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Lab: Flame Test by 1t94zsh0

VIEWS: 38 PAGES: 3

									                                           Lab: Flame Test

Introduction:
        The normal configuration of the electrons about the atom or ion of an element is called the
“ground state”. The electrons of this stable particle are at their lowest possible energy level. However,
when these stable particles are heated to very high temperatures, some of the electron leave their
ground state and become excited; that is, these electrons move to a higher energy level. They do so by
absorbing energy in the form of heat. This excited state is quite unstable and the tendency of these
“excited” electrons is to return to their ground state. As these electrons return to the ground state,
they release the absorbed heat energy in a different form. Some of this released energy is in the
visible region of the electromagnetic spectrum that we see as different colors of light. An excited
electron that falls from the 6th energy level down to the 2nd energy level emits a different color of light
than an electron falling from the 6th energy level to the 4th energy level. Thus, when heated, metals
can emit different colors based on where their ground state electrons “live”. Using colors to identify a
metal is called a flame test.

Materials: burner       24-well reaction plate
           Matches      modified jumbo pipets holding various known solutions
           7 Q-tips     unknown solution
           Cobalt sheet

Procedure:
1. Put on goggles and apron
2. Light your burner and properly adjust the flame
3. Soak the end of a Q-tip in a solution of sodium nitrate (NaNO3)
4. Without burning the cotton, passed the soaked end of the Q-tip slowly back and
       forth through the flame of the burner until there is a color changed in the
       flame (NOTE: this may require 6 or more passes through the flame)
5. Record the color (make yourself a chart!)
6. Repeat the process with the remaining solutions, including the unknown
7. Repeat steps 3-6 using the cobalt sheet

A word of caution….

The flame test’s effectiveness may be limited by contaminants in the solutions, particularly sodium.
Sodium is present in many compounds (even though it isn’t on the label!) and can color the flame.
Cobalt glass is often used to filter out the yellow color given off by sodium.
Name _______________________________

                               Lab Report Sheet for Flame Test Lab
                    (to answer these questions you will need to reference chapter 10 )

Pledge:

1. Why do the chemists use flame tests?



2. Describe, in terms of electrons, how light is produced in a flame test. Be sure to include the terms
ground state and excited state.




3. Is the flame color a test for the metal or for the anion in each solution? _____________.
Explain




4. List the ions and their color. Note the data given is not in ion form.
Solution ID                        Ion                   Flame Color

Copper(II)Nitrate                                       Green
Barium Nitrate                                          Yellow Green
Rubidium Nitrate                                        Deep Red
Sodium Nitrate                                          Intense yellow
Potassium Nitrate                                       Pale pink-violet
Indium nitrate                                          Blue

5a. Which metal ion produced color in the highest-energy range (do not ask Mr. Campbell or Ms.
Albach whether it is with or without the cobalt – you figure it out!)?



b. Which metal ion produced color in the lowest-energy range?




6. What letter is your unknown? ____A_______
     Color : Yellow Green



7. What metal is present in your unknown? _______________. How do you know?
8. What would happen to the color of the flame if you mixed 2 of the solutions together (would the 2
colors mix and become 1 new color or would the two colors be separate)? Explain.




9. Planck’s equation relates the energy of a photon of light to its frequency.
              E = hv
              Where Planck’s constant h =6.64 x 10-34 J∙s
              And frequency (v) is measured in 1/seconds or hertz(Hz)

   a. If the frequency of a red spectral line is 1.60 x 1014 Hz, how much energy does one photon of
      this light have?




   b. If the frequency of a violet spectra line is at 2.50 x 1014 Hz, how much energy does one photon
      of this light have?




10. The visible light spectrum represents only a small portion of the electromagnetic spectrum.
Infrared and ultraviolet radiations lie directly outside the visible range.
       * UV radiation is dangerous and is located just past the violet region on the spectrum.
       *IR radition is harmless and is located just before the red region on the spectrum.
Based on your calculations in question 9 above, explain why you think UV is more dangerous than IR.




13. If you had two metals that have the same flame color, what additional test could you do to
distinguish between them.

								
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