Hi Bayu,

I suggest you should,

1. Find concentration of (sample) X and (blank) Y first from the calibration curve.

2. Minus the concentration of the blank from the sample ( X - Y)

3. Multiply concentration of sample X by D.F

Please note that sometimes high results can mean there is contamination during analysis. If you suspect this, do a repeat.

Also, I advise you to run a known concentration (reference) sample throughout the whole analysis even during dilution steps; a leaf or soil reference would suffice.

Hello there,

The answer is simple. If the concentration of an analyte in the diluted sample is X (found from the calibration graph), then the concentration of analyte in the original sample is as follows:

concentration of analyte in sample= ((X- blank) * (100/25) * (250 / weight of sample)).

I hope this is clear, if not contact me again.

C is apparently what you should do,but you have to include the weight factor volume factor and dilution factor. First, i do not know the model of your AAS instrument?

some instruments like the one i use: SHIMADZU AA-7000,controlled by wZARD software has the ability to give you the actual conc. using the formular: Actual conc=DFXVFXCF/WT where:-DF-dilution factor,VF-volume factor,CF-correction factor (e.g 0.0001ppm = 100%) .....hope this helped

thank you so much for the answers

so we can conclude that I must determine the concentration of sample and blank, so the real concentration of analyte is the difference between them, right?

hmm what if the absorbance of blank is lower than the intercept of calibration curve?

should I go with A option?? why this option is not right?

i just want to know the right way to get the concentration of the analyte in the volumetric flask.

Hi there,

If the concentration of the blank is lower than the intecept, it just means that your blank is of high quality, so the concentration of analyte is not able to be detected since the detection limit (as given by the intercept) is higher than the analyte level in the blank. Remember the errors are very high at lower levels. In such cases I suggest you take the intercept as the blank level and subtract this value from the value of the diluted sample - it won't make much difference if you include errors in the analysis. Remember, operational standards, internal standards, replicates and international standards must be included in the analytical batch for quality assurance and accuracy of the analysis. i hope this helps.

Hi

It is very simple. there are two ways to run Blank in AAS. You can choose it as blank and run then you need not to bother about subtracting the blank or run blank as sample and subtract from the absorbance of sample. In your case you cannot subtract the blank from your diluted sample as the acid concentration is differing in both. u need to dilute your blank also same way as your sample and then you need to subtract. or while you dilute your sample you have to adjust the acid concentration in such a way that both are having same concentration of acid.

In your case you cannot subtract the blank from your diluted sample as the acid concentration is differing in both. u need to dilute your blank also same way as your sample and then you need to subtract. or while you dilute your sample you have to adjust the acid concentration in such a way that both are having same concentration of acid.

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This is a very important point !

Hello Bayu,

considering your additional remark on the very low signal of the blank solution I would add some general remarks on effects in AAS, which might hamper your results. On the one hand you can fit and calculate the signals of the analyte by use of an external calibration curve, taking in account the blank; this is the common method. As indicated by the colleagues the blank must be adjusted to the matrix composition of the analyte solution. Unfortunately, in AAS there are many element combinations, which will yield for the very same element concentrations a different signal in the analyte solution and the calibrations solution (e.g. depression of Sr (or Ca) signals by tiny admixtures of Al (or P, B); increase of signal by background noise due to high background concentrations). So the adaption of blank solutions or calibration solutions needs the knowledge of the matrix composition. To a certain degree you can avoid the problems by using the method of standard addition (which is an internal calibration). I have to admit: it is a rather cumbersome method; but in many cases a good advise. And you need a good quantity of analyte solution. Prepare 4 identical volumes of analyte. Take the first volume as it is; this volume has the conc “x”. To the second volume you add a tiny amount (“A”) of high concentration of the element you are interested; the solution has the conc “x + a”. Third volume: add two times “A” and you get “x + 2a”. To the next volume you add 3 “A” which yields “x + 3a”. Be sure that the resulting total concentrations do not exceed the calibration curve and, possibly, remain in the linear range (straight line; second order curve should be avoided for this method). After measuring you plot the signals in a X-Y-axis diagram. The signal of the first solution is placed on the Y-axis at X = zero. On the X-axis you plot in equal distances the increasing conc’s of “a” and plot the related AAS-signals as Y-values. All values should result in a straight line. Extend the line backwards until it intersects the X-axis (negative branch). Here you can read off the concentration of the analyte.

The method of (internal) standard addition circumvents many problems due to matrix effects. However, in case of high background concentrations you also should check the position of the zero-value by adjusting a blank to the background. This value would be the X-Y-axis intersection.

Best wishes

D. Zachmann

Oladele Osibanjo, Basel Convention Coordinating Centre for Africa for Training and Technology Transfer, University o Ibadan, Nigeria.

Hi Bayu,

Information on weight of sample taken is missing in your statement of the problem. This parameter is critically important in calculating the concentration of analyte in sample digested which l suspect is a solid. The unit of concentration is ug/g or mg/kg for solid sample . The corresponding concentration for liquid sample is ug/ml or mg/l.

The AAS instrument must have been calibrated with different known concentrations of standard analyte solution commercially purchased or prepared from pure metal or salt of the method according to recommended analytical methods.

It is also important to carry out analysis of blank (otherwise called procedure blank analysis) to care of impurities that may be present in solutions used to digest the sample and which would add to the actual concentration value of analyte.

The calculation of concentration of analyte in the sample digested is straight forward as follows :

Let the concentration of an analyte in the diluted sample be X (obtained from the calibration graph or automatically generated by the AAS previously calibrated), then the concentration of analyte in the original sample (where w is weight of sample taken in g) is as follows:

Concentration of analyte (ug/g or mg/kg) in sample= ((X- blank) * (100/25) * (250 / weight of sample in g)).

The expression can generally put as follows :

concentration of analyte (ug/g or mg/kg) in sample= ((X- blank) * Dilution Factor (DF) * (Total volume of digest solution / weight of sample in g)).

Please follow the following steps carefully.

1. Let us assume that you have taken W gm of sample, after digestion you made upto 250 ml (Let us say that it is your original sample solution).

2. You further diluted original sample solution to 4 times (DF = 4).

3. Absorbance of dil. sample solution is : x - y (after blank correction)

Find the conc. from the calibration plot already prepared w.r.t. abs. x-y.

Let us say that conc. : z mg/L

4. Conc. of your original sample solution: 4*z mg/L

5. Hence quantity of metal present in your sample: (4z*250)/1000 = W

I have difficulty with Parashuram Sahoo's solution. Unit of concentration is analyte weight per sample weight for solids (that is mg/kg or ug/g) OR analyte weight per volume of final solution after digestion, for solutions (that is mg/L or ug/mL). Hence quantity of metal present which is amount or W is different from concentration of metal in solution.

Oladele Osibanjo

Thank u Oladele for pointing out a small mistake in my solution. As per my solution the quantity of metal in W g of solid sample is (4z*250)/1000 (let it be W1 mg)

Hence conc. of metal expressed as mg/kg is : (W1*1000)/W

In my earlier answer I presented in terms of total quantity which should be W1 but not W. It is a typographical error. I hope now it is clear.

The first method is correct.

It is recomended to find the concentration with calibration curve first then multiply that by the DF.

First one is correct

Muhammad Aslam's suggestion is correct. Very simple, your AAS only "sees" your sample as is, it does not know anything about its history, dilution or whatsoever. As long as the previous dilution does not have any effect on the solubility, simply determine the concentration in your sample and then multiply the result by the DF.

C is correct

Answer C is correct. From calibration curve you can find a concentration of your sample (new sample in diluted form), and this concentration multiply by dilution factor.

Using the calibration curve (concentration vs absorbance) you calculate the concentration of both blank and sample, after that you apply the dilution factor or other corrections.

The correct answer is C. You said the absorbance of the original sample was out of range. Consequently, you diluted the sample. Therefore, first you find the concentration of the diluted sample and then the original sample using the dilution factor. There is only one choice (C)

Concentration = (sample concentration - blank concentration) * final volume * dilution factor/ sample weight

Your first dillution was in 250 mL.In a second step, you dilluted your sample 4 times.

It is easy make your calculation considering that your sample was dilluted in one liter.

Easy.

C is correct

Hi,

You would have calibrated the AAS before analyzing the metal.

In present case you have analyzed sample and blank :- to calculate the sample concentration, please follow these step to obtain correct results.Using the concentration obtained from the calibration curve you can proceed.

Concentration of samples X :- X1 ppm ( parts per million)

Concentration of Blank Y :- Y1 ppm ( parts per million)

A=(X1-Y1) gives you the exact concentration of the sample injected on AAS. Multiply the obtained results with Dilution factor =4, A=(X1-Y1)*4

To obtain final concentration of metal in the orginal sample,

Metal concentration in digested sample (ppm)

= ( A X 250 ) / sample wieght taken for digestion

ARUN

1- If blank sample solution is different from that of the calibration curve:

Take de absorbance, minus blank sample absorbance, find de concentration with de calibtration curve, then multiply by DF.

2- if both blank solutions are identical, no need to subtract the blank absorbance as it was already considered when setting the "zero" reference.

Hi, Bayu,

If you digested a sample in acid/s, the most accurate is to model (match) a gradually rising metal concentrations into the „matrix“ consisting of the same concentration/s of acid/s used (to take in consideration resulting acid/s concentration/s in final volume of digerate). So you get the calibration curve of target (to be determined) element. The first „point“ of this calibration curve (consisting only of this acid/s concentration/s – with zero metal concentration to be determined – is blank of calibration curve (Blcc). Simultaneously prepare 3-4 solutions with the same acid/s concentration/s used for sample digestion and make with these solutions the same operations as with a sample (the same volume in the same bottle at the same temperature, and fill in the same volume as in sample case). There are operation blanks (Blop). In optimal case is an absorbance (concentration) of Blcc the same or very close to mean of Blop values. If not, you should take into consideration the difference (Blop-Blcc). If you made the „points“ of your calibration curve in units mg/L (numerically, but not consistently the same as in units ug/mL), the final results in your case is:

mg/kg element in dry matter:

(conc Blop – concBlcc)*conc M*(250/n)*(100/25)*100/(100-w),

where

conc… concentration in units ug/mL determined from (linear part of) calibration curve; M… element to be determined; n… mass of sample (in g) with determined moisture w (%).

However, in many cases is not the acid/s concentration/s in Blop or in Blcc the same as the ones after sample digestion (due to evaporation, initial acid/s reduction due to reaction with organic matter, etc.).

Hence, it is recommended to determine actual concentration/s of acid/s after material digestion (e.g. volumetrically) and to match the real finish acid/s concentration/s for calibration “points”.The metal concentration (contamination) should be keep the same.

Karel.

The correct is C, if the signal of the blank is part of the calibration curve, you can directly interpolate the signal of the sample (diluted) and then the C option. If the blank signal is not part of the calibration curve first subtract the blank signal from de sample signal and then interpolate, at the end consider the factor dilution x 4.

Yes, the answer is 'C'. But the detailed discussion is given in my answer before a few days back.

easiest way to avoid confusion (however it is not at all confusing).

Put the blank solution in each of compartment and set it auto zero option.

Now,

replace one of blank solution (from analysing compartment) by the the test sample and take absorbance (ideally value 0.1 -0.9 ).

if absorbance is high then dilute it further (x/y/z times) unless and until its value comes in between 0.1-0.9.

Now calculate corrosponding concentration of diluted solution from standard curve.

Multiply it bye total dilution factor.

You would got concentration in the sample solution (you have)

A. take the absorbance of sample (X) minus blank absorbance (Y)

B. find the concentration with calibration curve

C. multiply that by the DF

'C' is the correct answer.

first, take the absorbance of sample (X) minus blank absorbance (Y)

second, find the concentration with calibration curve

then, multiply that by the DF

What was the mass of your sample?

I agree to Asaimonat Md Noushad Javed .

you should use the blank solution and set it zero in each measurement.

A is correct way of calculating metal conc ,ist minus the blank then multiply with dilution factor

The dilution factor must not be included in the interpolation, first the interpolation and at the end multiply the concentracion, not the signal.

you are taking the right steps.

First you determine  absorbance of diluted sample. With respect to that determine concentration of diluted sample using calibration plot. Multiply dilution factor to that concentration to get actual concentration of sample. In calibration plot blank value as zero concentration must have been used. Hence no need of doing blank again.

Dear, simply do this

concentration of respective metal in sample = A - B (Where B is concentration of sample * DF (Dilution Factor) * (Total volume of digest solution (1st time dilution)/ weight of sample in g).

I think this link is very useful, 

http://www.instanano.com/characterization/theoretical/concentration-calculation-from-uv-vis-absorbance/

Bayu, Dissolve/digest a known weight of metal in the acid and make the total solution 250ml. Calculate the concentration of the metal in solution.also run blank and make 250ml solution.Now measure absorption of analyte solution.If it shows good peak,good or have known dilution,till you get the good peak.Calculate Molar absorptivity.

UseLlambert law to calculate.You should be sure that the absorbance shown by analyte sample is due to metal ion and not due to solvent alone.

Dear colleague,

I proposed C method

C is correct as the instrument deal with sample only so use calibration curve first to know concentration the multiply by Dilution factor

Option C is the correct one.

as 0.2 is the gravimetric factor for Lead, Please  what is gravimetric factor for other metals lide Iron, Cadmium, Copper,Zinc, Potassium etc