Free Essay

Lightergas Rapport

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Submitted By litleblondie
Words 711
Pages 3
Kemirapport for
Lightergas forsøg

Lavet af Camilla Næblerød
Kemi HF
2014
Samarbejde med Tine

Formål
Formålet er at bestemme lightergas molarmasse og at foreslå en molekyleformel for gassen.
Teori
Ved at opsamle en portion gas i et omvendt måleglas anbragt under vand, kan man bestemme gassens volumen V. Kendes værdierne for trykket P og temperaturen T, kan vi ved hjælp af idealgasloven
P*V=n*R*T
Finde n, som er stofmængden af gassen:
P*V=n*R*T ↔ n=P*VR*T
P måles i bar, voluminet V i L og temperaturen T i Kelvingrader. R er Gaskonstanten 0,0831 L*bar/mol*K.
Kendes massen m af gassen, kan molarmassen M findes ved division:
M=mn
Apparatur og kemikalier
I øvelsen har vi brugt et måleglas på 205 ml, et termometer, en stor balje, en vægt og en engangslighter, Hårtørrer.

Fremgangsmåde
En engangslighter vejes med 0,1 milligrams nøjagtighed. Tag et 250 ml måleglas og et vandfad/balje. Fyld vandfadet/baljen 2/3 op og fyld måleglasset til renden. Læg nu din flade hånd på vandoverfladen i måleglasset så det slutter tæt. Hold måleglasset ind over vandfadet, vend bunden i vejret og før hurtigt mundingen ned under vandoverfladen. Tag forsigtigt hånden op. Kontroller, at der ikke er sluppet luftbobler ind i glasset. Hvis det er tilfælde, må du starte forfra. Spænd evt. måleglasset fast i et stativ så det ikke vælter. Før så lighteren ind under måleglasset og lad lightergassen boble op indtil du har fyldt ca. 200 ml lightergas i. Aflæs trykket og vandtemperaturen. Nu skal lighteren tørres. Ryst det overskydende vand af og tør den 2-3 minutter med hårtørreren. Vej lighteren og tør den atter 2-3 minutter. Vej den igen og sammenlign det forrige resultat. Fortsæt denne procedure indtil vægten ikke aftager mere (på nær små udsving).

Resultater og resultatbearbejdelse
Vægt på lighter før brug: 16,190 g
Vægt på lighteren efter brug: 15,760 g
Massen på lighter 16,190 g - 15,760 g = 0,43 g P | 1,0194 bar (DMI) | T | 307 Kelvin | V | 198 V |

T= 34℃ + 273 kelvin = 307 K V= 198 Volumen / 1000 = 0,198 Liter

1) C1H2*1+2 = CH4 Methan 2) C2H2*2+2 = C2H6 Ethan 3) C3H2*3+2 = C3H8 Propan 4) C4H2*4+2 = C4H10 Butan
Molarmassen
1) 12,01g/mol + 4*1,008g/mol = 16,042g/mol 2) 2*12,01g/mol + 6*1,008g/mol = 30,068g/mol 3) 3*12,01g/mol + 8*1,008g/mol = 44,094g/mol 4) 4*12,01g/mol + 10*1,008g/mol = 58,12g/mol

n=P*VR*T n=1,0194 bar*0,198 L0,0831L*barmol*K*307K = 0,0079 mol
Molarmasssen af lighteren
M= mn M= 0,43g0,0079mol = 54,43g/mol

Fejlkilder
Fejlkilderne i denne øvelse kunne være hvis der nu stadig var bobler i måleglasset, eller hvis man brugte forskellige vægte til at veje lighteren med. Måleglasset kunne også være beskadiget så gassen fra lighteren sivede ud. Trykket kunne også være en fejlkilde, da det blev taget om morgenen og ude ved Roskilde lufthavn over DMI`s hjemmeside.

Diskussion
Ud fra mit resultat af lighterens molarmasse, som blev 54,43g/mol og de resultater jeg fik i mine udregninger af molekyleformlerne, vil jeg sige at resultatet lænder sig meget op af at være Butan da den udregning blev 58,12g/mol og der derfor kun er en forskel på 3,69g/mol. Og med det er jeg blevet en hel del klogere på hvad der egentlig er inde i en lighter og det er Butan C4H10. Desværre er resultatet ikke fuldstændig identiske og det kan skydes nogle af de fejlkilder der er nævnt ovenfor, som bla trykket som jo ikke er målt fuldstændig sammen sted og er taget fra DMI´s hjemmeside. Er sikker på at hvis vi selv havde målt trykket i kemilokalet så var resultatet også blevet mere nøjagtigt.
Konklusion
Jeg er kommet frem til den konklusion at indholdet i en lighter m være butan, da det efter mine beregninger ligger tættest på de molekyleformler som jeg har udregnet. Udregningen af lighterens molarmasse blev 54,43g/mol og udregninger af molekyleformlerne blev henholdsvis 16,042g/mol (methan), 30,068g/mol (ethan), 44,094g/mol (propan) og 58,12g/mol (butan).

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