annisa nurul fix.pptx
Transcript of annisa nurul fix.pptx
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ANNISA NURUL AZZAHRA1111102000029
Tugas Analisis instrumen
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Halaman 255• 1. untuk analisis sampel semen,
seri dari standar telah disiapkan dan intensitas emisi untuk natrium dan kalium diatur pada 590 dan 768nm. Tiap standar larutan mengandung 6300μg/mL dari calsium sebagai CaO dalam pembacaan alkali. Hasil ditunjukan dalam tabel. Untuk setiap sampel semen 1.000g dilarutkan dalam 100ml. Hitung persen dari Na2O dan K2O.
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Kurva kalibrasi
0 20 40 60 80 100 1200
20
40
60
80
100
120
f(x) = 0.946153846153846 x + 13.5R² = 0.953797519656318
Kurva kalibrasi Na2O
konsentrasi
abso
rban
si
0 20 40 60 80 100 1200
20
40
60
80
100
120
f(x) = 0.986593406593 x + 4.914285714286R² = 0.991407996552159
Kurva kalibrasi K2O
Konsentrasi
Abs
orba
nsi
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Perhitungan kadar
%kadar = x100%Cs= konsentrasi sampelV= volumem= sampel Sampel : 1 g/100 ml = 10.000 µg/ml
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Kadar Na2OCement A (y = 28)
y = 0,946x + 13,5x = (y – 13,5): 0,946x = (28-13,5): 0,946x = 14,5 : 0,946 = 15,328 ppm
% kadar = 15,328/10.000 x 100% = 0,15328%
Cement B (y = 58)y = 0,946x + 13,5x = (y – 13,5): 0,946x = (58-13,5): 0,946x = 44,5 : 0,946 = 47,040 ppm
% kadar = 47,040/10.000 x 100% = 0,47040%
Cement C (y = 42)y = 0,946x + 13,5x = (y – 13,5): 0,946x = (42-13,5): 0,946x = 28,5 : 0,946 = 30,127 ppm
% kadar = 30,127/10.000 x 100% = 0,30127%
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Kadar K2OCement A (y = 69)
y = 0,986x + 4,914x = (y – 4,914): 0,986x = (69-4,914): 0,986x = 64,086 : 0,986 = 64,996 ppm
% kadar = 64,996/10.000 x 100% = 0,64996%
Cement B (y = 51)y = 0,986x + 4,914x = (y – 4,914): 0,986x = (51-4,914): 0,986x = 46,086 : 0,986 = 46,740 ppm
% kadar = 46,740 /10.000 x 100% = 0,46740%
Cement B (y = 63)y = 0,986x + 4,914x = (y – 4,914): 0,986x = (63-4,914): 0,986x = 58,086 : 0,986 = 58,911 ppm
% kadar = 58,911 /10.000 x 100% = 0,58911%
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3. Boron memberikan sebuah seri dari sebuah berkas fluktuasi yang seharusnya radikal BO2 yang terletak di green portion dari spektrum . Meskipun berkas yang overlap memberikan masalah dalam perhitungan, keterbatasan minimum diantara berkas dapat digunakan. Berikut hasilnya. Berapa konsentrasi boron dalam sampel ?
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Kurva kalibrasi
0 50 100 150 200 2500
1020304050607080
f(x) = 0.163 x + 35.9R² = 0.999887099202168
Kurva kalibrasi 518 nm
konsentrasi
Abs
orba
nsi
0 50 100 150 200 25005
101520253035404550
f(x) = 0.063 x + 32.9R² = 0.999244712990937
Kurva kalibrasi 505 nm
konsentrasi
abso
rban
si
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Konsentrasi boron dalam 518Sampel A (y = 45)
y = 0,163x + 35,9x = (y – 35,9): 0,163x = (45-35,9): 0,163x = 9,1 : 0,163 = 55,828 ppm
Sampel B (y = 85)y = 0,163x + 35,9x = (y – 35,9): 0,163x = (85-35,9): 0,163x = 49,1 : 0,163 = 301,227 ppm
Sampel C (y = 66)y = 0,163x + 35,9x = (y – 35,9): 0,163x = (66-35,9): 0,163x = 30,1 : 0,163 = 184,663 ppm
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Konsentrasi boron dalam 505Sampel A (y = 36,5)
y = 0,063x + 32,9x = (y – 32,9): 0,063x = (36,5-32,9): 0,063x = 3,6 : 0,063 = 57,143 ppm
Sampel B (y = 65)y = 0,063x + 32,9x = (y – 32,9): 0,063x = (65-32,9): 0,063x = 32,1 : 0,063 = 509,524 ppm
Sampel C (y = 50)y = 0,063x + 32,9x = (y – 32,9): 0,063x = (50-32,9): 0,063x = 17,1 : 0,063 = 271,429 ppm
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Mass Spectrometry
2.1 Pada uji suatu senyawa dengan Mass Spectrometry, didapatkan BM sebesar 212,0833. Apakah rumus molekul dari senyawa tersebut?
• JawabanRumus molekul yang didapatkan dari senyawa tersebut menurut Apendix A adalah C14H12O2 dengan BM sebesar 212,0837.
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2.2 Senyawa yang memiliki rumus molekul di Problem 2.1 memberikan kenaikan spektrum. Simpulkan struktur dari senyawa tersebut.
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• Jawaban
Rumus molekul C14H12O2 atau C15H16O
Base peak : m/z = 105
Berat Molekul : 212, sehingga m/z 212 – 105
= 107 Kesimpulan:
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2.3 Spektrum massa 2-butenal menunjukkan peak pada m/z 69 dengan 28,9%. Kemukakan paling tidak satu rute fragmentasi dan jelaskan kenapa fragmen ini stabil!
• JawabanRumus molekul : C4H6O, BM 70
Base peak : m/z 69 CH3CH=CHC=Om/z 70-69 = 1 H Kesimpulan : CH3-CH=CH-CHO
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2.4 Spektrum masa dari 3-butuna-2-ol menunjukkan base peak di m/z 55. Jelaskan kenapa fragmen yang memberi peningkatan pada peak tersebut stabil?
• JawabanRumus molekul : C4H6O, BM 70
Base peak : m/z 53 ke 55 (M+2 ) CH3-CC-CH2m/z 70-53 = 17 OHKesimpulan : CH3-CC-CH2-HO
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2.5 Tentukan struktur dari dua isomer dibawah ini, jika keduanya memiliki rumus C10H14 ?
Isomer A :BM : 134 -> rumus molekul : C10H14Base Peak :m/z 105 ->m/z 134-105= 29 -> C2H5
Kesimpulan : - C2H5
Isomer B• m/z 91 Ke 105 -> • m/z 134 – 105 = 29 -> C2H5
• Kesimpulan : C2H5
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2.7 Tentukan Strukturnya !
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Rumus Molekul : C9H10, BM : 118 (Kemungkinan 1) Base peak :
• m/z 77 (M-) -> C6H5 • m/z 118 – 77 = 41 -> C3H5
Kesimpulan : -CH=CH-CH3
(Kemungkinan 2) Base peak :• m/z 43 -> C3H7
• m/z 118 – 43 = 75-> (CH3O)2CH Kesimpulan : CH3-CH2-CH2-CH(OCH3)2
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Rumus molekul : C2H5I, BM : 156 - Base peak • m/z 156-127= 29 -> C2H5
• m/z 127 -> I
Kesimpulan :CH3-CH2-I
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Rumus molekul : C7H7Cl, BM : 126 Base peak m/z 91 -> m/z 126-91= 35 -> ClKesimpulan : - Cl
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- Rumus molekul : C7H7Br, BM : 170 Base peak • m/z 91 -> • m/z 170-91= 79-> Br• Kesimpulan : - Br
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Jawab :Rumus molekul : C7H8O, BM : 108- Base peak m/z 79 -> C6H5 + 2H m/z 108-79= 29 -> CHO
Kesimpulan : -> C6H5-CH2-OH
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Jawab :RM : C16H27O4 BM: 283 Jika BM 283, RM: C11H22O4-Base peak m/z 43 : CH=CH-OH m/z 283-43 = 240 (krn tdk ad di apendix) m/z 44 = COO- m/z 240-44= 196 : C14H28
-Kesimpulan : C14H28-COO-CH=CH-OH
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-Rumus molekul : C5H9O2I, BM : 229- Base peak m/z 127 -> I m/z 169-127= 42 -> C3H6 m/z 229-169= 60
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Infrared Spectrometry
• 3.2No. senyawa struktur IR Bands
1 Benzamide 3350(s) 3060(m), 1635(s)
2 Benzoic acid 3200-2400(s), 1685 (b,s,)705 (s)
3 Benzonitril 3080(w), nothing 3000-2800, 2230 (s) 1450(s) 760(s), 688(s)
4 Biphenyl 3030 (m), 730(s) 690(s)
5 Dioxane 2955(s) 2850(s) 1120 (s)
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6 Diphenyl sulfone 3080 (m), nothing 3000-2800, 1315 (s) 1300(s) 1155(s)
7 Formic acid 2900(bs) 1720 (b,s)
8 isobuthylamin 3380 (m), 3300(m), nothing 3200-3000, 2980 (s), 2870 (m) 1610(m), ~900 ~700
9 1-Nitropropane 2946(s) 2930(m) 1550(s) 1386(m)
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C-H aromatik No CC (rangkap tiga) Strech
C=C aromatik CH3 bond
OOP monosubstitusi
Jadi spektrum A : Diphenylacetylene
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Jadi spectrum B : 1,3 Cyclohexadiene
alkenaC-H alifatis
No cis C=C
Cis alkena OOP
CH2 bend
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Jadi, Spectrum C : 2 pentene
C=C strech
Sp3 Strech CH2 bendCis alkene disubstitusiCH2 bend
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Jadi, Spectrum D : 1-oktene
Sp2 CH StrechSp3 CH Strech
C=C Vinil alkena monosubstitusi
CH3 bend
CH2 bend
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