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Download Dear Students, Welcome to AP Chemistry, a little early. We will have
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AP Chemistry Summer Assignment Dr. Kalish Dear Students, Welcome to AP Chemistry, a little early. We will have a fabulous year together in which you will be taught everything you need for success in the course and on the AP Chemistry Exam in May 2015. I apologize for the summer assignment, but I want to have at least a month to review with you prior to the AP Chemistry Exam. The assignment is mostly review and NOT comprehensive of all material covered in honors chemistry. I would recommend that you start the assignment one week before school starts, as summer should be reserved for fun. Summer Assignment: 1. Read the Chapter 1 notes provided. These notes, as well as the two other packets, correspond with an older textbook and do not match the chapters in your book perfectly. Complete problems from Hwk 1.1 (the homework for the three chapters has been packaged as a unit) and Clwk 1.1. 2. Read the Chapter 2 notes provided. Complete problems from Hwk 1.2 and Clwk 1.2A and B (1.2B is NOT available on line). Note: Organic chemistry is no longer assessed on the APC Exam but is assessed on the SAT Chemistry Subject Test 2 so I have included some information on this concept. 3. Read the Chapter 3 notes provided. Complete problems from Hwk 1.3. We will review this section extensively, as stoichiometry is a critical component of APC! The advanced stoichiometry was not covered at the honors or pre-APC level. 4. Memorize the chemical symbols for elements on the periodic table (PT). For example, Mg represents Magnesium. You can use a PT on every assessment, but the name of the element is NOT listed. The periodic table that you will use all year and on the AP Chemistry Exam is provided in this packet. Also included are formula sheets (p. 160161), which are provided on the APC Exam, polyatomic ions, which are not, and other pertinent information. Please do not lose these sheets. You must memorize all of the polyatomic ions that have been provided. You will take Quiz 1.1 at the beginning of the second week of school. This quiz will assess material from the summer assignment (i.e. compound naming, formula writing, stoichiometry, empirical formula, etc.; NOT kinetics, equilibrium, etc.). If you have any questions or concerns, please email me at [email protected] Have a wonderful and safe summer. I look forward to working with all of you next year! Dr. Kalish I :;:. no NOT DETACH "CI FROM nOOK. n g !3 PERIODIC TABL~E iii' '".... OF THE ELElVlENTS '<I n H lie (lOX ::0 CD M >::: :::; '4 rt o ::l 10 I Li n C I N 0.9-1 HU:d I·tol 16.00 IC).OO l(l,IK II I 1 16 Ii' I' n Ar 'So· Na Mg AI 22,<)9 ::...!.~o 1(1(}X 19 20 21 ')~, 24 2:\ I 25 I 16 n 2H 29 Co Ni Cu 30 1 .\ 1 7 P ::'-:.(N .H ~C.97 :n As o :F s el 32.06 .q ':'lAS Yi rt ~ CfQ ~ ~ 0... Ca Sc Ti v - .... 9<1 -I7,<}() )().t)-I 'i~,O(; )-l9'" 6.1.55 6.'UtJ 79.90 X3.:':0 3H Jtj 40 41 .+~ .+3 44 45 .\.6 47 41\ (;9.12 .\.9 711.96 17 52 5.i 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Te I Xc X,'i..l7 1-:7.e:: Ml,.91 91,12 tJ2.tJI 95.')-1 l'>X) 101.1 102.91 lOll..!::' IO?iC 112.-11 11-1.-;2 IIX.71 121.75 127.60 12691 55 56 51 72 73 74 75 76 77 lX 71) XI) Xl X.2 &1 X..J It') Cs B~l I La I Ht' Au Hg TI Ph Hi Po At 1'291 X7 Fr 1i7 XX Ra Cr I Mn I Fe ...:...c....:. :....:....... +...:.. .:. . :. :.. :...............:5....:!!....:...6:.:.9. ..... OS Ir Pt nX.91 17H.-I9 ISO.Y.'i 1K1X'i IX()21 {\)(1.2 1950X H<J IOl 10:" 106 107 lOX 192.2 109 tAc Rf Bh Hs 22()J)21227.()1; (261) .'is CD 0 'iLantrmnidi! Scrr:s g Cc "" I Re Ta Ob 1~()2) Sg I (266 I (2b-l11 [277) I 1 10 Vlt Os (26~) (211) Zn. Ga 19IJ9~P()()591:[)-l.\X Ge Br Kr Se "07:l2(H91I!(2()t)) 2Jf)) III Rg 1(272) 59 nO (, I Pr Nd Pm ().:; 6..t (],'i M I 67 Mi (,9 71) } I 167 .26 16~,93 Lu 162'50 II ~J-l'B Yb I ¥1'1I l) I 17.1()-l 17·1.97 ')l\ t}9 l()() li)1 IU~ 1m Th Pa Cf Es ("m Md No Lr 2J2.11-l ::.H.O-1 (2:;1 L~52) (257) (2~X) (:.'5 Q I' 1.)0 gs Dv I Ho Er Tm ct C/] I' ~9.():'l 36 "'O.CK r-J (j 0 ~e K P> ~ I .N.IO 51 !:): :J s Q. 6 tv t...J Ul(!) III -" 22.', S g t Aetinide SCI' C'i ) ,2()2) UI ~9 X6 Rn I (~22) l".1 ~ !3 1:1 ~ .... ::l c STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25 C Hal f-reaction ) + 2 e + 1' 2.87 1.82 Au'''' + 31'Au(n 2CI !K)+2e 1.50 2 H,O(/) O:::(K) + 4H'" + 41' 1.23 2 Br Br2 (/) + 2 2 Hg 2 ... Hg2+ Hg:::+ + 21' + 2 e Hg(/) Ag(s) 0.80 Hg,:::++2e 2 Hg(/) +1' 0.53 curs) 0.52 + 21'Cuts) + 1' 0.34 1'- + 2 H+ + 0.15 0.15 Sn-+-+2e Sis) 0.79 21 + Cu 2 + 0.85 0.77 12(~)+2e Cu 2+ 1.07 0.92 +1' Cu+ 1.36 21' H 0.14 2H++2e H,(g) 0.00 Pb 2+ PhIs) 0.13 + 21' Sn 2-+2e Sn(s) 0.14 Ni 2+ + 21'Ni(s) + 2e Cots) Cd"+ + 21'Cd(s) Cr2+ Cr'+ + 1' Fe"+ + 21' Fe(s) 0.25 Co 2+ 0.28 Cr 3++3c Cr(s) Zn 2+ + 21' 0.40 0.41 0.44 0.74 Zn(s) 2H 2 0(/)+ 21' H 2 (g)+20H 0.76 Mn 2+ 1.18 + 21' :'vines) AI)+ + 31' Al(s) Be 2+ + 2c Be(s) Mg2+ + 21'- Na+ + e- + 2e Sr2+ + 2e Ba:::'" + 2e Rb+ + e Ca='- 0.83 1.66 1.70 Mgt.» Na(s) -2.37 Cats) -2.87 2.71 Sr(s) 2.89 Ba(s) 2.90 Rb(s) -2.92 K+ +e K(s) 2.92 + e Li+ + e Cs(s) 2.92 Lien -3.05 Cs+ GO ON TO THE NEXT PAGE. 3 AP Chemistry Course and Exam Description Appendix B: AP Chemistry Equations and Constants Throughout the test the following symbols have the definitions specified llnles, othemise nOled. L,mL g nm atm liter(s). milliliter(s) gram(s) nanometer( S) atmosphere(s) mm Hg J. kJ millimeters or men:ury jOllle(s). kilojoule(s) V \olt{s) mol mole(s) ATOMIC STRUCTURE E = energy E = Ill' \' = frequency /.= Planck's constant. II = 6.626 x 1 Speed of light. c = 2.9l)1{ x /\\ ogadro' s number = 6.022 Electron charge. e = Js 108 111 xI 1.602 A s"! mo\') 10')'1 coulomb EQUILIBRIUM [C]'[D]d K = . where a A + b B ~ c C + d D ( [A],lrBt K = (p, C l'tP K, (molar conccntrations) It! D pressure'» K" (weak acid) K1, (weak base) Kit (waler) Ki' (P")"(P13/' I' Equilibrium Constants K - fOWJlHB h - [B] K" = IWIIOWj pH = -log[WJ, pOH = 1.0 x 1O,j4 at 25°C [OWl 14 = pH + pOH _ [A] pH - pK" + log rHAl KINETICS InlAL InfAln = -kr _1_ =kl [Alo f ' _ 1/2 - rate constant r = time r Vl half,life k 0.693 -k- Retllrn to the Taj~e 0: Contents Appendix B GASES, LIQUIDS, AND SOLUTIONS PV P == pressure V = volume IlRT PA Plot,)1 T = temperature x X..... where X" = mole, A n = number of mole, III p .... + Pfl + Pc + ... 111 11 K JJ == mas" = molar mass D =- den,ily KE ;;; kinetic energy M \. = velocity °c + 273 A == absorbance = molar absorptivity h = path length (I D=!!.!.. V KE per molecule c = concentration 1 , = -::; 1111"Gas constant. R = 8.314 J mol-I Molarity, M A = moles of solute per liter of ,olution = abc = 0.08206 L atm mol-I K- 1 == 62.36 L torr mol-I K I atm = 760 111m == 760 tOIl' I STP == O.OOcC and 1.000 atm THERMOCHE:\USTRY/ ELECTROCHEMISTRY q lIlcf'::..T f'::..SC I so MIG I !:J.H~ products - t:,.G C products -ISo reactants I !:J.H! reactants I t:,.Gi products -I t:,.Gf reactants q heat 1/1 mas, c T specific heat capacity temperature So He qandard enthalpy GO ;,tandard free energy number of moles 11 E' O -RTln K 1=9.. q Faraday's constant. F t !volt Returr: to the Table of Corcter:ts 2013 The College Board. <;tamlan.l entrop: standard reduction potential current (amperes) charge (coulombs) lime (seconds) 96.485 coulombs per mole of electrons I joule [coulomb Chemistry Reference Tables Polvatomic Ions Activitv Series M~tal I . I Formula CH\COO' Name Acetatc Ammonium lOll Li Rb I Dr. Kalish i ~m; i I :,Sr I Arsenate Ca Na i ~Ig I Mil i Zn Cr Fe Cd Co Ni Sn Ph II, Sb Bi Cu Hg Ag I Azide Bromate Carbonate Chlorate Chlorite Chromate Cyanide Dichromate AsO.( N, i BrCh CO, CIO,' CIO, Cre}," CN Cr,O··· l\'ame Dihvdrogen phosphate Hydrogen carbonate or bicarbonate Hvdrogen sulfate Hydroxide Hvpochlorite Iodate Iodite l\1erc ury (l) Methvlammonilllll Monohydrogen phosphate Nitrate Nitrite lOll Formula H,P0 4 I-ICO., 1011 l\'ame Oxalate Perchlorate Formula C 20 4 U0 4 HSO,j' OH ClO Pennanganate Peroxide Phosphate Phosphite Sulfate Sulfite Thiocvanate 'fhiosull:1te Mn()j 0:" P0 4 ' 10,' 10:: Hg,· CH,NH, HPO.j·· NO., NO, Important Constants = 0.022 x 10:' atoms Speed oflight (c): cc- 2.991\ x ]0' Planck's constant: h 111 S 6.6262 x I J-scc Pt Au SO.j S( ), seN S::O," I un, ;'\lumber 1 2 ~ I mole Avogadro's constant: i I PO, . I Uranvl Prefix Mono· Oi ... Tri· f--.. T etraPentaHexa Hepta· I Octa· Nona Oeca I :1 I 4 I 5 6 7 11 9 10 Electroneaativities .,.,. Atomic Element [lectroIlegathit~ Atomic " limber Element Electronegativity -. Ni Cu 1.9 1.6 28 29 30 2.0 34 2.5 3.0 35 Se Hr Rb Ag Cd Sn Sb I Cs Ba Pt Au Hg Pb Bi Po ~umber It Li Be B C N I 0 F :-.la Mg AI Si 3.4 :17 47 4.0 4x 0.9 50 1.3 51 53 _. 16 P S 17 CI . ) .') 19 20 25 K O.X Ca Mn Fe Co 1.0 1.11 1.X 1.9 I :1 4 5 6 7 1\ 9 II 12 13 14 15 26 I n I 1.0 1.6 1.9 ') I 2.0 ' 55 56 78 79 80 fP ,- x3 84 Zn 1.9 1.6 '.6 :1.0 () .11 1.9 i 1.7 2.0 2.0 2.7 O.X 0.9 _. I i I 2.4 1.9 1.8 1.9 2.0 ! Reference Tables for 2 nd Semester: Table I: l!sefu I ('onstants ;\\,'gatiro' -; Numb;:r l illi\ ;:r,al G'l.-; C"lhtallt \' R R 6J)22 x 10" panicl;:, 0.01121 L-Jtlll ll1(lle-K or /I ..' 14 L-kPa lllok-K Planck', ":OlbWnt Sr;:..:d of light (in \iKUum) \lolar V"lull1e (STP) Ion rrodu..:t till' water ° Ta ble 2 : P ropertles fC,ommon S ~ 0 Ivents Solvent ktJ Til (0C) (nC-kg/mole) I Acctic acid Bell/elle Camphor Cvelohex<lnc Cvciohex<1nol Nitrohcll/ene Phefwl Wain I 17,l)O /10.100 20742 I ~O.725 2,530 2.53 5,611 2.75 - - 210,/1 5,24 3,MO 0,515 1/l1,~.W IOO,OO() Tr (0C) 16,66 5,533 17 /1, :5 ()54 25. i:; 5,7() 40,<)0 kr (OC-kg/mole) 6J,262, 10 ")-,,-,c 2,')9/1 x I ()' !1l sec I,,, = 22414 Lnwle f:. 1,0 X IOIJ Ii I 3,90 5,12 ~1, 7,7 2()() 39,':; (,1\52 740 1,1\53 O,()OO a)l l e.i S : ,peCIIC T TH eat 0 fC,ommon Substances at 20"(' Substance Specific Heat (Jig 0c) Table 4- Polvatomic Ions Air ~ Aluminum Carh,)ll (di,1I1Hlllti) Carb"11 il!rarhlte) Cill'holl dinxide Copper Ethyl alcohol Gold Granite Iron L\.'ad Parutlin Sihcr Stainless steel Water i I ,Of) (J,895 O.5()2 ! CaCO" CaU". CI1 1 •. · CJ I",. ('4 i II I<llel Formula CH 1COO NH.j Arsenate Azide Borate Bromate Carbonate Chlorate Chlorite Chromate AsO.j N, 80/ BrO, CO,-' CIO, C10, CrO.j . Cyanide Dichromate eN (,r::O-·' () 71 I (U132 O,3l\7 245 (),129 f),803 0,44/1 0,12:-\ 2.9 0,233 0,51 4,IX T a bl c..:; : Ent ha irlles ' 0 f FormatIOn Substance ~H n (kJ/mole) -45.9 '\IH, '" NlljCI, -314.4 NIl.jFI, -125 -3(l5,56 NH,NO" Br, Ion ;-';ame Acetate Ammonium (),on -1207,6 -6:'\4.9 -74.9 -104,7 -125.7 -300.9 -393,5 C1"H" " CO'. F'::c n.oo H> HBr:'J (l,On -36,29 HC) ·9:U Ion ~ame Dihydrogen phosphate Hydrogen carbonate or bicarbonate Hydrogen sulfate Hydroxide Hypochlorite Iodate Iodite Mercury (I) Methylammoniul11 Monohydrogen phosphate Nitrate Nitrite Substance Hr', , H:O," H-Ol\ H,O"" H2SO.j I FeO"1 Fc:O",. \lnO-, NcO," '\I:OJ," ~Ht (kJimole) 0: n.oo Na:O" Na-SO"" so' . -414.2 -110 I -2%,8 SO"e" -395,7 I Formula HePO, HC01 Ion :'-lame Oxalate Perchlorate HSO.j OH' CIO 10,' 10 2 Hg 2ClhNH, HP0 1 Pemlan!!anate Peroxide .. Phosphate Phosphite Sulfate Sulfite ThiocYanate Thiosu Ifate NO, NO, Lranvl Formula C 20 1 ClO. M1104 (), - PO+' PO," SO.jSO," SCN' S:O, LO, - -2:3 ..\ Table 6: Heats of Fusion and Vaporization for Some Common Substances Substance lIeat of Fusion Heat of (.J/g, \'aporization (Jig) 472(, i 205 Coppcr ,>1,79 Ell1\ I ,".koho! 109 ()4,5 Gold 157i\ Lead 24.7 X5S Sil\cr 88 2:'\00 \Vater 334 22<>0 -241.x2 -285,8 -187:-; -/\ J1,.9i\lJ -~15,5 -111:-\.4 -52(),() -il2.1 '9,16 i AP Chemi:-;try II and P: Chapter 1 Dr. Kalish Page I Chemistry: 'latter and Measurement I. Introduction: Chemistry enables us to design all kinds of materials: drugs (disease): pesticides: fertilizers: fuels: fihers (clothing): building materials: plastics: etc. A. Key Terms: 1. Chemistry: study of the composition. structure. properties. and reactions of matter a. Matter: anything that has mass and occupies space Examples: wood, sand. water. air. gold. ctc. 1) l\lass: quantity of matter in an ohject a) use a balance to measure mass: the unit on a balance is the gram. but the fundamental unit is the kilogram (kg) 2) Volume: space that object occupies a) use a ruler to measure a regular solid. and a graduated cylinder to measure an irregular solid (water displacement) or a liquid b) the units vary: em~. ml. L [1 ml = I em}] 2. Atoms: smallest distinctive units in a sample of matter (huilding blocks) 3. Molecules: larger units in which two or more atoms are joined together a. The way in which matter hehaves depends on the atoms present and the manner in which they are comhined 4. Composition: types of atoms and their relative proportions in a sample of matter B. Properties: 1. Physical property: characteristic displayed hy a sample of matter without it undergoing any change in its composition: what you can see or measure without altering the chemical nature of the material Examples: color. mass. density. state. Tm. Tr, Tb 2. Chemical property: characteristic displayed by a sample of matter as it undergoes a change in composition Examples: flammability. ability to react with acids C. Types of changes: 1. Physical change: change at the macroscopic level but no change in composition: the same substance must remain after the change Examples: Phase changes. dissolving AI' Chemistry [I and P: Chapter 1 Dr. Kalish Page .., 2. Chemical change or chemical reaction: change in composition and/or the structure of its molecules: one or more substances is altered--new substances arc formed Examples: cooking and spoiling of foods: burning, digestion. fcnnentation a. Reactants -7 Products h. Evidence of a Chemical Change 1) Evolution of a gas 2) Fonnation or a ppt. 3) Evolution or absorption of heat (exo- vs, endothennic reactions) 4) Emission of light 5) Color change D. Classifying :\:latter \1aterial Homogeneous IVlaterial: Substance Elements Homogeneous !vlix tures (Solutions) Heterogeneous Mixtures Compounds 1. Material: any specific type of matter a. Homogeneous Materials: unifonn matter b. Heterogeneous Materials: nonunifo1111 matter 2. Mixture: consists of two or more different atoms or compounds with no fixed composition; the atoms or compounds are mixed together physically a. Heterogeneous Mixture: variable composition and/or properties throughout: two or more distinct phases: interface Examples: blood, granite: Oil and water b. Homogenous Mixture or Solution: has the same composition and properties throughout Examples: salt water: sugar water 3. Substance: type of matter that has a definite or fixed composition that docs not vary from one sample to another Examples: Elements or compounds Dr. Kalish AP Chemistry Hand P: Chapter 1 Page -' a. Element: substance that cannot be broken down into other simpler substances by chemical reactions: substances composed of one type of atom; represented by a chemical symbol h. Compound: substance made up of atoms of two or more elements that combine in fixed propoltions or definite ratios: represented by a chemical fonnula Example: H 20 with 111 and 1 0 4. Key differences behveen mixtures and compounds a. The properties of a mixture ret1ect the properties of the substances it contains: the properties of a compound bear no resemblance to the properties of the elements that comprise the compound. h. Compounds have a definite composition by mass of their combining elements. while the components of a mixture may be present in varying propOitions. E. Scientific !\lethods: 1. Observation 2. Hypothesis: tentative prediction or explanation concerning some phenomenon 3. Experiment: procedure used to test a hypothesis a. Data 4. Scientific Laws: summary of patterns in a large collection of data 5. Theory: multi-tested and contlnned hypothesis II. Scientific Measurement: A. International System of l'nits 1. Length: the SI base unit is the meter '1 Mass: the quantity of matter in an object a. SI base unit is the kilogram 3. Time: SI base unit is the second 4. Temperature: property that tells us in what direction heat will t10w a. S I base unit is Kelvin (K) Seven Fundamental Units ill Sf Qllautity Length Mass Time Them10dvnamic Temperature Amount of Substance Electric Current Luminous Intensity I l"nit Symholfi)1' qlumti(r ! III r T 11 [ 1 I Abbreviation for unit I i i meter kilogram second kelvin mole ampere candela 111 kg s K mol A cd AP Chemistry Hand P: Chapter I Dr. Kalish Page 4 B. Den sity 1. mass per unit volume of a substance ~ - ./ Lowcst density d=m/V Density of water is 1.00 giml Problems ... Common Sf prefixes Prefix I Exponential:\1 ultiplier l'nit abbreviation I Jfea 11 illJ( i GiEaMega • KiloHecto Deca G lvl k h da . deci centi milli microi nano , pleo femto d 10" 10" 10 lO- i I 000 000 000 i t 000 000 1000 tOO . 1 10 10' 10" 10 1cr 1 1 ]0 1 I U() H)" 10'(' 1 1000 1 1 000000 n 10' I 1 000 000 000 p 10 10' , c m Jl f , 1 Greatest density I 1 000 000 000 000 1 I 000 000 000 000 000 C. Conversions within the same quantity vs. those behYeen different quantities I. Same Quantity 2. Different Quantities: Factor-Labell\lethod or Dimensional Analysis Eq utllities and Collversioll Factor.'" 2 cups== 1 pint 2.54 centimeters = I in 12 inches 1 foot 2 pints 1 quart 4 cups 1 quart 3 feet = 1 yard 1 liter = 1.06 quarts 5280 feet = 1 mile 16 ounces = I pound 1 meter = 39.37 inches I molc= 6.022 x 1O~i awm, I metcr 1.09 yards 3 1 ml 1 cm 1 km 0.62 miles Problems.... 28.3 grams·=; 1 ounce 1 kilogram = 2.2 pounds 453.6 g = J lb 28.35 g I ounce 3.785 L = 1 gal 29.57 ml I tl oz ' Dr. Kalish AP Chemistry Hand P: Chapter 1 Page .) D. Precision and Accuracy in Measurements I. Precision: how closely individual measurements agree with one another 2. Accuracy: closeness of the average of the set to the "correct" or most probable value **Precise measurements are 1\OT always accurate.** Example: Darts--Ifyou hit the same spot outside the bulls-eye five times, you have precision but not accuracy. You are accurate \vhen you hit thc bulls-eye. Percent Error =~(Measured Value -.Accepted VaIUC)}* I 00 1 Accepted Value Eo Significant Figures I. All digits known with certainty plus the tirst uncertain one (estimated) arc significant digits or significant figures 2. Significant figures reflect the precision of the measurement 3. D etermmmg ,S"Igm°fiIcant D"" 19lts . Number Digits to Count Example Number of Significant Digits i All , i\onzero Digits 3279 4 .) 11.2 ~ i Leading Zeroes (zeroes before an integer) None Captive Zeroes (zeroes between two integers) All Trailing Zeroes (zeroes , after the last integer) : 0'()O45 1 5.007 4 7 100 100. 100.0 0.0100 1.7 x 10'-1 1.30 x l((~ , 0.(0) x 10' All BCT be careful of "incolTcctly" written scicntitic notatilll1 Scientific Notation O.O()OOO5 6.f)OO.Om~ Counted only if the number contains a decimal point .., I 3 :4 ., .) 2 3 1 4. Rules: a. :vIultiplying and Dividing: Round the calculated result to the same number of significant figures as the measurement having the least number of significant figures. [carryall numbers through and then round oft] Example: 3.45 cm figures) * 2 4.5555 cm -7 15.7 cm (Answer is expressed in 3 significant !\P Chemistry Hand P: Chapter 1 Dr. Kalish Page 6 b. Addition and Subtraction: The answer can have NO more digits to the right of the decimal point than there are in the measurement with the smallest number of digits to the right of the decimal point. Example: 3.45 cm + 100,1 em ~ 103,6 em (Express answer to the tenth place) c, Rounding Fives: If the last significant digit before the five is odd. round up. If the last significant digit before the five is even (and therc are not any numbers other than zero after the tive). do NOT round up (leave it alone), Example: 3.15 ~ For two significant digits or to the tenth place. round to ],2 Example: 3.45 ~ For t\VO significant digits. round to 3.4 Example: 3.451 ~ For two signiticant digits. round to 3,5 28 Chapter 1 Chemistry: Matter and Measurement Matter is made up of atoms and molecules and can be subdivided into two broad cat egories: substances and mixtures. Substances have fixed compositions; they are either ele ments or compounds. Compounds can be broken down into their constituent elements through chemical reactions. but elements cannot be subdivided into simpler substances. Mixtures are either homogeneous or heterogeneous. Substances can be mixed in varying pro portions to produce homogeneous mixtures (also called solutions). The composition and properties are uniform throughout a solution. The composition and/or properties of a het erogeneous mixture vary from one part of the mixture to another. / Substances exhibit characteristics called physical properties without undergoing a change in composition. In displaying a chemical property, a substance undergoes a change in composition-new substances are formed. A physical change produces a change in the appearance of a sample of matter but no change in its microscopic structure and composi tion. In a chemical change, the composition and/or microscopic structure of matter changes. Four basic physical quantities of measurement are introduced in this chapter: mass, length, time, and temperature. In the SI system, measured quantities may be reported in the base unit or as multiples or submultiples of the base unit. Multiples and submultiples are based on powers of ten and reflected through prefixes in names and abbreviations. nanometer(nm) 10- 9 m micrometer (fLm) 1O-6 m millimeter (mm) 10- 3 m meter(m) m 1 1 kilometer (km) 103 m The SI base unit of temperature, the kelvin (K), is introduced in Chapter 5, but in this chapter two other temperature scales, Celsius and Fahrenheit, are considered and compared. 1 tF = 1.8 tc + 32 tc = -'-'----' 1.8 To indicate its precision, a measured quantity must be expressed with the proper num ber of significant figures. Furthermore, special attention must be paid to the concept of sig nificant figures in reporting calculated quantities. Calculations themselves frequently can be done by the unit-conversion method. The physical property of density also serves as an im portant conversion factor. The density of a material is its mass per unit volume: d = m/V. When the volume of a substance or homogeneous mixture (cm 3) is multiplied by its densi ty (glcm\ volume is converted to mass. When the mass of a substance or homogeneous mix ture (g) is multiplied by the inverse of density (cm3/g), mass is converted to volume. In this chapter and throughout the text, Examples and Exercises illustrate the ideas, methods, and techniques under current discussion. In addition, Estimation Examples and ac companying Exercises deal with means of obtaining estimated answers with a minimum of calculation. Conceptual Examples and accompanying Exercises apply fundamental con cepts to answer questions that are often of a qualitative nature. A inl yo pil pn de tic PI; AP Chemistry Summer Assignment Dr. Kalish Page 1 Homework Problems: Note: Numbers in the left margin correspond with book problems from the PH textbook and your answer key. 2) 11) 12) 13) 14) 15) 32) 36) 42) 46) 48) 50) 2) 8) 10) 12) 19) Hwk 1.1: Chapter 1 1) Which of the following are examples of matter? a) Iron c) The human body e) Gasoline b) Air d) Red light f) An idea 2) Which of the following is NOT a physical property? c) Natural gas burns. a) Solid iron melts at a temperature of 1535 oC. b) Solid sulfur as a yellow color. d) Diamond is extremely hard. 3) Which of the following describe a chemical change, and which a physical change? a) Sheep are sheared and the wool is spun into c) Milk sours when left out. yarn. d) Silkworms feed on mulberry leaves and b) A cake is baked from a mixture of flour, produce silk. baking powder, sugar, eggs, shortening, and e) An overgrown lawn is mowed. milk. 4) Which of the following represent elements? Explain. e) Na a) C c) Cl f) KI b) CO d) CaCl2 5) Which of the following are substances, and which are mixtures? Explain. a) Helium gas used to fill a balloon c) A premium red wine b) Juice squeezed from a lemon d) Salt used to de-ice roads 6) Indicate whether the mixture is homogeneous or heterogeneous. a) Gasoline c) Italian salad dressing b) Raisin pudding d) Coke 7) Convert the following quantities: a) 546 mm to meters c) 181 pm to µm e) 46.3 m3 to L (careful) f) 55 mi/h to km/min b) 87.6 mg to kg d) 1.00 h to µs 8) How many significant figures are there in each of the following quantities? e) 1.60 x 10-9 s a) 4051 m c) 0.0430 g 9 f) 0.0150 oC b) 0.0169 s d) 5.00 x 10 m 9) Perform the indicated operations, and provide answers in the indicated unit with the correct number of significant digits. a) 13.25 cm + 26 mm – 7.8 cm + 0.186 m (in cm) b) 48.834 g + 717 mg – 0.166 g + 1.0251 kg (in kg) 10) Calculate the density of a salt solution if 50.0 ml has a mass of 57.0 g. 11) A glass container has a mass of 48.462 g. A sample of 4.00 ml of antifreeze solution is added, and the container with the antifreeze has a mass of 54.513 g. Calculate the density of the antifreeze solution expressed in the correct number of significant figures. 12) A rectangular block of gold-colored material measures 3.00 cm x 1.25 cm x 1.50 cm and has a mass of 28.12 g. Can the material be gold if the density of Au is 19.3 g/cm3? Calculate the percent error. Hwk 1.2: Chapter 2 1) When 24.3 g of magnesium is burned in 16.0 g of oxygen, 40.3 g of magnesium oxide is formed. When 24.3 g of magnesium is burned in 80.0 g of oxygen, (a) What is the total mass of substances present after the reaction? (b) What mass of magnesium oxide is formed? (c) What law(s) is/are illustrated by this reaction? (d) If 48.6 g of magnesium is burned in 80.0 g of oxygen, what mass of magnesium oxide is formed? Explain. 2) What is the atomic nucleus? Which subatomic particle(s) is/are found in the nucleus? 3) Which of the following pairs of symbols represent isotopes? Which are isobars? a) 70 33 E and 70 34 E d) 7 3 b) 57 28 E and 66 28 E e) 22 11 c) 186 74 E and 48 E E and 44 22 E E and 186 74 E 4) What do atomic mass values represent? 5) What type of information is conveyed by each of the following representations of a molecule? AP Chemistry Summer Assignment 20) 24) 25) 26) 38) 47) 48) 50) 52) 54) 56) 58) 60) 62) 64) 68) Dr. Kalish Page 2 a) Empirical formula b) Molecular formula c) Structural formula 6) A substance has the molecular formula C4H8O2. (a) What is the empirical formula of this substance? (b) Can you write a structural formula from an empirical formula? Explain. 7) Are hexane and cyclohexane isomers? Explain. 8) For which of the following is the molecular formula alone enough to identify the type of compound? For which must you have the structural formulas? a) An organic compound c) An alcohol e) A carboxylic acid b) A hydrocarbon d) An alkane 9) Explain the difference in meaning between each pair of terms: a) A group and period on the periodic table c) An acid and a salt (P.T.) d) An isomer and an isotope b) An ion and ionic substance 10) Indicate the numbers of electrons and neutrons in the following atoms: a) B-11 c) Kr-81 b) Sm-153 d) Te-121 11) Europium in nature consists of two isotopes, Eu-151, with a mass of 150.92 amu and a fractional abundance of 0.478, and Eu-153, with a mass of 152.92 amu and a fractional abundance of 0.522. Calculate the weighted average atomic mass of Europium. 12) The two naturally occurring isotopes of nitrogen are N-14, with an atomic mass of 14.003074 amu, and N-15, with an atomic mass of 15.000108 amu. What are the percent natural abundances of these isotopes? {Hint: set one at x and the other at 1-x} 13) The two naturally occurring isotopes of rubidium are Rb-85, with an atomic mass of 84.91179 amu, and Rb-87, with an atomic mass of 86.90919 amu. What are the percent natural abundances of these isotopes? {Hint: set one at x and the other at 1-x} 14) Identify the elements represented by the following information. Indicate whether the element is a metal or nonmetal. a) Group 3A (13), period 4 d) Group 1A (1), period 2 b) Group 1B (3), period 4 e) Group 4A (14), period 2 c) Group 7A (17), period 5 f) Group 1B (3), period 4 15) Write the chemical symbol or a molecular formula for the following, whichever best represents how the element exists in the natural state. a) Chlorine c) Neon e) Sodium b) Sulfur d) Phosphorus 16) Which of the following are binary molecular compounds? a) Barium iodide c) Chlorofluorocarbons e) Sodium cyanide b) Hydrogen bromide d) Ammonia 17) Write the chemical formula or name the compound: d) Phosphorus f) Dinitrogen pentoxide a) PF3 b) I2O5 pentachloride c) P4S10 e) Sulfur hexafluoride 18) Write the chemical symbol or name for the following monatomic ions: e) Ba2+ a) Calcium ion c) Sulfide ion 3+ d) Fe f) Se2b) Cobalt(II) ion 19) Write the chemical formula or name for the following polyatomic ions: d) CrO42a) HSO4f) Dichromate ion b) NO3 e) Hydrogen phosphate g) Perchlorate ion c) MnO4ion h) Thiosulfate ion 20) Name the following ionic compounds: f) KOH k) K2Cr2O7 a) Li2S b) FeCl3 l) Ca(ClO2)2 g) NH4CN c) CaS h) Cr(NO3)3 9H2O m) CuI i) Mg(HCO3)2 d) Cr2O3 n) Mg(H2PO4)2 e) BaSO3 j) Na2S2O3 5H2O o) CaC2O4 H2O 21) Write the chemical formula for the following ionic compounds: a) Potassium sulfide c) Aluminum bromide d) Potassium sulfite b) Barium carbonate hexahydrate e) Copper(I) sulfide AP Chemistry Summer Assignment f) Magnesium nitride i) Potassium nitrite g) Cobalt(II) nitrate j) Zinc sulfate h) Magnesium dihydrogen heptahydrate phosphate 22) Name the following acids: a) HClO(aq) d) HF(aq) b) HCl(aq) e) HNO3(aq) c) HIO4(aq) f) H2SO4(aq) 23) Write the chemical formula for the following acids. a) Hydrobromic acid e) b) Chlorous acid f) c) Perchloric acid g) d) Nitrous acid h) 8) 22) 26) 36) 37) 40) 43) 44) 50) 56) 58) 64) 68) Dr. Kalish Page 3 k) Sodium hydrogen phosphate l) Iron(III) oxide g) H2SO3(aq) h) H2C2O4(aq) Acetic acid Phosphorous acid Hypoiodous acid Boric acid Hwk 1.3: Chapter 3 1) What are the empirical formulas of the compounds with the following molecular formulas? c) C10H8 a) H2O2 b) C6H16 d) C6H16O 2) Calculate the molecular or formula mass of the following. d) K3[Co(NO2)6] a) C2H5NO2 b) Na2S2O3 e) Chlorous acid c) Fe(NO3)3 9H2O f) Ammonium hydrogen phosphate 3) Calculate the mass, in g, of the following: c) 0.615 mol chromium(III) oxide a) 4.61 mol AlCl3 b) 0.314 mol HOCH2(CH2)4CH2OH 4) Calculate the mass percent nitrogen in the compound having the condensed structural formula, CH3CH2CH(CH3)CONH2. 5) Calculate the mass percent of beryllium in the mineral, Be3Al2Si6O18. Calculate the maximum mass of Be obtainable from 1.00 kg of Be. 6) The empirical formula of apigenin, a yellow dye for wool, is C3H2O. The molecular mass of the compound is 270 amu. What is the molecular formula? 7) Resorcinol, used in manufacturing resins, drugs, and other products, is 65.44 %C, 5.49 %H, and 29.06 %O by mass. Its molecular mass is 110. amu. What is the molecular formula? 8) Sodium tetrathionate, an ionic compound formed when sodium thiosulfate reacts with iodine is 17.01 % Na, 47.46 % S, and 35.52 % O by mass. The formula mass is 270 amu. What is its formula? 9) A 0.0989 g sample of an alcohol is burned in oxygen to yield 0.2160 g CO2 and 0.1194 g H2O. Calculate the mass percent composition and empirical formula of the compound. 10) Balance the following equations: e) Al2(SO4)3 + NaOH Al(OH)3 + Na2SO4 a) TiCl4 + H2O TiO2 + HCl b) WO3 + H2 W + H2O f) Ca3P2 + H2O Ca(OH)2 + PH3 c) C5H12 + O2 CO2 + H2O g) Cl2O7 + H2O HClO4 d) Al4C3 + H2O Al(OH)3 + CH4 h) MnO2 + HCl MnCl2 + Cl2 + H2O 11) Write a balanced chemical for each of the following: a) Decomposition of solid potassium chlorate upon heating to generate solid potassium chloride and oxygen gas b) Combustion of liquid 2-butanol c) Reaction of gaseous ammonia (NH3) and oxygen gas to generate nitrogen monoxide gas and water vapor. d) The reaction of chlorine gas, ammonia vapor, and aqueous sodium hydroxide to generate water and an aqueous solution containing sodium chloride and hydrazine (N2H4, a chemical used in the synthesis of pesticides). 12) Toluene and nitric acid are used in the production of trinitrotoluene (TNT), an explosive. ___C7H8 + ___HNO3 ___C7H5N3O6 + ___H2O a) What mass of nitric acid is required to react with 454 g of C7H8? b) What mass of TNT can be generated when 829 g of C7H8 reacts with excess nitric acid? 13) Acetaldehyde, CH3CHO (D = 0.789 g/ml), a liquid used in the manufacture of perfumes, flavors, dyes, and plastics, can be produced by the reaction of ethanol with oxygen gas. AP Chemistry Summer Assignment 74) 14) 82) 15) 84) 16) 89) 17) 90) 18) 92) 19) Dr. Kalish Page 4 ___CH3CH2OH + ___O2 ___ CH3CHO + ___H2O a) How many liters of liquid ethanol (D = 0.789 g/ml) must be consumed to generate 25.0 L acetaldehyde? Boron trifluoride reacts with water to produce boric acid and fluoroboric acid. 4BF3 + 3H2O H3BO3 + 3HBF4 a) If a reaction vessel contains 0.496 mol BF3 and 0.313 mol H2O, identify the limiting reactant. b) How many moles of HBF4 should be generated? A student needs 625 g of zinc sulfide, a white pigment, for an art project. He can synthesize it using the reaction, Na2S(aq) + Zn(NO3)2(aq) ZnS(s) + 2NaNO3(aq) a) What mass of zinc nitrate will he need if he can make the zinc sulfide in a 85.0 % yield? Calculate the molarity of each of the following aqueous solutions: a) 2.50 mol H2SO4 in 5.00 L solution b) 0.200 mol C2H5OH in 35.0 ml of solution c) 44.35 g KOH in 125.0 ml of solution d) 2.46 g oxalic acid in 750.0 ml of solution e) 22.00 ml triethylene glycol, (CH2OCH2CH2OH)2 (D = 1.127 g/ml) in 2.125 L of solution f) 15.0 ml isopropylamine, CH3CH(NH2)CH3, (D = 0.694 g/ml) in 225 ml of solution A stock bottle of nitric acid indicates that the solution is 67.0 % HNO3 by mass (67.0 g HNO3/100.0 g solution) and has a density of 1.40 g/ml. Calculate the molarity of the solution. A stock bottle of potassium hydroxide solution is 50.0 % KOH by mass (50.0 g KOH/100.0 g solution) and has a density of 1.52 g/ml. Calculate the molarity of the solution. If 50.00 ml of 19.1 M NaOH is diluted to 2.00 L, calculate the molarity of NaOH in the diluted solution. AP Chemistry Clwk 1.1 Dr. Kalish Page 1 Date: Period #: Matter--Substances vs. 1\1ixtures All matter can be classified as wither a substance (element or compound) or a mixture (heterogeneous or homogeneous). Directions: Classify each of the following as a s ubstance or a mixture. If it is a substance. write element or a compound in the substance column. Ifit is a mixture. write heterogeneous or homogeneous in the mixture column. Mixture cream Physical vs. Chemical Changes In a physical change. the original substance still exists. It has changed in form only. In contrast. a new substance is produced when a chemical change occurs. Energy always accompanies chemical changes. Directions: Classify each of the follO\ving as a chemical (C) or physical (P) change. I. Sodium hydroxide dissolves in water. 2. Hydrochloric acid reacts with potassium hydroxide to produce a salt. water. and heat. 3. A pellet of sodium is sliced in two. 4. Water is heated and changed to steam. 5. Potassium chlorate decomposes to potassium chloride and oxygen gas. 6. Iron rusts. 7. When placed in water. a sodium pellet catches on tire as hydrogen gas is liberated and sodium hydroxide forms. 8. Evaporation 9. Ice Melting 10. Milk sours. AP Chemistry Clwk 1.1 Dr. Kalish Page 2 11. Sugar dissolved in water. 12. Wood rotting 13. Pancakes cooking on a griddle 14. Grass growing in a lawn 15. A tire is intlated with air. 16. Food is digested in the stomach. 17. Water is absorbed by a paper towel. Physical vs. Chemical Properties A physical property is observed with the senses and can be determined without destroying the object. For example. color. shape. mass. length. and odor are all examples of physical properties. A chemical property indicates how a substance reacts with something else. The original substance is altered fundamentally when observing a chemical property. For example. iron reacts with oxygen to form rust. which is also known as iron oxide. Directions: Classify each of the following properties as either chemical or physical by denoting with a check mark. Physical Property I. 2. 3. 4. 5 - 6. 7. 8. 9. 10. _. II. 12. 13. 14. IS. Blue color DeI1~ity Flammability Soll:l~ili~y Reacts with acid to form He ... ?l:lPP?I}:~ combustion .... Sour taste ~J~I~il1g Point Reacts with water to form a gas Reacts with base to form water Hardness Boiling Point Can neutralize a base Luster Odor Chemical Property AI' Chemistry Ii and P: Chapter .2 Dr. Kalish Page 1 Atoms, 'lo1ecules~ and Ions I. La\\'s and Theories: A Brief Historical Introduction A. Laws of Chemical Combination I. Lavosier (1743-1794): The Law of Conservation of 1\1 ass a. The total mass remains constant during a chemical reaction Example: HgO ~ IIg -+ O 2 Mass of reactants = Mass of products ') Proust (1754-1826): The Law of Constant Composition or Definite Proportions a. All samples of a compound have the same composition or the same proportions by mass of the elements present Example: NaCI is 39.34 l'ia and 60.66 % CI Example: O:Mg in MgO is 0.6583: I. What mass ofMgO will faTIn when 2.000 g Mg is converted to MgO by buming in pure 02? 2.000 g Mg x 0.6583 0 1 Mg 1.317 gO 2.000 g Mg 1.317 gO 3.317 g MgO B. John Dalton (1766-1844) and the Atomic Theory of Matter (1803) 1. Law of :Vlultiple Proportions a. When two or more different compounds of the same two elements are compared. the masses of one element that combine with a fixed mass of a second element are in the ratio of small \vhole numbers Examples: CO vs. CO 2 : S02 \'s. SO, 2. Atomic Theory a. All matter is composed ofe.\'treme~v small, indivisible particles called atoms b. All atoms ofa given clement are alike in mass and other properties. but atoms of one clement differ from the atoms of every other element c. Compounds are fanned when atoms of different elements unite in fixed proportions d. A chcmical reaction involvcs a rearTangemcnt of atoms. l'io atoms are creatcd. destroyed or broken apart in a chemical reaction. Examples ... 3. Dalton used the Atomic Theory to restate the Lav. .' of Conservation of Mass: Atoms can neithcr be created nor destroyed in a chcmical rcaction. and as a consequence. the total mass remains unchangcd. AP Chemistry Hand P: Chapter ::: Dr. Kalish Page ::: C. The Divisible Atom I. Subatomic Particles a. Proton 1) Relative mass = 1 2) positive electrical charge b. Neutron I) Relative mass 1 (although slightly greater than a proton) 2) no charge == 0 c. Electron 1) mass = I! 1836 of the mass of a proton 2) negative electrical charge -I Particle I Symbol Proton Neutron Electron p n e Approximate Relative l\lass 1 L 0.000545 Relative Charge I 0 1 Location in Atom Inside nucleus Inside nucleus Outside nucleus An atom is neutral (has no net charge) because p = e-. :t The number of protons (Z) detennines the identity of the element. 4. Mass number (A)== protons + neutrons a. neutrons A - Z 1 Example: ticr Detennine the number ofp. e-. and n 5. Isotopes: atoms that have the same number of protons but different numbers of neutrons Examples: IH. 2H, 3H [or H-J. H-2. H-3J: 32S. "S: 5l)CO. 6!JCO 6. Isobars: atoms with the same mass number but different atomic numbers 1-1 H [ a. Examp Ie: C. N D. Atomic Masses 1. Dalton arbitrarily assigned a mass number to one atom (H-l) and detennined the masses of other atoms relative to it. 2. Current atomic mass standard is the pure isotope C-12. 3. Atomic mass unit (amu): l!12themassofC-12. 4. Atomic Mass: weighted average of the masses of the naturally occurring isotopes of that element Ne-20: 90.51 %. 19.99244 u a. Example: Ne-21: 0.27 %. 20.99395 u Ne-22: 9.22 ~O. 21.99138 u AP Chel1li~try II and P: Chapter :2 Dr. Kalish Page J E. The Periodic Table I. Dmitri Mendeleev's (1869) Periodic Table: a. Arranged elements in order of increasing atomic mass. from left to right in ro\\'s, and from top to bottom in groups b. Elements that most closely resemble each other are in the same vertical group (more important than increasing mass), c. The group similarity recurs periodically (once in each row) d. Gaps for missing clements: predict characteristics of yet to be discovered clements based on their placement 2. Modern Periodic Table a. Elements are placed according to increasing atomic number b. Groups or Families: vel1ical columns c. Periods: horizontal ro\VS d. Two series "pulled out" 1) Lanthanide and Actinide Series e. Classes 1) Most elements arc Metals, which are to the len (NOT touching) the stair-step line a) luster. good conductors of heat and electricity b) malleable (hammered into thin sheets or f()il). ductile (drawn into wires) c) Solids at room temperature (except mercury) 2) Nonmetals are to the right (NOT touching) of the stair-step line a) poor conductors of heat and electricity b) many are gases at RT 3) Metalloids: touch the vertical and or horizontal of the stair-step line (except Al and Po 11. Introduction to :Vlolecular and Ionic Compounds A. Key Terms 1. Chemical Symbols are used to represent clements 2. Chemical F0l111Ulas are used to represent compounds a. Subscripts indicate how many atoms of each element are present or the ratio of Ions B. Molecules and Molecular Compounds I. Molecule: group of two or more atoms held together in a definite spatial arrangement by covalent bonds ') Molecular Compound: molecules arc the smallest entities, and they detennine the propcI1ies of the substance 3. Empirical Formula: simplest fOl111Ula for a compound a. indicates the elements present in their smallest integral ratio Example: CH~O = 1 C: .2 H: 10 4. Molecular Formula: true fonnula for a compound {n = MFmass/EFmassl a. indicates the elements present and in their actual numbers Example: C6Hl~06 = Q C: 12 H: Q 0 5. Diatomic Elements: two-atom molecules, which don't exist as single atoms in nature a. Br~. h, N2, Ch. H> O 2• F~ 6. Polyatomic Elements: many-atom molecules Dr. Kalish AP Chemi,try H 3nd P: Chapter 2 Page --l a. Sg. P-l 7. Structural Formulas: shows the alTangement of atoms a. lines represent covalent bonds between atoms C. Writing Formulas and Names of Binary Molecular Compounds I. Binary Molecular Compounds: comprised of 1\\'0 elements. which are usually nonmctals a. The first element symbol is usually the element that lies farthest to the lcft of its period andlor lowest in its group (exceptions: Hand 0) [Figure 2.7] b. Molecular compounds contain prefixes far subscripts (exception: mono is not used far the first element) Prefix :".'umber c. The name consists of two \\'ords: mono 1 (prefix) element prefix~ide fonn I , * rule with oxide ! ditri tetra pcnta hexa hepta octa - 8 ! 110113 9 i dec a 10 3 i 4 :'\ f, 7 D. Ions and Ionic Compounds 1. Ion: charged particle due to the loss or gain of one or more electrons a. ::\1onatomic Ion: a single atom loses or gains one or more e 1) use the PT to predict charges 2) more than one ion can fann with transition elements b. Cation: positively charged ion [usually a metal] c. Anion: negatively charged ion [usually a nonmetal] d. Polyatomic Ion: a group of covalently bonded atoms loses or gains one or more e e. Ionic Compounds: comprised of oppositely attracted ions held together by electrostatic attractions; no identifiable small units 2. Formulas and Names for Binary Ionic Compounds a. Cation anion (~ide fonn) b. Cation (Roman Numeral) anion (-ide fann) 3. Polyatomic Ion: charged group of bonded atoms a. suftixes are often -ite (1 less 0) and ~ate b. prefixes arc often hypo- (1 less 0 than ~ite fonn) and per-( 1 more 0 than -~atc fann) Hypochlorite CIO' c. Example: CI0 2 Chlorite ClO, Chlorate CIO-l Perchlorate 4. Hydrates: ionic compounds in which the tonnula unit includes a fixed number of water molecules together with cations and anions a. Example: CaCI 2 ' 6H 2 0 Calcium chloride hexahydrate AP Chemistry II and P: Chapter .2 Dr. Kalish Page" b. Anhydrous: without water Acids. Bases, and Salts 1. Basic Characteristics of Acids and Bases when dissolved in water a. Acids: I) taste sour 2) sting or prick the skin 3) turn litmus paper red 4) react with many metals to produce ionic compounds and fl:: lgi 5) react with bases b. Bases 1) taste bitter 2) feel slippery or soapy 3) turn litmus paper blue 4) react with acids 2. The Arrhenius Concept ( 1887) a. Acid: molecular compound that ionizes in water to form a solution containing If and anions b. Base: compound that ionizes in water to t<mn a solution containing OH- and cations c. Neutralization: the essential reaction betv,·een and acid and a base. called neutralization. is the combination of H - and OH- ions to fonn \vater and a salt 1) Example: HCl NaOH -7 i\aCli- H:::O 3. Formulas and ~ames of Acids, Bases, and Salts a. Arrhenius Bases: cation hydroxide 1) Examples: NaOH = Sodium hydroxide KOH Potassium hydroxide Ca(OHb Calcium hydroxide b. '1olecular Bases: do not contain OH- but produce them when the base reacts with water 1) Example: NIh = Ammonia c. Binary Acids: H combincs with a nonmetal 1) Examples: HCl 1g1 = Hydrogen chloride HCl 1a4 ) = Hydrochloric acid HI 1gi = Hydrogen iodide HI1<141 = Hydroiodic acid H:,Slg) Hydrogen sulfide H:::S (aql Hydrosulfuric acid d. Ternary Acids: FI combines with two nonmetals 1) oxoacids: H combines with 0 and another nonmetal a) Examples: H:ypochlorous Acid HCIO Chlorous Acid 11CIO::: Chloric Acid HCI03 Perchloric Acid HCIO-1 Sulfurous Acid H2 S0 3 Sulfuric Acid H 2S04 b) ate-ic ite-ous AP Chemistry 1I and P: Chapter :2 Dr. Kalish Page 6 I II. Introduction to Organic Compounds (Carbon-based Compounds) A. Alkanes: Saturated Hydrocarbons (contain H and C) Stem Number I. molecules contain a maximum number of H Atoms ll1eth· I 2. Formula: C 1H2n-2 eth :2 a. Methane: CH 4 : prop 3 b. Ethane: C2H" but· 4 c. Propane: C.1H~ : pent 5 (, hex d. Butane: C4H 10 7 hept· 1) Two possible structural fOl11mlas R oct· --except methane. ethane, & propane 9 non· I 2) Compounds with the same molecular formula dec lO but different structural fOl11mlas are known as isomers, and they have di t1crent properties. B. Cyclic Alkanes 1. FOl11mla: CnH::n 2. prefix cyc1o C. Alkenes: unsaturated hydrocarbon 1. Formula: CJhn a. Ethene: C:,H4 b. Propene: C3H6 c. Butene: C4H~ D. Alkynes: unsaturated hydrocarbon 1. Fonnula: C nl1::n.2 a. Ethyne: C2H:~ b. Propyne: C3H4 c. Butvne: C.\H 6 E. Homology 1. a series of compounds \vhose fonnulas and structures vary in a regular manner also have properties that vary in a predictable manner a. Example: Both the densities and boiling points of the straight-chain alkanes increase in a continuous and regular fashion with increasing numbers ofC. ! ! ! F. Types of Organic Compounds 1. Functional Group: atom or group of atoms attached to or inserted in a hydrocarbon chain or ring that confers charactcristic properties to the molecule a. usually where most of the reactions of the molecule occur 2. Alcohols (R-OII) where R represcnt the hydrocarbon a. Examples: CH30H methanol CH3CH20H = ethanol CH3CH2CH20H = I-propanol Cf'hCH(OH)CH 3 2-propanol or isopropanol b. Not bases! 3. Ethers (R-O-R') where R' can represent a different hydrocarbon than R a. Example: CH,CH 20CH:-.CH, = Diethyl ether AP Chemistry If and P: Chapter 2 Dr. Kalish Page 7 4. Carboxylic Acids (R-COOH) a. Examples: HCOOH methanoic or formic acid CH 3COOH ethanoic or acetic acid b. the H of the COOH group is ionizable: the acid is classified as a weak acid 5. Esters (R'-COOR) a. Flavors and fragrances CH,COOCH:Crh = ethyl acetate b. Examples: CH 1COOCH:::CH 2CH:CH:::CH] pentyl acetate 6. Ketones (R-CO-R ') 7. Aldehydes (R-CO-H) 8. Amines (R-NH:::, R-NHR', R-NR'R") a. most common organic bases: related to ammonia b. one or more organic groups are substituted for the H in NH3 c. Examples: CH,NH::: = methyl amine CH 3CH:::Nlh ethyl amine 68 Chapter 2 Atoms, Molecules. and Ions TABLE 2.1 Some Classes of Organic Compounds and Their Functional Groups Class General Structural Formula" Example Name of Example Cross Reference Alkane R-H CH3CH2CH2CH2CH2CH3 hexane Section 2.9, 6.8, Chap. 23 Alkene " CH2=CHCH2CH 2CH3 I-pentene Section 9.10, Chap. 23 Alkyne -C=C CH3C==CCH2CH2CH2CH2CH3 2-octyne Section 9.10, Chap. 23 Alcohol R-OH CH3CH 2CH2CH 2OH I-butanol Section 2.10, Chap. 23 Alkyl halide R_Xb CH3CH2CH2CH2CH2CH2Br I-bromohexane Chap. 23 Ether R-O-R CH3-0-CH2CH2CH3 I-methoxypropane (methyl propyl ether)" l-aminopropane (propylamine)C Section 2.1 Amine / C=C / " R-NH2 CH 3CH2 CH2-NH2 II CH 3CH2CH 2C-H 0 0 Ketone R-C-R 0 II Carboxylic acid R-C-OH 0 II Ester R-C-OR 0 Amide Arene II R-C-NH z Ar-Hd Aryl halide Ar-Xb Phenol Ar-OH In or bo an cl~ Sections 2.10, 4.2, Chap. 15 by C3 2.1 na II R-C-H II ° 0 0 Aldehyde H' butanal (butyraldehyde)" Section 4.6, Chap. 23 3-hexanone (ethyl propyl ketone)C Section 4.6, Chap. 23 butanoic acid (butyric acid)" Sections 2.10, 4.2, Chap. 15,23 methyl butanoate (methyl butyrate)C Sections 2.10, 6.8 (fats) Chap. 23, Chap. 24 (polymers) butanamide (butyramide)" Section 11.6, Chap. 23, Chap. 24 (polymers) II CH3CH2CCH2CH2CH3 0 II CH 3CH 2CH2C-OH 0 II CH 3CH 2CH zC-OCH 3 0 II CH 3CH 2CH2C-NH 2 \(8j-CH 2CH 3 ethylbenzene o-B' bromobenzene Chap. 23 4-chlorophenol Section 9.10, Chap. 23 CI-Q-OH Section 10.8, Chap. 23 EI C( an R iar ie. co C: 11 "The functional group is shown in red. R stands for an alkyl group. stands for a halogen atom-F, Cl, Br, or I. C Common name. d Ar- stands for an aromatic (aryl) group such as the benzene ring. bX su Summary 71 o I R'-C-O-R or R'COOR where R' is the hydrocarbon portion of a carboxylic acid, and R is the hydrocar bon group of an alcohol. R and R' may be the same or different. Esters are named by indicating the part from the alcohol first and then naming the portion from the carboxylic acid with the name ending in -ate. For instance, o II CH3-C-O-CH2CH3 is ethyl acetate; it is made from ethyl alcohol and acetic acid. Many esters are noted for their pleasant odors, and some are used in flavors and fragrances. Pentyl acetate, CH3COOCH2CH2CH2CH2CH3' is responsible for most of the odor and flavor of ripe bananas. Many esters are used as flavorings in cakes, candies, and other foods and as ingredients in fragrances, especially those used to perfume household products. Some esters are also used as solvents. Ethyl acetate, for example, is used in some fingernail polish removers: It is a solvent for the resins in the polish. APPLICATION NOTE Butyric acid, CH)CH zCH 2COOH, is one of the most foul-smelling substances known, but turn it into the ester methyl butyrate, CH 3CH2CH 2COOCH 3 , and you get the aroma of apples. Amines The most common organic bases, the amines, are related to ammonia. Amines are compounds in which one or more organic groups are substituted for H atoms in NH3. In these two arnines, one of the H atoms has been replaced: H H I I I H-C-N-H H or CH3NH Z H APPLICATION NOTE H I I I H-C-C-N-H I I or CHFH 2NH 2 H H H Methylamine Ethylamine The replacement of two and three H atoms, respectively, is seen in dimethylarnine [(CH3hNH] and trimethylamine [(CH3hN]. In Chapters 4 and 15, we will see that mUch of what we learn about ammonia as a base applies as well to arnines. ~ummary The basic laws of chemical combination are the laws of conservation of mass, constant com position, and multiple proportions. Each played an important role in Dalton's development of the atomic theory. The three components of atoms of most concern to chemists are protons, neutrons, and electrons. Protons and neutrons make up the nucleus, and their combined number is the mass number, A, of the atom. The number of protons is the atomic number, Z. Electrons, found outside the nucleus, have negative charges equal to the positive charges of the pro tons. All atoms of an element have the same atomic number, but they may have different mass numbers, giving rise to isotopes. A chemical formula indicates the relative numbers of atoms of each type in a com POUnd. An empirical formula is the simplest that can be written, and a molecular formula ~fle~ts the actual composition of a molecule. Structural and condensed structural formulas ~. scnbe the arrangement of atoms within molecules. For example, for acetic acid: Amines with one or two carbon atoms per molecule smell much like ammonia. Higher homo logs smell like rotting fish. In fact, the foul odors of rotting flesh are due in large part toamines that are given off as the flesh decays. -..- - - -..... ... ---.--~ 72 Chapter 2 Atoms, Molecules, and Ions Key Terms H Acetic acid: acid (2.8) alcohol (2.10) alkane (2.9) amine (2.10) anion (2.7) atomic mass (2.4) atomic mass unit (2.4) atomic number (Z) (2.3) base (2.8) carboxylic acid (2.10) cation (2.7) chemical formula (p. 47) chemical nomenclature (p. 35) electron (2.3) empirical formula (2.6) ester (2.10) ether (2.10) formula unit (2.7) functional group (2.10) hydrate (2.7) ion (2.7) ionic compound (2.7) isomer (2.9) isotope (2.3) law of conservation of mass (2.1) law of constant composition (2.1) law of definite proportions (2.1) law of multiple proportions (2.2) mass number (A) (2.3) metal (2.5) metalloid (2.5) molecular compound (2.6) molecular formula (2.6) molecule (2.6) neutron (2.3) nonmetal (2.5) periodic table (2.5) poly atomic ion (2.7) proton (2.3) salt (2.8) structural formula (2.6) f 0 I II H-C-C-O-H I H Empirical formula Molecular formula Structural formula Condensed structural formula The periodic table is an arrangement of the elements by atomic number that places el ements with similar properties into the same vertical groups (families). The periodic table is an important aid in the writing of formulas and names of chemical compounds. A mole cular compound consists of molecules; in a binary molecular compound the molecules are made up of atoms of two different elements. In naming these compounds, the numbers of atoms in the molecules are denoted by prefixes; the names also feature -ide endings. Examples: NI3 nitrogen triiodide S2F4 = disulfur tetrafluoride Ions are formed by the loss or gain of electrons by single atoms or groups of atoms. Pos itive ions are known as cations and negative ions as anions. An ionic compound is made up of cations and anions, held together by electrostatic forces of attraction. Formulas of ionic compounds are based on an electrically neutral combination of cations and anions called a formula unit. The names of some monatomic cations include Roman numerals to designate the charge on the ion. The names of monatomic anions are those of the nonm~allic ele ments, modified to an -ide ending. For polyatornic anions, the prefixes hypo- and per- and the endings -ite and -ate are commonly found. Examples: MgF2 = magnesium fluoride copper(I) oxide CU20 Ca(CIOh calcium hypochlorite Li 2S = lithium sulfide CuO = copper (II) oxide KI04 = potassium periodate Many compounds are classified as acids, bases, or salts. According to the Arrhenius theory, an acid produces H+ in aqueous (water) solution, and a base produces OH-. A neu tralization reaction between an acid and a base fornls water and an ionic compound called a salt. Binary acids have hydrogen and a nonmetal as their constituent elements. Their names feature the prefix hydro- and the ending -ic attached to the stem of the name of the non metal. Ternary oxoacids have oxygen as an additional constituent element, and their names use prefixes (h)po- and per-) and endings (-ous and -ic) to indicate the number of 0 atoms per molecule. Examples: HI = hydroiodic acid HCI0 2 chlorous acid HI0 3 = iodic acid HCI0 4 = perchlonc acid Organic compounds are based on the element carbon. Hydrocarbons contain only the elements hydrogen and carbon. Alkanes have carbon atoms joined together by single bonds into chains or rings, with hydrogen atoms attached to the carbon atoms. Alkanes with four or more carbon atoms can exist as isomers: molecules with the same molecular formula but different structures and properties. Functional groups confer distinctive properties to an organic molecule when the groups are substituted for hydrogen atoms in a hydrocarbon. Alcohols feature the hydroxyl group, -OH, and ethers have two hydrocarbon groups joined to the same oxygen atom. Carboxylic acids have a carboxyl group, -COOH. An ester, R'COOR, is derived from a carboxylic acid (R'COOH) and an alcohol (ROH). Arnines are compounds in which organic groups are sub stituted for one or more of the H atoms in anmlonia, NH 3 . AP Chemistry Clwk 1.2A Dr. Kalish Page 1 Name: ________________________________.. ______.._____ Date: _ _ _ _ ___ Molecular Formula \\-'riting and Naming Name the following compounds: 1. .., SF 4 ~. R1Cl" 3. PBrs 4. NcO, 5. S, L) 6. SoO.) \Vrite the chemical formula for cach of the follO\ving compounds: 7 carbon dioxide R. sulfur hexafluoride 9. dinitrogen tetroxide 10. trisulfur heptaiodide 11. disulfur pentachloride 12. triphosphorus monoxide Ionic Formula Writing and Naming Directions: Name the following ionic compounds. 13. MgCh 14. NaF 15. NacO 16. AhOl 17. KI IR. AIF. 19. Mg 1N2 20. FeCh 21. MnO: 22. erN:, Compounds that include Polyatomic Ions: 23. Ca(OHh 24. (NH 4 hS 25. Al:(S04h 2A. H 1P0 4 ,.,!. ~ Ca(N01): AP Chemistry Clwk 1.2A Dr. Kalish Page 2 2R. CaCO, 29. 1\a cSO, 30. Co(CH,COOh 31. Cu c(S03h 32. Pb(OHh Directions: Write the correct formula for each of the following compounds. 1. Magnesium sultide 2. Calcium phosphide 3. Barium chloride 4. Potassium nitride 5. Aluminum sulfide 6. Magnesium oxide 7. Calcium fluoride R. Lithium fluoride 9. Barium iodide 10. Aluminum nitride II. Silver nitride* 12. Nickel(Il) bromide .--~-----~-----~-- 13. Lead(lV) phosphide 14. Tin(H) sulfide Compounds that include Polyatomic Ion~: 15. Aluminum phosphate IIi. Sodium bromate 17. Aluminum sulfite 18. Ammonium sulfate 19. Ammonium acetate 20. Magnesium chromate 21. Sodium dichromate 22. Zinc hydroxide* 23. Copper(Il) nitrite 24. Manganese(II) hydroxide 25. Iron(II) sulfate 26. lron(III) oxide --- .•..- - .•.. AI' Chemistry II and P: Chapter 3 Band S Ch ..\ fk Kabil Stoichiometr~": Chemical Calculations I. Stoichiometry of Chemical Compounds A. Molecular Masses and Formula Masses I. Molecular Mass: sum of the masses of the atoms represented in a molecular formula Example: Mass of CO 2 1 C = 1 x 12.0 amu l 0 = 2 x 16.0 amu mass of CO 2 =c 44.0 amu Fonnula Mass: sum of the masses of the atoms or ions represented in an ionic fonnula Example: Mass of BaCb 1Ba= I x D7.3al11u mass of BaCl 2 20X.3 amu lCl2x35.5amu B. The !\:1ole and Avogadro's Number I. Mole: amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of the C-12 isotope a. The elementary entities are atoms in elements. molecules in diatomic elements and compounds. and t(mnula units in ionic compounds. b. Avogadro's Number (1'\",) 6'()22 x 1O~' mor l 'I . I} I mole 6.022 x 10-- atoms. molecules. partIcles. etc. e. one mole of any element is equal to the mass of that element in grams I) For the diatomic elements. multiply the mass of the element by two. 2. Molar Mass: mass of one mole of the substance Example: Mass of BaCb 1moleBa=1 x D7.3g mass of 1 mole BaCI 2 20X.3 g I mole CI 2 x 35.5 g C. Mass Percent Composition from Chemical Fon11ulas I. Mass Percent Composition describes the prop0l1ions of the constituent elements in a compound as the number of grams of each element per 100 grams of the compound. Example: What is the % C in butanc (C..\H1n)? Mass ofCs x 100°;;) 4(!2.01g,) x 100°/0 MassofC..\H iIl 5S.14g D. Chemical Formulas from Mass Percent Composition I. Steps in the Detennination of Empirical FOI111Ula a. Change ~!O to grams. h. Convert mass of each elemcnt to moles. c. Detennine mole ratios. d. lfneccssary multiply mole ratios by a t~lctor to obtain positi\'c integers only. c. Write the empirical fonnula. /\P Chemistry II and P: Chapter 3 Band S CI1 4 Dr. K:lil~h Example: Cyclohexanol has the mass percent composition: 71.95% C. 12.08% H, and 15.97°0 O. Determine its empirical timl1ula. A compound has the mass percent composition as 1()llows: 36.33 C. 5.49 H, and 58.18 (~o S. Detcnninc its empirical t(mmtia . .., Relating Molecular F0ll11ulas to Empirical F0ll11ulas a. Integral Factor (n) Molecular Mass Empirical Fonnula Mass Example: Ethylene (M 28.0 u). cyclohexane (M 84.0 u). and I-pcntcnc (70.0 u) all have the empirical f0ll11ula CH:,. \Vhat is the molecular t(mnula of cach compound? II. Stoichiometry of Chemical Reactions A. Writing and Balancing Equations 1. chemical equation: shorthand description of a chemical reaction using symbols and formulas to represent clements and compounds. respectivcly. a. Reactants -7 Products h. C oefficicnts c. States: gas (g). liquid (1), solid (s). aqueous (aq) d. ll: Heat 2. Balancing Equations a. For an element. the same number of atoms must he on each side of thc equation. h. only coefticients can be changed 1) Balance the clement that only appears in one compound on each side of the equation first. 2) Balance any reactants or products that exist as the free clcment last. 3) Polyatomic ions should he treated as a group in most cases. Example: _SiCI 4 + __ H~O -7 _SiO~ . ~ _He) B. Stoichiometric Equivalence and Reaction Stoichiometry 1. Mole Ratios or stoichiometric factors Example: _SiCI.t + lH::O -7 _SiO:: ~·1HCI 2. Problems: a. Mole-tn-mole h. Mole-to-gram c. Gram-to-mole d. Gram-to-gram What is the mole ratio ofreactmts'! Dr. Kalish AP Chemistry If and P: Chapter 3 Band S Ch 4 C. Limiting Reactants I. Limiting reactant (LR) is consumed completely in a reaction and limits the amount of products t<:mned J To detenninc the LK compare moles. 3. Usc the LR to detennine theoretical yield. Example: FeS(s) + 2 HCI(aq) ~ FeCI 2(aq) H:,S(g) If 10.2 g HCI is added to 13 ' g FeS. what mass of H:,S can he formed'.' What is the mass of the excess reactant remaining? 10.2 g HCI x 1 mole HCI 36.46 g HCI 0.280 moles HCI [3,2 g FeS x 1 mole FeS 0.150 moles FeS LR 87.92 g FeS 0,280 moles HCI x 1 mole H2S :2 mole HCI 0.150 mole FeS - 0.140 mole FeS = x 34,()1.) g H,S1 mole H2S 4.n g I \.oS 0.010 mole FcS x 87.92 g FeS 1 mole FcS = (l~79 g FeS D. Yields of Chemical Reactions x 100 I. Percent Yield Actual Yield Theoretical '{ield J Actual Yield may be less than theoretical yield hecause of impurities. errors during experimentation. side reactions. etc, Example: If the actual yield of hydrogen sultidc was 3.56 g. calculate lhc percent yield, If the percent yield of hydrogen sulfide was 84.7 actual yield? (~o. what was the Solutions and Solution Stoichiometry I, Components of a Solution: a. Solute: substance being dissolving b. Solvent: substance doing the dissohing 1) water: universal solvent: solutions made \\i(h water as the sol\ent arc called aqueous solutions J Concentration: quantity ofsolutc in a given quantity ofsolwnt or solution a. ()i1ute: contains relati\cly little solutc with a large amount of solvcnt ,\P Chel1li~try 1I and P: Chapter.3 Rand S Ch -4 Dr. Kalish ['''ic'l' -+ b, Concentrated: contains a relatively large amount of solute in a given quantity solvent 3. Molarity or Molar Concentration Molarity moles solute Liters of solution or Example: Calculate the molarity of solution made by dissolving 2.00 moles NaCI in enough water to generate 4.00 L of solution. Molarity moles solute l.nO moles NaCI= O.SO() M 4.00 L Liters of solution Example: Calculate the molarity of solution made by dissolving 35.1 grams NaCI in enough water to generate 3.00 L of solution. 35.1 g :\aCI x I mole NaCI 58.44 g NaCI Molarity 0.60 I moles NaCI 0.601 moles NnCI 3.00 L moles solute Liters of solution f).lOI) M 4. Calculating the :\'lolarity of Ions and Atoms: Example: Calculate the molarity of Ca::' and chloride. Moles: cr in a 0.600 M solution of Calcium CaCI:: -7 Ca::- + leT I I ., 0.600 M CaCh x I mole Ca 2 ' I mole CaCI:: o.oon \1 Ca2 0.600 M CaCb x ., mole cr 1.20 M I mole CaCI:: cr Example: Calculate the molarity of C and H in 1.50 \1 propane. CJ-lx. Moles: C 311~ I "=" 3C 3 8H 8 1.50 M C,Hx x 3 mole C = 4.50 M C I mole C 3Hx 12.0 M H 1.50 M C;Hx x 8 mole H I mole C,H~ AI' Ch.:mistry II and P: Chapt.:r 3 Band S Ik Kalish eh 4 5. Dilution: the process by which dilute solutions are made by adding solvent to concentrated solutions a. the amount of solute (moles) remains the same. but thc solution concentration is altered b. M enllc x VWile = Moil X V cli1 Example: What is the concentration of a solution made by diluting sO.n 1111 of 1.00 M NaOH with 200. ml of water? III.Advanced Stoichiometry A. Allov..·s fiJr the conversion of grams of a compound to grams of an clement and composition to detem1ine empirical and molecular f0l111ula ° 0 Examples: t. A 0.1204 gram sample of a carboxylic acid is combusted to yield 0.2147 grams of CO-, and 0.0884 grams of water. a. Determine the percent composition and empirical ft)Jl11ula of the compound. Ansl\'cr: C!IIP: (';I/('(), b. If the molecular mass is .222 gimole. vvhat is the molecular tt)l'lnula'? c. Write the balanced chemical equation showing combustion of this compound. Dimethylhydrazine is a C-H-N compound used in rocket fuels. When burned completely in excess oxygen gas. a 0.312 g sample produces 0.458 g CO:, and 0.374 g H20. The nitrogen content of a separate 0.525 g sample is cOl1\clied to 0.244 g N a. What is the empirical t(mTIula of dimethylhydrazinc'? . 111'\\1 C},,( '//;.\ b. If the molecular mass is 150 gmo)c. what is the molecular ttmnula'? c. \Vrite the balanced chemical equation showing combustion of this compound. Empirical fonnulas can be detennined from indirect analyses In practice, a compound is seldom broken down completely to its elements in a quantitative analysis. Instead, the compound is changed into other compounds. The reactions separate the elements by capturing each one entirely (quantitatively) in a separate compound whose formula is known. In the following example, we illustrate an indirect analysis of a compound made entirely of carbon, hydrogen, and oxygen. Such compounds bum completely in pure oxygen-the reaction is called combustion-and the sole products are car bon dioxide and water. (This particular kind of indirect analysis is sometimes called a combustion analysis.) The complete combustion of methyl alcohol (CH30H), for example, occurs according to the following equation: 2CH30H + 30 z --- 2CO z + 4HzO The carbon dioxide and water can be separated and are individually weighed. No tice that all of the carbon atoms in the original compound end up among the COz molecules and all of the hydrogen atoms are in H 20 molecules. In this way at least two of the original elements, C and H, are entirely separated. We will calculate the mass of carbon in the CO 2 collected, which equal~ the mass of carbon in the original sample. Similarly, we will calculate the mass of hy drogen in the H 20 collected, which equals the mass of hydrogen in the original sample. When added together, the mass of C and mass of H are less than the total mass of the sample because part of the sample is composed of oxygen. By subtract ing the sum of the C and H masses from the original sample weight, we can obtain the mass of oxygen in the sample of the compound. A 0.5438 g sample of a liquid consisting of only C, H, and 0 was burned in pure oxy gen, and 1.039 g of CO 2 and 0.6369 g o( H 20 were obtained. What is the empirical formula of the compound? A N A LY SIS: There are two parts to this problem. For the first part, we will find the number of grams of C in the COz and the number of grams of H in the H 20. (This kind of calculation was illustrated in Example 4.10.) These values represent the number of grams of C and H in the original sample. Adding them together and sub tracting the sum from the mass of the original sample will give us the mass of oxygen in the sample. In short, we have the following series of calculations: grams CO 2 ----l> grams C grams H 20 ----l> grams H We find the mass of oxygen by difference. 0.5438 g sample - (g C + g H) g0 In the second half of the solution, we use the masses of C, H, and 0 to calculate the empirical formula, as in Example 4.14. SOLUTION: First we find the number of grams of C in the COz and of H ia the H 20. In 1 mol of CO2 (44.009 g) there are 12.011 g of C. Therefore, in 1.039 g of CO . 2 we have 12.011g gCO C 1.039 g CO2 X 44.009 "" 0.2836 g C 2 In 1 mol of H 20 (18.015 g) there are 2.0158 g of H. For the number of grams of H in 0.6369 g of H 2 0, 2.0158 g H 0.6369 g H 20 X 18.0]5 g H 0 0.07127 g H 2 The total mass of C and H is therefore the sum of these two quantities. total mass of C and H = 0.2836 g C + 0.07127 g H -c= 0.3549 g The difference between this total and the 0.5438 g in the original sample is the mass of oxygen (the only other element). mass of 0 0.5438 g - 0.3549 g = 0.1889 g 0 Now we can convert the masses of the elements to an empirical formula. ForC: ForH: For 0: 1 molC 0.2836 g C X 12.011 g C = 0.02361 mol C ImolH 0.07127 g H X 1.008 g H = 0.07070 mol H 1 malO 0.1889 g 0 X 15.999 gO = 0.01181 mol 0 Our preliminary empirical formula is thus C1J.02361H0.D707000.01I81' We divide all of these subscripts by the smallest number, 0.01181. CQ.02361 O.O1l81 HO.07070 O~ = 0.01181 0.01181 C1.999 H S.987 0 1 . The results are acceptably close to ~H60, the answer. Summary actual yield (3.10) Avogadro's number, NA (3.2) chemical equation (3.7) dilution (3.11) formula mass (3.1) limiting reactant (3.9) mass percent composition (3.4) molar concentration (3.11) molarity, M (3.11) molar mass (3.3) mole (3.2) molecular mass (3.1) percent yield (3.10) product (3.7) reactant (3.7) solute (3.11) solvent (3.11) stoichiometric coefficient (3.7) stoichiometric factor (3.8) stoichiometric proportions (3.9) stoichiometry (page 82) theoretical yield (3.10) Summary Molecular and formula masses relate to the masses of molecules and formula units. Mo lecular mass applies to molecular compounds, but only formula mass is appropriate for ionic compounds. A mole is an amount of substance containing a number of elementary entities equal to the number of atoms in exactly 12 g of carbon-12. This number, called Avogadro's number, is NA 6.022 X 1023 • The mass, in grams, of one mole of substance is called the molar mass and is numerically equal to an atomic, molecular, or formula mass. Conversions be tween number of moles and number of grams of a substance require molar mass as a con version factor; conversions between number of grams and number of moles require the inverse of molar mass. Other calculations involving volume, density, number of atoms or mo lecules, and so on may be required prior to or following the gram/mole conversion. That is, Molar mass Inverse of molar mass Formulas and molar masses can be used to calculate the mass percent compositions of compounds. And conversely, an empirical formula can be established from the mass percent composition of a compound;·to establish a molecular formula, we must also know the mo lecUlar mass. The mass percents of carbon, hydrogen, and oxygen in organic compounds can be determined by combustion analysis. A chemical equation uses symbols and formulas for the elements and/or compounds in VolVed in a reaction. Stoichiometric coefficients are used in the equation to reflect that a chemical reaction obeys the,law of conservation of mass. Calculations concerning reactions use conversion factors, called stoichiometric fac tors, that are based on stoichiometric coefficients in the balanced equation. Also required are ~lar masses and often other quantities such as volume, density, and percent composition. e general format of a reaction stoichiometry calculation is ~ 123 124 Chapter 3 Stoichiometry: Chemical Calculations no. mol B no. mol A no. mol A no. molB The limiting reactant determines the amounts of products in a reaction. The calculat ed quantity of a product is the theoretical yield of a reaction. The quantity obtained, called the actual yield, is often less. It is commonly expressed as a percentage of the theoretical yield, known as the percent yield. The relationship involving theoretical, actual, and percent yield is Percent yield actual X I0007 70 theoretical yield = The molarity of a solution is the number of moles of solute per liter of solution. Com mon calculations include relating an amount of solute to solution volume and molarity. So lutions of a desired concentration are often prepared from more concentrated solutions by dilution. The principle of dilution is that the volume of a solution increases as it is diluted, but the amount of solute is unchanged. As a consequence, the amount of solute per unit vol ume-the concentration-decreases. A useful equation describing the process of dilu tion is tv1conc X V cone = Mdil X V dil In addition to other conversion factors, stoichiometric calculations for reactions in solution use molarity or its inverse as a conversion factor. Review Questions 1. Explain the difference between the atomic mass of oxy gen and the molecular mass of oxygen. Explain how each is determined from data in the periodic table. 2. \\'hat is Avogadro's number, and how is it related to the quantity called one mole? 3. How many oxygen molecules and how many oxygen atoms are in 1.00 mol 0 2? 4. How many calcium ions and how many chloride ions are in 1.00 mol CaCh? 5. What is the molecular mass, and what is the molar mass of carbon dioxide? Explain how each is determined from the formula, CO 2 , 6. Describe how the mass percent composition of a com pound is established from its formula. 7. Describe how the empirical formula of a compound is de termined from its mass percent composition. S. \\'bat are the empirical formulas of the compounds with the following molecular formulas? (b) CgHl6 (e) CloHs (d) C6H 160 (a) HP2 9. Describe how the empirical formula of a compound that contains carbon, hydrogen, and oxygen is determined by combustion analysis. 10. \\'bat is the purpose of balancing a chemical equation? 11. Explain the meaning of the equation at the molecular level. Interpret the equation in terms of moles. State the mass relationships conveyed by the equation. 12. Translate the following chemical equations into words: (a) 2 Hig) + 02(g) (b) 2 KCl0 3 (s) ~ ~ 2 Hz0(l) 2 KCI(s) + 3 Gig) (e) 2 AI(s) + 6 HCI(aq) ~ 2 AICI 3(aq) + 3 H2(g) 13. Write balanced chemical equations to represent (a) the reaction of solid magnesium and gaseous oxygen to form solid magnesium oxide, (b) the decomposition of solid ammonium nitrate into dinitrogen monoxide gas and liq uid water, and (e) the combustion of liquid heptane, C 7H 16, in oxygen gas to produce carbon dioxide gas and liquid water as the sole products. 14. \\'bat is meant by the limiting reactant in a chemical re action? Under what circumstances might we say that a reaction has two limiting reactants? Explain. 15. \\'by are the actual yields of products often less than the theoretical yields? Can actual yields ever be greater than theoretical yields? Explain. 16. Define each of the following terms. (a) solution (d) molarity (b) solvent (e) dilute solution (e) solute (0 concentrated solution