U04L04-PercentCompandEmpiricalandMolecula:u04L04分比较和实证和分子

U04L04-PercentCompandEmpiricalandMolecula:u04L04分比较和实证和分子

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时间:2019-08-23

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1、%CompositionandEmpiricalandMolecularFormulasMr.ShieldsRegentsChemistryU04L041MolarCompositionWhenweseeachemicalformulaweimmediatelyknowhowManyandwhatkindofatomsareinthecompound.Butnumberofatomscanalsorepresent#ofmolesofatomsInthecompound.Forexample…InonemoleofN

2、aHCO3Ihave1molofNa,1molofH,1molofCand3molesofoxygenatomsAndin0.25molofNaHCO3IcancalculatethatIhave0.25molofNa,0.25molofH,0.25molofCand0.75molofOxygenatoms2BacktoHydratesNowtrythisproblem…Howmanymolesofwateraretherein0.67molofCoCl2.6H2OThinkofitthisway…IfIhad1mo

3、lofCoCl2.6H2OhowmanymolofwatermoleculeswouldIhaveifallthewaterwasdrivenoff?Now,IfIonlyhad0.67moleofCoCl2.6H2OhowmanymolofwaterwouldIhave?CoCl2.6H2OHEATCoCl2H20H20H20H20H20H20+0.67x6=4.02mol3Problem:Howmanymolesofwaterarein3molesofCaO∙8H20?Problem:Twomolesofwate

4、rweredrivenoffwhenasampleofCaO∙8H20washeatedleavingtheanhydroussaltbehind.Howmanymolesofhydratewereoriginallypresent?BacktoHydrates–molesofWaterIFIhad1moleofhydrateIwouldhave8molesofwaterbutIonlygot2molesofwatersoIonlyhad2/8=0.25moles3x8=24moles4Molesofatomsinh

5、ydratesHere’sanotherprobleminvolvingHydrates…A25gramsampleofahydrateisheateduntilallthewaterisdrivenoff.Themassoftheanhydrousproductremainingwas14grams.Whatisthe%waterinthehydrate?Ans:25gofhydrate–14gramsofremaininganhydrousProduct=11gramsofwater11gH20/25gHydra

6、te=44%waterWhenacompoundhasNOwaterofhydrationitissaidtobeAnhydrousandiscalledanAnhydride.5HydratesProblem:ifa15gsampleofahydrateisheateduntilallthewaterofhydrationisdrivenoff.Themassoftheanhydrousproductremainingis9.5grams.Whatisthepercentofwaterinthehydrate?An

7、swer:15g-9.5g=massofthewater=5.5g%water=(5.5g/15g)x100=36.67%6EmpiricalformulaDEFINITION:AchemicalformulathatstatesthesimplestwholenumberratiosofatomsinacompoundiscalledtheEMPIRICALFORMULAofthecompoundEmpiricalformula’sARENOTNECESSARILYthesameasthemolecularform

8、ula,buttheyCanbeEx.C8H16CH27EmpiricalformulaEx.C6H6(Benzene)isNOTanempiricalformulait’sthetrueMolecularFormula.Why?BUTConsiderMethane(CH4)It’smolecularformulaisalsoit’sempi

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