Flame Tests Identify Metals

GENERAL CHEMISTRY LABORATORY
FLAME TESTS
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CAUTION!!!CAUTION!!!
BE SURE TO WEAR SAFETY GLASSES AT
ALL
TIMES IN THE LABORATORY
NO EXCEPTIONS TO THIS RULE!

1. To explore the line spectra of a variety of elements by
heating them in a Bunsen burner flame.
2. The data gathered from the flame colors created by the
elements will be used to identify some elements present
in unknown solutions.
3. To get the idea about qualitative analysis.
4. Identification of Metal ion in compound.
FLAME TESTS

The Big Questions
 What is light?
 How is light emitted?
 What do electrons have to do with light?
 What are emission spectra?
 How do flame tests help identify metals?

The Electromagnetic Spectrum
• All light is part of the EM spectrum.
–Most is invisible:
•gamma, X-rays, UV, IR,
microwaves, radio waves
–Visible light: wavelength (w.l.) from
400 to 700 nm.

The EM Spectrum

Longer Wavelength
Higher Frequency and Energy

Longer Wavelength, Lower Energy

Flame Test for Cations
lithium sodium potassium copper

Flame Test
1. Electron absorbs
energy from the flame
goes to a higher energy
state.
2. Electron goes back down to
lower energy state and releases
the energy it absorbed as light.
Light Photon

EM Radiation
Light is a carrier of energy.
–Energy is proportional to frequency.
–Frequency is inversely proportional to
wavelength.
• Longer wavelength = lower frequency
= lower energy.
• Shorter wavelength = higher frequency
= greater energy.

Electrons and Quanta
• Ground state – the lowest energy position
an e-
can occupy.
• Excited state – a temporary high-energy
position.
• Quantum (pl. quanta) – the amount of
energy needed to move an e-
to a higher
energy level.

Electrons and Quanta
• If an atom absorbs exactly 1 quantum of
energy, an electron can be boosted from a
ground state to an excited state.
– The electron is only in the excited state for a
very short period of time.
• Soon the e-
returns to its ground state and emits
the quantum of energy as light.
– In some cases the emitted light is in the visible
spectrum.

Apparatus and materials
1. Bunsen burner
2. Watch glass dish
3. Platinum wire
4. Beaker
5. Salts
6. Hydrochloric acid solution, (dilute)
7. Be careful!

ProcedureProcedure
1. Dip the wire in a acid solution
2. Hold the wire in Bunsen flame (to check
if wire is clean)
3. Place some of the salt on the wire
4. Hold the salt in the flame
5. Note the colour imparted to the flame.
6. Repeat for other salts.
7. Record your results in flame test chart
(first table) , using different colours
8. Use the first table to fill in the blanks of
the second table.

The permitted energy levels of a hydrogen atom.
)
11
( 2
2
2
1 nn
R −=ν
Where, v = frequency
n = the quantum number
R = (Rydberg constant)
R = 3.29 ×1015
Hz
1 Hz = 1 s-1
Calculating the Balmer & Lyman Series
As noted earlier, the four bands
of light calculated by Balmer
could be simply calculated
using the Rydberg equation:

Wavelength (λ): Distance between
two consecutive peaks [unit: nm]
Frequency (ν): Number of waves
per second that pass a given point in
space [unit: s-1
(Hertz)]
λ ν = c
Where C is the speed of light
&
C = 2.9979×108
m/s
Recall that Frequency and Wavelength are related where
frequency times wavelength equals the speed of light.
Since the speed of light is a constant, as wavelength decreases,
then frequency must increase.https://www.facebook.com/fcoursesbd

 The wavelength and frequency of radiant energy are
inversely proportional to each other.
 Thus a radiation with short wavelength has high frequency.
 The wavelength, λ, frequency, ν, and speed, c, of light and
other forms of radiant energy are related as follows:
λν = c
Problem: What is the wavelength in nanometers of an x-ray
radiation that has a frequency of 9.4 x 1017
Hz? The speed of
light is 3 x 108
m s-1
.
Solution: The frequency of 9.4 x 1017
Hz = 9.4 x 1017
s-1
λ= c/ν [putting the value of c and ν]
λ= 3.2 x 10-10
m

Albert Einstine proposed that electromagnetic radiation can
be viewed as a stream of particles called PHOTONS. The
energy, E, of a photon of any form of radiation is proportional
to its frequency, and is given by the relationship
E=hν , where is an experimentally measured physical
constant, known as Planck’s constant (6.626x10-34
Js)

Violet Light 380-450 nm
Blue Light 475 nm
Green Light 495-570 nm
Yellow Light 570 nm
Orange Light 590 nm
Red Light 620-750 nm
400 nm
475 nm
510nm
570 nm
590 nm
650 nm
Range

The relationship among
1.Energy, E
2.Frequency, ν
3.Wavelength λ
4.Speed of light, c (where the value of c is 3 X 108
m s-1
)
5.Planck’s constant, h (where the value of h is 6.626 X 10-34
Js)
might be described as follows:
λ ν = c λ = c/ ν ν = c/ λ
E = hν λ = h c/ E

Problem: What is the wavelength in nanometers of a photon
whose energy, E, is 3.82 x10-19
J? In which spectral region can this
radiation be detected? [h= 6.626x10-34
Js and c= 3 x 108
m s-1
]
Solution: We know that,
E=hν
ν = E/h=(3.82 x10-19
J)/(6.626x10-34
Js)= 5.77x10-14
s-1
From another equation we know that,
λ= c/ν = (3 x 108
m s-1
)/(5.77x10-14
s-1
)
or, λ = 5.20x10-7
m
or, λ=5.20x10-7
m(109
nm/1 m) = 520 nm
The radiation with the wavelength of 520 nm falls in the visible
range, which is between 400 and 700 nm.

Exercise-1: What is the wavelength in nanometers of a photon
Js and c= 3 x 108
m s-1
]
Solution: 401 nm (Violet color)
Js and c= 3 x 108
m s-1
]
Solution: 510 nm (Green color)
Js and c= 3 x 108
m s-1
]
Solution: 662 nm (Red color)

Flame Tests Identify Metals

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