mol/L or M (mol ÷ vol)Energy E joule J (force x distance)

SI PrefixesName Symbol Meaning p.35tera T 1012 1 000 000 000 000giga G 109 1 000 000 000mega M 106 1 000 000kilo k 103 1 000hecto h 102 100deca da 101 10base 100 1deci d 10−1 0.1centi c 10−2 0.01milli m 10−3 0.001micro µ 10−6 0.000 000 1nano n 10−9 0.000 000 000 1pico p 10−12 0.000 000 000 000 1

base

T || G || M || k h da | d c m || µ || n || p

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Metric Conversion and Conversion Factors p.40To change the prefix from what you have to what you want, multiply it by a conversion factor tocancel what you have, and give you what you want.How many km in 30 000 m?

30 000 m = ______ km conversion factors: 1 km 1000 m 1000 m 1 km

30 000 m x 1 km = 30 000 m km = 30 km 1000 m 1000 m

How many mg in 6.25 kg?6.25 kg = ______ mg 6.25 kg x 1 000 000 mg = 6 250 000 mg

1 kgHow many seconds in 3.28 years?

3.28 y x 365 d = 1197.2 d x 24 h = 28732.8 h x 60 min = 1 y 1 d 1 h

1 723 968 min x 60 s = 103 438 080 s1 min

Laboratory Report FormatInvestigations in this class are intended to be exploratory and deductive (forming a conclusionfrom a set of data). In general, students will have to design their own experiments, and will notbe given a set of instructions. It is important to understand your experiment and all its parametersbefore commencing the lab.

Do not use a title page; put information on first page.Identifying Information: name, date, class, period, partnersTitle: GivenPurpose: GivenMaterials: Sometimes givenProcedure: Write a paragraph of instructions for others to follow if they were to repeat

your lab. Be thorough: include all directions and amounts.Data and Observations: Include all data recorded. Use a chart if possible.Calculations: Show all calculations.Conclusion: Provide a statement that answers the purpose.Reflection Questions: Answer all questions.Bonus Questions: If applicableGraphs: Graphs may be hand drawn or computer generated. (standard formatting)

Senses and TechnologyYour senses are easily fooled. We must therefore use technology to make our senses morereliable and to provide us with immediate, efficient, and prescriptive information. (comes fromsomething)

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Accuracy Vs. Precision p.44Every measured value inherently has a level of precision and accuracy.Precision describes how close a set of measurements come to each other. (taken same way)

3.247 cm is more precise than 3.2 cmAccuracy describes how close a measured value comes to its actual or accepted value.

if the actual value is 32.3 mL, then 32.0 mL is more accurate than 31.9 mL

Percent Error p.45Describes how close a value comes to an accepted value, as a percent of the accepted value

% error = accepted − experimental x 100 accepted

If the answer is positive, the experimental value is greater than the accepted value. If the answeris negative, the experimental value is less than the accepted value.

Significant Figures (s.f. are numbers that count for precision purposes) p.461. All non-zero numbers are significant. 615 has 3 sf

2. Zeros between real numbers are significant. 12 004 has 5 sf

3. For numbers greater than 1, (without a decimal) zeros at the end are not significant. 1 200 has 2 sf 60 000 has 1 sf

4. For numbers less than 1, zeros at the beginning are not significant. 0.0032 has 2 sf

5. For decimal numbers, zeros after non-zero numbers are significant. 0.0400 has 3 sf

6. For decimal numbers greater than 1, all zeros are significant.12.00 has 4 sf 50.0000 has 6 sf 9 000 000.0 has 8 sf

7. Exact numbers from conversion factors or from counting objects have an infinite number ofsignificant figures.

For addition and subtraction: add or subtract, then round off to the least number of DECIMAL PLACES. For multiplication and division: multiply or divide, then round off to the least number of SIGNIFICANT FIGURES.

Scientific Notation p.50Move the decimal to the place after the first non-zero number. The number of places it is movedbecomes the exponent of 10, which is multiplied.

65 000 → 6.5 x 104 0.000 0400 → 4.00 x 10−5 Numbers >1 have (+) exponents Numbers <1 have (−) exponents

Note: number of sig figs is the same in both notations

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Functions p.55~When one quantity depends on a second quantity, we say that the first is a function of the second.

The area of a circle depends on its radius:A = π r2 We say that A is a function of r.

When an object is dropped, the distance it falls (metres) depends on how long it falls or time(seconds).

d = 4.905t2 d is a function of t.what you do to x to get y ↑

In general, a quantity “y” is a function of “x” or y = f(x) ↓ ↓

We say “y is the value that f assigns to x” dependent independent variable variable

From above, A = f(r) = π r2 d = f(t) = 4.905t2

Recall: y = mx + b or y = m(x)

Graphing1. Always use a pencil.2. Decide witch variable is independent (the one that you are measuring with) and which isdependent. (the one that is measured)3. The independent variable goes on the x axis (horizontal) and the dependent goes on the y axis.4. Choose a scale for each axis. Use multiples of 0.1, 0.2, 0.5, 1, 2, 5, or 10 to make it easier towork with. Find the highest value in the data set, and figure out the best way to fit onto your axis.Don't forget to leave room for extrapolation, if required. 5. Label the axes with a name and units. ex) distance (m)6. Plot the data, use small points. Decide if the relationship is linear (straight line) or not. If it islinear, use a ruler to draw a line of best fit. If the graph is non linear, draw a smooth curve.Usually you do not connect the dots.7. Give the graph a meaningful title, incorporating the axis names and units.8. Put your name and the date in the upper right hand corner.

Components of Matter p.10In order to understand matter, we must examine its components. Some definitions:

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0 1 2 3 4 5 6

Time (s)

0

5

10

15

20

Dis

tanc

e (m

)

The Distance of a Runner (m) vs Time (s) name, date

Atom: a small "particle" that contains protons, neutrons, and electrons.

Molecule: particles that consist of 2 or more atoms covalently bonded together.

Element: a pure substance that cannot be decomposed into simpler substances. They are made ofidentical atoms. ex) gold, carbon, hydrogen

Compound: a pure substance made from the atoms of 2 or more elements that are chemicallybonded. ex) water - H2O ; sucrose - C12H22O11

��

HCl molecule↵ ��

�� ��

→

� Cl atom↵ � ��

�

+

� H atom↵ � �

� Element Element Compound (hydrogen) (chlorine) (hydrogen chloride)(note: both hydrogen and chlorine are actually diatomic, which will be explained later)

Chemical and Physical Changes p.11Physical change is matter changing in appearance without forming a new substance. It isbasically reversible. (rearrangement of particles) ex) phase change, dissolving, breaking a branch.recall: p.382

Gas Gas condensation� �vapourization � ||

Liquid sublimation deposition freezing� �melting || �

Solid Solid

Chemical change is a change in which a new substance is formed, having different properties. Itis not easily reversible. (new substance)

5 signs of a chemical reaction:1) change in colour 2) gives off light3) forms a gas (bubbles)4) forms a precipitate5) change in temperature

Elements and the Periodic Table p.20Elements are pure substances that cannot decompose chemically. Each element has characteristicproperties and can be grouped with similar elements on the periodic table. Vertical columns onthe table are called groups or families and share similar chemical properties. They are numbered1 to 18, left to right. Horizontal rows on the table are called periods. Properties of elementschange somewhat regularly across a period. They are numbered 1 to 7 from top to bottom.

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Metals and Nonmetals p.22Elements on the periodic table can classified as either metal or nonmetal. Elements to the right ofthe zigzag line are nonmetals and metals are to the left. (except hydrogen)

General Properties of Metals General Properties of Nonmetals-conduct heat and electricity -poor conductors of heat and electricity-solid at room temperature (higher m.p.) -gas at room temperature (lower m.p.)-grey and shiny in appearance -dull and varied in appearance-malleable (can be hammered into a sheet) -brittle, not malleable or ductile-ductile (can be drawn into a wire)

Metalloids p.23Some elements that border the line separating the metals and nonmetals are called metalloids.These include B, Si, Ge, As, Sb, and Te and have properties in between that of metals andnonmetals. They are also semiconductors of electricity.

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Unit 2: Naming and Organic Chemistry Chapters 7, 20 + 21Outline-ionic charges -acids -alkynes-polyatomic ions -alkanes -benzene and phenol-oxidation numbers -cycloalkanes -other functional groups-formulas from names -isomers-names from formulas -alkenes

Ionic Charges p.204Compounds will tend to form by gaining, losing, or sharing electrons so that the valence shell ofeach atom is satisfied. This is known as the octet rule, and even though there are manyexceptions, it is a useful concept.

Sodium has 1 valence electron, and will lose that electron in a reaction to become Na+. Noticethat Na+ has a full valence shell.Chlorine has 7 valence electrons and will gain 1 electron to complete its octet�� Cl�

n/a1�2�3�4±3+2+1+GroupCharge

NeFONCBBeLiFamily

Ions are named after their element, however anions have their suffix changed to "ide".Ca2+ calcium O2� oxideAg+ silver Cl� chloride

Many transition metals and metals to their right make more than one stable ion. In order to figureout which one is being used, you need to see a formula, or be given the type in the name. Forexample, copper makes 2 stable ions: Cu + and Cu 2+,

These ions are named copper(I) and copper(II) respectively.

Polyatomic Ions Ions can also be made of more than one atom: p.210

NH4+ ammonium SO4

2� sulfateClO3

� chlorate CO32� carbonate

NO3� nitrate PO4

3� phosphate OH� hydroxide HCl hydrochloric acidCH3COO� acetate (also C2H3O2

�) H2SO4 sulfuric acidHCO3

� hydrogen carbonate (bicarbonate) NH3 ammonia

Oxidation Numbers p.216The distribution of electrons in a molecule can be described by oxidation numbers, which areusually similar to the charge an atom will make. Oxidation numbers differ from charges in thatthey are not physically real, but only a mathematical explanation of the bonding.Some examples: oxidation states Li+1 , Ba+2 , S�2 , Cu+1 , Br��

charges Li� , Ba2+, S2�, Cu +, Br���

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Assigning Oxidation numbers 1. Atoms in pure elements have an OD# of 0. ex) Na, O2, P4, S8 all have OD# of 02. Elements are assigned the OD# equal to the charge they would have as an ion. The most andleast electronegative elements get their group charges. ex) NaOH Na is +1, O is −2 (H is +1)3. Fluorine is always −1.4. Oxygen is almost always −2. Exceptions H2O2 (O = −1), OF2 (O = +2)5. Hydrogen is +1 when bonded to a more electronegative element.6. OD#s add up to 0 in a neutral compound. In a polyatomic ion, they add up to equal the charge.7. These rules can also be applied to ionic compounds. ex) NaCl Na is +1, Cl is −1.Use these rules to assign OD#s to each element in a compound:

NaF +1NaF− 1 H2SO4

+1H2 S ? O4− 2 2(1) +___+ 4(−2) = 0 S is +6

PO43− ? PO4

− 2 ____ + 4(−2) = −3 P is +5

Formulas From Names p.206Ionic compounds (Metal + Nonmetal; Polyatomic Ions)Write the ions from the names. Then, using subscripts to balance charges, derive a formula.Recall: all compounds must have a net ionic charge of zero. The charges can be found by thegroup on the periodic table, from roman numerals, or from memory (ions).

barium fluoride LCM = 2 1(Ba2+) = +2

Ba2+ F� → BaF2 2(F1�) = �2 boron hydroxide 0 B3+ OH� → B(OH)3 (polyatomic ions need brackets if use more than 1)uranium (VI) oxide U6+ O2� → UO3

Molecular compounds (Nonmetal + Nonmetal) p.211If the compound name has a prefix(es), it is molecular. Simply turn the prefixes into subscriptsfor that element. If there is no prefix, it is single. DO NOT SWITCH SUBSCRIPTS.1-mono 5-penta 9-nona2-di 6-hexa 10-deca Prefixes for binary molecular compounds3-tri 7-hepta4-tetra 8-octa

dinitrogen trioxide → N2O3 iodine pentafluoride → IF5

tetraphosphorous decoxide → P4O10

Give the formula for the following compounds:1. sodium chloride 11. sodium chlorate 21. sulfur trioxide2. calcium fluoride 12. calcium nitrate 22. dinitrogen tetroxide3. magnesium sulfide 13. aluminum sulfate 23. carbon tetraiodide4. aluminum oxide 14. silver phosphate 24. diphosphorous pentachloride5. zinc oxide 15. ammonium chloride 25. carbon dioxide6. chromium(II) oxide 16. iron(II) hydroxide 26. beryllium nitrate7. copper(II) bromide 17. nickel(III) acetate 27. phosphorous(V) fluoride8. manganese(VII) oxide 18. titanium(IV) carbonate 28. cobalt(III) sulfate9. tin(IV) iodide 19. sodium bicarbonate 29. dihydrogen monoxide10. iron(III) oxide 20. uranium(III) oxide 30. potassium hydroxide

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Naming Compounds p.206Ionic Compounds (Metal + Nonmetal ; Polyatomic Ions) Name using the names of the ions.

MgCl2 → magnesium chloride Al2O3 → aluminum oxideIf there are polyatomic ions, name them in full.

Na2CO3 → sodium carbonate (NH4)2SO4 → ammonium sulfateIf there are metals that have more than one oxidation state, use roman numerals to representwhich charge was used. To find this, work back from the known anion. We always know theanion charge because the more electronegative element (on the right) gets its group charge.

Fe2O3 → iron( ) oxideoxide is O2� and there are 3�� anions total 6��� (3 x 2�)� the cations must total 6+take 6+ and divide by the number of Fe: 6�2 = 3+� each Fe is Fe3+

iron (III) oxide SnCl4 → Sn4+ Cl� → tin(IV) chlorideCu(C2H3O2)2 → Cu2+ C2H3O2

� → copper(II) acetateMnO2 → Mn4+ O2� → manganese(IV) oxide

Molecular Compounds (nonmetals + nonmetals) p.211There are 2 ways to name molecular compounds: prefix and stock1) the prefix method → give prefixes to show number of each atom in the molecule.

P2O5 → diphosphorous pentoxideAs2S3 → diarsenic trisulfide

Do not include “mono” on the first ion, but always on the second. CO2 → carbon dioxide N2O → dinitrogen monoxide

2) stock system → Give less electronegative element a roman numeral to show oxidation state, asdone with metals.

P2O5 → phosphorous(V) oxide CO2 → carbon(IV) oxideAs2S3 → arsenic(III) sulfide N2O → nitrogen(I) oxide

Name the following formulas:1. LiCl 11. Ba(OH)2 21. SO2

2. K2S 12. Na2SO4 22. CS2

3. Sr3N2 13. Cs3PO4 23. N2O4

4. Ga2O3 14. Sc(NO3)3 24. Cl2O5. ZrF4 15. Rb2S 25. P4O10

6. NiBr2 16. Mn(CH3COO)2 26. Co(C2H3O2)2

7. FeO 17. Cu(ClO3)2 27. NH4HCO3

8. Co2S3 18. Ti2(SO4)3 28. Zr3P4

9. Mn3N2 19. NH4F 29. H2SO4

10. CrO3 20. Fe2(CO3)3 30. BF3

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Acids p.214Binary acids contain hydrogen and one of the more electronegative elements. Oxyacids containhydrogen, oxygen and a third element, usually a nonmetal. Some common acids:Binary acids OxyacidsHF hydrofluoric acid CH3COOH acetic acidHCl hydrochloric acid H2CO3 carbonic acidHBr hydrobromic acid HNO3 nitric acidHI hydriodic acid H3PO4 phosphoric acidH2S hydrosulfuric acid H2SO4 sulfuric acid

The prefixes and suffixes refer to the number of oxygen atoms on the central atom:hypochlorous acid HClO hypo____ous means "2 less oxygen"chlorous acid HClO2 ____ous means "1 less oxygen"chloric acid HClO3 ____ic is the most common form or first discovered perchloric acid HClO4 per____ic means "1 more oxygen"

Organic ChemistryOrganic chemistry is the study of compounds containing carbon. (except carbonates and oxides)Organic compounds are named according to a different IUPAC system. (International Union ofPure and Applied Chemistry)

Saturated Hydrocarbons - Alkanes p.634Hydrocarbons are compounds with only hydrogen and carbon. Being saturated means that eachcarbon, which can make 4 bonds, has as much hydrogen on it as possible. Hydrocarbons withonly single bonds (saturated) are called alkanes, and use the following prefixes to tell the # of C.

meth_ = 1 hex_ = 6eth_ = 2 hept_ = 7prop_ = 3 oct_ = 8but_ = 4 non_ = 9pent_ = 5 dec_ = 10

Hmethane CH4 H-C-H H Hethane C2H6 � H H-C-C-Hpropane C3H8 �� H Hbutane C4H10 ���

pentane C5H12 ����

hexane C6H14 �����

heptane C7H16 ������

octane C8H18 �������

nonane C9H20 ��������

decane C10H22 ���������

alkanes CnH2n+2

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Adding groups p.636Alkyl groups are groups of atoms formed when 1 hydrogen is removed from an alkane. Prefixesare as before and suffixes are "_yl". ex) methyl, butyl, propyl, etc. ex) methypropane

Alkyl halides are hydrocarbons where one or more hydrogen atom replaced by a halogen. Theyreceive the appropriate prefix: fluoro_, chloro_, bromo_ etc. ex) chloroethane

Naming Alkanes p.6391. Name the longest continuous hydrocarbon chain (backbone). Give it a prefix to tell how manycarbons, add the suffix "ane"2. Add the names of the alkyl groups in front of the backbone name. If there is more than 1group, arrange them alphabetically. ex) ethyl methylhexaneIf there are 2 or more of the same group, use prefixes di (2), tri (3), and tetra (4).ex) diethyloctane Do this after they are in alphabetical order.3. Number the backbone to give the lowest numbers, and assign position numbers to groups.ex) 3-ethyl, 2,4,5-trimethyloctane4. Name halogens the same way as alkyl groups, but place them in front of them. (their higherpriority has no bearing on counting) ex) 4-bromo-2-methylhexane

Cycloalkanes p.635Cycloalkanes are alkanes where the first and last carbons form a bond, creating a ring. Theirempirical formula is CnH2n, because 2 hydrogen were lost to form the bond. Their name ispreceded by the prefix cyclo_. ex) cyclobutane � cyclopropane �

Isomers p.631Isomers are different structures of the same formula.

There are 3 isomers of pentane: n-pentane, 2-methylbutane, and 2,2-dimethylpropaneThere are 5 isomers of hexane:

Alkenes p.647Unsaturated hydrocarbons have double or triple bonds, and therefore do not have as manyhydrogen atoms as possible. Alkenes have at least one double bond, and their empirical formulais CnH2n. All the former naming rules still apply, with some additional considerations. ethene C2H4 ⁄⁄ propene C3H6 ⁄⁄� 1-butene C4H8 ⁄⁄�� trans-2-butene �\\�1-pentene C5H10 ⁄⁄���1-hexene C6H12 ⁄⁄���� cis-2-butene ��

Naming Alkenes p.6491. Name the longest backbone that contains a double bond. Use the suffix _ene, and if there ismore than 1 double bond, use _adiene, _atriene, and _atetraene. ex) hexadiene2. Number the backbone so that the double bonds have the lowest number.3. Indicate where the double bonds are with a position number in front of the backbone name.4. If the double bond can be identified as cis or trans configuration, label cis or trans in front ofits bond position number. ex) cis-2-trans-4-octadiene

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Alkynes p.651Alkynes have at least one triple bond, which is linear. As such, there is no cis or transconfigurations possible. The empirical formula for alkynes is CnH2n−2. The naming rules foralkenes apply to alkynes, except the suffix is _yne. ex) butyne, 1,4-hexadiyneethyne C2H2 ≡≡ propyne C3H4 ≡≡ 1-butyne C4H6 ≡≡�1-pentyne C5H8 ≡≡��1-hexyne C6H10 ≡≡���

Benzene and Phenol p.652Benzene is an important chemical because it is stable and is the foundation for many complexstructures. Benzene is an aromatic hydrocarbon, which means that it has 6 carbon atoms anddelocalized electrons. Its formula (1,3,5-hexatriene) would suggest 3 double bonds, but in fact allthe bonds are the same length. This suggests that the double bonds are shared over the wholemolecule as resonance hybrid bonds. Phenol is benzene with an alcohol group on carbon 1.

Three configurations are common when attaching 2 groups onto benzene: ex) with chlorine1,2-dichlorobenzene 1,3-dichlorobenzene 1,4-dichlorobenzeneortho-diclorobenzene meta-dichlorobenzene para-dichlorobenzene

Some Other Functional Groups p.663-679*Formula Name Example StructureR-OH alcohol 2-propanol (isopropyl alcohol)

R-O-R' ether diethyl ether (anesthetic ether) ¶R-C-H aldehyde formaldehyde (methanal) ¶R-C-R' ketone propanone (acetone) ¶R-C-OH carboxylic acid acetic acid (vinegar) ¶R-O-C-R' ester ethyl acetate

R-NH2 amine methylamine

note R means a hydrocarbon group and R' means a group also, but it may be a different one.also ¶ here means C=O, a carbonyl group

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Unit 3: The Mole chapters 3, 7 + 13Outline-relative mass -molar mass -molecular formulas-isotopes and avg atomic mass -m = n � mm -concentration-counting by mass -mult-step problems -n = c � V-the mole - % composition-#p = n�� 6.022 x 1023 -empirical formulas

Relative Mass p.78When calculating the mass of one object in terms of an other, take the mass of the object anddivide it by the mass of the other. For example, find the mass of a book relative to a pencil:

book = 1204 g book = 1204 g = 150.877... = 151 the book has a mass of 151pencil = 7.98 g 7.98 g "pencil mass units"

Atoms can be described in this way, and were originally massed according to hydrogen, being thesmallest. Later, this was revised so that 1 atomic mass unit (amu) was defined as 1/12 ofcarbon-12, an isotope of carbon.

Isotopes and Average Atomic Masses p.79Isotopes are atoms of the same element that have different numbers of neutrons, and thereforehave different masses. For example, hydrogen

(Protium) (Deuterium) (Tritium)Hydrogen-1 Hydrogen-2 Hydrogen-3

protons 1 1 1neutrons 0 1 2mass # 1 2 3abundance 99.985% 0.015% traceatomic mass 1.007825 amu 2.014102 amu 3.016049 amu

To find the average atomic mass for hydrogen, take the atomic mass of each isotope and multiplyit by its abundance (as a decimal) and add them together.

1.007825 x 0.99985 = 1.007673826252.014102 x 0.00015 = + 0.0003021153

1.00797594155 amu(In 1961, the average atomic mass of hydrogen was found to be 1.00797, but in 1981, this figurewas revised to be 1.00794 amu.)The periodic table displays the average atomic masses for the elements in amu.

Counting by Mass p.80An efficient way of counting large numbers of objects is to count by mass. For example, if youwere to count the number of paper dots in a bag of confetti, it would take a long time. Instead,you could find the mass of 1 dot, find the total mass of the dots and divide them. This would tellyou how many dots there are in the bag.

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The Mole p.81The mole is a unit for counting particles such as atoms and molecules. This is similar to countingeggs by the dozen, but much larger. A mole is defined as the number of atoms in 12 g ofcarbon-12, and is 6.022 x 1023. This number is called Avogadro's number in honour of AmedeoAvogadro, who's ideas were crucial in understanding this relationship. We can use this equation:

n = #p where n = number of moles in mol

6.022 x 1023 #p = number of particles

ex) How many atoms of gold are there in 3.25 mol of Au?n = 3.25 mol #p = n � 6.022 x 1023 = 3.25 � 6.022 x 1023 = 1.95715 x 1024

#p = ? = 1.96 x 1024 atoms

ex 2) How many atoms of oxygen are there in 6.29 mol of H2SO4?n = 6.29 mol #p = 6.29 mol � 6.022 x 1023 = 3.79 x 1024 molecules of H2SO4

#p = ? each molecule has 4 oxygen, so 4(3.79 x 1024) = 1.52 x 1025 atoms of oxygen

ex 3) How many moles of Na atoms are there in 5.21 x 1022 atoms of Na?n = ? n = #p = 5.21 x 1022 = 0.0865 mol#p = 5.21 x 1022 6.022 x 1023 6.022 x 1023

Molar Mass p.81Because of the definition of the mole, the atomic masses on the periodic table can also be viewedas the mass in grams of 1 mole of that element, or molar mass. From this, we get this equation:

n = m where n = number of moles in mol

mm m = mass in gmm = molar mass in g/mol

*When taking molar masses of elements from the periodic table, round to 2 decimal places.**The actual symbol for molar mass is M, but I prefer mm to reduce confusion*

ex) How many moles of atoms are there in 156 g of silver?m = 156 g n = m = 156 g = 1.45 moln = ? mm 107.87 g/molmm = 107.87 g/mol

ex 2) What is the mass of 0.54 mol of aluminum?m = ? m = n � mm = 0.54 mol � 26.98 g/mol = 15 g n = 0.54 molmm = 26.98 g/mol

Calculating Molar Mass of Compounds p.222When calculating the molar mass of compounds, find the total mass of each element and addthem together.

*In this class, round all molar masses to 2 decimal places.*

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ex) Find the molar mass of NaCl.Na + Cl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

ex 2) Find the molar mass of NH3.N + 3(H) = 14.01 g/mol + 3(1.01g/mol) = 14.01 + 3.03 = 17.04 g/mol

ex 3) Find the molar mass of H2SO4.2(H) + S + 4(O) 2(1.01) 2.02

32.07 32.074(16.00) + 64.00

98.09 g/molex 4) Find the molar mass of oxygen.Recall that H2, N2, O2, F2, Cl2, Br2, and I2 are all diatomic, along with P4 and S8.2(O) = 2(16.00) = 32.00 g/mol

Multi-step ProblemsWhen using 2 equations to solve a problem, or using calculated numbers, round to an extra sigfig through the problem. Round the final answer to the correct number of sig figs.

ex) What is the mass of 4.59 x 1024 molecules of carbon dioxide?m = ? n = #p = 4.59 x 1024 n = ? 6.022 x 1023 6.022 x 1023

mm = CO2 = 44.01 g/mol#p = 4.59 x 1024 = 7.622 mol (take an extra sig fig)

m = n � mm = 7.622 mol � 44.01 g/mol = 335.444...g

m = 335 g (the answer has 3 sig figs, as limited by the question)

Percent Composition p.226The percent composition tells you what percentage of the compound's total mass comes fromeach element. To find this, divide the mass of each component, by the total mass, or divide theiraccumulated molar masses by the overall molar mass. Don't forget to multiply by 100.

ex) Find the percent composition of a 42 g sample that contains 32.8 g iron and 9.2 g of oxygen.% Fe = 32.8 g = 78 % % O = 9.2g = 22 % (by mass)

42g 42 g

ex 2) Find the percent composition of methane. CH4 C H

First find the molar mass: 12.01 12.01 12.01 4.04 4(1.01) + 4.04 16.05 16.05

16.05 g/mol = 74.83 % = 25.17 %Divide each element's contribution by the molar mass, multiply by 100:��

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Empirical Formulas p.229An empirical formula is the lowest whole number ratio of bonding atoms in a compound. Forexample, C2H4, C5H10, and C100H200 all have the same empirical formula: CH2

To find an empirical formula, you must find the lowest whole number ratio of moles present.ex) A compound is analyzed and is found to contain 47.9 g of carbon and 127.7 g of oxygen.Find its empirical formula. CxOy

1. find the moles of each: n = 47.9 g 127.7g12.01 g/mol 16.00 g/mol

n = 3.988 mol : 7.9813 mol round to 1 extra sig fig2. divide each by the smaller: 3.988 7.9813

3.988 3.9883. this is the ratio of moles 1 : 2.0 *round to 1 decimal place*

so the empirical formula is CO2

*If the ratio is a whole number when rounded to 1 decimal place, use this ratio for your formula. If it rounds to 2.5 for example, find the lowest whole number multiple, and use that for yourformula. ex) 1 : 2.5 � 2 : 5, and the formula would be C2O5.

ex 2) Find the empirical formula of an iron-oxygen compound that is 70.0 % Fe and 30.0 % O bymass. Fe O1. assume you have 100 g n = 70.0 g 30.0 gthen treat your % as a mass 55.84 g/mol 16.00 g/mol

n = 1.254 mol 1.875 mol1.254 1.8751.254 1.254 1 : 1.5 next whole ratio is 2 : 3

Fe2O3

ex 3) Find the empirical formula of a compound that contains 6.61 g of hydrogen, 105 g of sulfurand 209 g of oxygen. H S O

n = 6.61 g 105 g 209 g1.01 g/mol 32.07 g/mol 16.00 g/mol

n = 6.545 mol 3.274 mol 13.06 mol6.545 3.274 13.063.274 3.274 3.2742.0 : 1 : 4.0

H2SO4

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Molecular Formulas p.232Once you have an empirical formula, you can determine the molecular formula, or the actualformula for that compound. It will be a multiple of the empirical formula, so you can comparethe molar masses to find the multiple. (the multiple will be a whole number like 1, 2, 3, etc.)

ex) Find the molecular formula of a compound if its molar mass is 42.09 g/mol and its empiricalformula is CH2. (CH2)x = molecular formulause molar masses (14.03 g/mol)x = 42.09 g/mol so 42.09 = 3 → the molecularto find how many 14.03 formula is 3times larger (CH2)3 = C3H6 times the empirical formula

ex 2) Find the molecular formula for a hydrocarbon if a 20.15 g sample contains 16.66 g ofcarbon, and its molar mass is 58.14 g/mol.first find the empirical C H 20.15 g − 16.66 g = 3.49 gformula and its mm n = 16.66 g 3.49 g

12.01 g /mol 1.01 g/mol = 1.3872 mol 3.4554 mol

1.3872 1.38721 : 2.5 � 2 : 5

empirical formula: C2H5 mm = 29.07 g/mol

(29.07 g/mol)x = 58.14 g/mol x = 2 (C2H5)2 = C4H10

Concentration p.412Concentration describes how many moles of solute there are dissolved per litre of solution.

c = n where: c = concentration in mol/L or M

V n = number of moles in molV = volume in L

ex) Find the concentration of a solution made by dissolving 0.291 mol of H2SO4 in enough waterto make 451 mL of solution. n = 0.291 mol c = n = 0.291 mol = 0.645 mol/Lc = ? V 0.451 L orV = 451 mL = 0.451 L = 0.645 M

ex 2) What mass of sodium chloride would you add to water in order to make 500.0 mL of a 1.00 M solution?m = ? n = c�� V = 1.00 mol/L � 0.5000 L = 0.5000 moln = ?mm = 22.99 + 35.45 = 58.44 g/mol m = n � mmc = 1.00 mol/L = 0.5000 mol � 58.44 g/molV = 500.0 mL = 0.5000 L = 29.2 g

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Unit 4: Chemical Reactions Chapter 8 + 17Outline-writing equations -types of reactions-equation symbols -activity series-diatomic elements -thermochemistry-law of conservation of mass -endo/exothermic reactions-balancing equations

Writing Equations p.242A chemical reaction is a change in which substances are converted into different substances.

magnesium and oxygen react to form magnesium oxide (word equation)

Chemical reactions are also represented by formula equations, which display the reactants andproducts as symbols or formulas. They are separated by an arrow, which indicates a chemicalreaction has taken place, as well as its direction.

"reacts with" "to produce"Mg(s) + O2(g) → MgO(s) (formula equation)

reactants products

The (s) and (g) refer to the state of the substance. For additional symbols, see p.246(s) solid (g) gaseous(�) liquid (aq) aqueous (dissolved in water)

Chemists often leave out the phases, as they don't affect the calculations. Also, recall that oxygenis a diatomic element, which exists in pairs in nature. The following form polyatomic molecules,which you need to remember:H2 N2 O2 F2 Cl2 Br2 I2 P4 S8

ex) Write the equation: solid aluminum carbide reacts with water to produce methane gas andsolid aluminum hydroxide.

Al4C3(s) + H2O(l) → CH4(g) + Al(OH)3(s)

The Law of Conservation of Mass p.243In normal chemical reactions, matter is neither created, nor destroyed and the total mass of thereactants must equal the total mass of the products. In other words, the number of atoms that gointo a reaction has to equal the number that come out. As a result of this law, we must balance allchemical equations using coefficients (large numbers in front of the symbol). You may notchange the formulas of the substances, just the ratio in which they react.

2Mg + O2 → 2MgO balanced (no "1" is used)

Coefficients multiply through the entire compound.2NaCl = 2 Na, 2 Cl3MgCl2 3 Mg, 6 Cl2H2SO4 4 H, 2 S, 8 O3Ca(NO3)2 3 Ca, 6 N, 18 O5Mg(CH3COO)2 5 Mg, 20 C, 30 H, 20 O

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Balancing Equations p.250� When balancing, look for polyatomic ions or elements that appear just once on each side

- balance those first.� You may need to change coefficients several times, so use a pencil.� If you run into an element that is difficult to balance, leave it and come back to it. Fractions

may be useful in some cases.� Do not include "1" as a coefficient.� Reduce coefficients to the lowest whole number ratio.� Double check by recounting.ex) balance these 3 equations:1) Zn + HCl → ZnCl2 + H2 3) C3H6 + O2 → CO2 + H2O

Zn + 2HCl → ZnCl2 + H2 C3H6 + O2 → 3CO2 + H2OC3H6 + O2 → 3CO2 + 3H2O

2) Al4C3 + H2O → CH4 + Al(OH)3 C3H6 + 9/2O2 → 3CO2 + 3H2OAl4C3 + H2O → CH4 + 4Al(OH)3 2C3H6 + 9O2 → 6CO2 + 6H2OAl4C3 + H2O → 3CH4 + 4Al(OH)3 Al4C3 + 12H2O → 3CH4 + 4Al(OH)3

Types of Chemical Reactions p.256 Activity SeriesWe will study 5 types of reactions: synthesis, decomposition, single replacement, Lidouble replacement, and combustion. Rb

K1) Synthesis Reaction: where 2 or more simple substances combine to form a more Bacomplex substance. A + B → AB Sr

2Na + Cl2 → 2NaCl sodium + chlorine → sodium chloride CaNa

2) Decomposition Reaction: where a complex substance breaks down into 2 or more Mgsimpler substances. AB → A + B Al

2NaCl → 2Na + Cl2 sodium chloride → sodium + chlorine MnZn

3) Single Replacement Reaction: (Single Displacement) where an uncombined element Crreplaces an element that is part of a compound. A + BX → AX + B Fe

Mg + 2HCl → MgCl2 + H2 Cdmagnesium + hydrochloric acid → magnesium chloride + hydrogen Co

Ni2Na + 2H2O → 2NaOH + H2 Snsodium + water → sodium hydroxide + hydrogen Pb

Note: single replacement reactions will only occur if the element doing the replacing H2

is more active (reactive) than the element being replaced. (activity series p.266) SbMg + 2HCl → MgCl2 + H2 Mg + NaCl → NR Bi

Cu4) Double Replacement Reaction: (Double Displacement) where different atoms in Hg2 different compounds replace each other, or when 2 compounds react to form 2 new Agcompounds. Double replacement reactions will always occur to some extent. PtAX + BY → AY + BX Au

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FeS + 2HCl → H2S + FeCl2

iron(II) sulfide + hydrochloric acid → hydrogen sulfide + iron(II) chloride(or dihydrogen monosulfide)

MgCO3 + 2HCl → MgCl2 + H2CO3 magnesium carbonate + hydrochloric acid → magnesium chloride + carbonic acid

(or hydrogen carbonate)

5) Combustion Reaction: where a substance combines with oxygen and released energy in theform of heat and light. CXHy + O2 → CO2 + H2O (for hydrocarbons)

C3H8 + 5O2 → 3CO2 + 4H2O propane + oxygen → carbon dioxide + water

Thermochemistry p.514-p.541All chemical reactions involve changes in energy. Some chemical reactions will release heat andget hot (exothermic), while others will absorb heat and get cold (endothermic). All chemicalreactions require energy to begin (activation energy); some need quite a bit, such as dynamite,while for others, there is enough thermal energy at room temperature to react. In an exothermicreaction, the amount of heat released is greater than the activation energy. In this case, chemicalpotential energy is converted into heat. Therefore, the products have less energy (chemicalpotential energy) than the reactants.

In an endothermic reaction, the amount of heat released is less than the heat input. The extra heatis converted into chemical potential energy. As a result, the products have more energy than thereactants. (fig.17-10, p.534) A catalyst is a substance that speeds up a reaction without beingconsumed. It is not a reactant, Pt

and is written above the arrow. NH3(g) + O2(g) → NO(g) + H2O(�)A catalyst increases reaction rates by lowering the amount of energy needed to break bonds(activation energy). (fig17-15, p.541)

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Unit 5: Stoichiometry Chapter 9Outline-stoichiometry problems: -limiting reactants

-moles to moles -amount in excess-mass to moles - percent yield-mass to mass - actual x 100-solution problems theoretical

Stoichiometry comes from the Greek words "stoicheion" (element) and "metron," (measure) andis a way of using the mole ratio to calculate many unknowns from given information. Forexample, if you are given 43 g of reactant, how many grams of product will you get? To solvethese problems, we use the ratio in which they react, but we must first convert to moles.

Moles to Moles p.280Use the mole ratio to predict unknowns.ex) When sodium reacts completely with 5.00 mol of chlorine, how many moles of sodiumchloride will you get?1. write a balanced equation 2Na + Cl2 → 2NaCl2. set up a mole ratio, n=5.00 mol n = ?cross multiply

1 = 2 1x = 2(5.00) x = 10.0 mol5.00 mol x

You will get 10.0 mol of NaCl.

ex 2) The complete combustion of 4.29 mol of octane would yield how many moles of carbondioxide?

2C8H18 + 25O2 → 16CO2 + 18H2O 2 = 16 n=4.29 mol n=? 4.29 mol x

2x = 68.64 x = 34.3 mol of CO2

Mass to Moles p.284If you are given the mass of a reactant (or product), first find out how many moles that is, thensolve as before.ex) Excess of calcium reacts with 127 g of oxygen to form how many moles of calcium oxide?place info on the equation:

m=127 g 1 = 2

2Ca + O2 → 2CaO 3.969 xn = m = 127 g n=3.969 mol n=? mm 32.00 g/mol mm=32.00 g/mol x = 7.94 mol

= 3.969 mol

Mass to Mass p.286Once you know the moles of one reactant or product, you can determine the mass of any other.ex) How many grams of oxygen and sulfur are produced when 298 g of sulfur trioxidedecomposes?

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m=298 g m=? m=?

8SO3 → 12O2 + S8n=3.722 mol n=5.583 mol n=0.4653 molmm=80.07 g/mol mm=32.00 g/mol mm=256.56 g/mol

n = 298 g 8 = 12 8 = 1 80.07 g/mol 3.722 x 3.722 x = 3.722 mol x = 5.583 mol x = 0.4653 mol

m = 5.583 mol�� 32.00 g/mol m = 0.4653 mol ��256.56 g/mol= 179 g = 119 g

SolutionsWe can perform stoichiometric calculations involving solutions; finding the # of moles is key.ex) What volume of 2.00 M AgNO3 is required to produce 4.00 L of 5.00 M Cu(NO3)2 whenreacted with copper?

Cu + 2AgNO3 → Cu(NO3)2 + 2Agn=40.00 mol n=20.00 mol

1 = 2 c=2.00 M c=5.00 M20.00 x V=? V=4.00 Lx = 40.00 mol

V = 40.00 mol n = 5.00 M ��4.00 LV = 20.0 L 2.00 M = 20.00 mol

Limiting Reactants p.288Up until now, we have examined ideal stoichiometric equations. In reality, one reactant is usuallyused up before the other, and will limit how much of the products you get. This reactant is calledthe limiting reactant, and the reactants that are left over are said to be in excess. It is important touse the limiting reactant when calculating yields.

To determine which reactant is limiting, you must be given the amounts of all of them. Take onereactant and predict how much of the other one(s) you would expect. If you have more of theother reactant than you expect, it is in excess; if you have less than you expect, it is limiting.ex) If 4.00 mol of Na react with 2.50 mol of Cl2, find the limiting reactant.

2Na + Cl2 → 2NaCl 2 = 1 n=4.00 mol n=2.50 mol 4.00 x

expect: 2.00 molLIMITING EXCESS

We can also determine the amount in excess: 2.50 − 2.00 = 0.50 mol excess

We can also determine the amount in excess in g: 0.50 mol � 70.90 g/mol = 35 g in excess

ex 2) Which reactant is limiting if 41.5 g of antimony react with 134 g of iodine? How manygrams of antimony(III) iodide would you expect?

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m=41.5 g m=134 g m=?

n = m 2Sb + 3I2 → 2SbI3 mm n=0.3408 mol n=0.5280 mol n=0.3408 mol

mm=121.76 g/mol mm=253.80 g/mol mm=502.46 g/mol= 41.5 g expect: 0.5112 mol m = n � mm121.76 g/mol LIMITING 2 = 3 EXC 2 = 2 = 0.3408 � 502.46= 0.3408 mol 0.3408 x 0.3408 x = 171 g

x = 0.5112 x = 0.3408

ex 3) When 58 g of zinc reacts with 640 mL of 1.0 M hydrochloric acid, what mass of zincchloride is formed. Find how many grams of the excess reactant is left over.

m=58 g m=

Zn + 2HCl → ZnCl2 + H2 2 = 1n=0.887 mol n=0.640 mol n=0.32 mol 0.640 xmm=65.39 g/mol c=1.0 M mm=136.29 g/mol x = 0.32 molexpect: 0.32 mol V=640 mL =0.640 L 2 = 1 LIM m = 44 g0.640 xx = 0.32EXCESS by 0.567 mol or (0.567 mol � 65.39 g/mol) = 37 g

Percent Yield p.293So far we have been dealing with the theoretical yield, or the most product possibly produced.Usually, the actual yield is less than this ideal due to side reactions and product lost in transfer orin purification. We can calculate the efficiency of a reaction using this equation:

% yield = actual yield x 100theoretical yield

*You must be given the actual yield in the question, you find the theoretical throughstoichiometry.*When given an actual yield, do not put that into the stoichiometric equation.

ex) When 59.7 g of propene is burned with 45.2 g of oxygen, 39.4 g of carbon dioxide is actuallyproduced. Calculate the percent yield of CO2. (first find the theoretical yield)

m=59.7 g m=45.2 g m=?

2C3H6 + 9O2 → 6CO2 + 6H2On=1.418 mol n=1.413 mol n=0.942 molmm=42.09 g/mol mm=32.00 g/mol mm=44.01 g/molEXCESS expect: 6.381 mol

n = 59.7 g 2 = 9 9 = 6 m = 0.942 � 44.01 42.09 g/mol 1.418 x 1.413 x = 41.5 g= 1.418 mol x = 6.381 x = 0.942 *For theoretical mass

LIMITING round to correct sf.

% yield = actual x 100 = 39.4 g x 100 = 94.9 %theoretical 41.5 g

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Unit 6: The Atom Chapters 3, 4 + 5Outline-Democritus' model -E = hv -Aufbau principle, Hund's rule and-laws of composition -Spectra Pauli exclusion principle-Dalton's model -Bohr's model -noble gas notation-J. J. Thompson's model -the quantum model -periodic table and e- configuration-Rutherford's model -quantum numbers -valence electrons-light as waves + particles -orbital notation -periodic trends: atomic radii, IE, and- c = λν -electron configuration electronegativity

Democritus p.65Democritus, a Greek philosopher, is credited for first formalizing a particle theory of matteraround 400 BC. He believed all matter to made of small, indivisible particles he called atoms. (fr. Gk atomos, indivisible) His theory had 6 points:� atoms are particles in the void� atoms are always in motion� atoms are indivisible� atoms come in different sizes, shapes, and are arranged in different ways� atoms make life predictable� the human soul is made of soul atoms

The idea of atoms was dismissed a generation later, by the more influential thinkers Plato andAristotle. For almost 2000 years the concept of the atom was forgotten.

Laws of Composition p.66With the arrival of better technology, certain laws were established that helped understand matterLaw of Conservation of Mass - states that matter can not be created nor destroyed in a chemical

reaction.Law of Definite Proportions - states that a chemical compound contains the same elements in the

same proportions regardless of the size or source of the sample. For ex) water is always 2 hydrogen and 1 oxygen: H2O (H2O2 is a different compound)

Law of Multiple Proportions - states that compounds are made from small, whole number ratios of elements. For ex) CO and CO2 (carbon dioxide is never C1/3H2/5)

Dalton p.66John Dalton, a teacher from England, re-established an atomic theory in 1808, based on hisexperiments. He explained the law of conservation of mass using the laws of definite andmultiple proportions. Dalton's atomic theory:1. All matter is composed of small particles called atoms.2. Atoms of the same element are identical, atoms from different elements differ in size, mass

and other properties.3. Atoms can not be divided, created, or destroyed.4. Atoms of different elements combine in small whole-number ratios to form compounds.5. In chemical reactions, atoms are combined, separated, or rearranged.

J. J. Thompson p.70By 1897, scientists such as William Crookes had been experimenting with cathode ray tubes.These were glass tubes with a gas at low pressure that glowed when subjected to high voltages.

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The experiments by many scientists, including Thompson, found that� the glow was made of particles streaming from the negative terminal to the positive� the particles could push objects� the particles were deflected by a magnetic field� the particles were deflected away from a negatively charged objectThompson was able to calculate a charge to mass ratio for these particles and found them to bethe same for all materials. In 1903, he proposed his "plum pudding" model of the atom: atomswere made from positive material, laced with small negative particles. The negative particleswere later named electrons. (electric + _on, fundamental particle)

MilikanRobert Milikan used more accurate equipment in his oil drop experiment. He found the mass ofan electron to be 1837 times smaller than the smallest atom. (hydrogen) Here are the masses:

proton (p+) 1.673 x 10-24 gneutron (no) 1.675 x 10-24 g these more accurate numbers were calculated laterelectron (e−) 9.109 x 10-28 g

Rutherford p.72Ernest Rutherford was a New Zealander who did some of his work in Canada. In 1911, He andsome colleagues bombarded thin gold foil with alpha particles. The foil was about 400 atomsthick, and alpha particles are fast, positively charged particles with a mass of 4 times that ofhydrogen. If the atom was uniformly dense, as was believed, the particles should have passedthrough will only small deflections.

This was the case, however, 1 in every 8 000 particles was redirected toward the source.Rutherford eventually concluded that the centre of an atom must be dense and positive, whilemost of the atom was empty space. He named the centre the nucleus, and didn't know where theelectrons were, but hypothesized that they orbited the nucleus like planets orbit the sun. Protonsin the nucleus should repel each other, but are held together by short-range attractive forcescalled nuclear forces.

The neutron was later discovered by Irène and Frederic Joliot-Curie in 1932 and was named byJames Chadwick.

Light as Waves and Particles p.91Before 1900, light was believed to behave as a wave. Waves have characteristics such aswavelength (λ), amplitude (ψ), and frequency (ν). For all electromagnetic radiation, wavelengthand frequency are inversely proportional in the equation

where: c = the speed of light 3.00 x 108 m/sc = λν λ = wavelength in m

ν = frequency in Hz or cycles per secondLater, Einstein proposed that light also behaved as a particle, and the energy of a photon can becalculated using Plank's idea of the quantum or amount of energy.

where: E = the energy of a photon in JE = hν h = Planck's constant = 6.6262 x 10−34 Js

ν = frequency in HzIn other words, photons at a given frequency will possess a known amount of energy.

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Bohr p.94When hydrogen gas in a tube is energized using electricity, its electrons are excited to a higherenergy state. When they fall back to their ground state, they give off a photon with energy equalto that gained. Its colour can be observed and calculated using E = hν. When adding moreelectricity, we should see a continuous colour change, but in fact we only see 4 distinct colours.

In 1913, Niels Bohr proposed that electrons can occupy only certain energy levels. The electronorbital lowest in energy (ground state) is closest to the nucleus, and higher energy levels aresuccessively further away. This model of the atom explained the hydrogen spectrum, but failed toaccount for many electron atoms. (see energy series fig. 4-8, 4-9, p.96)

The Quantum Model p. 98In order to understand the quantum nature of the atom, scientists had to change their model of theelectron. In 1924, Louis de Broglie proposed that electrons could behave as both particles andwaves. In order to detect where electrons are in the atom, if we use light or photons, they willknock an electron off course. Heisenberg's uncertainty principle states that the more certain youare of an electron's position, the less certain you are of its speed, and vice versa. So instead oftalking about where electrons are, we use electron probability clouds, or areas where you arelikely to find an electron.

This dual nature of the electron led Erwin Schrödinger to develop the Schrödinger waveequation. This is the foundation for quantum theory.

where: ψ = Schrödinger wave functiond 2 ψ + 8 π2 m (E −V) ψ = 0 x = positiond x2 h2 h = Planck's constant

E = total energyV = potential energy

Quantum Numbers p.101From this equation, we receive 4 quantum numbers to describe electrons and orbitals:

1. Principle quantum number ( n ): describes the energy level of the electron. It also determineshow many orbitals you have. There are n2 solutions to the Schrödinger equation, or n2 orbitals.ex) n = 1, n = 2, n = 3...

2. Angular momentum quantum number ( l ): describes the type of orbitalwhen l = 0 s orbital see shapes, p.102

l = 1 p orbitall = 2 d orbitall = 3 f orbitall = 4 g orbital

3. Magnetic quantum number ( ml ): describes the orientation or shape of the orbital. (x, y, z) It ranges from −l to +l. ex) if l = 2, ml = −2, −1, 0, +1, +2 5 orientations

4. Magnetic spin quantum number ( ms ): describes the spin of the electron. If there are there aretwo electrons in the same orbital, they must spin in opposite directions: + ½ or − ½. There canonly be 2 electrons per orbital. Spin is also shown with arrows: ↑ or ↓.

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Electron Configuration: Orbital Notation p.106According to the Aufbau principle, an electron will occupy the lowest-energy that will accept it.ex) hydrogen has 1 electron ↑ -energy level 1 (the lowest)

1s -s orbital (first to fill)

ex 2) nitrogen has 7 electrons ↑↓ ↑↓ ↑ ↑ ↑ 1s 2s 2px 2py 2pz

-1s fills, then 2s, then the 2p (there are 3 of them)-2 electrons in the same orbital have opposite spin (↑ vs. ↓)-according to Hund's rule, orbitals of equal energy levels are each occupied by one electronbefore 2 electrons are placed in the same orbital.

ex 3) nickel has 28 electrons↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ 1s 2s 2p 3s 3p 4s 3d

Electron configuration notation is a shorthand way of writing this:1s2 2s2 2p6 3s2 3p6 4s2 3d8 or more correctly 1s22s22p63s23p63d84s2

note: order according to energy level

The filling order for orbitals is1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 ...

-note each energy level has 1 s orbital, 3 p orbitals, 5 d orbitals, and 7 f orbitals-s fill first, then p, then d, then f -all you have to remember is the filling order of the energy levels:

1 2 2 3 3 4 3 4 5 4 5 6 4

then fill in orbital types:1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f

Pauli Exclusion PrincipleThe Pauli exclusion principle states that no 2 electrons (in the same atom) have the same 4quantum numbers:ex) neon (10 e-) ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↓

1s 2s 2px 2py 2pz

draw a line from theelectron to the set n = 2 n = 2 n = 2 n = 2of quantum numbers l = 0 l = 1 l = 1 l = 1

ml = 0 ml = +1 ml = −1 ml = −1ms = +½ ms = +½ ms = −½ ms = +½

Noble Gas Notation p.111Note that for nickel, the closest noble gas that has fewer electrons is Ar = 1s2 2s2 2p6 3s2 3p6

so another way of writing nickel is [Ar]3d84s2

ex 2) iodine (53 e-) is [Kr] 4d105s25p5

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The Periodic Table p.123Dmitri Mendeleev is regarded as the father of the periodic table. He noticed that when elementswere arranged according to increasing atomic mass, similar chemical properties recur at regularintervals. For example, Li, Na, and K have similar properties. His work led to the periodic law:when elements are arranged in order of increasing atomic number, elements with similarchemical and physical properties appear at regular intervals. Review periodic tableGroup 1: Alkali metals - react with water to form base, low m.p., very reactive

2: Alkaline earth metals - also form base in water, slightly higher m.p., reactive3-12: Transition metals - a transition from reactive metals to nonmetals 13: Boron family - metals and metalloid, not very reactive14: Carbon family - not very reactive �15: Nitrogen family - not very reactive � mix of metals, metalloids, and nonmetals16: Oxygen family - more reactive �17: Fluorine family - very reactive, form halogenated compounds18: Noble gases - full valence shells, will not react under normal circumstancesLanthanides and Actinides: transition metals, elements 84+ are radioactive, elements 93+ are synthetic

Periodic Table and Electron Configuration p.128Understanding electron configuration brings new perspective to the periodic table. For example,there are 2 columns for the s-block elements because s orbitals can hold 2 electrons; there are 10transition elements because d orbitals hold 10 electrons. Therefore:Groups 1 + 2 are the s-block elementsGroups 3-12 are the d-block elements see fig. 5-5, p.129Groups 13-18 are the p-block elementsLanthanides and Actinides are the f-block elements

Valence ElectronsValence electrons are the outermost s and p electrons. Although you elements will never havemore than 8 valence electrons, (s2 + p6) energy levels hold the following electrons: energy level electrons configuration

1 2 1s2 2 8 2s2 2p6 3 18 3s2 3p6 3d10 4 32 4s2 4p6 4d10 4f14

5 32 5s2 5p6 5d10 5f14

6 18 6s2 6p6 6d10

7 8 7s2 7p6

(valence)

Periodic Trends: Atomic Radii p.140The size of an atom (atomic radius) is determined by the charge of the nucleus and the number ofenergy levels. As we move down a group, we notice atoms increase in size. This makes sense,because electrons are filling higher energy levels, which are further from the nucleus. Innerelectrons also shield the outer electrons somewhat from the nucleus. However, as we move left toright across a period, atoms decrease in size. Although the number of electrons is increasing, theyare on the same energy level, and are pulled closer to a more positive nucleus.

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Periodic Trends: Ionization Energy p.143Ionization energy (IE) is the amount of energy required to remove an electron, to form an ion. Aswe move down a group, we notice that IE decreases. This is because the electrons are fartherfrom the nucleus and shielded from the positive nucleus by the middle electrons. As we moveacross a period, IE generally increases because there is a greater positive charge attracting eachelectron on the same energy level. (which is also closer to the nucleus)

Periodic Trends: Electronegativity p.151Electronegativity is the measure of the ability of an atom to attract electrons. It is measured on ascale of 0 - 4.0 with fluorine, the most electronegative element, arbitrarily assigned 4.0. As wemove down a group, electronegativity slightly decreases due to increased size and shielding. Aswe move across a period left to right, electronegativity generally increases due to strongernuclear charge and smaller size.

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Unit 7: Chemical Bonding Chapter 6Outline-nonpolar-covalent bonds -Lewis structures -hybridization-polar covalent bonds -resonance structures -intermolecular forces-ionic bonds -ionic bonds -dipole-dipole forces-covalent bonds -metallic bonds -hydrogen bonding-octet rule -VSEPR theory geometry -London dispersion forces

A chemical bond is the mutual attraction between the nucleus of an atom and the valenceelectrons of another atom. Bonds occur because the particles have a lower potential energybonded to each other than they do separately.

Types of Bonds p.161Chemical bonds can be classified into 1 of 3 categories: nonpolar-covalent, polar-covalent, andionic. In general, metals tend to lose electrons, and have lower electronegativities thannonmetals, which tend to gain electrons. We can identify the bond type based on the difference ofelectronegativity.

Nonpolar-covalent BondsCovalent means sharing valence electrons, and this occurs when two atoms have similarelectronegativity values. A non-polar bond is a one where the electrons are shared equally, andno partial charges occur. These bonds occur between atoms when their electronegativitydifference is between 0 and 0.3. (for electronegativity values see p.151)ex) H2 H − H

(e-neg) 2.1 2.1 e- are shared exactly equallydifference = 0.0

Polar-covalent BondsPolar-covalent means that there are charged ends to the molecules, and although they are sharingvalence electrons, they are not shared equally. These bonds occur between atoms when theirelectronegativity difference is between 0.3 and 1.7.ex) HCl e- are shared, but not equally; partial charges occur

δ+H − Clδ−because of uneven electron distribution: e- are more

2.1 3.0 strongly attracted to the Cl difference = 0.9

Ionic BondsIonic bonds are so polar that electrons are actually transferred from one atom to another, creatingions. These oppositely charged particles strongly attract each other. These bonds occur betweenatoms when their electronegativity difference is 1.8 or greater.ex) NaCl Note that NaCl is not a molecule!

Na+ − Cl−

0.9 3.0 (e-neg diff) � difference = 2.1 0 0.3 1.7 4.0

|______|_____________|____________|a helpful chart: NPC PC I

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Covalent Bonds p.165To form a bond, atoms must release energy; to break it, the same amount must be added. Bondenergy is the energy required to break a bond to form neutral atoms, and is measured in kJ/mol.The higher the bond energy, the stronger the bond. Bonding can be represented through orbitalnotation:ex) fluorine can make 1 bond with itself: F2

F ↑↓ ↑↓ ↑↓ ↑↓ ↓ 1s 2s 2p covalent bond (overlapping orbitals)

F ↑↓ ↑↓ ↑↓ ↑↓ ↑ 1s 2s 2p

ex 2) Nitrogen can make 3 bonds with hydrogen: NH3

3 Hs ↓ ↓ ↓1s 1s 1s

3 covalent bondsN ↑↓ ↑↓ ↑ ↑ ↑

1s 2s 2p

Octet Rule p.168The octet rule states that chemical compounds tend to form so that each atom (by gaining, losing,or sharing electrons) has an octet of electrons in its highest occupied energy level. Although thereare many exceptions to the octet rule, (for example H2, BF3, or expanded d-orbitals) it is a usefulway to view covalent bonding.

Electron Dot Notation and Lewis Structures p.170Electron dot notation shows only valence electrons around an elemental symbol. Begin with theelement's symbol, and place dots around it to show the number of valence electrons. Begin at thetop and add one dot clockwise until you run out of dots. Each of the 4 sides can only have 2 dots,at least initially. For example � ��� �

Be � �� O :��������

Lewisdiagram

NeFONCBBeLiFamily

Lewis structures show molecules and their valence electrons. ex) H2O � �Drawing Lewis Structures H : O : H� Count the total number of valence electrons. ���� �

� Draw a skeleton structure using shared pairs of electrons to show bonding.� Use carbon for the central atom, otherwise use the least electronegative element.� Add unshared pairs of electrons so that each atom (that will accept them) is surrounded by 8 e-.� Multiple pairs of electrons may be shared between 2 atoms, forming double and triple bonds,

to satisfy the octet rule.� If there is more than 1 possible location for a double bond, use resonance structures to show all

possibilities; separate resonance structures with double arrows.

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For examples see p.171 − 175

Ionic Bonds p.176Ionic bonds form between cations and anions as a crystal lattice. This means that each cation issurrounded by anions and vice versa. This is why ionic crystals are brittle, because if the lattice isshifted, strong repulsive forces cause it to shatter.

Metallic Bonds p.181Metallic bonds form between metal atoms, and are different from covalent or ionic bonds.Metallic bonds do not follow the octet rule, and vacant s and p orbitals overlap. This enableselectrons to move freely throughout the entire structure. This is why metals are good conductorsof heat and electricity. The bond structure is the same in all directions, which means that you canbend and reshape the material without stressing the structure. (recall that metals are malleableand ductile) Metals can absorb a wide range of light frequencies, which excites their electrons.These electrons returning to their ground state gives off light, which accounts for the shinyappearance of metals.

VSEPR Theory and Molecular Geometry p.183VSEPR stands for valence shell electron pair repulsion, and this theory states that the repulsionbetween valence electrons in molecules causes their atoms to be oriented as far apart as possible. See p.186ex) In BeF2, the 2 fluorines must be placed as far apart as possible, which creates a 180° bond.Bond Shape Drawing Bond Angle(°) Empirical Formula Example

Linear 180 AB2 BeF2

Bent or angular ~117 AB2E SnCl2

Trigonal-planar 120 AB3 BF3

Tetrahedral 109.5 AB4 CH4

Trigonal-pyramidal ~107 AB3E NH3

Bent or angular ~105 AB2E2 H2O

Trigonal-bipyramidal 90 + 120 AB5 PCl5

Octahedral 90 AB6 SF6

Note: A is the central atom, B are bonded atoms, and E are electron pairs.Hybridization p.187Hybrid orbitals are orbitals of equal energy produced by combining 2 or more orbitals on thesame atom. Orbitals to combine Hybrid orbitals Shape Bond angle (°)

s + p sp Linear 180s + p + p sp2 Trigonal-planar 120s + p + p + p sp3 Tetrahdral 109.5

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Hybridization explains why carbon can make 4 bonds:

C ↑↓ ↑↓ ↑ ↑ Carbon should make 2 bonds... 1s 2s 2p

but the s orbital and the 3 p orbitals combineC ↑↓ ↑ ↑ ↑ ↑ to make 4 sp3 orbitals, which are 1 part s orbital,

1s 2sp3 2sp3 2sp3 2sp3 and 3 parts p orbital.

Intermolecular Forces p.189Intermolecular forces are forces of attraction between molecules. The forces that act within atomsto hold them together are called intramolecular forces. Forces between molecules are generallyweaker than intramolecular or ionic attractive forces. We can tell how strong intermolecularforces are by the melting point of the substance:

Bond typesMetallic ↑ Increasing strength ; Increasing melting pointsIonic ↑Polar-covalent ↑ � Intermolecular forcesNonpolar-covalent ↑ �

There are 2 types of intermolecular forces1. Dipole - Dipole Forces p.190A polar molecule has an end that is negative and one that is positive. This occurs when moleculeshave lone pairs of electrons or the electronegativities do not balance in 3 dimensions.(symmetrical molecules are nonpolar)

H−Cl positive end δ+ −|→ δ− negative end (more electrons)

ex) H2O and NH3 are polar, while CCl4 and CO2 are nonpolar molecules See p.191

Dipole-dipole forces are the strongest intermolecular forces and occur between 2 polarmolecules. A dipole can also induce a dipole in a nonpolar molecule by temporarily attractingsome of its electrons. This attraction is weaker than a dipole-dipole force.

1b. Hydrogen Bonding p.192A particularly strong category of dipole-dipole bonding is hydrogen bonding. This only occurswhere there are H-F, H-O and H-N bonds. The high difference in electronegativity between Hand F, O, and N make very polar molecules, and the positive H is attracted strongly to thenegative end of another molecule.

δ+H−Fδ−

- - -δ+H−F

δ−

−|→ −|→This is why H2O (has hydrogen bonding) boils at 100°C, while a similar compound H2S, (no hydrogen bonding) boils at −60°C.

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2. London Dispersion Forces p.93London dispersion forces are weak attractive forces caused by instantaneous dipoles whenelectrons happen to end up on the same side of a molecule. They occur in all atoms andmolecules, even noble gases. (also in molecules with dipole-dipole forces) London forcesincrease with mass because there are more electrons, as illustrated by boiling points of:

H2 −253°CO2 −183°CCl2 −34°CBr2 + 59°C

End of Course Notes

Lab: Copper(II) Chloride and Aluminum FoilPurpose: To investigate and understand the reaction between copper(II) chloride and aluminum.Materials: 150 mL beaker, Al foil, thermometer, balance, other equipment if desiredProcedure:Data/Observations:Calculations:Conclusion: answer reflection questionsReflection Questions:

1. What did you do (or what could you have done) to ensure that your senses were not fooled?2. What was the substance formed during the reaction?3. Where did the aluminum go? How can you be sure?

Bonus Questions:1. Of what were the bubbles made?2. Write a balanced chemical equation for the reaction.

Lab: Mass and Volume of SubstancesPurpose: To determine the relationship between mass and volume for methanol - CH3OH, water - H2O, ethylene glycol - C2H6O2.Materials: balance, 3 substances, 50 mL beaker, micropipet, 10 mL graduated cylinder Procedure: Data/Observations: empty cylinder dry ________volume methanol water ethylene glycol1.00 mL ______ ______ ______2.00 mL ______ ______ ______etc.Calculations: net mass, slopeConclusion: State the constant relationship (a number) between mass and volume for each

substance. Hint: you need the graph.Reflection Questions:

1. What is the mass of 1.000 x 103 L of each substance?2. Write the relationship between mass and volume as a function for all 3 substances.

3. Give the % error for each of your 3 constants. Provide an explanation for any error over 5 % with sources of error.

Graph: mass vs. volume for the 3 substances

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Lab: The Thickness of Aluminum FoilPurpose: To determine the thickness of aluminum foil in mm.Materials: sheets of aluminum foil, ruler, balanceProcedure:Data/Observations:Calculations: given: DAl = 2.70 g/mLConclusion: The thickness of aluminum foil is ____ mm.Reflection Question:

1. If your sheet were made of tin, what would be its mass?

Lab: The Height of an Oleic Acid MoleculePurpose: To determine the height of an oleic acid molecule in mm.Materials: 0.1 % oleic acid in methanol, cafeteria tray, overhead sheet, marker, lycopodium

powder, micropipet, graph paper, rulerProcedure:Data/Observations:Calculations: -area of the "hole"

-volume of 1 drop-volume of oleic acid-height of oleic acid

Conclusion: The height of an oleic acid molecule is _____ mm.Bonus Question:

1. What is the mass of an oleic acid molecule if we assume that the molecule's length and width are both 1/10th of its height, and its density is 1.00 g/mL?

Lab: Identification of UnknownsPurpose: To identify two unknown organic samples.Materials: 2 samples, thermometer, 250 mL beaker, table of unknownsProcedure: (use cooling curves to identify)Data/Observations: -chart of temperatures and time

-graph of temperature vs. time (cooling curves) Conclusion: State the identity of the two unknownsReflection Question:

1. Draw the structure of 10 substances from the table of unknowns

Lab: Finding the FormulaPurpose: To determine the formula of the hydrated crystal of copper(II) sulfate. Materials: balance, test tube, copper(II) sulfate, Bunsen burner, test tube holder.Procedure: Data/Observations:Calculations:Conclusion:Reflection Question:

1. How many atoms of hydrogen are there in 253 g of copper(II) sulfate?

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Lab: Endothermic / Exothermic LabPurpose: To determine the change in temperature per mole for each reaction.Materials: Ba(OH)� 8H2O, NaOH, (NH4)SCN, thermometer, 50 mL graduated cylinder,

2 -150 mL beakers, balanceProcedure: reaction 1: 1.00 g of NaOH in 50.0 mL of waterreaction 2: 2.00 g of Ba(OH)� 8H2O with 1.00 g of (NH4)SCNData/Observations:Calculations:Conclusion: State the temperature change per mole for each reactionReflection Questions:

1. Write balanced equations for these 2 reactions.Bonus Question:

1. To 2 sig figs, determine the amount of heat released per mole for each reaction. For thespecific heat capacity of the 2 solids, you use a reasonable estimation if you reason it out.

Lab: Percent YieldPurpose: To calculate the percent yield of a productMaterials: 100 mL beaker, 250 mL Erlenmeyer flask, funnel, filter paper, stirring rod, watch

glass, 1M CuSO4, iron fillings, methanolProcedure: React 1.00 g of iron with 30 mL of CuSO4 and calculate the % yield, assuming one of

the products is iron(II) sulfate.Data/Observations:Calculations: show all workConclusion: State the % yield copper for your reaction.Reflection Questions:

1. Write the balanced equation for the reaction.2. How pure was your sample of copper? What did you do (or could you have done) to find out?

Bonus Question:1. Calculate the % yield if the products were copper and Fe2(SO4)3 .

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