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8/3/2019 Report Kimfis Lapindo
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Group 2
Ahmad Farizzi Artyani Batari Thomy Dj Vollmer
Faculty of Engineering
University of Indonesia
Analytical Chemistry
PBL II: Lapido Mud
FlowAtomic and Molecular Spectroscopy
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Table of Contents
PBL Mind Map ............................................................................................................ 1
Table of Contents ...................................................................................................... 2
Problem Definition .................................................................................................. 3
Theory .................................................................................................................................... 3
Lapindo Mud Flow ................................................................................................ 3
Analysis Method ................................................................................................................. 8
Infrared spectrometry ................................................................................................... 10
Answer of Questions .................................................................................... 19References............................................................................................................. 6
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PROBLEM DEFINITION
- Lapindo flow muds analysis using atomic and molecular spectroscopy- Atomic Absorption Spectrometry (AAS) and IR spectroscopy
THEORY
LAPINDO FLOW MUD
The Sidoarjo mud flow or Lapindo mud, also informally abbreviated as Lusiis a mud volcano in the sub
district ofPorong, Sidoarjo in East Java, Indonesia that has been in eruption since May 2006. This biggest
mud volcano in the world was created by the blowout of a natural gas well drilled by PT Lapindo
Brantas, although company officials contend that it was caused by a distant earthquake.
At its peak Lusi was spewing up to 180,000 m of mud per day. In mid August 2011, mud was being
discharged at a rate of 10,000 cubic meters per day, with 15 bubbles around the gush point. This was a
significant decline from a year previous, when mud was being discharged at a rate of 100,000 cubic
meters per day with 320 bubbles around the gush point. It is expected that the flow will continue for the
next 25 to 30 years. Although the Sidoarjo mud flow has been contained by levees since November
2008, resultant flooding regularly disrupts local highways and villages, and further breakouts of mud are
still possible
http://en.wikipedia.org/wiki/Poronghttp://en.wikipedia.org/wiki/Sidoarjohttp://en.wikipedia.org/wiki/East_Javahttp://en.wikipedia.org/wiki/Indonesiahttp://en.wikipedia.org/wiki/Blowout_(well_drilling)http://en.wikipedia.org/wiki/PT_Lapindo_Brantashttp://en.wikipedia.org/wiki/PT_Lapindo_Brantashttp://en.wikipedia.org/wiki/Leveehttp://en.wikipedia.org/wiki/Leveehttp://en.wikipedia.org/wiki/PT_Lapindo_Brantashttp://en.wikipedia.org/wiki/PT_Lapindo_Brantashttp://en.wikipedia.org/wiki/Blowout_(well_drilling)http://en.wikipedia.org/wiki/Indonesiahttp://en.wikipedia.org/wiki/East_Javahttp://en.wikipedia.org/wiki/Sidoarjohttp://en.wikipedia.org/wiki/Porong -
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The content of lapindo flow mud
Test Results
Parameter Result marksStandart Quality
(PP no. 18/1999)
Arsen 0,045 Mg/L 5 Mg/L
Barium 1,066 Mg/L 100 Mg/L
Boron 5,097 Mg/L 500 Mg/L
Pb 0,05 Mg/L 5 Mg/L
Hg 0,004 Mg/L 0,2 Mg/L
Si 0,02 Mg/L 20 Mg/L
Trichlorophenol 0,017 Mg/L 2 Mg/L (2,4,6 Trichlorophenol)
400 Mg/L (2,4,4 Trichlorophenol)
http://id.wikipedia.org/wiki/Arsenhttp://id.wikipedia.org/wiki/Bariumhttp://id.wikipedia.org/wiki/Boronhttp://id.wikipedia.org/wiki/Timbalhttp://id.wikipedia.org/wiki/Raksahttp://id.wikipedia.org/wiki/Raksahttp://id.wikipedia.org/wiki/Timbalhttp://id.wikipedia.org/wiki/Boronhttp://id.wikipedia.org/wiki/Bariumhttp://id.wikipedia.org/wiki/Arsen -
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Effect of lapindo flow mud
- decrease the gas sources- water pipe ofSurabayas PDAM was broke because of the surface is going down- 600 hectare of field was submerged- 16 Housing residents, schools and another public place were flooded- PERTAMINAs gas pipe was explode because of the high pressure mud flow- 30 factories are flooded with lusi- Etc
Cause of lapindo flow mud
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Mud eruption chronology
On May 28, 2006, PT Lapindo Brantas targeted gas in the Kujung Formation carbonates in the Brantas
PSC area by drilling a borehole named the 'Banjar-Panji 1 exploration well'. In the first stage of drilling
the drill string first went through a thick clay seam (5001,300 m deep), then sands, shales, volcanic
debris and finally into permeable carbonate rocks. At this stage the borehole was surrounded by a steel
casing to help stabilize it. At 5:00 a.m. local time (UTC+7) a second stage of drilling began and the drill
string went deeper, to about 2,834 m (9,298 ft), this time without a protective casing, after which water,
steam and a small amount of gas erupted at a location about 200 m southwest of the well. Two further
eruptions occurred on the second and the third of June about 8001000 m northwest of the well, but
these stopped on June 5, 2006. During these eruptions, gas was released and local villagers observed
hot mud, thought to be at a temperature of around 60 C (140 F).
A magnitude of 6.3 earthquake occurred in Yogyakarta at ~06:00 local time 27 May 2006, approximately
250 kilometers South West from Sidoarjo. Seven minutes after the earthquake a mud loss problem in
the well was noted. After two major aftershocks, the well suffered a complete loss of circulation. A loss
of circulation is when drilling mud that is pumped down a shaft does not return to the surface but is lost
into some opening or a fault system. This mud loss problem was finally stopped when a loss circulation
material was pumped into the well, a standard practice in drilling an oil and gas well. A day later the well
suffered a kick, an influx of formation fluid into the well bore. The kick appears to have been killed
within three hours. The next day, 29 May 2006, steam, water and mud began erupting 200 meters away
from the well, a phenomenon that is now known as the Lusi mud volcano.
Hypotheses on the possible causes
The birth of Lusi was a major disaster for the general population that lives nearby, the loss of their
houses, belongings as well as their livelihood. For the scientific community, however, it was a chance to
study the evolving geological process of a mud volcano. In the past, mud vulcanologists could only study
existing or ancient mud volcanoes during dormant periods. This is a rare occasion and a unique
opportunity to conduct scientific experiments to further our understanding. Lusi also offered
opportunities to study the down hole condition of a mud volcano from the neighboring Banjar-Panji
exploration well lithologies.
To explain what triggered the mud volcano, three hypotheses have been suggested, though none has
won universal support:
Hydro-fracturing of the formation, hence a drilling related problem
From a model developed by geologists working in the UK, the drilling pipe penetrated the overpressured
limestone, causing entrainment of mud by water. The influx of water to the well bore caused a
hydrofracture, but the steam and water did not enter the borehole; they penetrated the
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surrounding overburden and pressuredstrata. The extra pressure formed fractures around the borehole
that propagated 12 km to the surface and emerged 200 m away from the well. The most likely cause of
these hydraulic fractures was the unprotected drill string in the second stage of drilling. Though steel
casing is used to protect the well bore in oil or gas exploration, this can only be applied in stages after
each new section of the hole is drilled; see drilling for oil.
The relatively small distance, around 600 feet (180 m), between the Lusi mud volcano and the well being
drilled by Lapindo (the Banjarpanji well) may not be a coincidence, as less than a day before the start of
the mud flow the well suffered a kick. Their analysis suggests that the well has a low resistance to a
kick.Similarly, a NE-SW crack in the surface in the drill site may be evidence of an underground blowout.
The well may have suffered an underground blowout that resulted in a surface breach. Also, a likely
contributor is the dissociation of methane hydrates. These hydrates are formed when fresh water used
in fracking seeps into methane beds and at depth and pressure form crystal like rock which contain up
to 205 C (401 F) in each molecule. These hydrates cannot be formed in salt water. They are endemic
wherever drilling for oil and gas have used fresh water as part of the slurry that pressurizes wells on
land, permafrost, and under the earth crust along the continental shelves.
Solution of it
- Stop the flow mud with snubbing unit, it is use for pushing down the borehole which left, then itwill cover up the mud flow hole.
- Sidetracking to avoid the borehole which is left- putting some sort of heavy metal object shaped like balls to stop the eruption. It costs a fortune
and as it is nature, my personal opinion is that is not going to work. Its science: there is
pressure, there is a leak, you cover the leak, the pressure will go somewhere else.
Unfortunately, the only thing is to let the mother nature to reach its equilibrium.
http://en.wikipedia.org/wiki/Stratumhttp://en.wikipedia.org/wiki/Stratum -
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ANALYSIS METHOD
Atomic Absorption Spectrhophotometry (AAS)
Introduction
Atomic-absorption (AA) spectroscopy uses the absorption of light to measure the concentration of gas-
phase atoms. Since samples are usually liquids or solids, the analyte atoms or ions must be vaporized in
a flame or graphite furnace. The atoms absorb ultraviolet or visible light and make transitions to higher
electronic energy levels. The analyte concentration is determined from the amount of absorption.
Applying the Beer-Lambert law directly in AA spectroscopy is difficult due to variations in the
atomization efficiency from the sample matrix, and nonuniformity of concentration and path length of
analyte atoms (in graphite furnace AA). Concentration measurements are usually determined from aworking curve after calibrating the instrument with standards of known concentration.
Schematic of an atomic-absorption experiment
Instrumentation
Light source
The light source is usually a hollow-cathode lamp of the element that is being measured. Lasers are also
used in research instruments. Since lasers are intense enough to excite atoms to higher energy levels,
they allow AA and atomic fluorescence measurements in a single instrument. The disadvantage of these
narrow-band light sources is that only one element is measurable at a time.
http://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/data/wcurve.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/optics/sources/lamps.htm#hollow-cathodehttp://elchem.kaist.ac.kr/vt/chem-ed/optics/sources/lasers.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/afs.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/graphics/aa-expt.gifhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/afs.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/optics/sources/lasers.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/optics/sources/lamps.htm#hollow-cathodehttp://elchem.kaist.ac.kr/vt/chem-ed/data/wcurve.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htm -
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Atomizer
AA spectroscopy requires that the analyte atoms be in the gas phase. Ions or atoms in a sample must
undergo desolvation and vaporization in a high-temperature source such as a flame or graphite furnace.
Flame AA can only analyze solutions, while graphite furnace AA can accept solutions, slurries, or solid
samples.
Flame AA uses a slot type burner to increase the path length, and therefore to increase the total
absorbance (see Beer-Lambert law). Sample solutions are usually aspirated with the gas flow into a
nebulizing/mixing chamber to form small droplets before entering the flame.
The graphite furnace has several advantages over a flame. It is a much more efficient atomizer than a
flame and it can directly accept very small absolute quantities of sample. It also provides a reducing
environment for easily oxidized elements. Samples are placed directly in the graphite furnace and the
furnace is electrically heated in several steps to dry the sample, ash organic matter, and vaporize the
analyte atoms.
Light separation and detection
AA spectrometers use monochromators and detectors for uv and visible light. The main purpose of the
monochromator is to isolate the absorption line from background light due to interferences. Simple
dedicated AA instruments often replace the monochromator with a bandpass interference filter.
Photomultiplier tubes are the most common detectors for AA spectroscopy.
Picture of a flame atomic-absorption spectrometer:
http://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/optics/detector/pmt.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/graphics/aa.jpghttp://elchem.kaist.ac.kr/vt/chem-ed/optics/detector/pmt.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htm -
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Infrared spectrometry
Theory
Introduction
The term "infra red" covers the range of the electromagnetic spectrum between 0.78 and 1000 mm. In
the context of infra red spectroscopy, wavelength is measured in "wavenumbers", which have the units
cm-1.
wavenumber = 1 / wavelength in centimeters
It is useful to divide the infra red region into three sections; near, midand farinfra red;
Region Wavelength range (mm) Wavenumber range (cm-1
)
Near 0.78 - 2.5 12800 4000
Middle 2.5 50 4000 200
Far 50 -1000 200 10
The most useful I.R. region lies between 4000 - 670cm-1.
Theory of infra red absorption
IR radiation does not have enough energy to induce electronic transitions as seen with UV. Absorption
of IR is restricted to compounds with small energy differences in the possible vibrational and rotational
states.
For a molecule to absorb IR, the vibrations or rotations within a molecule must cause a net change in the
dipole moment of the molecule. The alternating electrical field of the radiation (remember that
electromagnetic radation consists of an oscillating electrical field and an oscillating magnetic field,
perpendicular to each other) interacts with fluctuations in the dipole moment of the molecule. If the
frequency of the radiation matches the vibrational frequency of the molecule then radiation will be
absorbed, causing a change in the amplitude of molecular vibration.
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Molecular vibrations
The positions of atoms in a molecules are not fixed; they are subject to a number of different vibrations.
Vibrations fall into the two main catagories ofstretching and bending.
Stretching: Change in inter-atomic distance along bond axis
Bending: Change in angle between two bonds. There are four types of bend:
Rocking Scissoring Wagging Twisting
Vibrational coupling
In addition to the vibrations mentioned above, interaction between vibrations can occur (coupling) if the
vibrating bonds are joined to a single, central atom. Vibrational coupling is influenced by a number of
factors;
Strong coupling of stretching vibrations occurs when there is a common atom between the twovibrating bonds
Coupling of bending vibrations occurs when there is a common bond between vibrating groups Coupling between a stretching vibration and a bending vibration occurs if the stretching bond is
one side of an angle varied by bending vibration Coupling is greatest when the coupled groups have approximately equal energies No coupling is seen between groups separated by two or more bonds
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Instrumentation
Sources
An inert solid is electrically heated to a temperature in the range 1500-2000 K. The heated material will
then emit infra red radiation.
The Nernst gloweris a cylinder (1-2 mm diameter, approximately 20 mm long) of rare earth oxides.
Platinum wires are sealed to the ends, and a current passed through the cylinder. The Nernst glower can
reach temperatures of 2200 K.
The Globar source is a silicon carbide rod (5mm diameter, 50mm long) which is electrically heated to
about 1500 K. Water cooling of the electrical contacts is needed to prevent arcing. The spectral output is
comparable with the Nernst glower, execept at short wavelengths (less than 5 mm) where it's output
becomes larger.
The incandescent wire source is a tightly wound coil of nichrome wire, electrically heated to 1100 K. It
produces a lower intensity of radiation than the Nernst or Globar sources, but has a longer working life.
Detectors
There are three catagories of detector;
Thermal Pyroelectric Photoconducting
Thermocouples consist of a pair of junctions of different metals; for example, two pieces of bismuth
fused to either end of a piece of antimony. The potential difference (voltage) between the junctions
changes according to the difference in temperature between the junctions
Pyroelectric detectors are made from a single crystalline wafer of a pyroelectric material, such as
triglycerine sulphate. The properties of a pyroelectric material are such that when an electric field is
applied across it, electric polarisation occurs (this happens in any dielectric material). In a pyroelectric
material, when the field is removed, the polarisation persists. The degree of polarisation is temperature
dependant. So, by sandwiching the pyroelectric material between two electrodes, a temperature
dependant capacitor is made. The heating effect of incident IR radiation causes a change in the
capacitance of the material. Pyroelectric detectors have a fast response time. They are used in most
Fourier transform IR instruments.
Photoelectric detectors such as the mercury cadmium telluride detector comprise a film of
semiconducting material deposited on a glass surface, sealed in an evacuated envelope. Absorption of IRpromotes nonconducting valence electrons to a higher, conducting, state. The electrical resistance of
the semiconductor decreases. These detectors have better response characteristics than pyroelectric
detectors and are used in FT-IR instruments - particularly in GC - FT-IR.
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IR Spectroscopy Tutorial: How to analyze IR spectra
Mono-functional molecules there are so many IR bands that it is not feasible to assign every band in an
IR spectrum. Instead, look for tell-tale bands -- the region from 4000-1300 cm-1 is particularly useful for
determining the presence of specific functional groups.
3500-3300 cm-1 NH stretch 1&Mac251;, 2&Mac251; amines
3500-3200 cm-1 OH stretch alcohols, a broad, strong band
3100-3000 cm-1 CH stretch alkenes
3000-2850 cm-1 CH stretch alkanes
1760-1665 cm-1 C=O stretch ketones, aldehydes, esters
1680-1640 cm-1 C=C stretch alkenes
looking in the region from 4000-1300. Look at the CH stretching bands around 3000:
Indicates:
Are any or all to the right of 3000? alkyl groups (present in most organic molecules)
Are any or all to the left of 3000? a C=C bond or aromatic group in the molecule
Look for a carbonyl in the region 1760-1690. If there is such a band:
Indicates:
Is an OH band also present? a carboxylic acid group
Is a CO band also present? an ester
Is an aldehydic CH band also present? an aldehyde
Is an NH band also present? an amide
Are none of the above present? a ketone
(also check the exact position of the carbonyl band for clues as to the type of carbonyl compound it is)
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Look for a broad OH band in the region 3500-3200 cm-1. If there is such a band:
Indicates:
Is an OH band present? an alcohol or phenol
Look for a single or double sharp NH band in the region 3400-3250 cm-1. If there is such a band:
Indicates:
Are there two bands? a primary amine
Is there only one band? a secondary amine
Other structural features to check for:
Indicates:
Are there CO stretches? an ether (or an ester if there is a carbonyl band too)
Is there a C=C stretching band? an alkene
Are there aromatic stretching bands? an aromatic
Is there a CC band? an alkyne
Are there -NO2 bands? a nitro compound
If there is an absence of major functional group bands in the region 4000-1300 cm-1 (other than CH
stretches), the compound is probably a strict hydrocarbon.
Also check the region from 900-650 cm-1. Aromatics, alkyl halides, carboxylic acids, amines, and amides
show moderate or strong absorption bands (bending vibrations) in this region.
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Group Frequencies
Typical Infrared Absorption Frequencies
Stretching Vibrations Bending Vibrations
Functional Class Range (cm-1
)Inten
sityAssignment
Range
(cm-1
)
Inten
sityAssignment
Alkanes 2850-3000 Str CH3, CH2 & CH
2 or 3 bands
1350-
1470
1370-
1390
720-
725
med
med
wk
CH2 &
CH3 deformati
on
CH3 deformati
on
CH2 rocking
Alkenes 3020-31001630-1680
1900-2000
medvar
str
=C-H & =CH2 (usuallysharp)
C=C (symmetry
reduces intensity)
C=C asymmetric
stretch
880-995
780-
850
675-
730
strmed
med
=C-H & =CH2(out-of-plane
bending)
cis-RCH=CHR
Alkynes 3300
2100-2250
str
var
C-H (usually sharp)
CC (symmetry
reduces intensity)
600-
700
str C-H
deformation
Arenes 3030
1600 & 1500
var
med-
wk
C-H (may be several
bands)
C=C (in ring) (2
bands)
(3 if conjugated)
690-
900
str-
med
C-H bending &
ring puckering
Alcohols &
Phenols
3580-3650
3200-3550
970-1250
var
str
str
O-H (free), usually
sharp
O-H (H-bonded),usually broad
C-O
1330-
1430
650-770
med
var-
wk
O-H bending
(in-plane)
O-H bend(out-of-plane)
Amines 3400-3500 (dil. soln.)
3300-3400 (dil. soln.)
wk
wk
N-H (1-amines), 2
bands
N-H (2-amines)
1550-
1650
660-
med-
str
NH2 scissoring
(1-amines)
NH2 & N-H
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir2http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir3http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4chttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4chttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir3http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir2 -
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1000-1250 med C-N 900 var wagging
(shifts on H-
bonding)
Aldehydes &
Ketones
2690-2840(2 bands)
1720-1740
1710-1720
1690
1675
1745
1780
med
str
str
str
str
str
str
C-H (aldehyde C-H)
C=O (saturated
aldehyde)
C=O (saturated
ketone)
aryl ketone
, -unsaturation
cyclopentanone
cyclobutanone
1350-
1360
1400-
1450
1100
str
str
med
-CH3 bending
-CH2 bending
C-C-C bending
Carboxylic
Acids&Derivative
s
2500-3300 (acids)
overlap C-H
1705-1720 (acids)
1210-1320 (acids)
1785-1815 (
acyl halides)
1750 & 1820
(anhydrides)
1040-1100
1735-1750
(esters)
1000-1300
1630-
1695(amides)
str
str
med-
str
str
str
str
str
strstr
O-H (very broad)
C=O (H-bonded)
O-C (sometimes 2-
peaks)
C=O
C=O (2-bands)
O-C
C=O
O-C (2-bands)C=O (amide I band)
1395-
1440
1590-1650
1500-
1560
med
med
med
C-O-H bending
N-H (1-
amide) II bandN-H (2-
amide) II band
Nitriles
Isocyanates,Isothi
ocyanates,
Diimides, Azides
& Ketenes
2240-2260
2100-2270
med
med
CN (sharp)
-N=C=O, -N=C=S
-N=C=N-, -N3, C=C=O
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5 -
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INTERPRETING AN INFRA-RED SPECTRUM
Example :
The infra-red spectrum for a simple carboxylic acid
Ethanoic acid
Ethanoic acid has the structure:
You will see that it contains the following bonds:
carbon-oxygen double, C=O
carbon-oxygen single, C-O
oxygen-hydrogen, O-H
carbon-hydrogen, C-H
carbon-carbon single, C-C
The carbon-carbon bond has absorptions which occur over a wide range of wavenumbers in the
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fingerprint region - that makes it very difficult to pick out on an infra-red spectrum.
The carbon-oxygen single bond also has an absorbtion in the fingerprint region, varying between
1000 and 1300 cm-1depending on the molecule it is in. You have to be very wary about picking out
a particular trough as being due to a C-O bond.
The other bonds in ethanoic acid have easily recognised absorptions outside the fingerprint region.
The C-H bond (where the hydrogen is attached to a carbon which is singly-bonded to everything
else) absorbs somewhere in the range from 2853 - 2962 cm-1. Because that bond is present in most
organic compounds, that's not terribly useful! What it means is that you can ignore a trough just
under 3000 cm-1, because that is probably just due to C-H bonds.
The carbon-oxygen double bond, C=O, is one of the really useful absorptions, found in the range
1680 - 1750 cm-1. Its position varies slightly depending on what sort of compound it is in.
The other really useful bond is the O-H bond. This absorbs differently depending on its
environment. It is easily recognised in an acid because it produces a very broad trough in the range2500 - 3300 cm-1.
The infra-red spectrum for ethanoic acid looks like this:
The possible absorption due to the C-O single bond is queried because it lies in the fingerprint
region. You couldn't be sure that this trough wasn't caused by something else.
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SOLUTION
Assignment 1 :
1. Are the above-ground basins built to containment of the mud a sustainable solution? How doyou think to solve this as a long term solution?
Answer:
No, because it only can make the lapindo mud not expand to another area and its not a best or
permanent way, that because the above-ground basins can collapse.
Geologist: just wait until it reaches the equilibrium by itself. Equilibrium will happen if the
ground runs out the overpressure gas or when theres ground erosion so the ground will cover
up the hole of overpressure gas.
2. What do you think will happen if releasing the mud into an aquatic environment?Answer:
It would dangerous for aquatic biota. Because the heavy metals or another dangerous metals
will poisoning the aquatic biota. Besides that it will contaminated water. And we all know that
we need water in our life every day and every activity.
3. Will you describe any organic and inorganic might be found in the mud?Answer:
Test Results
Parameter Result marksStandart Quality
(PP no. 18/1999)
Arsenic 0,045 Mg/L 5 Mg/L
Ba 1,066 Mg/L 100 Mg/L
B 5,097 Mg/L 500 Mg/L
Pb 0,05 Mg/L 5 Mg/L
Hg 0,004 Mg/L 0,2 Mg/L
FreeSi 0,02 Mg/L 20 Mg/L
Trichlorophenol 0,017 Mg/L 2 Mg/L (2,4,6 Trichlorophenol)
400 Mg/L (2,4,4 Trichlorophenol)
http://id.wikipedia.org/wiki/Sianidahttp://id.wikipedia.org/wiki/Sianidahttp://id.wikipedia.org/wiki/Sianidahttp://id.wikipedia.org/wiki/Sianida -
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4. What analysis methods can be used for analyzing heavy metals and organic compounds inpolluted soil such as lapindo mud?
Answer:
The best analysis method for analyzing heavy metals and organic compounds in polluted soils
(such as Lapido Muds) are the Atomic and Molecular Spectroscopy, which uses the absorption of
light to measure the concentration of gas-phase atoms. What makes it accurate is, that the
liquids or solids have to be vaporized and then analyzed in a flame or graphite furnace.
Example:
- Atomic-Absorption for Atomic Spectroscopy
- IR Spectra for Molecular Spectroscopy
Assigment 2:
1. What you know about Atomic Absorption Specthrophotometry (AAS)? Can you describe thefundamental theory and instrumentation in this method of analysis?
Answer:
Atomic-absorption (AA) spectroscopy uses the absorption of light to measure the concentration
of gas-phase atoms. Since samples are usually liquids or solids, the analyte atoms or ions must
be vaporized in a flame or graphite furnace. The atoms absorb ultraviolet or visible light and
make transitions to higher electronic energy levels. The analyte concentration is determined
from the amount of absorption.
Applying the Beer-Lambert law directly in AA spectroscopy is difficult due to variations in the
atomization efficiency from the sample matrix, and nonuniformity of concentration and pathlength of analyte atoms (in graphite furnace AA). Concentration measurements are usually
determined from a working curve after calibrating the instrument with standards of known
concentration.
2. How do you describe the procedure for performing analysis of heavy metal in soil/mud usingAAS?
Answer:
In order to analyze a sample for its atomic constituents, it has to be atomized. The atomizersmost commonly used nowadays are flames and electrothermal (graphite tube) atomizers. The
atoms should then be irradiated by optical radiation, and the radiation source could be an
element-specific line radiation source or a continuum radiation source. The radiation then
passes through a monochromator in order to separate the element-specific radiation from any
other radiation emitted by the radiation source, which is finally measured by a detector.
http://elchem.kaist.ac.kr/vt/chem-ed/quantum/at-lvls.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/data/wcurve.htmhttp://en.wikipedia.org/wiki/Monochromatorhttp://en.wikipedia.org/wiki/Monochromatorhttp://elchem.kaist.ac.kr/vt/chem-ed/data/wcurve.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/quantum/at-lvls.htm -
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3. After running series of standard chromium solution and the unknown sample, the following datais collected. You are preparing a callibation curve by plotting absorbance versus concentration
for standard solutions. How do you calculate the concentration of Cr in mud, if unknown sample
is obtained from 2 gram of dry mud after you dried it in oven at 105 C? it is dissolved in 100ml
water
chromium (ppm) absorption
0.1 0.006
0.5 0.033
1 0.064
2 0.121
3 0.174
4 0.232
Answer:
unknown sample absorption = 0.105
0.105 = x 0.0572 + 0.039
http://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/graphics/aa-expt.gif -
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x = 1.77 ppm/100ml
Assignment 3:
1. How do you describe the fundamental theory and instrumentation for IR Spectroscopy analysis?Answer:- Theory
IR radiation does not have enough energy to induce electronic transitions as seen with
UV. Absorption of IR is restricted to compounds with small energy differences in the
possible vibrational and rotational states.
For a molecule to absorb IR, the vibrations or rotations within a molecule must cause a
net change in the dipole moment of the molecule. The alternating electrical field of the
radiation (remember that electromagnetic radation consists of an oscillating electrical
field and an oscillating magnetic field, perpendicular to each other) interacts with
fluctuations in the dipole moment of the molecule. If the frequency of the radiationmatches the vibrational frequency of the molecule then radiation will be absorbed,
causing a change in the amplitude of molecular vibration.
----
y = 0.0572x + 0.0039
0
0.05
0.1
0.15
0.2
0.25
0 1 2 3 4 5
absorption
absorption
Linear (absorption)
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- InstrumentationSources
An inert solid is electrically heated to a temperature in the range 1500-2000 K. The
heated material will then emit infra red radiation.
The Nernst gloweris a cylinder (1-2 mm diameter, approximately 20 mm long) of rare
earth oxides. Platinum wires are sealed to the ends, and a current passed through the
cylinder. The Nernst glower can reach temperatures of 2200 K.
The Globar source is a silicon carbide rod (5mm diameter, 50mm long) which is
electrically heated to about 1500 K. Water cooling of the electrical contacts is needed to
prevent arcing. The spectral output is comparable with the Nernst glower, execept at
short wavelengths (less than 5 mm) where it's output becomes larger.
The incandescent wire source is a tightly wound coil of nichrome wire, electrically
heated to 1100 K. It produces a lower intensity of radiation than the Nernst or Globar
sources, but has a longer working life.
Detectors
There are three catagories of detector;
Thermal
Pyroelectric
Photoconducting
2. What is the characteristic of functional group absorption peaks of phenol spectra. If you havepropanoic acid with OH and CO functional groups will the IR spectra be different from IR spectra
of phenol?
Answer:
Phenol, also known as carbolic acid, phenic acid, is an organic compound with the
chemical formula C6H5OH. It is a white crystalline solid. The molecule consists of a phenyl (-
C6H5), bonded to a hydroxyl (-OH) group.
Propanoic acid (from 'propane', and also known as propionic acid) is a naturallyoccurring carboxylic acid conneted to (-OH) group with chemical formula CH3CH2COOH.
Alcohol/Phenol O-H Stretch 3550 - 3200 (broad, s)
Carboxylic Acid O-H Stretch 3000 - 2500 (broad, v)
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3. If you have phenol in mud, what sample preparation techniques can be applied for IRspectroscopy analysis?
Answer:
First you have to get the phenol in the mud as pure, which you can achieve by the
Purification Method, and then you can get the pure evaporated phenol.
Another method would be the other way round, just to freeze the mud, because of the different
Melting Points; there will be substances that will freeze earlier and some that will freeze later, in
this steps we can separate the Phenol from the mud and so have the prepared.
If you have solids you have heat it up with a heat of 1500K-2000K, this melting then is ready to
be analyzed.
4. The following is the IR spectra of compound with the molecular formula of C6H12O, can yougive prediction what compound is? Give explanation based on functional group absorption
peaks.
Answer:
3342 is O-H as the functional Group Alcohol
2935 and 2858 are double peaks which combine =C-H as the functional Group Alkana
1453 isC-H as the functional Group Alkane
1070 is C-O as the functional Group alcohol
971 is =C-H as the functional Group Alkene
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REFERENCES
http://www.wpi.edu/Academics/Depts/Chemistry/Courses/General/infrared.html
http://www.atsdr.cdc.gov/toxprofiles/tp17-c7.pdf
http://orgchem.colorado.edu/hndbksupport/irtutor/analyzeir.htmlhttp://www.chem.ucla.edu/~webspectra/irtable.html