Data Analysis and Presentation

Data Analysis and PresentationChapter 5- Calibration Methods and Quality AssuranceEXCEL How To Do 1- least squares and linear calibration curve/function (chapter 4)2- standard addition (chapter 5)3-internal standard (addition) (chapter 5)

CHAPTER 05: Opener B

CHAPTER 05: Unnumbered Figure 5.1

CHAPTER 05: Table 5.1

Calibration functionQ: How the selected instrument responds to the change in quantity (concentration, M ppm ) of the measured analyte?A: Measure the instrument responses (peak hights, peak areas etc.) from known but different concentrations of the analyte (say 0, 5,10, 15 20 ppm standards). Then construct the response diagram figure (response vs. concentration) with the response or calibration function through the points.

If one assumes that the calibration curve is linear there are physical reasons for that. There is a linear function that can best represent the response of the instrument.If points are out of line: errorThere are methods based on the propagation of error that can help us calculate the best fit: linear regression of the straight-line calibration curvesOLD SLIDE: Caffeine again: first step calibration

Liner vs. nonlinear

CALIBRATION: Instrument used to measure signals must be calibrated, must have the calibration relation. Example (linear): y = m [x ]+ b (general algebraic form for linear equation)Often given in other similar forms:Smeasured(total signal) =k nA (concentration) + Sreag ( signal from reagents, from blank) Calibration is the determination of that relation: it is used for the determination of analyte by using standards and blanks to determine a relationship (function) between concentrations and assay responses. Validation implies that other labs have approved the analytical method of analysis and produced similar results.Standards are materials containing accurately known concentrations of the desired analyte. Standardization is the process of determining relationship between the measured signal and the amount of analyte (fittting for k in linear relations).

Blanks are solutions containing all added reagents except the sample analyte.(method, reagent and field blanks)Controls can be an alternate sample, in which the contents are well known.

Primary and Secondary reagents

similar expressions that you could find:SA(signal due to analyte) =k nA (moles of analyte)Smeasured(total signal) =k nA +Sreag ( signal from reagents, from blank)Smeasured(total signal) =k nA +Sreag + SI( signal from interferents)Total analysis technique & Concentration techniquesSA(signal due to analyte) =k CA (concentration of analyte)

Calibration Language:-Accuracy, precision, sensitivity, selectivity, detection limits, noise-Robust and rugged method-Scale of operation (analytes classification )

OK , if physics of the response process for that instrument is linear, that is the more analyte you have the higher the signal, how can we get that calibration function:Answer 1: One point standard:

Answer 2: Rather than do a one point determination, we run a series of standards (of known concentration) and then relate their signals to the signal of the unknown sample to find its concentration.Example: We have an analytical instrument available for a chemical analysis. We prepare a blank and a series of external analyte standards (solutions of known analyte concentration). We monitor and record their instrumental signal response. From this information, we prepare a calibration curve plot. Next, we run the unknown sample and record its signal response. Finally we use the calibration curve to find the analyte's concentration Note: External standards are not added to our unknown samples!!!

CalibrationAnswer 1:Function from single point standardization -and its domain: not the best !!!

Calibration Data:Answer 2

FYI: Uncertainty: Finding the best calibration function from the data: linear regression

If one assumes that the calibration curve is linear there are physical reasons for that- there is linear function that can best represent the response of the instrument.If points are out of line: errorThere are methods how to calculate the best fit: linear regression of straight-line calibration curves

We can run a series of standards (of known concentration) and then find the relationship We can use that relationship to relate the signal of an unknown sample to find its concentration. We can use EXCEL program for that.

Example 1 We are trying to determine the concentration of an unknown Fe+2 solution. We have prepared several Fe+2 solutions of known concentration to be used as known standards. We record and plot the concentration and absorbance data for all Fe+2 solutions and unknown solution. We zeroed the standards with the blank to get the net absorbance.# Absorbtion S measured(units) Concentration C (ppm)10.0161.0020.0562.0030.1103.5040.1607.5050.43010.00unknown reading = 0.119 (What is the concentration of this solution??)(1) EXTERNAL STANDARD (standard is not added to the sample)

MFYI: Method of Least Squares/Linear regressionThe method of least squares assumes that the errors in the y values are substantially greater than the errors in the x values. A second assumption is that the uncertainties in all of the y values are similar.

Derivation of the Least Squares Method(Equation for a straight line)y = mx + b Vvertical deviation = dii = yiy = yi(mxi+b)SSome of the deviations are positive and some are negative. To minimize the magnitude of the deviations irrespective of their signs, we square the deviations to create positive numbers.

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di2 = (yiy)2 = (yi mxi b)2

Least-squares slope:m = EQ \f(n \i\su(,, )(xiyi) \i\su(,, )xi \i\su(,, )yi,D)

Least-squares intercept:b = EQ \f(\i\su(,, )(xi2) \i\su(,, )yi \i\su(,, )(xiyi) \i\su(,, )xi,D)

D = n EQ \i\su(,, )(xi2) EQ \B(\i\su(,, )(xi)) 2

How Reliable are Least-Squares Parameters?

Standard deviation of slope:

m = y EQ \r(\f(n,D))

Standard deviation of intercept:b = y EQ \r(\f(\i\su(,, )(xi2),D))

y = EQ \r(\f(\i\su(,, )(di2),n2))

We will see that the error behavior of individual points is also important!

Back to the example 1 NOTE : There is an outliner this is due to the systematic error, maybe the dilution of the standard stock solution was not prepared well in this flaskTAKE IT OUT!!! How to get the correlation function using least square based linear regression: Use Excel!

Excel: this works!!

Using Exel trend line

Sheet1

00STANDARD

19031.7

41068.3

58096.6

Samples

164.3386.043843843864.3

263.5381.23903903963.5

364.5387.24504504564.5

463.3380.037837837863.3

Sheet1

Concentration (ppm)

Instrument response (absorption)

external standard

Sheet2

Calculated concentration (ppm)

Instrument respone (measured)

Results of assay on replicate samples

Sheet3

A spreadsheet for Least Squares: A more detailed analysis in ExcelExample LSQ fit

CHAPTER 04: Unnumbered Table 4.3

CHAPTER 04: Figure 4.12

CHAPTER 04: Table 4.7

CHAPTER 04: Equation 4.27

Is such external standardization the only way to calibrate and acquire the accurate and precise values? No, there are also other methods that include addition of analyte or some other material into the sample (aliquot)

Possible problems with external standardization

SpikeSometimes (often) the response to analyte is affected by something else in the sample, which we call MATRIX.SPIKE is a known quantity of analyte added directly to the sample to verify if the response to analyte is the same as that expected from pure sample observed in the calibration curve.

2. Standard additionSStandard addition: Known quantities of analyte are added to the unknown, and the increased signal lets us deduce how much analyte was in the original unknown. Typically we use the method of standard addition when unknown sample matrix is sufficiently complex. The blank and standards are not representative of the unknown sample and lead to analysis error. This method requires a linear response to analyte.For a single spike, one trial [X]i / [X]f + [S]f = Ix / I s+x [X]i = unknown initial concentration of analyteIx = signal from first solution[S]f = concentration of standard in second solution[X]f = diluted concentration of analyteI s+x = signal from second solution

EExample 2 A blood serum sample containing sodium ion gives a signal of 4.27 mV on a light intensity meter in an atomic emission experiment. The Na+ concentration in the serum is then increased by 0.104 M by a standard addition, without significantly diluting the sample. This "spiked" serum sample gives a signal of 7.98 mV in atomic emission. Find the original concentration of Na+ in the serum.[X]i = unknown initial concentration of Na+Ix = 4.27 mV[S]f = 0.104 M[X]f = diluted concentration of analyte = [X]i I s+x = 7.98 mV [X]i = 0.120 M

FYI: (2) Sampledirect additionabS sample = S spikea b

(1)Sampleseparate aliquots (portions)abS sample = S spikea bFYI: Accurate description Ssample = signal sampleCS= concentration spikedCA= concentration analyte

WWe use standard addition methods when our sample has matrix effects which are hard to duplicate in a blank. Let's again find the amount of Pb+2 in a blood sample. A 5.00 mL blood sample containing lead yielded a signal of 0.712 units. Afterwards, the sample was spiked with 5.00 L of 1560 ppb Pb+2 standard. (note: a small addition) This spiked solution gave a reading of 1.546 units. Find the concentration of Pb+2.[X]i / [X]f + [S]f = Ix / I s+x or [C]i / [C]f + [Cs]f = Sx / S s+x

Be careful with volume! Example 3. CA=1.33ppb

from previous multipoint figure : many points fitGraphic treatment (standard additions plot), alternatively we can perform a series of additions and plot the signal response verses spiked concentration and extrapolate to find the initial concentration. Note, we can plot the x-axis as concentration of spike or as the concentration of spike with dilution.

FYI: Standard addition: How to work with many points: FIT -EXCEL

FYI: Similar: multipont,For standard addition(s)

FYI: Standard addition - Example 4A 5.00 mL sample solution containing an unknown amount of analyte was analyzed and then a series of spikes was added to the sample. The unknown solution was analyzed after each addition. The concentration of the standard used for the spikes was 400.0 ppm. The readings and respective spikes are shown:

so for y=0, x=.1212/1.077=0.1125mL (x axis is volume in mL)

By using, Conc of unknown => (0.1125mL)X(400 ppm) = (5.00 mL)(Conc?)

answer = 9.003 ppm

Chart1

0.119

0.231

0.339

0.442

volume of spike

Signal

Std additions plot

prob1&2

prob#1

00.06prob#115.1206896552

51.48

102.283.67

153.98prob#21.0769230769

204.610.256

Equation of best fit line is y = 0.232X + 0.162

Letting Y = 3.67

solving for the concentration X for unknown = 15.1

prob1&2

conc

signal

prob3

#3For three trials

for the 10 ppmblankdiff in reading and blank

0.1890.0010.188

0.1960.0020.194

0.1930.0020.191

Average0.1930.0020.191

0.191

Unknowndiff in reading and blank

0.1120.0010.111You must consider the blank(10*0.112)/0.191

0.1130.0020.111Correct Unknown Conc is =>5.864

0.1170.0020.115

Wrong answer(10*0.114)/0.1927

Average0.1140.112Unknown Conc if blank ignored5.916

prob4

#4sample wt12.983

sample dilution volume, L0.25

aliquot tested, mL50

final sample dilution volume100

conc of spike std ppm1

volume added, mL10

signal w/o spike0.345

signal w/ spike0.546

Use

[Xi] / [Xi] + [Sf] = Ix / Ix+s

and

(50/100) [Xi] = 0.345

(50/100) [Xi] + (10/100) [1ppm] = 0.546

solve for [Xi] =>

(0.5[Xi])(0.546)=(0.5[Xi] + (0.1 ppm)(0.345)

[Xi] of unkn in ppm =>0.343

%Hg = 0.343 mg Hg/L X 0.25 L X 1 gram/1000mg X 100/12.983 g

%Hg =>0.00066

prob7

volume of stdsignalconc of spikevolume of aliquots =>50 mL

010840

0.11844(0.1*0.531)/(50+0.1)

0.22473(0.2*0.531)/(50+0.2)stand conc0.531M

0.33266(0.3*0.531)/(50+0.3)

0.44010(0.4*0.531)/(50+0.4)

conc of spikesignalConc x =>0.00156molar

0.000001084or

0.0010618441.561mmolar

0.002122473

0.003173266

0.004214010

prob7

Conc of spike

signal

std add plot

prob6

#6

mL addedreadings

00.119y =>zerovolume of sample5mL

0.10.231conc of stand400ppm

0.20.339from graphx =>0.1125348189mL

0.30.442

Conc =>9.0027855153ppm

Conc of unknown => (0.1125mL)X(400 ppm) = (5.00 mL)(Conc?)

prob6

volume of spike

Signal

Std additions plot

prob5

#5

standardscon, ppmsignal

[X]known200.115

[S]known100.321

unknown mixture

[X]unknown?0.266

[S]unknown150.275

use ([X]unk/[S]unk)/([X]kn/[S]kn) = (signal ratio of Unknown/signal ratio of known)

([X]unk/15)/(20/10) = (0.266/0.275)/(0.115/0.321)

[X]unknown found =>80.9985770751

example

conc(Cfinal)unitsmLs addedVfinalppm(added)ppm added/Vf

0.000.1050.000

0.040.2120.1050.120.0399201597

0.080.4090.2050.240.0796812749

0.120.6710.3050.360.1192842942

0.200.890.5050.5100.198019802

x = 0.021 ppm in unknconc of std =20ppm

example

conc [mg/L]

abs units

Std Add plot

FYI: Standard additions example 5 : (the x-axis as the concentration of spike with dilution)This time we have enough sample solution to make several standard addition solutions.We remove five 50 mL aliquots, we read one without a spike. Then we add successive amounts of 20.0 ppm standard to the rest and determine their signal. We plot signal verses the final concentration of standard solutions.

Solve for X intercept! X= 0.021 ppm X=.0903/4.201

Chart1

0.105

0.212

0.409

0.671

0.89

conc [mg/L]

abs units

Std Add plot

Sheet1

00.1190000.119

0.10.231405.17.84313725490.231

0.20.339805.315.09433962260.339

0.30.4421205.621.42857142860.442

Sheet1

0

0

0

0

Sheet2

0.000.105

0.040.212

0.080.409

0.120.671

0.200.89

-0.0200

Sheet2

0

0

0

0

0

0

Sheet3

conc(Cfinal)unitsmLs addedVfinalppm(added)ppm added/Vf

0.000.1050.000

0.040.2121.0050.120.0399201597

0.080.4092.0050.240.0796812749

0.120.6713.0050.360.1192842942

0.200.895.0050.5100.198019802

x = 0.021 ppm in unknconc of std =20ppm

Sheet3

0

0

0

0

0

conc [mg/L]

abs units

Std Add plot

3. Internal StandardIn addition to the analyte you measure add another similar compound (not analyte ) that has similar response ( sensitivity ) as the analyte.

Internal Standard is sometimes added to an unknown sample. The reason may be to verify the signal response in situations where instrument response varies slightly from run to run. For example, an analysis is preformed on different days or different instruments or under different operating conditions. Typically, the internal standard resembles the analyte. Let's say we are separating isomers of octane and determining their concentration on a gas chromatograph (GC). To verify how much is present we might add a known amount of cyclohexane as an internal standard

Internal StandardAn internal standard is a known amount of a compound, different from the analyte, that is added to the unknown sample. A Standard mixture of analyte and standard is prepared before hand. Internal standards are desirable whenever losses of sample are likely to occur during handling or analysis. They are also used to calibrate the instrument when same analysis is done on different days. Use the following relation:

Internal StandardSA(signal due to analyte) =k A CASIS(signal due to int. stand) =kIS CIS

Example Example on internal standard: Note we are not diluting the sample significantly!A lab uses Cu+2 as a internal standard for a Pb+2 analysis in blood. Known solution standards for Pb+2 (1.75 ppm) and Cu+2 (2.25 ppm) give readings such that the signal ratio of Pb to Cu is 2.37. A blood sample was spiked with the Cu standard (2.25 ppm). We assume that the blood sample volume does not change significantly. The ratio of the readings for Pb to Cu in the unknown spiked solution is 1.80. Based on the response for the known amount of Cu+2 added to a blood sample determine, the amount of Pb+2 present.SA(signal due to analyte) = k A CASIS(signal due to int. stand) = kIS CISCACIS= KK=3.05CA=1.33ppb Pb 2+CA=CISSASISKFirst measurementSecond measurement

Uncertainty and instruments: how to use information and make decisions

Accuracy, precision Accuracy vs. precision

Microanalysis - limits of trace analysisThere are detection limits associated with every analytical method. As samples and their analyte content become smaller, there is a real concern how good is the analysis. This is especially true when we get in the ~ parts per million range and beyond. Here are several factors that affect trace analysis:1) the nature of the analyte2) matrix effect of the component, interferences3) reagent quality and sample preparation4) type of assay ( instrumental or lab bench)5) pre-concentration of analyte

-Scale of operation (analytes classification

Scale of operation

The detection limit is the concentration level to which a method can reliably give a signal statistically different from the background signal (noise).Often for an instrumental method, it is the ratio of the average signal amplitude / the average noise amplitude. In this instance the detection limit is defined as when the ratio (S/N = 2).For a wet analysis it may be defined as the Analyte Concentration which produces a signal three times the Std. Dev. of the blank.Sensitivity is the method's ability to discriminate the differences between two trace concentration levels. Stated a little different:Sensitivity is ratio for the change in signal output to the change in concentration of analyte.Sensitivity = I (signal) / C (analyte concentration)

Sensitivity, selectivity, detection limits, noise

Selectivity: A method is selective if its signal is a function of only the amount of analyte present in the sample, not the interferents etc..

NoiseTThree types of noise in electrical instruments: White noise, Drift noise, and Line noise.i) White noise (thermal) is always present. It involves the motion of electrons in the circuits.iii) Drift noise (sometimes called flicker or 1/f ) is not well understood but is known to be inversely proportional to frequency of the signal. It is significant at frequencies lower than ~100 mHz.

iii) Line noise arises from environmental noise (power lines, TV & radio signals, lighting, etc.).Line and drift noise can be reduced by using frequency modulation, filters, and insulation.

Signal Averaging can be used to improve signal to noise ratios in certain instrumental techniques. This is true due to the digital storage of signal data.To improve the signal-to-noise ratio by a factor of n requires averaging n2 spectra.(S/N ratio improvement, Fellgett, 1958)The S/N ratio improvement through signal averaging stems from the fact that the noise, N, will grow by a factor of n (where n is number of times the spectrum is recorded) where as the signal, S, should grow by a factor of n. As seen in the figure above; 4 scans vs. 256 scans, the signal becomes enhanced due to S/N improvement.

There is a working range for an analysis. It should be such that the analyte concentration is at least ~ 10 times the Detectin .Limit and small enough to produce a linear signal response. If performing an established method of analysis, the information about D.L. and linear range for calibration curves can be found.

(*if the analyte concentration is too high(concentrated) such that the signal response is not linear to concentration we can DILUTE. If opposite we can concentrate

Constant sample error stays the same regardless of sample size.Proportional sample error varies with variation in sample size.

HW and practice problemsAll Examples in the Chapter with solutionsExercises (solutions at the end): 5-a, 5-b, 5-c


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