' +JJJJ ?\>m0M='+l> /+l   d]@ŵLҦ]]LF L}BBL] X  ` 鷎귭෍ᷩ췩緈JJJJx Lȿ L8ᷭ緍췩 緍i 8 `巬 췌`x (`(8`I`B` ``>J>J>VU)?`8'x0|&HhHh VY)'&Y)xꪽ)' `Hh`V0^*^*>&` aI꽌ɪVɭ&Y&&Y& 꽌ɪ\8`&&꽌ɪɖ'*&%&,E'з꽌ɪФ`+*xS&x'8*3Ixix&& 8  '  & x)*++`FG8`0($ p,&"ųųೳŪŪųųij  !"#$%&'()*+,-./0123456789:;<=>?   1 '" *"( (9"1 ( ,.(0# 2  /#0/#0 *?'#07#00/0/'#07#0:"4<*55/**5/*%5/)1/)1/)1/)'#0/#0*5/*75/**5/*:5//#0/#0'#07#0:::*::'#07#0"):$(%"%:$(%"%$$2%4%$$2%4%$(2()!)E(!8b $!H(+ "@H !D)"E` @ $ C ` DQ &J80^݌Hh ü ü݌ ռ ռ ռA ļD ļ? ļAEDE?HJ>h Լ ռ ռ ռ`HJ>݌h Hh݌`ɃlXLȔЖȔЖȠHIHHHHhHH݌hHhHh݌H6 VDP (ED Z $0x8x D- ܸDD# H8`?E Vk *f???0xE Hh D#-EEE8` D ܸx D - ܸx8`-0ݩ?ʥD EEE`   LDcpq` [` ~  LӜu`".Q`pNФbptťܥm2<(-Py0\|e<6e<g< JJJJj귍hI  aUL@ kU8  L  Q^R(jQ0l^l\  wUuW ԧ H h@ [_ /QSIRb_L`LLLL`ª`LQLYLeLXLeLee ўQH\(h0L& Ꝥ$`( R \ZLl8 ўR HH\`\Z[YS6`LxQɿu3'RͲʎRʎ]]]ɍuL͟ɍ}RLRɍg^H8 ^hZLɍR LͲɊRR% QLܤͲ Z@ -^ ş\[Z QY\[Z8`l6Lş_Ȍb_Ͳ] )Y h( ֭ͲLɍ [LLĦ__ ^ 9 LҦ3 9 a   0LjLY u< (_9 ˭ɠuɠK_9 ?LˆʎõĵL õ ĵµ aµ`` L̦µ_bJLuLz`  ȟ QL߼J̥KlV  ȟ QlV eօ3L e3L &RL &QL d L4 Ne)n `@-eff L f`L . tQLѤ LҦL` OPu d L Ne)noon 8ɍ` ^f\õL ^NR  RΩLҦ)\Z ʽ LHv 3h`0h8` [L NС õ`A@` ŵL^Lõ`  \ 濭0 \  ȟ Q ^\lZl^?cqH şch`fhjõĵ@OAP`u@`@&`QR`E Ls  @DAE@u`8` %@ @A@`@`@A`Mµ ) LЦ`8@AWc@8@-@HAȑ@hHȑ@ȑ@hHȑ@Ȋ@ch8&ȑ@Hȑ@Ah@LHȑ@ȑ@ htphso`hMhL`9V8U897T6S67`INILOASAVRUCHAIDELETLISUNLOCCLOSREAEXEWRITPOSITIOOPEAPPENRENAMCATA_OMONOMOPRINMAXFILEFINBSAVBLOABRUVERIF!pppp p p@p p`" t""#x"p0p@p@@@p@!y q q p@  LANGUAGE NOT AVAILABLRANGE ERROWRITE PROTECTEEND OF DATFILE NOT FOUNVOLUME MISMATCI/O ERRODISK FULFILE LOCKESYNTAX ERRONO BUFFERS AVAILABLFILE TYPE MISMATCPROGRAM TOO LARGNOT DIRECT COMMANč$3>L[dmx-(0 0Ϡ@跻~!Wo*9~~~~ɬƬ~_ j ʪHɪH`Lc (L ܫ㵮赎 ɱ^_ J QL_Ls贩紎 DǴҵԵƴѵӵµȴ 7 ַ :ŵƴѵǴҵȴµ納贍﵎ٵ്ᵭⳍڵL^ѵ-I `  4 ò-յ!  8صٵ紭ﵝ 7L (0+BC  7L HH`LgL{0 HH` õL H hBL BH [ h`Lo õ ڬL B ڬ LʬH hB@ յյ [L (ȴ) ȴ 7L L ( L (ȴL{ƴѵ洩ƴǴҵ 7 ^* B0 HȱBh ӵԵ 8 L8 ݲ` ܫ  / / ED B / / ]ƴS0Jȴ ȴ)  紅D贅E B ƴ  / 0L Ν `HD٤DEEhiHLGh ` ŵBѵ-` ѵB-` ܫ XI볩쳢8 DH E𳈈췍Ȍ X0 · JLǵBȵC`,յp` 䯩 R-յյ`յ0` K R-յյ`ɵʵӵԵ` 4 K ( ѵҵLBȱBL8` DBHBH : ַ޵BȭߵBhhӵԵ RBܵmڵ޵ȱBݵm۵ߵ` 䯩LR˵̵ֵ׵`êĪLR E( 8` R` ELRŪƪ`췌 յյI뷭鷭귭ⵍ㵍跬ª 뷰` Lf ݵܵߵ޵ ^`8ܵ i B8` 4L ֵȱB׵ ܯ䵍൭嵍 ` DȑB׵Bֵ  ַ յյ`굎뵎쵬 뵎쵌``õĵBCõĵ`µµ`L õBĵCصص Qƴ0"Bƴ 󮜳` 0۰ϬBƴ8`i#`ЗLw!0>ﵭ` m ﳐ 7i볍 8 ЉLw`H h ݲL~ `浍국䵍뵩嵠Jm赍嵊mjnnn浈ۭm浍浭m䵍䵩m嵍`"L ŵ8ŵH ~(` i d ֠z# u`computer's attentionEraseEdit responseEscape to indexAnswerBack"j- 2:I6114312:10,I260,I::3x2 360:100ESCAPE":Gl 4:10:"You can 'ESCAPE' to the mainindex of a lesson by pressingthe key marked ESC.":360 N:2:8:"@c5@4@F@0SUMMARY@1":G:7:5:"KEY":20:7:"FUNCTION" 9:3:"RETURN"(31)""(29)"ESCCTRL ACTRL B"E( "@P":9:12:"Get rd'C2:I3014012:20,I252,I::36H360:1150X N:2:13:"@c5@4@F@0HELP":G 3:10:"To see the answer to a questionhold down 'CTRL' and press 'A'.":17:"@x2@c5CTRL-A@x1@c7 gives you the @x2@c5a@x1@c7nswer.":360  N:2:9:"@c5@4@F@0f Markup Symbols@C6":1:3:B40:G:5:4:"=X"(5)(4)""(19)""(20)""(24)""(23)""(22)""(16)>5:13:"wrong letterno letter hereinverted lettersinsert beforeinsert aftershould be lower caseshould be upper casemove word leftinsert wor anexplanation of these symbols.":300:15:1:E40,415:1:"@Pwrong letter":17:8:"inverted order":19:18:"missing letter":15:25:"extra letter" 2:20,11095,75:90,125120,75:164,140164,75:230,110200,75:3:360h4N:5:"Summary omesymbols under your answer.":360N:"If you typed sadimu chlrides insteadof sodium chloride this is how youranswer will be marked."ZA$"sadimu chlrides":8:12:"@:"ZA$A"sodium chloride":9:11:ZM$615:4:"Press @X2@C5RETURN@ fossonhold down 'CTRL' and press 'B'.":17:"@x2@c5CTRL-B@x1@c7 allows you to go @x2@c5b@x1@c7ack.":360:2500#N:6:"If you have made a typing or spellingerror in your answer the computerwill try to help you locate thespecific problem by putting sosponse bypressing the key marked ";(29)".Pressing ";(29);" once erases yourresponse.If you press ";(29);" again your answer";"will return one word at a time.":360N:2:12:"@c5@4@F@0BACK":Gm2:10:"To go back a display in a leresponse, press RETURN to have your answer judged.":360:2000f~N:2:16:"@c5@4@F@0@J@1":BK2:G3:10:"@PThe key marked @J will erase whatyou have typed on the screen.":360N:2:15:"@c5@4@F@0@K@1":G8:3:"@PYou can edit your reanswer to aquestion, press RETURN to have yourresponse evaluated by the computer.":360LN:16:2:"@4@F@0@c5"(1)"@1":BK6:G:8:1:"This symbol "(1)" will appear on thescreen when you are to respondto a question."?`14:"After you type your RETURN@x1@c7 to go on.":300:N:9:"@c5@4@F@0RETURN@1":BK1:G:7:1:"@PPressing the key marked RETURN getsthe computer's attention."11:"Press RETURN when you have finishedreading and are ready to go on."Y15:"After you have typed an A"N":ZA%Ĺ770,96:771,0:772,0:"Sound effects will @x2@c5NOT@x1@c7 be used.":360:100~14:4:"Type: y or n":R:220,P:ZK%2350Z1283006ZK%27100@J100^1:BK100,1100,1150,1200,1250,1000hG:23:5:"@1Press @x2@c510:2:"@#@c3@4@F@0SOUND":Gi8:4:"Do you want sound effects whileworking these programs? (y,n) 12:10:ZL%1:I:14:4:E14:12:10:E4A"Y":16:4:ZA%Ĺ770,173:771,48:772,192:"Sound effects @x2@c6will@ be used.":360:100[s for Using These Programs2. Sound3. See Index of Lessons@c6":B0,46,278,192,4rxP:AZK%48:A1A3120A1000,200,130N:1:B:4:9:"@F@0@4ONE MOMENT, PLEASE..."::G:(13);(4);"RUN ROUTER"N:1:B0,0,278,49:B0,49,278,192,4:tworth N.H. 03282"e <17:10:"(603) 764-5831":22:5:"Press @x2@c6RETURN@x1@c7 to continue.":300 dN:B0,0,278,45:G:Y4:5:"@2@FPERCENT COMPOSITION EMPIRICAL FORMULAS":G:3:8:"@F@0@x2@c6Choose topic by number.":GUn7:12:"1. Instruction(7:22:"(c) 1983 by Smith and Kean.":B2,177,276,189:9:Y180:"@x2PRESS RETURN.":P 2N:1:9:"@F@x2@0@c6Distributed by@x1":5:B1,20,276,80,12,6:6:Y37:"@F@4@C5@0COMPress":G 713:6:"A Division of Wadsworth, Inc.":15:3:"P.O. Box 102, Wen,72,964 I09:A:912I,A:: ONERR GOTO FIXUPt N:"@0@P@1@C7@v4":GS:6:B0,0,278,191,2,2:G:B2,2,276,45 #Y9:5:"@F@4@C1CHEMISTRY":G:8:5:"@F@2PERCENT COMPOSITION EMPIRICAL FORMULAS":G &G:17:5:"Stanley Smith Elizabeth KeanW KEY TABLE,A"((24584)256(24585)):D$"BLOAD WHITE SET,A"((24580)256(24581)) WS(24580)256(24581)1029:D$"BLOAD BOLD TABLE,A"WS:H(WS256):LWS256H:24582,L:24583,H D$;"BLOAD BIN.STRING.EXP.794"  104,168,104,166,223,154,72,152X 782,208:783,245:784,174:785,0:786,3:787,76:788,2:789,3:790,96:791,0:792,0j 793,0:800,0 ::10:8:"ONE MOMENT PLEASE...":12:6:"LOADING ENBASIC CAI SYSTEM":22:16:"D5/V3" (13);(4)"BRUN ENBASIC" D$(4)Q D$"BLOAD z : HELLO 11/2/834 (C) SMITH KEAN 1983?38000H30T 1012,0 104,168,104,166,223,154,72,152,72,96 I09:A:912I,A:770,173:771,48:772,192:773,136:774,208:775,5:776,206:777,1:778,3:779,240:780,9:781,202            a i yLvh \hgQMEưikmjln[lX `ĠԤ{ ~21;27BwQr+|yg_htpq}WDhBjJ01~l8*9n*yL Pu5 UČJFGaI2 Dӝ[f'&;:(@ ّSb?:Jvs]J>?<;8HUTW N8Q4Z""8?>>> ?DDDD      EF$"K2SO3":PC(1)49.4:PC(2)20.3:PC(3)30.3:AW$"Atomic Weights: K=39.1 S=32.1 O=16.0":i20.3:PC(3)30.3:AW$"Atomic Weights: K=39.1 S=32.1 O=16.0":" N=14.0 O=16.0":3 NE3:E$(1)"K":E$(2)"S":E$(3)"O":AW(1)39.1:AW(2)32.1:AW(3)16.0:EF$"K2SO4":PC(1)44.9:PC(2)18.4:PC(3)36.7:AW$"Atomic Weights: K=39.1 S=32.1 O=16.0":[4*NE3:E$(1)"K":E$(2)"S":E$(3)"O":AW(1)39.1:AW(2)32.1:AW(3)16.0:14.0:AW(3)16.0:EF$"NaNO2":PC(1)33.3:PC(2)20.3:PC(3)46.4:AW$"Atomic Weights: Na=23.0 N=14.0 O=16.0":3NE3:E$(1)"Na":E$(2)"N":E$(3)"O":AW(1)23.0:AW(2)14.0:AW(3)16.0:EF$"NaNO3":PC(1)27.1:PC(2)16.5:PC(3)56.4:AW$"Atomic Weights: Na=23.0C(1)40.0:PC(2)6.7:PC(3)53.3:AW$"Atomic Weights: C=12.0 H=1.0 O=16.0":1NE2:E$(1)"B":E$(2)"F":AW(1)10.8:AW(2)19.0:EF$"BF3":PC(1)15.9:PC(2)84.1:AW$"Atomic Weights: B=10.8 F=19.0":m2 NE3:E$(1)"Na":E$(2)"N":E$(3)"O":AW(1)23.0:AW(2) O=16.0":0NE3:E$(1)"C":E$(2)"H":E$(3)"O":AW(1)12.0:AW(2)1.0:AW(3)16.0:EF$"CH4O":PC(1)37.5:PC(2)12.5:PC(3)50.0:AW$"Atomic Weights: C=12.0 H=1.0 O=16.0":L1NE3:E$(1)"C":E$(2)"H":E$(3)"O":AW(1)12.0:AW(2)1.0:AW(3)16.0:EF$"CH2O":Pts: S=32.1 O=16.0":/NE2:E$(1)"N":E$(2)"O":AW(1)14.0:AW(2)16.0:EF$"NO2":PC(1)30.4:PC(2)69.6:AW$"Atomic Weights: N=14.0 O=16.0": 0NE2:E$(1)"N":E$(2)"O":AW(1)14.0:AW(2)16.0:EF$"NO":PC(1)46.7:PC(2)53.3:AW$"Atomic Weights: N=14.0 mic Weights: P=31.0 Cl=35.5":.NE2:E$(1)"S":E$(2)"O":AW(1)32.1:AW(2)16.0:EF$"SO2":PC(1)50.1:PC(2)49.9:AW$"Atomic Weights: S=32.1 O=16.0":/NE2:E$(1)"S":E$(2)"O":AW(1)32.1:AW(2)16.0:EF$"SO3":PC(1)40.1:PC(2)59.9:AW$"Atomic Weigh"Atomic Weights: C=12.0 O=16.0":-NE2:E$(1)"P":E$(2)"Cl":AW(1)31.0:AW(2)35.5:EF$"PCl3":PC(1)22.5:PC(2)77.5:AW$"Atomic Weights: P=31.0 Cl=35.5": .NE2:E$(1)"P":E$(2)"Cl":AW(1)31.0:AW(2)35.5:EF$"PCl5":PC(1)14.9:PC(2)85.1:AW$"Ato55.7:AW$"Atomic Weights: Cu=63.5 Br=79.9":,NE2:E$(1)"C":E$(2)"O":AW(1)12.0:AW(2)16.0:EF$"CO2":PC(1)27.3:PC(2)72.7:AW$"Atomic Weights: C=12.0 O=16.0":$-NE2:E$(1)"C":E$(2)"O":AW(1)12.0:AW(2)16.0:EF$"CO":PC(1)42.8:PC(2)57.2:AW$"Atomic Weights: Cu = 63.5 Br = 79.9":+NE2:E$(1)"Cu":E$(2)"Br":AW(1)63.5:AW(2)79.9:EF$"CuBr2":PC(1)28.5:PC(2)71.5:AW$"Atomic Weights: Cu = 63.5 Br = 79.9":/,NE2:E$(1)"Cu":E$(2)"Br":AW(1)63.5:AW(2)79.9:EF$"CuBr":PC(1)44.3:PC(2)*vNE3:E$(1)"Na":E$(2)"C":E$(3)"O":AW(1)23.0:AW(2)12.0:AW(3)16.0:EF$"Na2CO3":PC(1)43.4:PC(2)11.3:PC(3)45.3:AW$"Atomic Weights: Na=23.0 C=12.0 O=16.0":*+NE2:E$(1)"Cu":E$(2)"Br":AW(1)63.5:AW(2)79.9:EF$"CuBr":PC(1)44.3:PC(2)55.6:AW$:E$(1)"Fe":E$(2)"Cl":AW(1)55.8:AW(2)35.5:EF$"FeCl2":PC(1)44.0:PC(2)56:AW$"Atomic Weights: Fe = 55.8 Cl = 35.5":)lNE2:E$(1)"Fe":E$(2)"Cl":AW(1)55.8:AW(2)35.5:EF$"FeCl3":PC(1)34.3:PC(2)65.6:AW$"Atomic Weights: Fe = 55.8 Cl = 35.5":E2:VT:CX:E2:VT:CX:60501(C27S32:6050F(C64A1$(CS)a(C48C58ĺ(15);"D";(A1$;:A$A$A1$:S0:CXCX1:(A$)20ī6050((8ZK%0:NTNT1:ZA%L20:I55455:T1,I:T2,L:CL::VT:CX2:OK$:(BVT:CX2:NO$:z)bNE2"@:";:ES0:HP0:P:A1$(ZK%):CZK%:C13ZA$A$:6160J'ZK%2ġ:100Y'C127C8t'PS0:C8(A$)06050'C17ı'C216040'C1NTNT1:ZC%ZC%1:ZA$A$:HP1:6160'C8(A$)16040(C8A$(A$,(A$)1):CXCX1:VT1:CX:he shift key for UPPER case"P&z24:4:"Press @X2@C5CTRL-B@ for the Index.":\&"@@@$"w&PS1ĢVT:CX2:E4:K&PS1ZK%13CZK%:A1$(ZK%):VT:CX:6070&A$"":S0:HTHC:HT:VT1:E20:HT:VT:E20:HT:VT:(2):HTHT2:CXHT:HT:VT8'14 %5030% X0:Y128::E1,8:Y100:E40,3:Y120:E40::Y104:X30:"Press CTRL-Q for a calculator.":24:2:"Press @X2@C5CTRL-B@ for the Index."%100% %p12:1:(24591)0ĺ"Lower case: press ESC and then a letter"&u(24591)Ė3:"Use t$912:X10:Y(222):"Cannot evaluate";I$Y69ĺ" - TOO LARGE":5814j$Y16ĺ"-Syntax Error":5814$Y133ĺ"-Division by zero":5814$Y191ĺ" - Too complex":5814$Y107ĺ" - Bad Subscript":5814%Y224ĺ" - Undefined Function":58L:EX$EX$L$%#PP1:P(ZA$)5500/#5070}#|794:J1(((((EX))2.3)2)):EX((EX10J.5)):EXEX10J:X4:"= ";#EX1000EX.099ĺEX:5030#EX1000JJ2:EX10J"x10@+"J"@D":5030#EX.099JJ2:EX10J"x10@+"J"@D":5030))+*-/^ only.":5030""P15300F"L405100: CHECK FOR "("c"L141EX$EX$"*":5300"L147L158EX$EX$"*":5300: NO IMPLIED MULT"L425300: "*""L142EX$(EX$,(EX$)1):L94:L$(94):5300: "**" TO "^" #L1:E39:YTBLF:YY:X0:E39:A!R:X0:YTB:ZL%30:I:ZK%25900T!ZK%27ġ:100e!ZA$""5040n!X10!TYTBLF:YTY!EX$"":C1:P1!L$(ZA$,P,C):L$"x"L$"*":5300!L$"="L$" ""L(L$):L62L94L69:"Use numbers and (Type a numerical expression and press RETURN. Press CTRL-B when finished. "^ 24:1::E40: X1:Y101:5:X,Y1X3971,Y1X3971,Y2113X,Y2113X,Y1:3 Y1107:LF11:TBY1 TBTB22:TB180TBY122!YTB:X0::E39:YTB4:YY:X01:6200:ZA%Ģ12:3500Qf "":VT1:HC1:ZM$:16:1:"The formula is: ";:3600_k ZK%3410np 250:3430 290:NT10:2010I1(EF$):A$(EF$,I,1):(A$)48(A$)57ĺ"@D";A$;::290CA1:CC1:5800L Y104:X3:E38:"@Z@P<K (24591)ĺ"Use the shift key for Upper andlower case."nM NE2Ģ10:10:R(1)" "E$(1)" "R(2)" "E$(2)N NE3Ģ10:6:R(1)" "E$(1)" "R(2)" "E$(2)" "R(3)" "E$(3)R VT13:HC12:6020:ZK%17įR:7:14:E30:5000:R:3410\ A"":PSumber by "M(L)" gives theratio:":15:8:R(1)" "E$(1)" "R(2)" "E$(2)" "R(3)" "E$(3)b> 290H N:BK8:4:"Type the empirical formula of thecompound which has this mole ratio:"J 20:1:(24591)0ĺ"For lower case press ESC and thena letter"such as 1, 2, and 3 are wholenumbers so the answer is no. Try: no.":ZK%3340^ 250:3350|* N1000:I13:M(I)LLI/ 2 R(1)(100M(1)M(L).005)100:R(2)(100M(2)M(L).005)100:R(3)(100M(3)M(L).005)100Y4 12:1:"Dividing each nmallest number":5:0,10279,10:3o 1:3:""M(1)" moles "E$(1)" "M(2)" moles "E$(2)" "M(3)" moles "E$(3) 6:1:"Is the ratio "M(1)" : "M(2)" : "M(3)" in lowestwhole number terms?" 9:14:500:A"":600:ZA%3370O 12:1:"Numbers 0 250:3250' L2:M(1)M(2)L1i R(1)(100M(1)M(L).005)100:R(2)(100M(2)M(L).005)100 12:1:"Dividing each number by "M(L)" gives theratio:":15:10:R(1)" "E$(1)" "R(2)" "E$(2):290:3400" N:BK7:"Step 3: Divide by the s" moles "E$(1)" "M(2)" moles "E$(2)r 6:1:"Is the ratio: "M(1)" : "M(2);" in lowest wholenumber terms?" 9:14:500:A"":600:ZA%3270 12:1:"Numbers such as 1, 2, and 3 are wholenumbers so the answer is no. Try: no.":ZK%324 l ZK%3150v 250:3160 11:1:" Press RETURN to continue.":M(HE)(100A)100:300:HEHE1:HENEĢ5:1::E40,5:11:1:E40::3140 NE33310 N:BK6:"Step 3: Divide by the smallest number":5:0,10279,10:3( 1:3:M(1)8N 8:12:500:ZK%17įR:8:12::E30::5000:R:3150fS 360:A"":590:ZA%3200X 10:1:N1ĺ"Your number is right, but please includethe units of moles in your answer."b N0ĺPC(HE)"grams/"AW(HE)" g per mole = "ML" moles."ert grams to moles":5:0,10279,10:3:420o: I1NE:Y18:X6055(NE3)70(I1):XX:PC(I)"g "E$(I)::HE1D 5:1:"How many moles of "E$(HE)" are in "PC(HE)" grams?":Y42:X10:AW$:RI MLPC(HE)AW(HE):ML(ML100)100:MXML.05ML:MNML.05ML:MNPC(HE).05C 45HE:14:500:360:A"":590:ZA%3100 1:75HE:N0ĺPC(HE)"% "E$(HE)" is "PC(HE)" grams per 100 grams.Try: "PC(HE)" grams." 430:ZK%3030 250:3040 HEHE1:HENE3020& 290+0 N:BK5:"Step 2: Conv250:2120V N:BK4:"Step 1: Obtain weights of elements in g.":5:0,10279,10:3 I1NE:Y20:X8030(NE3)70(I1):XX:PC(I)"% "E$(I)::HE1 15HE:1:"How many grams of "E$(HE)" are there in 100grams of the compound?":R:MXPC(HE).05"E$(I)" ";::Y25:X1508(AW$)2:X1X1D2XX:AW$:6000:4204VT7:HC14:6020:ZK%17įR:7:14:E30:5000:R:21009ZK%13000>A"":PS1:6200:ZA%Ģ12:2040HVT1:HC1:ZM$:10:8:"For help press CTRL-A"\ZK%2100 f:10:3:"Press @X2@C6RETURN@ to continue.":300:2010P(RN)7010,7020,7030,7040,7050,7060,7070,7080,7090,7100,7110,7120,7130,7140,7150,7160,7170,7190,7200,7210:2090*N:BK3:"Type the empirical formula of thiscompound: ";,/I1NE:PC(I)"% one step at atime.":290:D(1)1:100QMP20:BK3:I1MP:P(I)I::NT0::NPMPNT1P(RN)P(NP):NPNP1:PRPR1:PR5D(2)1:100NP02000RN(NP(1)1):NT0:PS0205071:12::E40::PR(NT1)" right";:27:MRPR(NT1)" to go"4:"In this program you will be given thepercent of each element in a compound.From these data you are to calculatethe empirical formula."%12:"If you have trouble working a problempress CTRL-A and the computer will leadyou through the solution N2İ400"ZL%1ZL%208I:ZK%27ġ:100FZK%17ıXZK%2ġ:350rZK%1ZC%ZC%1:530A"":ZA%1ZL%1505ZK%0:NTNT1:NN1ZA%0:ZW%1XM:ZK%0:ZA%L20:I55455:T1,I:T2,L:CL::bN:BK2:F3):|ZA%0:ZW%1:S"You have to many numbers afterthe decimal point.":"Your number is right, but please includethe units of grams in your answer.":2:1,99276,99:3:Y104:X30:"Press CTRL-Q for a calculator.":N1İ410 00,2090,3000,3120,3210,3310- hZA$""N0: jA$"":C$"":N0:SF0:DM0:I1(ZA$):B$(ZA$,I,1):(B$)45(B$)58A$A$B$:C$C$" ":SFSF(DM1) m(B$)57(B$)46C$C$B$ n(B$)46DM1r:ZA$C$:A(A$):AMXAMNZA%1:ZW%0:N1(Sndex.":(13);(4);"RUN ROUTER"C I7510025:T1,I:T2,30:CL:T T:ZK%260g ZK%27ġ:100y ZK%2ġ:350  "G:24:4:"Press @X2@C6RETURN@ to continue.":ZK%0:ZS0:1 ,P:ZK%27ġ:100 6ZK%2ġ:350 @ ^BK200,100,1blems - Empirical Formulas@R@R3. Return to the MAIN Index"s x22:4:"Press @X2@C6ESC@ to return to this index. I12:D(I)1ĢY2(I1):2:"*" 300:AZK%48:A0A4150 A1000,2000,200 N:8:"@3@F@C1@GReturning tothe Main IS>"( 2OK$"@X2@C1OK@":NO$"@X2@C5NO@ dN:BK1:1:15:"INDEX":3:6:"@2@FEmpirical Formulas@1@S":6:7:"@x2@c5@FREVIEW PROBLEMS":G:7:1:U$ i10:3:"@x2@c6@F@0Select topic by number.":G:PR0:MR5; nY13:Y:4:"1. Introduction@R@R2. Pro  I09:A:912I,A:R V$"" #Y$"" %N$"" (GM$"":590:ZA%2440{!t 14:1:"If you divide a number by itself youget 1. So try: "MM:ZK%2410!~ 250:2420! 12:8:MM:14:A:13:14:"moles Mg = 1.00":49,9988,99" 16:8:MC:18:A:17:14:"moles of Cl = "(100MCs of Mg and Cl byany number we want as long as we doit to both of them.":290 [ N:BK12:6:2:""MM" moles Mg : "MC" moles Cl" ` 1:5:"What number should we divide by toconvert "MM" moles of Mg to 1 moleof Mg?":MXMN.01:MNMM.1+!j 10:14:50z #    Ԥ Ϡ  ԠԠ8ӠӠ̠Ӡ8ǠԠ"Ӡ. ĠŠ Ϡ!Ԥ! à#٠Š̠Ӡ8 ŠԠήǮЮҠles Mg : "MC" moles Cl"aB 16:"Now we must reduce this to the lowestwhole number ratio.":290L N:BK11:6:2:""MM" moles Mg : "MC" moles Cl"O V 6:1:"Since an empirical formula gives theratio of the atoms in the formula, wecan divide the mole"S. 20:1:"Next we must convert this compositioninto a chemical formula.":2908 N:BK10:"The product of our reaction of Mg withHCl contains "MM" moles of Mg and "MC"moles of chlorine.":8:"So the ratio of atoms in the formula is: "MM" mo000:MC(1000WC35.5)1000:MM$(MM):MC$(MC) 1:10:WM$" g Mg x":9:13:"1 mole Mg":11:"24.3 g Mg":91,75159,75:10:22:"= "MM$" moles Mg"$ 1:16:WC" g Cl x":15:13:"1 mole Cl":17:"35.5 g Cl":91,123159,123:16:22:"= "MC$" moles ClWM". Try: "WPWM" grams."-430:ZK%2250<250:2270 N:BK9:"The product contains "WM$" grams ofmagnesium and "WC" grams of chlorine.These weights may be converted tomoles, a unit which counts the numberof atoms."/ MM(1000WM24.3)1WM.001:MNWPWM.001L7:14:500:ZK%17įR:7:14:E30:5000:R:2250360:A"":590:ZA%Ģ24:1:E40:290:23101:9:N0ZC%1ĺ"Subtract the weight of the magnesiumfrom the weight of the product."N0ZC%1ĺWP"- "WM$" = "WPr weighed "WB$"grams so the product of our reactionweighs "WP" grams.":290N:BK8:"The product of the reaction of "WM"grams of magnesium with HCl weighs"WP" grams. What is the weight of thechlorine (Cl) in the product in grams?":420MXWP has been boiled off.Let's let the beaker cool and thenreweigh it.":290jN:BK7:X110:Y10:2:204092,44100,49166,49174,44:135,50135,54:84,54182,54182,8884,8884,54:32119,47M9:15:""WBWP"Grams":16:3:"The empty beake1:F3:S2:2160:ZS3:300:B0:F4:2160:290<24:1:E4012:1:E40,5:20:E38:7132,81:16:1:"Next we will boil off the aqueous acid.":0:I2545:111,I155,I:ZS.2:300:J4112,46:3:2119,47:7132,81:16:1:E40:13:"All of the acid290Hf12:5:E30,3:24:1:E40:4112,47:I02:5111,457I::2170upIBFS:61195I,44:61195I,35::,zB0:F4:S1:2160:12:2:"Bubbles are forming as the magnesiummetal dissolves. Let's wait until thereaction is over.":ZS3:300:B0:500:360:A"":590:ZA%Ģ24:1:E40:290:2140v>22:1:N0ĺWBWM"- "WB" = "WM" grams. Try: "WM" grams"H430:ZK%2100R250:2110\N:BK6:X110:Y10:2:2040:3:2119,47:12:6:"Next let's add some aqueoushydrochloric acid.":some magnesium metal.":290*16:1:E40,3:13:"The empty beaker weighs "WB$" grams,and the beaker plus magnesium is "WT$"grams. How much does the magnesiumweigh in grams?",2119,47:9:15:E7:WT$/MXWM.01:MNWM.01:24:1:E408414:2l";7"calculate the weight of the magnesium.":290W N:BK5:X110:Y10:2:204092,44100,49166,49174,44:134,50134,54:84,54182,54182,8884,8884,54:3 9:15:""WB$"Grams":16:1:"The beaker weighs "WB$" grams. Next,we will add ,Y35X3,Y38X43,Y38X46,Y35X46,YX48,Y2:"We will run the experiment in thisbeaker.First we will weigh the beaker, then wewill add some magnesium and weigh thebeaker containing the magnesium. Fromthe difference in the weights we wil the product.":290X(5(1)):WB10.100X:X(5(1)):WBWBX10:X(5(1)):WM.25X100:WPWM95.324.3:WP(1000WP)1000:WB$(WB)"00":WM$(WM)"0"WTWMWB:WT$(WT)"0":WCWPWMN:BK4:X110:Y20:2040:11:1:20504X2,Y2X,YXnesium chloride and hence itsempirical formula.":290:D(1)1:100N:BK3:5:"Treatment of magnesium metal withhydrochloric acid forms a compoundwhich contains both magnesium andchlorine.Let's run the experiment and thencalculate the formula ofterminethe empirical formula for magnesiumchloride.A weighed sample of magnesium metalwill be allowed to react with hydro-chloric acid to form magnesium chloride.B"From the increase in weight you willbe able to calculate the compositionof mag:ZK%27ġ:100ZK%17ı/ZK%2ġ:350UZK%1ZC%ZC%1:SS1:NO$"":530oA"":ZA%1ZL%1505{ZK%0:NN1ZA%0:ZW%1XM:ZK%0:ZA%L20:I55455:T1,I:T2,L:CL::bN:BK2:2:"In this experiment you are to des after thedecimal point.":r"Your number is right, but please includethe units of grams in your answer.":2:1,99276,99:3:Y104:X30:"Press CTRL-Q for a calculator.":N1İ410N2İ400ZL%1ZL%20SS0:I:NO$NN$:C$"":N0:SF0:DM0:I1(ZA$):B$(ZA$,I,1):(B$)45(B$)58A$A$B$:C$C$" ":SFSF(DM1) m(B$)57(B$)46C$C$B$ n(B$)46DM1 r:ZA$C$:A(A$):AMXAMNZA%1:ZW%0:N1(SF3): |ZA%0:ZW%1:"You have too many number260 ZK%27ġ:100( ZK%2ġ:350. o "G:24:4:"Press @X2@C6RETURN@ to continue.":ZK%0:ZS0:1 ,PZS:ZK%27ġ:100 6ZK%2ġ:350 @ ^BK200,100,100,2000,2030,2060,2140,2200,2230,2310,2360,2380,100 hZA$""N0:a jA$""122(I1):4:"*" H WT((100(1)))1001:NH0:T(0)0:W(0)WTg 300:AZK%48:A0A3150| A1000,2000,200 N:8:"@G@3@C1@FReturning tothe Main Index.":(13);(4);"RUN ROUTER" SS0āI7510025:T1,I:T2,30:CL: T:ZK%:"@2@FMg-HCl Experiment@1@S":6:1:U$:9:3:"@X2@C6@F@0Select topic by number.":G:NO$NN$ n12:6:"1. Introduction@R@R2. Mg - HCl Experiment@R@R3. Return to the Main Index" x20:4:"@X2@C6ESC@ returns you to this index." I13:D(I)1Q #Y$"" %N$"" (GM$"" 2OK$"@X2@C1OK@":NO$"@X2@C5NO@":NN$NO$] dN:BK1:1:15:"INDEX":3:5( : MG EXPERIMENT 3/23/83 11/83Q: COPYRIGHT SMITH, CHABAY,KEAN 1983[100f36950 T1768:T2769:CL770:CK16336:U$""D(5),P(5),T(10),W(10)V$""          BVT:CX2:NO$:A$)20ī6050+>o>8:ZK%0:ZA%L20:I55455:T1,I:T2,L:CL::VT:CX2:OK$:>BVT:CX2:NO$:ANO$: :CL::VT:CX2:OK$:>BVT:CX2:NO$:(A$,(A$)1):CXCX1:VT1:CX:E2:VT:CX:E2:VT:CX:6050S=C27S32:6050h=C64A1$(CS)=C48C58ĺ(15);"D";=A1$;:A$A$A1$:S0:CXCX1:(A$)20ī6050==8:ZK%0:ZA%L20:I55455:T1,I:T2,L:CL::VT:CX2:OK$:>$NN$:A1$(ZK%):CZK%:C13ZA$A$:6160L<ZK%1SS1:NO$"":ZC%ZC%1:^<ZK%2ġ:350q<ZK%27ġ:100<C127C8<PS0:C8(A$)06050<C216040<C1NTNT1:ZC%ZC%1:ZA$A$:HP1:NO$"":6160<C8(A$)16040>=C8A$30:"Press CTRL-Q for a calculator.".; 1004;@;"@H@$"[;PS1ĢVT:CX2:E2:K;PS1ZK%13CZK%:A1$(ZK%):VT:CX:6070;A$"":S0:HTHC:HT:VT1:E20:HT:VT:E20:HT:VT:(2):HTHT2:CXHT:HT:VT)<"@:";:ES0:HP0:SS0:P:NO:5814):Y16ĺ" - Syntax Error":5814O:Y133ĺ"-Division by zero":5814r:Y191ĺ" - Too complex":5814:Y107ĺ" - Bad Subscript":5814:Y224ĺ" - Undefined Function":5814:5030$; X0:Y128:E1,8:Y100:E40,3:Y120:E40:Y104:X((EX))2.3)2)):EX((EX10J.5)):EXEX10J:X4:"= ";Y9EX1000EX.099ĺEX:50309EX1000JJ2:EX10J"x10@+"J"@D":50309EX.099JJ2:EX10J"x10@+"J"@D":50309912:X10:Y(222):"Cannot evaluate";:Y69ĺ" - TOO LARGE"XY69ĺ" - TOO LARGE":5814V8L147L158EX$EX$"*":5300: NO IMPLIED MULTq8L425300: "*"8L142EX$(EX$,(EX$)1):L94:L$(94):5300: "**" TO "^"8L1L:EX$EX$L$8PP1:P(ZA$)550085070:9|794:J1(((271007ZA$""5040*7TYTBLF:YTY=7EX$"":C1:P1d7L$(ZA$,P,C):L$"x"L$"*":5300w7L$"="L$" "7L(L$):L62L94L69:"Use numbers and ()+*-/^ only.":50307P153007L405100: CHECK FOR "("8L141EX$EL-B when finished."^6X1:Y101:5:X,Y1X3971,Y1X3971,Y2113X,Y2113X,Y1:3u6Y1107:LF11:TBY16TBTB22:TB180TBY1226YTB:X0:E39:YTB4:YY:X0:E39:YTBLF:YY:X0:E396R:X0:YTB:ZL%30:I:1:ZK%259007ZK%e molecular weight by@: @Pthe empirical formula weight3. Multiply subscripts in the empirical@: @Pformula by the ratio from 2."5 290:D(3)1:1005CA1:CC1:58006Y104:X3:E38:"Type a numerical expression and press RETURN. Press CTRts so the molecular formula is C@D3H@D6.":2904 N:BK23:12:"@F@C5@2SUMMARY":G:4:6:"Finding Molecular Formulas from Empirical Formulas":5:1,50278,50278,511,51:35 9:1:"1. Find the weight of the empirical@: @Pformula2. Divide thle?">3 17:14:500:A"":600:ZA%32603 19:1:"The molecular weight is 3 times theempirical formula weight so there mustbe 3 of them. Type: 3":ZK%32303 250:3240.4 19:1:"Right, the molecular formula has 3 CH@D2uni formula by the weight ratio.":5:0,23279,23:32 5:1:"We have found:":7:6:"Empirical formula: CH@d2Empirical formula weight: 14Molecular weight: 42Ratio of weights: 42/14 = 33 14:1:"How many CH@D2 units are there in theactual molecu:4:"Right! Press RETURN to continue.":300:32001l 10:1:ZC%1ĺ"42/14 = 3 which means there are 3empirical formula units in the molecule.1q ZC%1ĺ"Try: 31t ZK%31701v 250:318012 N:BK20:"Step 3: Multiply subscripts of theempirical":5:0,22279,22:30X Y26:X1:"The molecular weight is 42 and theempirical formula weight is 14. How manyCH@D2 units are there in the compound?":4200b 9:12:500:ZK%17Ģ9:12:E20:R:5000:R:317031g A"":600:ZA%Ģ11tinue.":300:3150W/ 1:10:N0ZC%1ĺ"Sum the atomic weights of all of theatoms."/ N0ZC%1ĺ"1x12.0 + 2x1.0 = 14. Try: 14"/& ZK%3070/0 250:30900N N:BK22:"STEP 2: Divide molecular weight by the empirical formula weight":5:0,10279,10:3~. 3:3:"What is the formula weight of CH@D2?Atomic weights: C=12.0 H=1.0":420:MX14.01:MN13.9999. 14:7:500:ZK%17įR:6:14:E20:5000:R:3070/ 360:A"":590:ZA%Ģ10:3:"Right, press RETURN to conet's calculate the MOLECULAR formulaof a compound which has the EMPIRICALformula CH@d2 and a molecular weight of42."- "We will start by calculating the weightof the empirical formula.":290. N:BK21:"1. Calculate weight of empirical formula@C5MOLECULAR@ formula tells usexactly how many moles of each elementthere are in a mole of the compound.", 16:"To convert an empirical formula intoa molecular formula, you must know themolecular weight of the compound.":290z- N:BK20:6:"@pLlest@: @Pnumber to obtain subscripts"+ "4. If necessary, multiply subscripts@: @Pby a factor to get whole numbers.":290:D(2)1:100+ N:BK19:2:"The @X2@C5EMPIRICAL@ formula tells usthe ratio of the atoms in a compound.l, 8:"The @X2ormula is: Fe@D2O@D3":290s* N:BK18:9:"@G@3@F@C5SUMMARY":G:4:5:"Calculating Empirical Formulas":6:1:U$* 9:"1. Obtain the weights of all elements in the compound.")+ "2. Convert weights to moles@R3. Divide all moles by the smal450m)t 22:1:"If we multiply both 1 and 1.5 by 2 we get2 and 3 which are whole numbers. Try: 2":ZK%2410|)~ 250:2420) 15:5:E35,3:20:16:E20* 16:4:"2 x (1 mole Fe : 1.5 mole O) = 2 mole Fe : 3 mole O":20:1:"So, the empirical f`(L N:BK17:"STEP 4: Multiply by factor":5:1,12279,12:3:3:6:" 1 mole Fe : 1.5 moles O(` 8:1:"What is the smallest number that whenmultiplied by 1 and 1.5 will change themto whole numbers?")j 13:14:500:A"":600:ZA%2:"STEP 4: Multiply by factor":5:1,12279,12:3:3:6:" 1 mole Fe : 1.5 moles O(B 10:1:"1 is a whole number, but 1.5 is not.We need to multiply both 1 and 1.5 bysome number such that the product willbe whole numbers such as 2, 3, 4, etc.":290.88 moles/1.25 moles = 1.5 O":16:1:"Our ratio is now 1 : 1.5. Is this inlowest whole number terms?":24:1:E40& 19:14:500:A"":600:ZA%İ290:2360&$ 21:1:"1.5 is not a whole number. Answer: NO":ZK%2330&. 250:2340S'8 N:BK163: Divide by smaller number":5:0,12279,12:3:6:3:"1.25 moles Fe : 1.88 moles O"% 1:6:"A good way to begin to reduce thisratio to whole numbers is to divideeach number by the smaller quantity.":290t& 12:6:"1.25 moles/1.25 moles = 1 Fe1 that theratio of moles is: 1.25 moles Fe and 1.88 moles O"$14:"Since chemical formulas use numberslike 1,2,3 and not numbers such as 1.25or 1.88, we need to convert the ratio 1.25 : 1.88 to a whole number ratio.":290V% N:BK15:"STEP les"B#N0ZC%1ĺ"Divide grams by atomic weight to getmoles."#N0ZC%1ĺ"30 grams O x(1 mole O/16.0 g O) =1.87 moles. Try: 1.87 moles"#ZK%2220#250:2240B$N:BK14:4:"We have found that for our compoundwhich is 70% Fe and 30% Oof O? (Atomic weight of O is 16.0)":420:MX1.88:MN1.869p"12:6:500:ZK%17įR:6:12:E20:5000:R:2220"360:A"":590:ZA%Ģ10:1:"Right, press RETURN to continue.":300:2290#1:8:N1ĺ"Please include the unit: mo weight to getmoles."m!zN0ZC%1ĺ"70 grams Fe x (1 mole Fe/55.8 g Fe) = 1.25 moles. Type: 1.25 moles"{!ZK%2130!250:2150!N:BK13:"Step 2: Convert grams to moles":5:0,10279,10:3:"3:2:"How many moles of O are there in 30 g1.2499= R12:6:500:ZK%17įR:6:12:E20:5000:R:2130 \360:A"":590:ZA%Ģ10:1:"Right, press RETURN to continue step 2.":300:2200 f1:8:N1ĺ"Please include the unit: moles"!pN0ZC%1ĺ"Divide grams by atomiccontain 70 grams Fe and 30 grams O.Next we need to convert grams to moles.":290>N:BK12:"Step 2: Convert grams to moles":5:0,10279,10:3 H3:1:"How many moles of Fe are there in 70 gof Fe? (Atomic weight of Fe is 55.8)":420:MX1.256:MN1:MN69.9999= 14:10:500:360:A"":590:ZA%21001:14:N0ĺ"70% Fe in a compound means that thereare 70 g of iron in 100 g of compound.Type: 70 g" 430:ZK%2060*250:2070R414:1:"Right! 100 grams of our compound would calculate thenumber of grams of each element.":290N:BK11:"STEP 1: Calculate the number of gramsof each element in 100 g of compound.":5:0,22279,22:3 6:1:"How many grams of Fe are there in 100 gof a compound which is 70% Fe?":MX70.0he empirical formulafor a compound which is 70.0% iron and30.0% oxygen."8:"To do this, we must determine the ratioof the number of moles of each elementin the compound."514:"It is convenient to start with 100 gramsof the compound and theny: HO"(ZK%1290#2250:1300/<P:100N:BK9:7:"Chemical analysis can tell you thepercent composition of a compound.Let's see how to calculate the empiricalformula from the percent composition.":290JN:BK10:2:"Let's calculate troxide is H@d2O@d2.What is the empirical formula?"w 14:12:ZF$"@H":500:A"":600:ZA%İ290:D(1)1:1001:16:ZC%1ĺ"The ratio of atoms in H@D2O@D2 is 1:1"ZC%1ĺ"Since the ratio of atoms is 1:1 theempirical formula is HO. Tr empirical formula.WHC14:VT12:EM0:6020:A"":6200:ZA%İ290:128016:1:"The ratio of atoms is 1:2 so theempirical formula must be: CH@D2.":ZK%1250250:PS1:12604N:BK8:4:2:"The molecular formula for hydrogenpe40=16:1:ZC%1ĺ"The ratio of atoms in C@d6H@d6 is 1:1"ZC%1ĺ"Since the ratio of atoms is 1:1 theempirical formula is CH. Try: CH"ZK%1190250:1200N:BK7:6:"The molecular formula for cyclohexaneis C@D6H@D1@D2. Type thecal formula just tells usthe ratio of the atoms.":290N:BK6:13:"@2@FC@D6H@D6@1@S":6:1:"The molecular formula for benzene isC@D6H@D6. What is the empirical formula?":5:0,25260,25:312:11:ZF$"@H":500:A"":600:ZA%İ290:125:2:10:"@2@FGLUCOSE@1@S":6:1:" MOLECULAR FORMULA: C@D6H@D1@D2O@D6":8:" EMPIRICAL FORMULA: CH@D2O"U$12:"The molecular formula tells us exactlyhow many moles of each element arepresent in a mole of the compound."718:"The empiriH2O1":ZA%EM1=BA"":6200:ZA%1140L19:1:EM0ĺ"The ratio of atoms is 1:2:1 so theempirical formula must be: CH@D2O."VEM1ĺ"The subscript 1 should not be written."`ZK%1080j250:PS1:1100t290s~N:BKa tells only theratio of the number of atoms of theelements in a compound.":3:1:U$.11:1:"The molecular formula for glucose isC@D6H@D1@D2O@D6. Type its empirical formula.(Remember the ratio of atoms is 1:2:1)"8HC14:VT16:EM0:6020:A"C1otice the subscripts 6 12 6 havethe ratio 1 : 2 : 1.":290N:U$:6:6:"Formula:":5:18:"@2@FC@D6H@D1@D2O@D6@1@S":1:11:"Ratio of Atoms:":10:20:"@2@F1 2 1@1@S":290W$N:BK4:5:"@2@FEMPIRICAL FORMULA@1@S":1:5:"The EMPIRICAL formul formula indicates 6 carbons in eachglucose molecule. So, 1 mole of glucosemolecules would contain 6 moles ofcarbon atoms. Type: 6":ZK%1020250:1030>11:1:"Right, 1 mole of glucose contains 6moles of C, 12 moles of H, and 6 molesof O.NN:BK3:2:2:"Glucose has the formula C@D6H@D1@D2O@D6.How many moles of carbon are in 1mole of glucose?":MX6:MN68:14:500:A"":ZA%İ600:1050360:A"":590:ZA%105014:1:"TheM:ZK%0:ZA%L20:I55455:T1,I:T2,L:CL::7bN:BK2:6:"The molecular formula tells you howmany moles of each element there arein 1 mole of the compound.Let's see how the molecular formuladiffers from the empirical formula.":290vPress CTRL-Q for a calculator.":/N1İ410=N2İ400CUZL%1ZL%20xSS0:I:NO$NN$:ZK%27ġ:100ZK%17ıZK%2ġ:350ZK%1ZC%ZC%1:SS1:NO$"":530A"":ZA%1ZL%1505ZK%0:NN1ZA%0:ZW%11Xr:ZA$C$:A(A$):AMXAMNZA%1:ZW%0:N1(SF3):J|ZA%0:ZW%1:"You have too many numbers after thedecimal point.":"Your number is right, but please includethe units of grams in your answer.":!2:1,99276,99:3:Y104:X30:"1240,100,2000,2010,2040,2110,2200,2290,2310,2360,2380,100,3000,3030,3060,3150_ hZA$""N0: jA$"":C$"":N0:SF0:DM0:I1(ZA$):B$(ZA$,I,1):(B$)45(B$)58A$A$B$:C$C$" ":SFSF(DM1) m(B$)57(B$)46C$C$B$ n(B$)46DM18 SS0āI7510025:T1,I:T2,30:CL:9 T:ZK%260L ZK%27ġ:100^ ZK%2ġ:350d  "G:24:4:"Press @X2@C6RETURN@ to continue.":ZK%0:ZS0:1 ,G:P:ZK%27ġ:100 6ZK%2ġ:350 @M ^BK200,100,1000,1010,1060,1150,1180, to return to this index."A I13:D(I)1Ģ112(I1):3:"*"G v WT((100(1)))1001:NH0:T(0)0:W(0)WT 300:AZK%48:A1A4150 A1000,2000,3000,200 N:8:1:"@G@f@3@c1Returning tothe Main Index.":(13);(4);"RUN ROUTER"(2@FEmpirical Formulas@1@S":5:1:U$:8:3:"@X2@F@0@C6Select topic by number.":G:NO$NN$ n11:5:"1. Empirical Formulas@R@R2. Calculating Empirical Formulas@R@R3. Molecular Formulas@R" x"4. Return to the Main Index":22:3:"Press @X2@C6ESC@Y$"" %N$"" (GM$"" 2OK$"@X2@C1OK@":NO$"@X2@C5NO@":NN$NO$Z dN:BK1:1:14:"INDEX":3:5:"@6% : EMPIRICAL FORMULAS 1/11/83M: COPYRIGHT 1983 SMITH,CHABAY,KEANX36950b100 T1768:T2769:CL770:CK16336:U$""D(5),P(5),T(10),W(10)V$""M #                    XI[------------??????????g)-------- $$$$$$$$$$$566666666666%$$$$$$$$$$$566666666666 $$$$$$$$$$$$ed 6$  66%??MI$$$$$$$$$$$$?7666666.66666%$$$$$$$$$$$766666&<< $  I"- $,(-6$%66v---- dd$$$$$??????7------->??????7------->??????w------>?????w----->????w---- ???-------------------%----------------------????? X0:Y130:E,7:E2,7:Y100:E40,3:Y120:E40:E,8:Y104:X30:"@PPress CTRL-Q for a calculator."v>100|> 00> O$=Y107ĺ" - Bad Subscript":5814L=Y224ĺ"-Undefined Function":5814V=5030= X0:Y130:E1,7:Y100:E40,3:Y120:E40:Y104:X30:"Press CTRL-Q for a calculator."=100= Tscript":5814=Y224ĺ"-Undefined Function":5814>10J"x10@+"J"@D":5030H<EX.099JJ2:EX10J"x10@+"J"@D":5030s<912:X10:Y(222):"Cannot evaluate";<Y69ĺ" - TOO LARGE":5814<Y16ĺ" - Syntax Error":5814<Y133ĺ"-Division by zero":5814<Y191ĺ" - Too complex":5814C;L142EX$(EX$,(EX$)1):L94:L$(94):5300: "**" TO "^"W;L1L:EX$EX$L$p;PP1:P(ZA$)5500z;5070;|;:794:J1(((((EX))2.3)2)):EX((EX10J.5)):EXEX10J:"= ";;EX1000EX.099ĺEX:5030<EX1000JJ2:EX$" "L:L(L$):L62L94L69:"Use numbers and ()+*-/^ only.":5030`:L37L$"*.01"n:P15300:L405100: CHECK FOR "(":L141EX$EX$"*":5300:L147L158EX$EX$"*":5300: NO IMPLIED MULT;L425300: "*"Y1!9TBTB22:TB180TBY122\9YTB:X0:E39:YTB4:YY:X0:E39:YTBLF:YY:X0:E399R:X0:YTB:ZL%30:I:1:ZK%259009ZK%271009ZA$""50409TYTBLF:YTY9EX$"":C1:P19L$(ZA$,P,C):L$"x"L$"*":5300:L$"="L 100 ="WT" %"8p ZK%3410+8z 250:3430@8CA1:CC1:58008Y104:X3:E38:"Type a numerical expression and press RETURN. Press CTRL-B when finished."8X1:Y101:5:X,Y1X3971,Y1X3971,Y2113X,Y2113X,Y1:39Y1108:LF11:TB"What is the weight percent "EL$" in "M$"Atomic weights: "AW$:420z7R 7:12:500:ZK%17įR:7:12:E30:5000:R:34107\ MXWT.1M:MNWT.1M:360:A"":590:ZA%Ģ12:30408f 9:1:N0ĺ"("EW" g "EL$"/ "MW" g "M$") x,Ag,silver>":v64 M$"NaClO@D4":WT18.8:EL$"Na":MW122.5:EW23.0:AW$"Na=23.0 Cl=35.5 O=16.0":V$"":6> M$"K@D2CO@D3":WT56.6:EL$"K":MW138.2:EW78.2:AW$"K=39.1 C=12.0 O=16.0":V$"":D7H N:14:"@2@F"M$"@1@S":4:1:GM$>":590:ZA%Ģ12:3040Z5 9:1:N0ĺWT" g x (100 g cpd/"PC" grams "M$")= "M" grams"m5 430:ZK%3310|5 250:33305 PP(3(1)1):PP3370,3380,3390:34006* M$"AgNO@D3":WT63.5:EL$"Ag":MW169.9:EW107.9:AW$"Ag=107.9 N=14 O=16.0":V$"":590:ZA%Ģ12:3040r3 9:1:N0İ800:M" grams "M$"x("PC" g "M$"/100 g compound)= "(ZZWT)ZZ" grams"3 430:ZK%32103 250:32303 PP(3(1)1):PP3270,3280,3290:33003 M$"P":PC22:3 M$"S":PC22.6:3 M$"Ni":PC45.3:2:V$"":2 N:M(5(1)2):"How many grams of "M$" are containedin "M" grams of a compound which is"PC" % "M$"?":420:WTMPC1002 6:12:500:ZK%17įR:6:12:E30:5000:R:32103 MXWT.1WT:MNWT.1WT:360:A"":1l M$"Cu":PC28.4:V$"":2v M$"Ag":PC75.tomic Wts: "AW$:420J0 7:12:500:ZK%17įR:7:12:E30:5000:R:31000& MXWT.1WT:MNWT.1WT:360:A"":590:ZA%Ģ12:304010 9:1:N0ZC%1ĺ"Divide the gram atomic weight of "EL$"by the gram formula weight 27.1:EL$"Na":MW85:EW23.0:AW$"Na=23 N=14 O=16":V$"":/ M$"Na@D2CO@D3":WT11.3:EL$"C":MW83:EW12.0:AW$"Na=23 C=12 O=16":V$"":0 N:17(M$)2:"@2@F"M$"@1@S":4:1:"What is the weight percent of "EL$" in"M$"? A050,3160,3260,3360^. 1:MPNP(NT1)" right";:27:NP(NT1)" to go":24:1:E40:290:3010. PP(3(1)1):PP3060,3070,3080:3090. M$"ZnCl@D2":WT47.9:EL$"Zn":MW136.4:EW65.4:AW$"Zn = 65.4 Cl = 35.5":V$"":H/ M$"NaNO@D3":WTnext section you can practice:":10:6:"finding weight percentsusing weight percents asconversion factors":290:D(2)1:100- MP4:BK18:I1MP:P(I)I::NT0:NPMP- NT1P(RN)P(NP):NPNP1:NP0D(3)1:100- RN(NP(1)1):NT0. P(RN)3 soyou need conversion factor B. Type: B":ZK%2500C, 250:2510, 19:14:E20:20:4:"50 g C x":19:13:"100 g CO@D2":21:"27.3 g C":93,156164,156:20:23:"= 183 g CO@D2", 5:45,16469,147:131,173155,158:3:290- N:BK17:6:"In the grams of CO@D2?":13:6:"A"+ 9:14:"27.3 g C":16:"100 g CO@D2":Y116:65,Y126,Y:21:13:"B":24:14:"100 g CO@D2":16:"27.3 g C":177,Y245,Y+ 15:19:ZF$"@H":500:A"B":600:ZA%25304, 21:1:"You want to end up with grams of CO@D2cel leaving grams O.":300* N:BK16:"CO@D2 is 27.3% carbon by weight. Let'scalculate the number of grams of CO@D2needed to supply 50 grams of carbon."+ 7:"Which conversion factor, A or B, shouldwe multiply times 50 grams to give thenumber of2O")t ZC%1ĺ"TRY: A"&)~ ZK%24005) 250:2410) 290:11:1:E40,3:14:7:E29,8:14:7:E25:6:16:"88.9 grams O":8:"100 grams H@D2O":110,52200,52:14:12:"= 44.5 grams O":* 5:21,6470,40:120,72182,53:3:16:1:"The grams H@D2O can9 g O":18:"100 g H@D2O":20:15:"B":22:16:"100 g H@D2O":18:" 88.9 g O"q(V Y131:66,Y130,Y:164,Y230,Y(` 20:14:ZF$"@H":500:A"A":600:ZA%2450)j 22:1:ZC%1ĺ"You need g H@D2O in the denominator so itwill cancel the units on 50 g H@Dfactor.":290'B N:BK15:3:"Let's calculate how many grams ofO there are in 50 grams of water.":7:1:"50 grams H@D2O x (CONVERSION FACTOR)":11:1:"Which conversion factor, A or B, shouldwe use?'G 2:1,25278,25:3N(L 7:15:"A":16:9:" 88.360:A"":590:ZA%2360x& 12:1:N0ĺ"88.9% means there are 88.9 g foreach 100 g of water. Try: 88.9 g"&$ 430:ZK%2320&. 250:2330 '8 13:1:"Right, we can use the relationship: 88.9 g O per 100 g H@D2Oas a conversion ound to besure they add up to 100%.":290%N:BK12:15:"@2@FH@D2O@1@S":4:6:"H = 11.1 % O = 88.9 %":2:1,35278,35:3% 6:2:"Water is 88.9% O by weight. How manygrams of O are there in 100 grams ofwater?":MX89:MN88.8!& 15:10:500:"x100 = 88.9 %":107,140145,140:290$N:BK13:14:"@c6@4@FH@D2O":G:7:12:"@:H = 11.1 %O = 88.9 %":120,71169,71:11:15:"100.0 %"*%15:1:"@PTo check the accuracy of our work wecan sum the weight percents of all ofthe elements in the compthere must be 16.0 grams of oxygenper mole. Try: 16 grams"O#430:ZK%2220^#250:2230#11:1:"The gram-molecular weight of H@D2O is 18grams. Therefore the weight % oxygenis:"%$18:4:"Weight % O =":17:15:"16 g":19:"18 g":18:20:"2:1,25273,25:3V"5:1:"How many grams of O are there in onemole of H@D2O?""13:9:MX16.02:MN15.99:500:360:A"":590:ZA%2260<#12:1:N0ĺ"There is 1 oxygen atom per mole of H@D2OSince the atomic weight of oxygen is 1.000 = 11.1 %":107,140145,140:290!N:BK11:9:3:"We calculated that water is 11.1%H by weight. Let's calculate theweight percent O in water.":290!N:BK12:"Calculation of the weight percent O inH@D2O. Atomic weights: H = 1.0 O = 16.0" of H@D2O?":R:MX18.02:MN18Q \14:13:500:360:A"":590:ZA%2180 f16:1:N0ĺ"2x1.0 + 16.0 = 18.0 grams. Try: 18 grams" p430:ZK%2140 z250:2150"!18:1:"3.":18:4:"Weight % H =":17:15:" 2 g":19:"18 g":18:20:"x1"There are 2 moles of hydrogen atomsper mole of H@D2O. Since the atomicweight of hydrogen is 1.0 there mustbe 2x1.0 or 2 grams of hydrogen permole. Try: 2 grams">430:ZK%2090H250:2100 R11:1:"2. What is the gram-molecular weight 's calculate the weight percent H inH@D2O. Atomic weights: H = 1.0 O = 16.0":2:1,25273,25:3 5:1:"1. How many grams of H are there in one mole of H@D2O?"*13:9:MX2.02:MN2.0:500:360:A"":590:ZA%2130412:1:N0ĺ:"44.0g":5:171,110220,110:3:30017:1:"3. Divide the grams of C by the gram-molecular weight and multiply by 100:" 21:1:"Weight % C =":20:13:" 12 g C":22:"44 g CO@D2":90,163150,163:21:21:"x 100 = 27.3%":300cN:BK10:"Letweights: C = 12.0 O = 16.0)":5:0,34279,34:36:1:"1. Find the weight of C in a mole of CO@D2":8:6:"1x12.0 g C/mole = 12.0 grams":290%10:1:"2. Find the total weight of 1 mole of CO@D2@:":12:14:"1x12.0 = 12.0g2x16.0 = 32.0g";:23ibutes to the totalweight of the compound.":16:"Let's see how to calculate the per-cent composition - the percent ofthe total weight contributed byeach element.":2901N:BK9:"Here is how to calculate the weightpercent carbon in CO@D2.(Atomic percent ofan element in a compound from thechemical formula.":290:D(1)1:100^300:100N:BK8:"@2@F PERCENT COMPOSITION@1@S "U$:2:7:"The chemical formula tells us thenumber of each type of atom in acompound."12:"Each atom contr40250:125014:1:"Right, so the weight percent oxygen inKClO@D3 is:":18:3:" "M"@grams O":20:WT" grams KClO@D3":19:18:"x 100 = 39.1%":16,148115,148:290PN:BK7:8:2:"In the next section we will see howto calculate the weight htis "M" grams.":7:1:"What is the weight of oxygen in "WT"grams of KClO@D3?":MXM.02:MNM.0211:14:500:360:A"":590:ZA%128013:1:N0ĺ"The loss in weight equals the weight ofthe oxygen. Try: "M" grams."430:ZK%12y again.":300:NH0:1020.T20NH71230j"@x2@c5PRESS@":9:"1 To continue heating2 to go on300:AZK%48:A1A21210A1050,1230cN:BK6:MWTW(NH):"Heating "WT" grams of KClO@D3 yields"W(NH)" grams of KCl. The loss in weig:XX:T(I):YY10::Y78:I0NH:X1807(W(I)1):XX:YY:W(I):YY10:20:4:T20NH7:"Let's heat some more to be sureall of the oxygen is gone.":290:1050NH7T20Ė1:22:"You should heat longer betweenweighings. Press RETURN to trW2)100:NHNH1:T(NH)T:W(NH)W2t2:X90:Y25:X,YX100,YX100,Y30X,Y30X,Y:X110:Y16:X,YX10,Y5X50,Y5X60,Y:140,25140,22:3:8131,9~15:6:""W2" grams":9:4:"Time, Minutes Weight, Grams"HY78:I0NH:YY:X847(T(I)10)24:1:E40:1100B1070L20:4:E35,3:3:"Let's let the crucible cool a fewminutes.":0:7132,80:10131,64:3:ZS3:PZS:290`N:BK5:T20W2.60848WT:1135eT0W2WT:1140jW2WT(T20)(WT.6085WT):W2(100W2)100 oW2(100,445:10131,64:3u$16:4:"Heating time in minutes:":7:20:"Press RETURN when youwant to stop heating.".28:16:E6:28:16:T:ZS.4:PZS:ZK%1311003ZK%271008TT1:T30Ģ20:1:E40,3:4:"You have heated long enough.":290:otassium chlorate havebeen put in this crucible.":8119,72:14:4:"Next we will heat the crucible todrive oxygen out of the salt.":290:T0N:BK4:8126,32:7132,80:2:168,95168,97112,97112,95168,95:160,95160,16:162,95162,16:118,44164, it loses oxygen. Let's weighsome KClO@D3, heat it until all of theoxygen has been lost, and then re-weigh the sample."15:"From the loss in weight we willcalculate the weight percent oxygenin KClO@D3.":290N:BK3:4:1:""WT" Grams of pL20:I55455:T1,I:T2,L:CL::+b9N1İ410vN2ĺ"Your number is right, but the units arepercent."| WT10ZZ100:*WT100ZZ10:4WT100ZZ1:>ZZ10:|N:BK2:5:"When potassium chlorate, KClO@D3, isheatedalculator.":N1İ410)N2İ400/AZL%1ZL%20dNS0:I:NO$NN$:ZK%27ġ:100rZK%17ıZK%2ġ:350ZK%1ZC%ZC%1:NS1:NO$"":530A"":ZA%1ZL%1505ZK%0:NTNT1:NN1ZA%0:ZW%1%XM:ZK%0:ZA%AMXAMNZA%1:ZW%0:N1(SF2):7|ZA%0:ZW%1:r"You have to many numbers after thedecimal point.":"Your number is right, but please includethe units of grams in your answer.": 2:1,99276,99:3:Y104:X30:"Press CTRL-Q for a c,1230,100,2000,2020,2070,2190,2200,2280,2300,2370,2470,100L hZA$""N0: jA$"":C$"":N0:SF0:DM0:I1(ZA$):B$(ZA$,I,1):(B$)45(B$)58A$A$B$:C$C$" ":SFSF(DM1) m(B$)57(B$)46C$C$B$ n(B$)46DM1%r:ZA$C$:A(A$):R", NS0āI7510025:T1,I:T2,30:CL:= T:ZK%260P ZK%27ġ:100b ZK%2ġ:350h  "G:24:4:"Press @X2@C6RETURN@ to continue.@":ZK%0:ZS0:1 ,G:P:ZK%27ġ:100 6ZK%2ġ:350 @: ^BK200,100,1000,1020,1050,1120ESC@ to return to this index."E I13:D(I)1ĢY2(I1):2:"*"K z WT((100(1)))1001:NH0:T(0)0:W(0)WT 300:AZK%48:A0A4150 A1000,2000,3000,200 N:8:"@3@f@G@c1Returning tothe Main Index.":G:(13);(4);"RUN ROUTEMPOSITION@1@S":5:1:U$:8:3:"@f@x2@c6Select topic by number.":G:NO$NN$ nY11:Y:4:"@:1.@P KClO@D3 Experiment@R@R2. Calculating Percent Composition@R@R3. Percent Composition Problems@R" x"4. Return to the Main Index":22:3:"Press @X2@C6N,NO,NOT,ISN'T,CAN'T>"} %N$"" (GM$"" 2OK$"@X2@C1OK@":NO$"@X2@C5NO@":NN$NO$L dN:BK1:1:15:"INDEX":3:5:"@2@FPERCENT CO5$ :WT PERCENT 1/13/83 11/83M:COPYRIGHT SMITH, CHABAY, KEAN 1983X36950b100 T1768:T2769:CL770:U$""D(5),P(5),T(10),W(10)V$""C #Y$" ND$(13)(4):ST34474j O232,150:233,140:0:1:793,A:ST35990 PA1000,200,300,400,500,600,700 D$"BLOAD FORMULAS SHAPES,A"ST:D$;"RUN WT PERCENT":P ,D$"BLOAD FORMULAS SHAPES,A"ST:D$;"RUN EMPIRICAL ber.":Ga 28:13:"1. Introduction@R@R2. Percent Composition@R@R3. Empirical Formulas@R@R"; 7"4. Mg-HCl Experiment@R@R5. Empirical Formula Problems 8(793)0Ė6:112(793):(29) <P:AZK%48:A1A560 =A11000# KN:2:6:"  : ROUTER 3/16/83= : (C) SMITH CHABAY KEAN 1983H 37000Q20N:15:2:"INDEX":4:5:"@2@FPercent Composition Empirical Formulas":GB0,0,278,20:6:B0,0,278,191,4:B4,18,274,20:G (3:11:"@F@X2@C2@0Choose topic by numuEX$EXE J#$ ߠȱȱʈ lթ gݩ! ߪ + ?#$` D0>3>33300>333>3?633>0333300033373cwcc333333333>33>00>063333333 cck63 333? ?&30 33?33333333333333>;3>333?33300003333?cwccc337?;333333333>ccc{3n333303? 333333ccccc6ccckwc33 33333 ?0 ?  3336666Cc0 fc   c66c ?  ?0 337;330 ?00008<6300?0033?0 333333>0    CH``cc~<>cc``c#> ??c66c08<000UUUUUUUU8<>?><8 ? ? ? 88 ?      X1:VT1:CX:E2:VT:CX:E2:VT:CX:6050@=C27S32:6050U=C64A1$(CS)p=C48C58ĺ(15);"D";=A1$;:A$A$A1$:S0:CXCX1:(A$)20ī6050==8:ZK%0:ZA%L20:I55455:T1,I:T2,L:CL::VT:CX2:OK$:=BVT:CX2:NO$:ZK%:C13ZA$A$:61609<ZK%1SS1:NO$"":ZC%ZC%1:K<ZK%2ġ:350^<ZK%27ġ:100m<C127C8<PS0:C8(A$)06050<C216040<C1NTNT1:ZC%ZC%1:ZA$A$:HP1:NO$"":6160<C8(A$)16040+=C8A$(A$,(A$)1):CXCor a calculator."; 100!;-;"@H@$"H;PS1ĢVT:CX2:E2:Ky;PS1ZK%13CZK%:A1$(ZK%):VT:CX:6070;A$"":S0:HTHC:HT:VT1:E20:HT:VT:E20:HT:VT:(2):HTHT2:CXHT:HT:VT<"@:";:ES0:HP0:SS0:P:NO$NN$:A1$(ZK%):C - Syntax Error":5814<:Y133ĺ"-Division by zero":5814_:Y191ĺ" - Too complex":5814:Y107ĺ" - Bad Subscript":5814:Y224ĺ" - Undefined Function":5814:5030; X0:Y128:E1,8:Y100:E40,3:Y120:E40:Y104:X30:"Press CTRL-Q f((EX10J.5)):EXEX10J:X4:"= ";F9EX1000EX.099ĺEX:5030w9EX1000JJ2:EX10J"x10@+"J"@D":50309EX.099JJ2:EX10J"x10@+"J"@D":50309912:X10:Y(222):"Cannot evaluate";9Y69ĺ" - TOO LARGE":5814:Y16ĺ"RGE":5814C8L147L158EX$EX$"*":5300: NO IMPLIED MULT^8L425300: "*"8L142EX$(EX$,(EX$)1):L94:L$(94):5300: "**" TO "^"8L1L:EX$EX$L$8PP1:P(ZA$)550085070'9|794:J1(((((EX))2.3)2)):EX50407TYTBLF:YTY*7EX$"":C1:P1Q7L$(ZA$,P,C):L$"x"L$"*":5300d7L$"="L$" "7L(L$):L62L94L69:"Use numbers and ()+*-/^ only.":50307P153007L405100: CHECK FOR "(" 8L141EX$EXY69ĺ" - TOO LAK6X1:Y101:5:X,Y1X3971,Y1X3971,Y2113X,Y2113X,Y1:3b6Y1107:LF11:TBY16TBTB22:TB180TBY1226YTB:X0:E39:YTB4:YY:X0:E39:YTBLF:YY:X0:E396R:X0:YTB:ZL%30:I:1:ZK%259006ZK%271007ZA$""by@: @Pthe empirical formula weight3. Multiply subscripts in the empirical@: @Pformula by the ratio from 2."5 290:D(3)1:1005CA1:CC1:58006Y104:X3:E38:"Type a numerical expression and press RETURN. Press CTRL-B when finished." formula is C@D3H@D6.":2904 N:BK23:12:"@F@C5@2SUMMARY":G:4:6:"Finding Molecular Formulas from Empirical Formulas":5:1,50278,50278,511,51:3u5 9:1:"1. Find the weight of the empirical@: @Pformula2. Divide the molecular weight 00:A"":600:ZA%32603 19:1:"The molecular weight is 3 times theempirical formula weight so there mustbe 3 of them. Type: 3":ZK%32303 250:32404 19:1:"Right, the molecular formula has 3 CH@D2units so the molecularght ratio.":5:0,23279,23:32 5:1:"We have found:":7:6:"Empirical formula: CH@d2Empirical formula weight: 14Molecular weight: 42Ratio of weights: 42/14 = 32 14:1:"How many CH@D2 units are there in theactual molecule?"+3 17:14:5 RETURN to continue.":300:32001l 10:1:ZC%1ĺ"42/14 = 3 which means there are 3empirical formula units in the molecule.1q ZC%1ĺ"Try: 31t ZK%31701v 250:31802 N:BK20:"Step 3: Multiply subscripts of theempirical formula by the wei30X Y26:X1:"The molecular weight is 42 and theempirical formula weight is 14. How manyCH@D2 units are there in the compound?":4200b 9:12:500:ZK%17Ģ9:12:E20:R:5000:R:3170 1g A"":600:ZA%Ģ11:4:"Right! PressD/ 1:10:N0ZC%1ĺ"Sum the atomic weights of all of theatoms."t/ N0ZC%1ĺ"1x12.0 + 2x1.0 = 14. Try: 14"/& ZK%3070/0 250:30900N N:BK22:"STEP 2: Divide molecular weight by the empirical formula weight":5:0,22279,22:3k. 3:3:"What is the formula weight of CH@D2?Atomic weights: C=12.0 H=1.0":420:MX14.01:MN13.9999. 14:7:500:ZK%17įR:6:14:E20:5000:R:3070. 360:A"":590:ZA%Ģ10:3:"Right, press RETURN to continue.":300:3150MOLECULAR formulaof a compound which has the EMPIRICALformula CH@d2 and a molecular weight of42."- "We will start by calculating the weightof the empirical formula.":290. N:BK21:"1. Calculate weight of empirical formula":5:0,10279,10:ula tells usexactly how many moles of each elementthere are in a mole of the compound.", 16:"To convert an empirical formula intoa molecular formula, you must know themolecular weight of the compound.":290g- N:BK20:6:"@pLet's calculate the to obtain subscripts"}+ "4. If necessary, multiply subscripts@: @Pby a factor to get whole numbers.":290:D(2)1:100+ N:BK19:2:"The @X2@C5EMPIRICAL@ formula tells usthe ratio of the atoms in a compound.Y, 8:"The @X2@C5MOLECULAR@ formO@D3":290`* N:BK18:11:"@2@FSUMMARY@1@S":4:5:"Calculating Empirical Formulas":6:1:U$* 9:"1. Obtain the weights of all elements in the compound."+ "2. Convert weights to moles@R3. Divide all moles by the smallest@: @Pnumber "If we multiply both 1 and 1.5 by 2 we get2 and 3 which are whole numbers. Try: 2":ZK%2410l)~ 250:2420) 15:5:E35,3:20:16:E20 * 16:4:"2 x (1 mole Fe : 1.5 mole O) = 2 mole Fe : 3 mole O":20:1:"So, the empirical formula is: Fe@D2STEP 4: Multiply by factor":5:1,12279,12:3:3:6:" 1 mole Fe : 1.5 moles O(` 8:1:"What is the smallest number that whenmultiplied by 1 and 1.5 will change themto whole numbers?"(j 13:14:500:A"":600:ZA%2450])t 22:1:iply by factor":5:1,12279,12:3:3:6:" 1 mole Fe : 1.5 moles O'B 10:1:"1 is a whole number, but 1.5 is not.We need to multiply both 1 and 1.5 bysome number such that the product willbe whole numbers such as 2, 3, 4, etc.":290P(L N:BK17:"oles = 1.5 O":16:1:"Our ratio is now 1 : 1.5. Is this inlowest whole number terms?":24:1:E40& 19:14:500:A"":600:ZA%İ290:2360&$ 21:1:"1.5 is not a whole number. Answer: NO":ZK%2330&. 250:2340C'8 N:BK16:"STEP 4: Multaller number":5:0,12279,12:3:6:3:"1.25 moles Fe : 1.88 moles O"% 1:6:"A good way to begin to reduce thisratio to whole numbers is to divideeach number by the smaller quantity.":290d& 12:6:"1.25 moles/1.25 moles = 1 Fe1.88 moles/1.25 mof moles is: 1.25 moles Fe and 1.88 moles O"$14:"Since chemical formulas use numberslike 1,2,3 and not numbers such as 1.25or 1.88, we need to convert the ratio 1.25 : 1.88 to a whole number ratio.":290F% N:BK15:"STEP 3: Divide by sm%1ĺ"Divide grams by atomic weight to getmoles."#N0ZC%1ĺ"30 grams O x(1 mole O/16.0 g O) =1.87 moles. Try: 1.87 moles"#ZK%2220#250:22402$N:BK14:4:"We have found that for our compoundwhich is 70% Fe and 30% O that theratio eight of O is 16.0)":420:MX1.88:MN1.869`"12:6:500:ZK%17įR:6:12:E20:5000:R:2220"360:A"":590:ZA%Ģ10:1:"Right, press RETURN to continue.":300:2290"1:8:N1ĺ"Please include the unit: moles"2#N0ZColes."]!zN0ZC%1ĺ"70 grams Fe x (1 mole Fe/55.8 g Fe) = 1.25 moles. Type: 1.25 moles"k!ZK%2130z!250:2150!N:BK13:"Step 2: Convert grams to moles":5:0,10279,10:3*"3:2:"How many moles of O are there in 30 gof O? (Atomic w12:6:500:ZK%17įR:6:12:E20:5000:R:2130 \360:A"":590:ZA%Ģ10:1:"Right, press RETURN to continue step 2.":300:2200 f1:8:N1ĺ"Please include the unit: moles"!pN0ZC%1ĺ"Divide grams by atomic weight to getmrams Fe and 30 grams O.Next we need to convert grams to moles.":290>N:BK12:"Step 2: Convert grams to moles":5:0,10279,10:3H3:1:"How many moles of Fe are there in 70 gof Fe? (Atomic weight of Fe is 55.8)":420:MX1.256:MN1.24991 R1 14:10:500:360:A"":590:ZA%21001:14:N0ĺ"70% Fe in a compound means that thereare 70 g of iron in 100 g of compound.Type: 70 g" 430:ZK%2060*250:2070F414:1:"Right! 100 grams of our compound wouldcontain 70 ghenumber of grams of each element.":290N:BK11:"STEP 1: Calculate the number of gramsof each element in 100 g of compound.":5:0,22279,22:36:1:"How many grams of Fe are there in 100 gof a compound which is 70% Fe?":MX70.01:MN69.9999 formulafor a compound which is 70.0% iron and30.0% oxygen."8:"To do this, we must determine the ratioof the number of moles of each elementin the compound.")14:"It is convenient to start with 100 gramsof the compound and then calculate tZK%12902250:1300#<P:100N:BK9:7:"Chemical analysis can tell you thepercent composition of a compound.Let's see how to calculate the empiricalformula from the percent composition.":290>N:BK10:2:"Let's calculate the empiricald2O@d2.What is the empirical formula?"k 14:12:ZF$"@H":500:A"":600:ZA%İ290:D(1)1:1001:16:ZC%1ĺ"The ratio of atoms in H@D2O@D2 is 1:1"ZC%1ĺ"Since the ratio of atoms is 1:1 theempirical formula is HO. Try: HO"(l formula.KHC14:VT12:EM0:6020:A"":6200:ZA%İ290:128016:1:"The ratio of atoms is 1:2 so theempirical formula must be: CH@D2.":ZK%1250250:PS1:1260(N:BK8:4:2:"The molecular formula for hydrogenperoxide is H@6:1:ZC%1ĺ"The ratio of atoms in C@d6H@d6 is 1:1"ZC%1ĺ"Since the ratio of atoms is 1:1 theempirical formula is CH. Try: CH"ZK%1190250:1200 N:BK7:6:"The molecular formula for cyclohexaneis C@D6H@D1@D2. Type the empiricala just tells usthe ratio of the atoms.":290N:BK6:13:"@2@FC@D6H@D6@1@S":6:1:"The molecular formula for benzene isC@D6H@D6. What is the empirical formula?":5:0,25260,25:312:11:ZF$"@H":500:A"":600:ZA%İ290:124041:"@2@FGLUCOSE@1@S":6:1:" MOLECULAR FORMULA: C@D6H@D1@D2O@D6":8:" EMPIRICAL FORMULA: CH@D2O"U$12:"The molecular formula tells us exactlyhow many moles of each element arepresent in a mole of the compound.".18:"The empirical formuZA%EM14BA"":6200:ZA%1140L19:1:EM0ĺ"The ratio of atoms is 1:2:1 so theempirical formula must be: CH@D2O."VEM1ĺ"The subscript 1 should not be written."`ZK%1080j250:PS1:1100t290j~N:BK5:2:10s only theratio of the number of atoms of theelements in a compound.":3:1:U$.11:3:"The molecular formula for glucose isC@D6H@D1@D2O@D6. Type its empirical formula.(Remember the ratio of atoms is 1:2:1)" 8HC14:VT16:EM0:6020:A"C1H2O1":the subscripts 6 12 6 havethe ratio 1 : 2 : 1.":290N:U$:6:6:"Formula:":5:18:"@2@FC@D6H@D1@D2O@D6@1@S":1:11:"Ratio of Atoms:":10:20:"@2@F1 2 1@1@S":290Q$N:BK4:5:"@2@FEMPIRICAL FORMULA@1@S":3:5:"The EMPIRICAL formula tellla indicates 6 carbons in eachglucose molecule. So, 1 mole of glucosemolecules would contain 6 moles ofcarbon atoms. Type: 6":ZK%1020250:1030811:1:"Right, 1 mole of glucose contains 6moles of C, 12 moles of H, and 6 molesof O.Notice N:BK3:2:2:"Glucose has the formula C@D6H@D1@D2O@D6.How many moles of carbon are in 1mole of glucose?":MX6:MN68:14:500:A"":ZA%İ600:1050360:A"":590:ZA%105014:1:"The formuM:ZK%0:ZA%L20:I55455:T1,I:T2,L:CL::9bN:BK2:6:"The molecular formula tells you howmany moles of each element there arein 1 mole of the compound.Let's see how the molecular formuladiffers from the empirical formula.":290x"Press CTRL-Q for a calculator.":1N1İ410?N2İ400EWZL%1ZL%20zSS0:I:NO$NN$:ZK%27ġ:100ZK%17ıZK%2ġ:350ZK%1ZC%ZC%1:SS1:NO$"":530A"":ZA%1ZL%1505ZK%0:NN1ZA%0:ZW%13X:r:ZA$C$:A(A$):AMXAMNZA%1:ZW%0:N1(SF3):L|ZA%0:ZW%1:"You have too many numbers after thedecimal point.":"Your number is right, but please includethe units of grams in your answer.":#2:1,99276,99:3:Y104:X30:0,1240,100,2000,2010,2040,2110,2200,2290,2310,2360,2380,100,3000,3030,3060,3150a hZA$""N0: jA$"":C$"":N0:SF0:DM0:I1(ZA$):B$(ZA$,I,1):(B$)45(B$)58A$A$B$:C$C$" ":SFSF(DM1) m(B$)57(B$)46C$C$B$ n(B$)46DM1* SS0āI7510025:T1,I:T2,30:CL:; T:ZK%260N ZK%27ġ:100` ZK%2ġ:350f  "G:24:4:"Press @X2@C6RETURN@ to continue.":ZK%0:ZS0:1 ,G:P:ZK%27ġ:100 6ZK%2ġ:350 @O ^BK200,100,1000,1010,1060,1150,118 to return to this index."A I13:D(I)1Ģ112(I1):3:"*"G v WT((100(1)))1001:NH0:T(0)0:W(0)WT 300:AZK%48:A1A4150 A1000,2000,3000,200 N:8:3:"@F@C1@3@0Returning tothe Main Index...":(13);(4);"RUN ROUTER"2@FEmpirical Formulas@1@S":5:1:U$:8:3:"@X2@F@0@C6Select topic by number.":G:NO$NN$ n11:5:"1. Empirical Formulas@R@R2. Calculating Empirical Formulas@R@R3. Molecular Formulas@R" x"4. Return to the Main Index":22:3:"Press @X2@C6ESC@Y$"" %N$"" (GM$"" 2OK$"@X2@C1OK@":NO$"@X2@C5NO@":NN$NO$Z dN:BK1:1:14:"INDEX":3:5:"@6% : EMPIRICAL FORMULAS 1/11/83M: COPYRIGHT 1983 SMITH,CHABAY,KEANX36950b100 T1768:T2769:CL770:CK16336:U$""D(5),P(5),T(10),W(10)V$""M #               d7CT|`_\@[]  ~   ~  L ?EE(轄x -v8塅&J 8 i轄x%I sw &%E&'i') ȥ8娅` yzz`y䍱zzy`, Oz,̠BX gyx` `zi`|z``8򅛥8ҥˆ yz0 𳦍J xJ J H pnhzĥ` g  _NO tgH  8 h pn MzIz`zH pnhz` pgi rgŜ$H z h pnƜ`zzz` 4z y 莩z wL MzHzH 8hzhze`yЎ``8`Fz`z0a m z0 'mH)JJJiJhh΄`zOzHHƥz ɀ쭧zH Kz ~hz  h ~ hzzzz` K  ~ Kzz٬zzzьz0ГɀНz  ~L{ĝ񝐁zz pnz枦LV|ĝ͹ ,ɩ  } }`NO`0"a0 b0z4z/z0*z,)a {~ { } ,~z, g;   g gL(} ~ΥL$} |ş | {z y  H 4h` 3~IRz =x v x w0Lp ȱA` ) y w` Ln.A& $< >,.?: ; ` 8```m`m``e` H!) 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