YEAR 12 Week 2 Answers

QUALITATIVE ANALYSIS

1. Compare and contrast the techniques of paper chromatography and thin layer chromatography by listing the advantages and disadvantages of each.

	Paper Chromatography	TLC
Stationary Phase	Absorbent Paper	Silica Fused to Glass or Plastic Backing
Solvent	No difference	No difference
Expense	Cheap	Relatively expensive ($2-5 a plate)
Preparation	Little	Just slightly more
Corrosive materials?	No ... dissolves paper	Yes
		Better Separation
		Better for smaller amounts
	Easy to handle & store	Very Fragile
	Slow	Faster
	More suited to polar & water soluble compounds	Better separation for less polar compounds

2. The dyes in a red jelly bean were extracted and separated using paper chromatography. Two spots were measured. One of these spots moved 0.9cm while the other moved 3.7cm. The solvent front moved 7.2cm.
(a) Draw the chromatogram for the jelly bean

(b) What are the likely dyes present in the bean?
R_f of Spot 1 = 0.9/7.2 = 0.13 ... this is close to the R_f of Green S (0.11) so spot 1 is possibly Green S
R_f of Spot 2 = 3.7/7.2 = 0.51 ... this is close to 0.48, so spot 2 is likely to be erythrosine

(c) What possible reasons may there be for small differences in the R_f values?
Difficulty in locating centre of spot, variations in conditions (thickness of stationary phase, temp etc)

(d) Why is the R_f value never more than 1?
It is impossible for the dye to be pushed ahead of the solvent ... ie it must be dissolved (desorbed) into the solvent for separation to take place.

(e) What is meant by the terms mobile phase, stationary phase, adsorption and solvent?
Mobile Phase: This is the liquid or gas that sweeps (and dissolves) the substance through the separating material
Stationary Phase: This is the solid which separates the components by adsorption.
Adsorption: Sticking to the surface. This is done by means of bonds between the molecules at the solid's surface and the molecules in the components of the sample.
Solvent: A liquid which dissolves a solid, liquid or gas.

3. The amino acids in a sample of fruit juice can be determined using paper chromatography.
(a) Calculate the R_f value of each spot on the chromatogram
This is measured directly from the TLC using a mm ruler. Spots were measured to the estimated midpoint.
Solvent front moved 6.3 cm

R_f of A = 5.2/6.3 = 0.83

R_f of B = 3.2/6.3 = 0.51

R_f of small spot in C = 4.5/6.3 = 0.71

R_f of large spot in C = 1.8/6.3 = 0.29

(b) Attempt to identify the amino acids in each juice

Spot A's R_f is very close to that of Isoleucine (0.81)

Spot B's R_f is similar to that of Alanine (0.56)

Spot C (small) R_f is similar to that of Valine (0.75)

Spot C (large) R_f is similar to both Serine (0.27) and Taurine (0.33) - no firm identification can be made in this instance.

(c) Which amino acid was least strongly adsorbed onto the paper?
Spot A (Isoleucine) as it had the greatest R_f

(d) Which amino acid was most strongly adsorbed to the paper?
The larger of the Sample C spots as it had the lowest R_f.

(e) Which amino acid showed the strongest attraction to the mobile phase?
Spot A (Isoleucine) as it had the greatest R_f

(f) Do any of the spots show signs of incomplete separation of 2 components? Explain.
Although it is not clear in this case, it is possible that the larger of the Sample C spots does as it is spread out, and has a R_f value in between both Serine (0.27) and Taurine (0.33). However it has the same "colour".

(g) Why is the origin marked with pencil not biro?
Biro has a number of solvent soluble dyes which may separate during the chromatography. This would interfere with the chromatography by producing extra spots and would also reduce the accuracy in measuring the origin.
Pencil is made from graphite, which does not separate in the chromatography as it is firmly adsorbed to the stationary phase.

(h) Why is the solvent level kept below the origin?
Otherwise the sample would dissolve into the solvent and not be separated into it's components.

4. How can samples of potassium, sodium, barium and calcium be quickly distinguished? How accurate is this?
Exciting the atoms in a flame would produce a characteristic colour of light

Potassium	Lilac (light purple)
Sodium	Yellow
Barium	Orange
Calcium	Red

This is not very accurate as it is hard to tell between the different colours by sight alone.

5. In a flame test, both calcium and lithium produce red light:
(a) What is the origin of this light?
When an excited electron returns to a lower energy level, it emits a photon with a characteristic wavelength l. The energy of the photon is inversely proportional to this energy ie

DE =C/l (where DE= difference in energy levels, l = wavelength, C = constant)

ie the greater the difference in energy levels, the lower the wavelength (alternately, the greater the frequency of the light).

(b) How could we use this light to distinguish between the two elements?
If the emitted light is passed through a prism, we will separate the light into the specific wavelengths emitted by the excited sample.
By comparing the spectrograph of the sample with that for Ca and Li, we can identify the identity of the unknown from the number, position & colour of the lines.

(c) How is this better than flame tests?
This is simply more specific, as we are comparing the specific lines, not overall colours. The lines correspond to the transitions between energy levels in the atom, and as these are unique for each element, the identity of the metal can be quickly established.

6. Below is a representation of some of the possible energy transitions in a given atom. Each return to a lower energy level is accompanied by the emission of a photon of light.
(a) Are these the only possible transitions for this atom? Explain?
No ... there are a number of energy transitions possible for the atom
If an electron is promoted into an excited state, then there are a number of possible pathways it can take to the ground state:

eg E₆ ->E₁

E₆ ->E₄ ->E₁

E₆ ->E₅ ->E₃ ->E₁ etc.

In other words, each possible transition will be made by a few electrons

(b) Why are there so many lines in an emission spectrum?
Simply because there are so many possible transitions:

From E₂ there is one possible transition

From E₃ there are two possible transitions

From E₄ there are three possible transitions

From E₅ there are four possible transitions

From E₆ there are five possible transitions

So just in the 6 levels shown, there would be 15 possible transitions. As there are tens of energy levels (shells) beyond the E₆ level, we will find hundreds of transitions for each element.

(c) Which transition would have the highest frequency?
The highest frequency has the lowest wavelength (and so the greatest energy difference). This would be the E₆ to E₁ transition.

(d) Which transition is most likely to be red? Explain
From DE =C/l (where DE= difference in energy levels, l = wavelength, C = constant), ie the greater the difference in energy levels, the lower the wavelength (alternately, the greater the frequency of the light).
As red has a wavelength of 7 x 10^-6m and violet has a wavelength of 4 x 10^-6 m, this tells us that red must be a small change in energy. From the diagram, the smallest transition is E₅ to E₃, though the E₆ to E₅ would be more likely to produce red.

(e) Is it possible for light to be absorbed by atoms? Explain.
If the energy of the light corresponded exactly to the difference between the energy levels, then there would be a promotion of the electron from ground state to an excited state. This is the reverse of the emission spectrum.
Thus when white light is passed through atoms, characteristic wavelengths of light would be absorbed when electrons are promoted. This is the reason why solutions of Cu²⁺are blue, and why rust is red. Because there are fewer possible transitions.