Please Help-Chem Questions.

[bec]

Thanks for that. I'm using heinemann - the only thing about it i can find (according to the index at the back) is a reference in a table....very brief information.
but it's ok because my teacher's notes are pretty thorough.

can anyone help me understand this:
"Excited electrons are unstable and will eventually jump back to a lower energy level provided there is an available space in the lower energy level."
In what circumstances wouldn't there be an available space?
Thanks

[Collin Li]

Colorimetry is a subset of UV-visible spectroscopy. It is simply the "visible" part of UV-visible spectroscopy. Questions about colorimetry usually examine what colour of light you would pick in colorimetry to observe the concentration of some coloured solution (you pick the complementary colour, as it is most strongly absorbed).

Thanks for that. I'm using heinemann - the only thing about it i can find (according to the index at the back) is a reference in a table....very brief information.
but it's ok because my teacher's notes are pretty thorough.

can anyone help me understand this:
"Excited electrons are unstable and will eventually jump back to a lower energy level provided there is an available space in the lower energy level."
In what circumstances wouldn't there be an available space?
Thanks

You're reading into that too much. There is never going to be a case where there is no available space. By definition an excited electron is one that vacates a lower energy state for a higher energy state. Nothing will fill it up in the meantime. I guess it included that in just in case people are wondering why the 3s electrons aren't jumping down to the 1s orbital (which is obviously full for any element with an atomic number larger than 2).

[bec]

Thanks - that makes sense (both things)

Something else i don't really get - why is AES used only for group I and II elements?

[Collin Li]

Something else i don't really get - why is AES used only for group I and II elements?

I never knew that!

A possible explanation is that group I and II elements are not very electronegative. Spectroscopy is often a field that is most effective with metallic atoms because of low electronegativity. It means it is easier to excite electrons, and hence the application of light to do so is much more effective.

[bec]

well if you never knew that, i don't need to know it either!
but what you said to explain it makes sense - same reasoning behind the fact that flame tests work with cations and not anions, isn't it?

[Collin Li]

Yep.

[bec]

Ok, I really don't understand colorimetry/uv-visible spectroscopy...

my understanding of colorimetry is that it's the same principal as guessing which glass of cordial has more cordial in it, based on which is a darker colour. only, in this case, instruments are used and they make precise measurements on wavelengths emitted (and therefore, the "colour" of a solution) which are compared with readings from solutions of known concentrations to determine a more exact value.

i was under the impression that it was totally different from AES and AAS in that it had nothing to do with providing energy to excite electrons...but when i asked my teacher if that was correct, she told me that it wasn't (and then couldn't really explain much more than that)

why?

what am i getting wrong?

[Toothpaste]

I'll attempt to explain it a bit ... but er ... open your Heinemann book to page 85/86 while you read (page 10 of the Heinemann workbook may help too):

You see how the front cover of your text book is purple? Well it's reflecting that colour (purple), which it does not absorb. The colour you observe is the colour that is being reflected from that object. This means that it's absorbing the rest of the colours in the visible spectrum (just not purple).

UV-visible spectroscopy is both quantitative and qualitative.

The wavelengths absorbed vary between different compounds, and we can use this property to identify what something is (qualitative).
We can also measure the absorptions of a "known" concentration (standard solution) and compare it with your "unknown" sample - at the same wavelength - using a calibration graph. Hence find how much is absorbed : how much of the element present can be determined (quantitative).


In UV-visible spectroscopy there is a light source (see diagram) providing energy of all wavelengths.
You would pick the wavelength of light that allows for maximum absorbance by your sample solution.

*The light supplies the energy to promote electrons to higher levels and the energy needed to do this differs in different substances. Similar to AAS atoms will absorb light that will promote an electron from ground state to a higher energy level.

On page 85, have a look at table 7.4.

Let's say your sample solution is green. (fourth instrument in figure 7.15 page 86)
Using the monochromator, you would pick the wavelength your solution would absorb. Since it is green(observed) you would pick purple (see table 7.4).

If it was red you would pick a blue-green wavelength. See it?

Alright now for measurements... look at the diagram 7.15 on page 86 again.
Generalising - let's say the "narrow beam" was 100 wavelength. If the light detector detects 30 wavelengths ... then that would mean your sample solution absorbed 70; as 30 went through and the other 70 was absorbed.

I personally learn through examples, so I'll do some questions for you.

Now for reading graphs:
Hmm on page 88 look at question 10.
Pick out the point on the graph where there is the lowest absorbance. It is between 400 - 500 nm and it corresponds to blue on the absorption spectrum below it. This means the dye absorbs blue the least. Therefore it is reflecting blue and this is the colour of the dye we see. To be specific it's violet-blue.

Question 12 I would highlight all the numerical values to make it easier to read chunks of text like these.
(a) It says there is an absorbance of 0.17. Grab a ruler, something straight, and find 0.17 on the absorbance axis. Draw a line across till it hits the other line, and from there draw a line down till it touches the concentration axis. Work out what concentration it is. It should be around 32mg/L (the answer).

(b)The concentration of P is 32 mg/L. We have to find the mass of it in 250mL (the amount of solution we have).
Change it to mL and let x denote the mass we want to find.




in

Our sample of detergent is 0.250g (dissolved in our 250mL).
We want the percentage mass of P in this.



% mass of P in the detergent = %

(c) Using table 7.4 on page 85 600nm corresponds to orange absorbed and blue observed.
The blob of text says that they converted the phosphate content to a blue-coloured molybdenum phosphate compound. Orange would provide the highest absorbance for the blue and has a wavelength of ~600nm.
The direct answer would be: "Orange light of wavelength 600 nm would be strongly absorbed by a blue solution."

*I think that was the only line relating to your question. Gees ... I go off on tangents instead of being straightfoward.
coblin could answer it better I bet

[Mao]

The theory behind these is, light is not "emitted" by these molecules, it is rather"reflected"

In ambient white light (every colour), these molecules in the solution absorbs all except some frequencies, which are reflected, and which are what we see in our eyes.

colorimetry is based on this principle, the amount of reflection is detected by our eyes, and we take make an estimate

Now, it is really hard to measure how much reflection there is, as that will have to be related to the ambient luminocity of the environment it is in, plus the reflected frequency will get mixed up with everything else and etc etc... all very complicated and crap

but if this is done in a dark environment, and the frequency that these molecules best absorb is shone, then the concentration can be measured by how much is absorbed. this is the principle behind UV vis spec.

Before a sample is actually tested, a spectrum of UV is shone at the sample, and the frequency that absorbs best is chosen. and the absorbance for standards are done and a calibration curve, etc are drawn and the shibang and u get ur results

but the principle is, these molecules absorbs most frequencies, and reflect some.

[Collin Li]

I think the preferred term is 'transmit' instead of 'reflect.' I think reflection is when light bounces off something, which hasn't happened in this case. Instead, light just passes right through (transmits), because the discrete nature of electron energy levels do not accept these other wavelengths of light.

I'm not really understanding what your confusion is though, bec, but AES has nothing to do with AAS!

[bec]

thank you so much - toothpick, how long did that take you!

Ok I'll try and make my questions clearer this time

The theory behind these is, light is not "emitted" by these molecules, it is rather"reflected"

(or "transmitted" coblin said)

Question 1: When you say light is not "emitted", what do you mean? Are you saying that the electrons remain at ground state and don't release any energy in the form of light?

...light just passes right through (transmits), because the discrete nature of electron energy levels do not accept these other wavelengths of light.

Question 2: So, in colorimetry, we're passing white light through a substance and measuring which wavelengths are absorbed. The light that the instrument at the end measures has not been absorbed "because the discrete nature of electron energy levels do not accept these wavelengths of light". Does this mean that ever time I see a colour, I'm seeing the energy that is emitted by excited electrons as they return to ground state? I'm wearing blue shorts, are the electrons in the dye emitting wavelengths that correspond to 'blue' radiation?

*Similar to AAS atoms will absorb light that will promote an electron from ground state to a higher energy level.

Question 3: Is this all just exactly the same principal as AAS? Is the only difference the fact that instead of using a specific hollow source cathode lamp, we have the whole spectrum of visible light available to us (+ UV, in UV-visible spec) and hence can use this technique qualitatively?

I can answer questions in my text book about this, since they're all pretty simple, but I'd really like to understand WHY and HOW all this works as well as just the fact that it just DOES. So thanks for the help!

[Mao]

The theory behind these is, light is not "emitted" by these molecules, it is rather"reflected"

(or "transmitted" coblin said)

Question 1: When you say light is not "emitted", what do you mean? Are you saying that the electrons remain at ground state and don't release any energy in the form of light?

In ambient light, the electrons arent actually excited (there's not enough energy). Imagine blowing a fan at a fishing net, for example (i should think of a better one). The wind experienced on the other side of the net is like the light you see, the net itself doesnt actually produce any wind, its the fan behind it that does, however, it allows some wind to pass through

putting that into presepctive, ambient light is a broad spectrum of wavelength. Elements/compounds absorb certain frequencies, and leaves the rest alone (transmit). when you see red wine, for example, the redness is not emitted by the wine molecules, but rather, every colour is absorbed by the wine except red, which is allowed to transmit through it.

With the knowledge of this, we can measure concentration from the amount of transmittence (colorimetry), or the amount of absorption (UV vis)


ps coblin, I think its a bit of both (transmit and reflect), in bec's question 2 scenario, you dont suppose frequencies are "transmitted" through bec's legs, do u?



Question 2
Colorimetry is pretty inaccurate, as the detector is our eyes. but the basic principle is, all samples are compared under the same level of ambient lighting, and the amount of transmittence is measured.
Simply put, comparing a gradient of colours in the same light...

and as for your blue shorts, the energy levels of electron orbits are discrete (due to the unique distance between electrons from each other and the nucleus and how much charge there are... etc), this means that each electron can only take up certain "quanta" of energy. EM radiation also have a definitive energy proportionate to their wavelength. This means that only certain EM radiation which energy exactly match the difference in energy levels of the electron orbits can be absorbed. The ones that dont exactly match simply passes on (or collide with electrons on the way and are deflected). This deflection is the blue that you see. The dye absorbs practically everything except the blue frequencies, and the blue frequencies are deflected.

If you find it hard to visualise EM radiation deflecting, think of the radiation as photons (tiny subatomic particles that is light). and even though your blue shorts is made up of atoms that are mainly empty space, there are so darn many that some of this tiny photons will collide with electrons (or nucleus). think of it as... gazillions of billard balls?


question 3
The similarity toothpick was referring to is that in AAS, it's not how much the atoms emitted (that's AES), but rather, how much the atoms absorbed, or the amount of energy used by the atoms so the electrons could be promoted.

In UV vis, you are also measuring the amount of absorbance. The way they do it is, before a sample is actually tested for absorbance, a broad spectrum is shone at it, and the frequency that this compount absorbs most is used (similar principle to why each metal have a specific cathode lamp in AAS). and then the amount of absorbance is measured.

Measuring emission is only useful for metals, as their low electronegativity means small amount of energy can excite the electrons and cause the promotion and emission. However, most abundant elements that we encounter everyday are have higher electronegativity (hence requiring more energy for promotion of electron), making the measurements of emission too inefficient and costly (energy wise).

The key thing to remember when trying to sort out the confusion is:
the colour you see is what the compound doesnt absorb, emission of light is only when the compound is tremendously excited (and doesnt exactly happen often, except in flames and lights, which are both extremely hot at the centre).

[Collin Li]

Mao, you are correct with the "pants" scenario. In spectroscopy, it is based on transmittance. Transmittance is required for reflection, so the only reason why you'd see blue pants is because blue light is transmitted and reflected.

I have answered these questions my way, because, sorry Mao, I think your answers are not very direct or relevant to the Chemistry course.

Question 1: When you say light is not "emitted", what do you mean? Are you saying that the electrons remain at ground state and don't release any energy in the form of light?

The electrons absorb the wavelengths of light. This means that the atom becomes excited, and yes, it will re-emit that absorbed wavelength of light. You assume that the wavelength of light that is re-emitted travels in all different directions. Even though some of the re-emitted light will go back towards the detector, only a small percentage of it will, so the absorbance reading at the detector will still proportionally change with the concentration, so this technique is still viable to determine the concentration of a solution.


Question 2: So, in colorimetry, we're passing white light through a substance and measuring which wavelengths are absorbed. The light that the instrument at the end measures has not been absorbed "because the discrete nature of electron energy levels do not accept these wavelengths of light". Does this mean that ever time I see a colour, I'm seeing the energy that  is  emitted by excited electrons as they return to ground state? I'm wearing blue shorts, are the electrons in the dye emitting wavelengths that correspond to 'blue' radiation?

No, we're passing coloured light through the sample. There should be a coloured filter in between the white light source and the sample. Colorimetry is essentially just the same as UV-visible, except it is restricted to the "visible" part of "UV-visible." Colorimetry isn't even explicitly stated on the study design. It is strictly a subset of UV-visible spectroscopy, and it is often used as a stepping stone in learning to demonstrate how UV-visible works (same principle of choosing a wavelength of light to be absorbed optimally by the solution).

The second part of your question: when you look at blue shorts under white light, the colour that is you see is due to mainly blue light being transmitted (and then reflected towards your eyes). Other colours of light are absorbed. The re-emitted light scatters and is not important.


Question 3: Is this all just exactly the same principal as AAS? Is the only difference the fact that instead of using a specific hollow source cathode lamp, we have the whole spectrum of visible light available to us (+ UV, in UV-visible spec) and hence can use this technique qualitatively?

UV-visible, AAS, and colorimetry have the same basic concept! They are just different versions of the same idea.

AAS has a specialised light-source, which is much more accurate and specific than UV-visible and colorimetry (which just uses white light, a prism and a slit to separate the light into a spectrum, and select the required wavelength). Colorimetry may use a coloured filter instead of a prism (monochromator).

[Mao]

I have answered these questions my way, because, sorry Mao, I think your answers are not very direct or relevant to the Chemistry course.

hehe, i like your answers better anyways

[bec]

thank you so much!