Final Exam: EXPT 2 - Atomic Absorption Spectroscopy of Sodium

Layout of the AAS (Components)

Light source → Sample Cell → Monochromator → Photodetector → Read out

General Principle of absorbance spectroscopy

Using Atomic Absorption Spectroscopy we will vaporize and atomize a sample of chips and use electronic absorption of the resulting gas-phase atoms measured using specific wavelengths to determine the amount of sodium in our sample.

AAS is very sensitive, high in selectivity, and its widely applicable.

Wavelength, frequency, and energy

Wave nature: c = λ \* v

Particle nature: E = hv = c / λ)

Planck’s equation:

Planck’s constant, h = 6.626 *10^-34 J*\*S

Energy transitions are quantized

E atom = h V = E photon

Atomic Emission Spectra

Observing photons of light emitted at specific wavelengths as the excited electrons return to the ground state

Atomic Absorption Spectra

Measures the energy absorbed in the ground state at specific wavelengths that excite the electrons and result in emission of light.

Beer’s Law

The amount of energy absorbed can be quantitated

A = e b c

e = molar absorptivity

b = sample cell length

c = concentration

Light Source

Typically a hollow cathode lamp, this case its a sodium cathode

Sample Cell

is the flame chamber onto which sample is aspirated into

Monochromator

is still necessary to filter out other interfering wavelengths from combustion process

Calibration Curve

x - concentration (ppm Na)

y - Absorbance

Standard stock solutions are prepared via 100 ppm stock solution

M1V1 = M2V2

(0.2, 0.5, 1.0, 2.0)

V1 = (2.0 ppm)(100 mL)/ 100 ppm

ppm

= mass solute / mass solution \* 10^6

1 mg / L = 1 ppm

1 ug / mL

1/1000 g = 1 mg = 1\*10^-3 g

1 ug = 1\*10 ^ -6 g = 1/1000000

% volume

= vol solute/ vol solution \* 100

% mass

mass solute/ mass solution \* 100

Dilution Factors

Since very difficult to weight out a small mass of chips ( 0.0243g) accurately, this is the reason to make up 100 x solution (2.43g) and then dilute to 1/100 for AAS sample!

Weight % of an aqueous solution containing 10. grams of NaCl per liter?

( 10. grams NaCl ) / (1000 g soln) x 100 = 1.0% (assuming d = 1.0 g/ mL)

Weight % of an aqueous solution containing 10. mg of NaCl per liter?

PPM of the solution in #2 instead?

( 0.010 g NaCl ) / (1000 g soln) x 100 = 0.001%

\
( 0.010 g NaCl ) / (1000 g soln) x 10^6 = 10 ppm

“Based on the manufacturer’s claim of 115 mg Na per serving and a 28-g serving size, calculate the proper chip sample weight that will give you a 100-mL solution whose Na content is 1.0 ppm (which is in the middle of the desired 0.2-2.0 ppm range of your calibration curve).”

If 1.0 ppm is 1.0 ug /mL times 100 mL → 100 ug Na in 100 mL = 1 ppm

\
115 mg in a 28 g serving size → 115 ug in 28 mg chips (0.028 g)

\
(100 ug / 115 ug ) x 28.0 mg = 24.3 mg

What is the purpose of the flame in an AAS?

To vaporize/atomize the sample into the path of the light

What type of sample is AAS used to analyze?

Typically solution containing metals

In AAS, what quantities are plotted on the x and y axes to obtain a straight-line calibration curve?

x-axis - time

y-axis - absorbance

What is Planck’s equation?

E = hv

What is meant by the term “quantization”?

To set things into discreet or specific levels – in this case discreet energy levels or packets of energy.

Which kind of light has the greatest (highest) wavelength – visible, ultraviolet, x-rays, or infrared?

Which has the greatest frequency?

Which has the greatest energy?

\

Infrared

X-rays

X-rays

Calculate the volume of a 50-ppm Na+ stock solution needed to make standard Na+ solutions of 3.00 ppm, 2.00 ppm, 1.00 ppm, and 0.200 ppm in 100 mL volumetric flasks (not by serial dilution).

M1V1 = M2V2

6\.0 mL, 4.0 mL, 2.0 mL and 0.4 mL

Based on the manufacturer’s claim of 115 mg Na per serving and a 28-g serving size, calculate the proper chip sample weight that will give you a 100-mL solution whose Na content is 1.0 ppm (which is in the middle of the desired 0.2-2.0 ppm range of your calibration curve).

If 1.0 ppm is 1.0 ug /mL times 100 mL →

100 ug Na in 100 mL = 1 ppm (or 0.100 mg Na)

0\.100 mg Na sample / 115 mg Na serving = x mass sample / 28.0 g serving size

x mass chip sample = 0.0243 g