DP Chemistry:
Term 01: Practical Task 01
Metal Salt Colour Test
And
Hydrogen Spectroscopy Test
Write Up
Part One: Metal Salt Colour Test
Aim:
To determine the chemical reactions that occur during the exposure of selected metal
salts to a blue Bunsen flames. Then using the results collected to determine the nature
of two unknown elements.
Hypothesis:
Energy, such as heat when applied to an element can cause electrons to “jump” to
higher energy shells the resulting energy if of an identical level to visible light will cause a
visible reaction. As every element is unique in its composition the energy given off
during a reaction due to heat varies from element to element therefore as visible light is
a range a range of reactions will be observed that can be attributed to a specific
element.
Apparatus:
-Bunsen burner
-Lab coat
-Safety Glasses
-Matches
-Nichrome wire in cork
- Amounts of Sodium Chloride (NaCl), Potassium Chloride (KCl), Strontium Chloride
(SrCl2),
Calcium Chloride (CaCl2), Copper (II) Chloride (CuCl2), Iron (III) Chloride (FeCl3),
Magnesium Chloride (MgCl2), Barium Chloride (BaCl2) and Lithium Chloride (LiCl)
-Amounts of Unknown element A and Unknown element B
-Hydrochloric Acid (HCl)
-Twelve watch glasses
-Heatproof mat
-Spatula
Method:
1. Rinse watch glasses and label them with the element names so as to not
confuse them.
2. Collect approximately a two spatulas of each element, cleaning the spatula
between each collection so as not to contaminate elements with one
another. Place elements into their respective watch glasses.
3. Fill the remaining watchglass with HCl.
4. Place heat proof mat and Bunsen burner on workspace.
5. Set up Bunsen burner, closing air hole.
6. Light Bunsen burner.
7. Sterilize the nichrome wire by submerging the end into the watchglass of
Hydrochloric acid.
8. Get rid of the excess acid on the wire by exposing the submerged tip to the
Bunsen flame.
9. Dip wire into HCl briefly (to make sure the metal salts adhere to the wire) dip
the wire into the watch glass containing NaCl.
10. Place the wire into the Bunsen flame.
11. Observe any changes in colour or any other physical reactions, noting them.
12. Clean the wire after all reactions between the metal salt and flame have
ceased by again dipping it into the acid.
13. Repeat steps 8 to 12 but with differing salts until all elements including the
unknowns have been tested. (If clarification is needed or the reaction is
missed repeat with the same element until satisfied)
14. Tabulate and process data, discuss findings and identify unknowns.
Results:
Tabulated data of visible reactions between elements and blue Bunsen flame:
Element
Colour and other observed characteristics
Sodium Chloride (NaCl)
Bright Orange
Potassium Chloride (KCl)
Sparks, Bright Orange
Strontium Chloride (SrCl2)
Vivid Red
Calcium Chloride (CaCl2)
Dark Orange
Copper (II) Chloride (CuCl2)
Light blue/green. White orange flames in
centre. Pinkish flames at top.
Iron (III) Chloride (FeCl3)
Bright yellow sparks, followed by
purple/dark blue/green flame, then a
orange flame/ green streaks
Magnesium Chloride (MgCl2)
Sparks
Barium Chloride (BaCl2)
Faint green with orange
Lithium Chloride (LiCl)
Red orange
Tabulated data of visible reactions between unknown elements and blue Bunsen
flame:
Element
Colour and other observed characteristics
Unknown A
Sparks, Bright Orange
Unknown B
A combination of orange, yellow and
green.
From comparison between the results of the known elements the unknown elements
can be named:
Unknown A: Potassium Chloride
Unknown B: Barium Chloride
Discussion:
An oddity in the results above would be in the complexity of some observed reactions
and the comparatively simple and short nature of others. I would attribute this to
systematic error and/or random human error, namely two things; the mixing of elements
when they were being collected (the spatula might have contained traces of another
element) causing a systematic error right from the onset or the nichrome wire being not
completely sterilized after each trial, due to random human error, this in turn causing
traces of another element to remain on the wire and thus react simultaneously with the
element being tested.
The idea that the wire was not properly sterilized can be supported by the fact that the
earlier results are, for the most part, the simpler ones, i.e. the residue built up after
multiple tests, which is plausible. Though, however the latter two observations do
debunk this idea quite a bit.
In order to preserve experimental validity it would be prudent to, in future cases, use a
certain spatula for each element during sample collection and a specific wire for each
element as well eliminating not only the chance of random error but having to sterilize
the wire after each trial. Of course if using multiple wires is not an option simply ensuring
that all residues are thoroughly removed using HCl and immersion of the wire tip in the
Bunsen flame would suffice.
Evaluation:
Despite the confusion and possible mixing of elements the main characteristics of each
element could still be deduced, to clarify, there were no insurmountable obstacles in
terms of identifying the characteristics of the specific metal salts, confusion perhaps but
not undue difficulty.
Part Two: Hydrogen Spectroscopy Test
Aim:
To ascertain via the use of a spectrometer the wavelength produced by “excited”
hydrogen gas and from this working out the energy values produced by hydrogen gas
in this state.
Apparatus:
-Power pack
-Hydrogen gas discharge tube set up
-Spectrometer
Method:
1. Turn on the Hydrogen gas discharge tube (not for extended periods, e.g. over
half a minute otherwise a burnout may occur).
2. Using the spectrometer observe the, and take note of the coloured lines
produced.
3. Compile the resulting data into a table showing both quantitative and
qualitative data (meaning colour and the wavelength in Angstroms).
4. Process data.
Results:
Tabulated data of wavelengths calculated via spectroscope of Hydrogen gas tube
Colour
Wavelength (in Angstroms)
Violet
4600
Green
5400
Orange
5800
Red
6700
The two equations used to convert Wavelength to Energy are as follows:
Energy (in Joules or Kilojoules) = Planck’s constant (4.00 x 10^-13 kJ.s.mol^-1) x Frequency (In Hertz
or 1 second ^ -1)
Speed of light (3.00 x 10^8 m.s ^ -1) = Wavelength (Nanometres, Centimetres or Angstroms) x
Frequency
As we need to find the frequency using the wavelength, which is a known value the
equation: Speed of light = Wavelength x Frequency will be used, albeit after it has been
rearranged to make the frequency the subject of the equation.
It can be rearranged as thus: Frequency = Speed of light
Wavelength
Using the frequency acquired using the above equation we can then use the next
equation, which requires the frequency to be a known value,
To find the energy values; Energy = Planck’s constant x Frequency
The working for all four collected data values are below:
Violet (4600 Angstroms):
Frequency = 3.00 x 10 ^8 m.s ^ -1
4600 Angstroms x 10 ^ -10 metres = 4.6 ^ -7
4.6 ^ -7 m
Frequency = 6.52173913 x 10 ^14 Hertz
Energy = 4.00 x 10 ^-13 kJ.s.mol ^-1(Planck’s constant) x 6.52173913 x 10 ^14 Hertz
Energy = 260.8695652 Kj.mol ^-1
Green (5400 Angstroms):
Frequency = 3.00 x 10 ^8 m.s ^ -1
5400 Angstroms x 10 ^ -10 metres = 5.4 ^ -7
5.4 ^ -7 m
Frequency = 5.555… x 10 ^14 Hertz
Energy = 4.00 x 10 ^-13 kJ.s.mol ^-1(Planck’s constant) x 4.01677563 x 10 ^14 Hertz
Energy = 222.222… Kj.mol ^-1
Orange (5800 Angstroms):
Frequency = 3.00 x 10 ^8 m.s ^ -1
5800 Angstroms x 10 ^ -10 metres = 5.8 ^ -7
5.8 ^ -7 m
Frequency = 5.172413793 x 10 ^14 Hertz
Energy = 4.00 x 10 ^-13 kJ.s.mol ^-1(Planck’s constant) x 6.52173913 x 10 ^14 Hertz
Energy = 206.8965517 Kj.mol ^-1
Red (6700 Angstroms):
Frequency = 3.00 x 10 ^8 m.s ^ -1
6700 Angstroms x 10 ^ -10 metres = 6.7 ^ -7
6.7 ^ -7 m
Frequency = 4.47761194 x 10 ^14 Hertz
Energy = 4.00 x 10 ^-13 kJ.s.mol ^-1(Planck’s constant) x 6.52173913 x 10 ^14 Hertz
Energy = 179.1044776 Kj.mol ^-1
Percentage Uncertainty:
Violet: 1.086956522%
Green: 0.925925925…%
Orange: 0.862068965%
Red: 0.746268656%
Colour
Energy
Red
179.1
Yellow
206.9
Green
222.2
Violet
260.9
(Kj.mol ^-1),
Correct to 1
decimal place
Discussion:
The percentage uncertainties were used instead of the absolutes for two reasons; firstly,
if absolute uncertainty were used it would have to be expressed in Angstroms and that
is no good indicator of uncertainty as the final answers are not in Angstroms. Secondly,
the values are being multiplied and divided therefore the percentage uncertainties are
needed, the reason of them not changing is that there were no other uncertainty
values to alter them, there being only one piece of measured data used, the rest were
given constants (id est the speed of light and Planck’s constant).