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Atomic Absorption Spectroscopy (AAS) Tutorial

Key Concepts

Atomic Absorption Spectroscopy (AAS) was developed by CSIRO Scientist Dr Alan Walsh in the 1950s.

The steps involved in using atomic absorption spectroscopy (AAS) data to determine the concentration of a species in a solution are:

Atomic Absorption Spectroscopy can be used to measure the concentration of metals in :

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Example : Undiluted Sample

Atomic Absorption Spectroscopy (AAS) can be used to determine the lead concentration in soil collected from the side of a road.

A student prepared standard lead solutions for comparison and the aborbance of each solution was measured.

A road-side soil sample was also prepared.

The results are shown in the table below.

SampleConcentration (ppm)Absorbance
Blank 0.00 0.00
Standard 1 1.00 0.17
Standard 2 2.00 0.34
Standard 3 3.00 0.48
Standard 4 4.00 0.65
Standard 5 5.00 0.83
Sample ? 0.58

What was the concentration of lead in the soil sample?

Step 1: Draw a calibration curve using the data:

  1. Plot the calibration curve using the concentrations and absorbances of the standard solutions (shown as red x's on the graph)
  2. Draw a line of best fit through the plotted points (shown as a red line on the graph)

Step 2: Use the calibration curve to find the concentration of lead in the sample:

  1. Mark the position of the 0.58 absorbance of the sample being investigated (shown on the graph as a blue x)
    (Draw a horizontal line, parallel to the x axis, from absorbance 0.58 until it meets the line drawn on the graph. Shown on the graph as a dotted blue line.)
  2. Read off the concentration of lead in the sample from the graph, 3.50 ppm
    (From the blue x on the line in the graph, draw a line vertically down to meet the x axis. Shown on the graph as a dotted blue line.)

The concentration of lead in the sample was 3.50 ppm.
(Check that your answer is sensible. The absorbance of the sample lies between the absorbance for standards 3 and 4, therefore the concentration of lead in the sample must be between 3.00 and 4.00 ppm)

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Example : Diluted Sample

Samples are often diluted before being analysed.
When this occurs, you will need to take this into account when you calculate the concentration of the original sample.

A student prepared standard lead solutions for comparison and the aborbance of each solution was measured.

A road-side soil sample was also prepared.

10 mL of this prepared soil sample was placed in a 100 mL volumetric flask and enough water was added to make it up to the mark.

The absorbance of this diluted soil sample was recorded.

The results are shown in the table below.

SampleConcentration (ppm)Absorbance
Blank 0.00 0.00
Standard 1 1.00 0.17
Standard 2 2.00 0.34
Standard 3 3.00 0.48
Standard 4 4.00 0.65
Standard 5 5.00 0.83
Sample ? 0.26

What was the concentration of lead in the original (undiluted) soil sample?

Step 1: Draw the calibration curve

  1. Plot the calibration curve using the concentrations and absorbances of the standard solutions (shown as red x's on the graph)
  2. Draw a line of best fit through the plotted points (shown as a red line on the graph)

Step 2: Use the calibration curve to determine the concentration of lead in the diluted sample

  1. Mark the position of the 0.26 absorbance of the sample being investigated (shown on the graph as a blue x)
    (Draw a horizontal line, parallel to the x axis, from absorbance 0.26 until it meets the line drawn on the graph. Shown on the graph as a dotted blue line.)
  2. Read off the concentration of lead in the diluted sample from the graph, 1.60 ppm
    (From the blue x on the line in the graph, draw a line vertically down to meet the x axis. Shown on the graph as a dotted blue line.)
  3. Concentration of lead in the diluted sample was 1.60 ppm
    (Check that your answer is sensible. The absorbance of the sample lies between the absorbance for standards 1 and 2, therefore the concentration of lead in the sample must be between 1.00 and 2.00 ppm)

Step 3: Calculate the concentration of lead in the original, undiluted sample.

Using logic:

  1. Concentration of lead in the diluted sample was 1.60 ppm (see above calculations)
        Since 1 ppm = 1 mg L-1
        1.60 ppm = 1.60 mg L-1
        Concentration of lead in diluted sample = 1.60 mg L-1
  2. Therefore the concentration of lead in the 100 mL volumetric flask used for the dilution was 1.60 ppm or 1.60 mg L-1
  3. The mass of lead in the 100 mL volumetric flask = concentration (mg/L) x volume (L)
        concentration = 1.60 mg/L
        volume = 100 mL = 100/1000 L = 0.100 L
    mass of lead in 100 mL volumetric flask = 1.60 mg/L x 0.100 L = 0.16 mg
  4. All the lead in the volumetric flask came from the 10 mL sample of the original solution.
    Therefore the 10 mL undiluted sample contained 0.16 mg of lead.
  5. Concentration of lead in the 10 mL undiluted sample = mass (mg) ÷ volume (L)
        mass (mg) = 0.16 mg
        volume (L) = 10 mL = 10/1000 L = 0.010 L
    concentration of lead in undiluted sample = 0.16 mg ÷ 0.010 L = 16 mg L-1
        Since 1 mg L-1 = 1 ppm
        16 mg L-1 = 16 ppm
  6. Lead concentration in the sample of road-side soil was 16 ppm (or 16 mg L-1).
    (Check that your answer is sensible. Since the original sample has been diluted, the concentration of the original undiluted sample must be greater than the concentration of the diluted sample used in the AAS experiment)

Using dilution equation (formula):

  1. c1V1 = c2V2
        c1 = lead concentration in undiluted sample = ? ppm
        V1 = volume of undiluted sample in litres = 10/1000 = 0.010 L
        c2 = lead concentration in diluted sample = 1.60 ppm
        V2 = volume of diluted sample in litres = 100/1000 = 0.100 L
  2. Substitute the values into the equation and solve:
        c1 x 0.010 = 1.60 x 0.100
        c1 x 0.010 = 0.16
        c1 = 0.16 ÷ 0.010
        c1 = 16 ppm
  3. Lead concentration in the sample of road-side soil was 16 ppm (or 16 mg L-1).
    (Check that your answer is sensible. Since the original sample has been diluted, the concentration of the original undiluted sample must be greater than the concentration of the diluted sample used in the AAS experiment)

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