Chem2_CH2 - Chapter 2 Atom Mole s and Ions s, cule First...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 2 Atom Mole s and Ions s, cule First Things First: Don't worry about history (nam s/date nationalitie of scie e s/ s ntists). C ntrateon thescie the discove d. once nce y re Le thenam s and sym of e m nts 1 through 20, 26 and 29. arn e bols le e Also, le theions the form arn y . Le Table 2.3, 2.4*, 2.5*, 2.6, 2.7 & 2.8 (m arn s onatom ions, ic pre s, e fixe tc.) * Befam with Table 2.4 & 2.5 (nam s of typeI I cations and iliar s e polyatom anions) ic Atoms and Subatomic Particles alle xhibits the An atomis thesm st particlethat e characte ristic prope s of an e m nt. rtie le e nte us At thece r of an atomis thenucle containing protons (p) and ne utrons (n). us d ore le Thenucle is surrounde by oneor m e ctrons (e). Subatomic particles Nam / sym e bol e ctron le e proton P ne utron n Mass (actual) charge(re lative ) 9.11 x 10-31 kg -27 -1 +1 1.67 x 10 kg 1.67 x 10 kg 0 Note Theactual chargeof thee is -1.60219 x 10 C : Atomic Structure C loud of e ctrons. In a ne le utral atomthe# e = # p - Z: Atom Num r = ic be # of protons in nucle us Z de s what theatomis (i.e fine . e m nt) le e A: Mass num r = be # p + # n in nucle us X: S bol for thee m nt ym le e X Z Isotopic notation Ions Whe # e don't e n qual # p; thecharge don't s balance Moree ctrons than protons le Ne gativecharge Anion Moreprotons than e ctrons le Positivecharge Cation Chemical compounds I onic solids (NaI ) + Mole s cule (C 2C 2) H l Na I- Me tals (Fe ) Chemical compounds Mole s areform d whe atom sharee to form cule e n s covale bonds. Typically, bonds be e non-m tal nt twe n e atom s. I onic solids areform d whe the is a com te e n re ple transfe of e be e atom Typically, bonds r - twe n s. be e m tal and non-m tal atom twe n e e s. Nonmetals are found to left of the the metalloids. Metals are found to thethe right of metalloids 11.1 Relative Atomic and Molecular Masses H 1.01 Al 26.98 P 30.97 H2 2.02 2 O 32.00 Question What is them cular m of ole ass m thane C 4? e , H Atom m s of: ic asse C arbon, C= 12.01 g/m ol Hydroge H = 1.01 g/m n, ol Answer Total atom m sin C 4: ic asse H 1(12.01g/m = 12.01 g/m ol.) ol H = 4(1.01 g/m = 4.04 g/m ol) ol Them cular m of C 4 is the ole ass H sumof thetotal atom m s of: ic asse C 12.01 g/m ol H + 4.04 g/m ol C 4 16.05 g/m H ol C= Question What is them cular m of ole ass S odiumsulfate Na2S 4? , O Atom m s of: ic asse S odium Na = 22.99 g/m , ol S ulfur, S= 32.07 g/m ol Oxyge O = 16.00 g/m n, ol Answer Total atom m s: ic asse Na = 2(22.99g/m = 45.98 g/m ol) ol S= 1(32.07 g/m = 32.07 g/m ol) ol O= 4(16.00 g/m = 64.00 g/m ol) ol Mole cular m of Na2S 4: ass O 142.05 g/m ol A "dimensional analysis " calculation How m wate m cule doe it taketo com te any r ole s s ple ly fill a 1.00 L volum tric flask? e Dimensional analysis question S gy: trate 1. Writedown thede d quantity. sire 2. Writedown give quantitie n s. 3. Writedown ne de conve e d rsion factors. 4. Unit conve rsion: L m g Mol. Mole s. L cule 5. Use"line se to obtain de d unit(s) and calculate ar" tup sire answe r. How many water molecules does it take to completely fill a 1.00 L volumetric flask? S p: te 1. De d quantity: sire H2O Mole s cule 1. Give quantity: n 1.00 L H2O 3. C rsion factors: onve 1L = 1,000 m L De nsity of H2O = 1.00 g/m L MW of H2O = 18.02 g/m ol. Avogadro's # = 6.022 x 10 23 units/m ol. 1. Unit conve rsion: L m g Mol. Mole s. L cule Dimensional analysis answer 25 2 3.34 x 10 H O m cule ole s Some Other useful Conversion Factors 1 M = 100 cm 1M/100cm 1 M = 1,000 m 1M/1,000m m m 1 inch = 2.54 cm 1in./2.54cm 3 3 1 cm = 1 m 1cm /1m L L 1 L = 1,000 m L 1L/1,000m L 1 gal = 3.785 L 1gal/3.785L The periodic table Groups (or Families) Periods Groups and Periods Ele e in thesam group have m nts e sim che ical prope s (e NaC ilar m rtie .g. l, KC RbC e l, l, tc.). Ele e in thesam pe do not m nts e riod havesim che ical prope s (e ilar m rtie .g. NaC MgC2, AlC3). l, l l +1 +2 +3 -3 -2 -1 Atomic Mass Isotopes and Atomic mass s le e s le e Isotope of an e m nt areatom of thee m nt which havethesam num r of protons, but have e be diffe nt num rs of ne re be utronsin the nucle ir us. re utron count cause isotope of s s This diffe ncein ne an e m nt to havediffe nt m s (i.e isotopic le e re asse . m ass). Isotopic Notation A = Mass num r be (sumof protons and ne utrons in nucle us) X Z X = S bol of ym thee m nt. le e Z = Atom num r ic be (# protons in nucle us) Theatom m of an e m nt is theave isotopic m ic ass le e rage ass of all of its isotope s. Two isotopes of Sodium Question I f you only know thenum r of protons in be a ne utral e m nt, which of thefollowing le e can you de rm ? te ine A) Thenum r of ne be utrons in thene utral e m nt le e B) Thenum r of e ctrons in thene be le utral e m nt le e C Thenam of thee m nt ) e le e Answer I f you only know thenum r of protons in be a ne utral e m nt, which of thefollowing le e can you de rm ? te ine A) Thenum r of ne be utrons in thene utral e m nt No-- le e m ultipleisotope e s xist B) Thenum r of e ctrons in thene be le utral e m nt Ye le e s-- e quals num r of protons be C Thenam of thee m nt Ye ) e le e s--num r of protons be de s thee m nt fine le e Question C hlorinehas two m ajor isotope 35C and 37C s, l l. Without using anything e pt thepe xce riodic table , which isotopeis m ore abundant? Solution S theave atom ince rage ic m of chlorineis35.45, ass the m bem 35C re ust ore l than 37C l. To calculatetheave atom m m rage ic ass, ultiply theatom m ic ass of e isotopeby its pe nt abundanceand add there ach rce sults. I sotope I sotopic m ass Abundance(% Ave atom ) rage ic (am u) m (am ass u) 62.9298 64.9278 69.09 30.91 63 29 65 29 Cu Cu 63.55 (62.9298 amu) 0.6909 = 43.48 amu (64.9278 amu) 0.3091 = 20.07 amu 63.55 amu Ion Formation e n utral atomgainsor lose s An ion is form d whe a ne oneor m e ctrons. ore le le e Anion gain of an e ctron(s); typically nonm tals ation loss of an e ctron(s); typically m tals. le e C Cation Formation Anion Formation Formation of cations Na Na + e + (S odium S odiumion) Mg Mg + 2e +2 - (Magne sium Magne sium ion) Ple notethee of anions in theaboveche ical ase ation ctrical ne le utrality m Form e quations. C +e C l l Metals forming Multiple Cations +2 +3 Iron: Fe , Fe C r: C +, C +2 oppe u u Le Pb+2, Pb+4 ad: Question (a) How m protons and how m any any +2 e ctrons arepre nt in oneFe cation le se (Fehas atom num r 26)? ic be (b) How m protons and how m any any e ctrons arepre nt in oneI - anion le se (I has atom num r 53)? ic be 26 protons 24 e ctrons le 53 protons 54 e ctrons le The Law of Conservation of Mass In a che ical re m action, m is ne r ass ithe cre d nor de ate stroye d. A "what do you think" question An unknown solid we ighing 250.0 g is place in a se d glass containe d ale r full of air, and se on a balance A t . le is use to focus sunlight, or lase ns d r light, onto thesam to igniteit. ple 250.0 g Whe thesam has burne com te will thebalance n ple d ple ly, re 250.0 g, m or le than 250.0 g? Assum a close ad ore ss e d syste . m Answer 250.0 g Question Is this answe consiste with the r nt law of conse rvation of m ass? Answer Ye s! Thetotal m of re ass acting spe s cie in a che ical re m action e quals the total m of thespe sproduce ass cie d. Mass is ne r cre d nor de ithe ate stroye d. This is thee nceof TheLaw of C rvation of sse onse Mass The Law of Multiple Proportions Atom of two or m e m nts can s ore le e com in diffe nt ratios to produce bine re m than onecom ore pound. C and C O O2 Question C onside two com r pounds, A and B, that both contain only carbon and oxyge n. 100.0 g of A contains 27.2 g of carbon 100.0 g of B contains 42.9 g of carbon Arethe com se positions consiste with theLaw of nt MultipleProportions (i.e CH4, C H6, C H8, e . tc)? 2 3 Answer A has 27.2 g of carbon and (100.0 - 27.2) = 72.8 g of oxyge n B has 42.9 g of carbon and (100.0 - 42.9) = 57.1 g of oxyge n oxyge n:carbon ratio in A = 72.8/27.2 = 2.67 ( i.e . CO2) oxyge n:carbon ratio in B = 57.1/42.9 = 1.33 ( i.e . CO) *2; ye s Molecular Formulas Them cular form te you the ole ula lls S toichiom tric re e lationship be e twe n atom in a com s pound (i.e atom and . s m s). ole H2S 4 O H2S H2O H2O2 S ulfuric Acid Dihydroge S n ulfide Wate r Hydroge Pe n roxide Question How m Na atom arein 14.0 g any s of Na2S 4? O Answer x Na atom 14.0 g Na2S 4 x s O 1m ol/142.05 g x 2 m Na/1 m Na S ol ol O 2 4 x 6.022 x 1023 Na atom ol = s/m Question Theform for wate H2O, te you that: ula r, lls A) Each wate m culehas twiceas m hydroge as r ole uch n oxyge by m n ass B) Each wate m culehas two hydroge atom and one r ole n s oxyge atom n C) Each wate m culehas twiceas m oxyge as r ole uch n hydroge by m n ass D) Each wate m culehas two oxyge atom and one r ole n s hydroge atom n Answer Theform for wate H2O, te you that: ula r, lls A) Each wate m culehas twiceas m hydroge as r ole uch n oxyge by m No n ass B) Each wate m culehas two hydroge atom and one r ole n s oxyge atom Ye n s C) Each wate m culehas twiceas m oxyge as r ole uch n hydroge by m No n ass D) Each wate m culehas two oxyge atom and one r ole n s hydroge atom No n Nomenclature Therule aregive in Zum s n dahl (pp. 34-44) Le m rial in Table 2.3, 2.6, 2.7, and 2.8. Befam arn ate s iliar with Table 2.4 and 2.5. s What's theform of potassiumoxide ula ? K2O What is theform of sulfuric acid? ula H2S 4 (Table2.8) O +1 +2 +3 -3 -2 -1 K O isneutral 2(+1) + 1(-2) = 0 2 Answer Question What would bethecorre che ical ct m form whe thefollowing se of ula n ts atom or ions com ? s bine Na and S Al and S +2 2 C and C a l Fe and C l Na S 3 Mg and PO4-3 C l aC 2 2 3 2 42 Al S Fe l C Nam binary com ing pounds Binary com pounds contain only two diffe nt re e m nts. le e Courte of Wile and S I nc. sy y ons, Formulas from Names - Question Givetheche ical form for: m ulas A) sodiumoxide B) silicon te trachloride C) le ad(IV) oxide D) galliumarse nide E) zinc sulfide F) nitrous acid Formulas from Names - Answers 2 A) sodiumoxideNa O (group 1 cation with group 6 anion) 4 B) silicon te trachlorideS l (te = 4) iC tra 2 C le V) oxide PbO (I V = +4 cation with group 6 anion) ) ad(I D) galliumarse GaAs (group 3 cation with group 5 anion) nide E) zinc sulfideZnS(know Zn is 2+, Table2.3) How come... Why is Ba(NO3)2 calle barium d nitrate but Fe 3)2 is calle , (NO d iron(II) nitrate ? i.e why isn't it barium ) nitrateor . (II iron nitrate ? Formulas from Names - Question Givetheche ical form for: m ulas A) Nitroge dioxide n B) S odiumsulfide C) I ron(III) oxide D) C alciumhydroxide E) C r(I) nitrate oppe F) Dinitroge te n troxide Formulas from names - Answers 2 A) Nitroge dioxideNO (usepre s whe non-m tals are n fixe n e bonde to e othe d ach r) 2 + B) S odiumsulfideNa S(no Rom num ral, Na ) an e 2 3 C I ron(I I I ) oxide FeO (I I I = +3 cation with m ) ultiplepossible charge +2 & +3) s: 2 Formulas from Names - Question Givetheche ical form for: m ulas A) Hydroge chloride n B) Hydrobrom acid ic C) Hydroge sulfide n D) Hydrosulfuric acid Formulas from Names - Answer A) Hydroge chlorideHCl (gas/solid phase n ) B) Hydrobrom acid HBr (aque phase ic ous ) 2 C) Hydroge sulfideH S(gas/solid phase n ) 2 D) Hydrosulfuric acid H S(aque phase ous ) Naming Oxy-Acids of Chlorine HC 4 lO HC 3 lO HC 2 lO HC lO HC l Pe chloric Acid r C ic Acid hlor C hlorousAcid Hypochlorous Acid Hydrochloric Acid +1 oxyge n Root Acid - 1 oxyge n -2 oxyge ns No oxyge n* * Ple noteHC is not an oxy-acid ase l Naming Oxy-Anions of Chlorine C 4lO C 3lO C 2lO C lO C l- Pe chlorate r C ate hlor C ite hlor Hypochlorite Chloride +1 oxyge n Root Base - 1 oxyge n -2 oxyge ns No oxyge n* Courte of Wile and S I nc. sy y ons, ...
View Full Document

This note was uploaded on 04/27/2009 for the course CHEM chem1A taught by Professor Hooker during the Spring '09 term at UCSB.

Ask a homework question - tutors are online