Post by: Jimbo1 on February 04, 2008, 07:25:31 pm
1) A bottle contains NaOCl labelled at 153ppm. 750mL of this bottle will contain:
A-9.3 x 10*20 Na+ ions
B-1.2 x 10*21 molecules of NaOCl
C-1.5 mol of OCl- ions
D-204 mg of NaOCl
2)Which of the following contains the greatest number of sulfur atoms:
A- 68.9g of CuSO4.5H20
B-29.4L of SO2 at SLC
C-2.0L of 0.40M H2SO4 solution
D-3.0 x 10*24 molecules of H2S
3) Glucose undergoes a combustion reaction to produce Co2 and H20.
a) Write a balanced equation to represent this reaction
b)What volume of C02 at 37*C and 101.3kPa would be produced when 10.0g of the glucose reacts.
c) What is the name given to this reaction
4) 125 mL of 0.250M copeer(II) nitrate solution is added to 90.0mL of 0.560M sodium hydroxide solution resulting in the formation of a precipitate of Calcium Hydroxide.
a) Write a balanced chemical equation for this reaction
b) Calculate amount in mol of both copper(II) nitrate and sodium hydroxide present at the start of the reaction and hence determine which reactant is in excess.
c) Calculate the mass of precipitate formed.
Please help.
Thank You :)
Post by: Toothpaste on February 04, 2008, 10:30:49 pm
Shot-gun the biology-like one.
3) Glucose undergoes a combustion reaction to produce Co2 and H20.
a) Write a balanced equation to represent this reaction
b)What volume of C02 at 37*C and 101.3kPa would be produced when 10.0g of the glucose reacts.
c) What is the name given to this reaction
a) C6H12O6 (aq) + 6O2 (g) --> 6CO2 (g) + 6H2O (l)
b) n(C6H12O6) = 10g / 180g
= .0555556 mol
=> n(CO2) = 6 * n(C6H12O6)
= 0.333 mol
37 degrees C = 310 K
PV = nRT
V(CO2) = (0.333mol * 8.31
= 8.48 L
c) Cellular (aerobic) respiration
Post by: Collin Li on February 04, 2008, 10:35:07 pm
Post by: midas_touch on February 05, 2008, 12:00:31 am
4) 125 mL of 0.250M copeer(II) nitrate solution is added to 90.0mL of 0.560M sodium hydroxide solution resulting in the formation of a precipitate of Calcium Hydroxide.
a) Write a balanced chemical equation for this reaction
b) Calculate amount in mol of both copper(II) nitrate and sodium hydroxide present at the start of the reaction and hence determine which reactant is in excess.
c) Calculate the mass of precipitate formed.
a) Cu(NO3)2 + 2NaOH -> 2NaNO3 + Cu(OH)2
b) Use n = cv which is a formula you must be familiar with come the unit 3 exam.
Cu(NO3)2:
n = 0.125*0.25
= 0.03125 mol
NaOH:
n = 0.09*0.56
= 0.0504 mol
However, you need to consider mole ratios of the reactants:
n(NaOH) reacting/n(Cu(NO3)2) reacting = 2
But, n(NaOH) present/n(Cu(NO3)2) present = 0.0504/0.03125 = 1.6128
Hence NaOH will run out first, making it the limiting reactant, and the Cu(NO3)2 in excess
c) Now we know which reactant is the limiting reactant, we can calculate the amount of the precipitate formed (Cu(OH)2):
n(NaOH): 0.0504 mol
Once again consider mole ratios in the equation
n(NaOH) reacted/n(Ca(OH)2) formed = 2
Hence n(Ca(OH)2) formed = 0.0504/2 = 0.0252 mol
Use m = nM (Another forumla you must learn to use) to figure out the mass of Ca(OH)2 formed
m = 0.0252*74.1
= 1.87g (correct to 3 sig figs)
Post by: cara.mel on February 05, 2008, 08:05:30 am
Disclaimer: I am doing these on 5 hours of sleep so sorry if I skip lines in working or whatever
1. 153ppm = 153mg/L = 0.153g/L => in 3/4 of a litre there is .153*.75=0.115g of NaOCl
n(NaOCl) = m/Mr = 0.115/74.5 = 1.54*10^-3mol
Then unfortunately it's just a matter of going through each option to see if they work
a) n(Na+) = n(NaOCl) = 1.54*10^-3mol
number of ions = n(Na+)*6.023*10^23 (avogadro's constant) = 9.3*10^20 ions => a is right
Confirming the other 3 are wrong:
b) number of NaOCl molecules = number of Na+ ions => wrong
c) n(OCl-) = n(NaOCl) => wrong
d) worked out there was 115mg of it earlier when working out the mole.
First step you do in any question you are unsure of is try and find the mole of something, it will usually take you where you want to go
2) Although the question says number of atoms, by looking at the question it would take less steps overall to get them all into moles than # of atoms (and seeing as they are directly proportional, we can do that). So we want to find the one with the higest amount (in mol) of S
a) n(CuSO4.5H2O) = m/Mr = 68.9/249.5 = 0.276mol
n(S) = n(CuSO4) = 0.276mol
b) At SLC, 1 mol = 24.5L
=> n(SO2) = 29.4/24.5 = 1.2mol
n(S) = n(SO2) = 1.2mol
c) n(H2SO4) = cV = 2*.4=.8mol
n(S) = n(H2SO4) = 0.8mol
d) n(H2S) = 3*10^24/6.023*10^23 = 5mol
n(S) = n(H2S) = 5mol
Therefore d is the biggest => answer
Post by: J.Z-M on February 05, 2008, 10:24:00 am
1.B
(Probably wrong though)
Post by: J.Z-M on February 05, 2008, 10:53:14 am
Post by: midas_touch on February 05, 2008, 01:48:33 pm
What's wrong with the first two? :P
Disclaimer: I am doing these on 5 hours of sleep so sorry if I skip lines in working or whatever
1. 153ppm = 153mg/L = 0.153g/L => in 3/4 of a litre there is .153*.75=0.115g of NaOCl
n(NaOCl) = m/Mr = 0.115/74.5 = 1.54*10^-3mol
Then unfortunately it's just a matter of going through each option to see if they work
a) n(Na+) = n(NaOCl) = 1.54*10^-3mol
number of ions = n(Na+)*6.023*10^23 (avogadro's constant) = 9.3*10^20 ions => a is right
Confirming the other 3 are wrong:
b) number of NaOCl molecules = number of Na+ ions => wrong
c) n(OCl-) = n(NaOCl) => wrong
d) worked out there was 115mg of it earlier when working out the mole.
First step you do in any question you are unsure of is try and find the mole of something, it will usually take you where you want to go
2) Although the question says number of atoms, by looking at the question it would take less steps overall to get them all into moles than # of atoms (and seeing as they are directly proportional, we can do that). So we want to find the one with the higest amount (in mol) of S
a) n(CuSO4.5H2O) = m/Mr = 68.9/249.5 = 0.276mol
n(S) = n(CuSO4) = 0.276mol
b) At SLC, 1 mol = 24.5L
=> n(SO2) = 29.4/24.5 = 1.2mol
n(S) = n(SO2) = 1.2mol
c) n(H2SO4) = cV = 2*.4=.8mol
n(S) = n(H2SO4) = 0.8mol
d) n(H2S) = 3*10^24/6.023*10^23 = 5mol
n(S) = n(H2S) = 5mol
Therefore d is the biggest => answer
I forgot how to do ppm :(, I was trying to do it on a mole basis.
Post by: Ninox on February 05, 2008, 07:50:31 pm
Good luck! :D
and impeccable work guys!
Post by: bec on February 09, 2008, 09:51:19 am
In each of the following equations, identify the acids and bases and name the conjugate acid-base pairs
a) HSO4- + H2O --> H3O+ + SO42-
b) NH4+ + S2- --> NH3+HS-
c) CH3COO- + H3O+ --> H2O +CH3COOH
I've done it but despite correctly identifying the acids/bases, some of my conjugate acid-base pairs were the wrong way around. I said:
a) HSO4-/SO42- (the book agrees with me here) and H3O+ /H2O (book has it other way around)
b) The pair I got wrong was HS- / S2-
c) I said CH3COOH[/b] / CH3COO-
I've always been taught to list them acid/base (ie, not base/acid) so did I get these wrong because I'm identifying acids and bases wrong, or does order just not matter?
Edit: just realised the book lists them in the order they appear in the original equation...which is correct, that or ensuring they're in the order acid/base?
Post by: Collin Li on February 09, 2008, 09:58:27 am
Post by: bec on February 19, 2008, 08:27:52 pm
can anyone explain this to me: "The decomposition of a compound into its elements must be a redox reaction."
I think i get it but i need confirmation....is it because in a compound, for the oxidation numbers to total to zero there must be at least one element that has a positive oxidation number and >1 element with a negative oxidation number - then, when the compound decomposes into its elements, each product will have the oxidation number of zero.... meaning that both oxidation and reduction have occurred?
like if CO2 --> C + O2 (if that can even happen, i wouldn't know)
even writing this has sort of solidified it in my mind actually (despite the rambling way i've written it...)
but a yay or nay would be good anyway so i know i'm on the right track!
thanks!
Post by: Collin Li on February 19, 2008, 08:34:14 pm
Post by: bec on February 22, 2008, 09:48:02 pm
in a compound containing elements more electronegative than oxgen, the more electronegative element "gets the negative oxidation number" (i quote from the notes my teacher gave us)
can someone rephrase that?
I'm assuming that means that in F2O, F has oxidation no. -1, O is +2? and in ClO, Ox. no of Cl = -1, O=+1?
is that right?
Post by: Collin Li on February 22, 2008, 09:53:16 pm
Post by: bec on February 29, 2008, 08:02:45 pm
A: The energy required to promote an electron in a chlorine atom is much higher than for the potassium atom. As a consequence, the flame of a Bunsen burner excites relatively few chlorine atoms. Furthermore, energies emitted as electrons in the excited atoms return to their lowest energy levels are mainly outside the energies of visible light.
in english, anyone? i don't understand a word
Post by: midas_touch on February 29, 2008, 08:22:13 pm
Post by: bec on February 29, 2008, 08:51:11 pm
why though? what is it about the electrons in chloride ions (can i generalise and say all anions?) that makes them require more energy to jump to higher energy states?
is it to do with the electron configuration?
Post by: Collin Li on February 29, 2008, 08:58:14 pm
The high energy wavelengths that are emitted will be in the ultraviolet region of the UV-visible spectrum, and that is why you cannot perform flame tests on non-metals.
Post by: bec on February 29, 2008, 09:32:50 pm
Post by: bec on March 04, 2008, 06:30:00 pm
Q: Name a substance, other than copper sulfate solution, whose concentration can be determined directly by UV–visible spectrometry.
A: Any coloured solution can be chosen, for example chlorophyll solution.
Why any coloured solution? I thought the whole thing with UV-visible spectrometry was that solutions didn't have to absorb wavelengths in the visible region?
Can the concentration of chlorophyll (which i'm guessing is green?) be determined using regular atomic absorption spectrometry?
Post by: Collin Li on March 04, 2008, 06:32:49 pm
AAS is based on the cathode lamps (restricted to metallic atoms), so unless chlorophyll contains some metallic atoms in it, I doubt it is possible.
Post by: bec on March 04, 2008, 06:50:57 pm
Atomic emission spectroscopy (AES) is regarded as superior to flame tests because, in AES, more elements produce emission spectra in the hotter flame, and a much smaller quantity of an element needs to be present in order for it to be detected. Using AES, the amount of the element present in a sample can be accurately determined, even in the presence of larger amounts of other elements that would mask the flame colour of the element to the naked eye.
Ok, so i understand that aes is better than flame tests because
1) a smaller quantity of an element is needed
2) can be used for quantitative analysis
3) uses a machine, therefore more accurate than naked eye thinking that something is "kind of bluey-green"
but....what's this "hotter flame" they're talking about? isn't the flame in aes only there so the substance can be atomised?
thanksss
Post by: iamdan08 on March 04, 2008, 07:22:50 pm
Post by: bec on March 04, 2008, 07:47:45 pm
how do we know which light comes from electrons excited in the flame, and which comes from electrons excited because of the light derived from the cathode lamp?
does that make sense?
Post by: Mao on March 04, 2008, 08:04:29 pm
the measurements are done on specific frequencies.
much like each atom has its own emission spectrum (back in unit 1), the complementary spectrum to this is their absorption spectrum.
Each cathode lamp is calibrated to a particular frequency for a particular metal (an AAS have a lamp for each type of metal, thats why they cost so darn much), and only that metal will absorb that frequency.
if that made sense?
Post by: Collin Li on March 04, 2008, 08:38:10 pm
AES is superior to flame tests in every way. Flame tests only give you a macroscopic "average" view of the emission spectra, and plus the flame may be too weak - even if it is strong enough, the emitted wavelengths will be invisible to the naked eye (UV - the high energy wavelengths of light).
For AAS, the flame in the atomiser is supposed to bring the atoms to their ground state (not 100% sure how, but you're supposed to assume this happens). I haven't seen an official explanation against why the emission caused by the flame does not affect the final results in the AAS, but I suspect this is due to the monochromator placed after the absorption, which selects a particular wavelength of light (the one that is being absorbed), ignoring the other emitted wavelengths (I'm aware this isn't a complete explanation). It's a very good question that my friend asked TSFX, who didn't have an answer for it. You probably don't have to worry about it.
Post by: bec on March 04, 2008, 08:53:26 pm
thanks...more to come i'm sure!
Post by: bec on March 05, 2008, 07:27:20 pm
Post by: iamdan08 on March 05, 2008, 07:53:53 pm
Quote
-Discuss the principles of colorimetry, and the relationship between concentration and absorption
-Use second-hand colorimetry data to construct a calibration curve and determine the concentration of an ingredient in a consumer product, concentration of phosphate ions in laundry products
-perform a colorimetric analysis such as the phosphate content of washing powders
So it looks like it is.
Post by: bec on March 05, 2008, 08:08:44 pm
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
Post by: Collin Li on March 05, 2008, 08:52:24 pm
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).
Post by: bec on March 05, 2008, 09:31:20 pm
Something else i don't really get - why is AES used only for group I and II elements?
Post by: Collin Li on March 05, 2008, 09:56:20 pm
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.
Post by: bec on March 06, 2008, 08:00:21 pm
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?
Post by: Collin Li on March 06, 2008, 08:28:56 pm
Post by: bec on March 08, 2008, 04:25:33 pm
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?
Post by: Toothpaste on March 08, 2008, 06:32:42 pm
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.
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
Post by: Mao on March 08, 2008, 10:50:17 pm
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.
Post by: Collin Li on March 09, 2008, 12:04:48 am
I'm not really understanding what your confusion is though, bec, but AES has nothing to do with AAS!
Post by: bec on March 09, 2008, 09:07:09 am
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!
Post by: Mao on March 09, 2008, 11:50:12 am
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).
Post by: Collin Li on March 09, 2008, 04:57:51 pm
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).
Post by: Mao on March 09, 2008, 05:11:35 pm
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 ;D
Post by: bec on March 10, 2008, 07:35:52 pm
Post by: bec on March 23, 2008, 08:51:29 pm
Why can NMR only be used to detect substances with an odd mass number? isn't it just the protons that are affected by the magnetic field? so, provided the number of protons is odd, why is it relevant how many neutrons there are?
Post by: Collin Li on March 23, 2008, 10:11:59 pm
Post by: bec on March 24, 2008, 08:06:21 am
eg. Could you use NMR to analyse nitrogen, since even though its mass number is 14, there are 7 electrons/protons...
Post by: Collin Li on March 24, 2008, 01:38:54 pm
Post by: bec on March 25, 2008, 05:38:37 pm
Post by: bec on March 25, 2008, 07:51:48 pm
1. Why is AAS more accurate than AES: couldn’t you use the detecting instrument, and the same setup, in AES?
2. UV-Vis Can be used for atoms, ions and molecules…………why is that special? Don’t the others? If not, why not?
3. In IR-spec, the equipment includes a “sample and reference cell or disc made out of NaCl, KBr or similar" – why? What’s “similar” to NaCl?
4. Do we need to know what sp2, sp3 stretches are?
Post by: Collin Li on March 26, 2008, 02:45:42 am
2. Where does it say it is "special"? AAS is restricted to metals only (because of the cathode lamp, which requires metal cathodes to operate it). GC and HPLC are better suited to molecules. Molecules are most susceptible to chromatography because of polarity and non-polarity. In that way, I guess UV-visible is special, since it can deal with anything - as long as it absorbs any wavelength of light in the UV-visible spectrum.
3. Similar compounds to NaCl are just relatively inert salts. If you remember from electrochemistry, these ions are weak reductants and oxidants, which means they will not react very readily. They are stable compounds, similar to the types of compounds that you would also use in a salt bridge.
4. Possibly, but not by those names. You should know what a "stretch" is, but you definitely do not need to know about the "sp" part. That is something you will learn in first year university.
Post by: bec on March 26, 2008, 07:37:47 am
You're right, I always get confused with AAS and AES...
Is this pretty much right?
AES:
- Like a flame test but with hotter flame, slit (NOT wavelength selector), prism that splits emitted energies into emission spectrum (coloured lines, black background)
- Only qualitative (for VCE...i checked in the textbook)
AAS:
- Hollow source cathode lamp (made of same metal as analyte) "beams" through a flame
- Sample sprayed into flame --> atomic vapour
- Monochromator and slit select WAVELENGTH (as opposed to just the slit in AES)
- Machine measures how much of that wavelength was absorbed
- Analyse sample solutions, draw calibration curve, plot point of sample to find concentration
Post by: Collin Li on March 26, 2008, 11:37:39 am
Post by: bec on March 26, 2008, 06:02:23 pm
Sounds pretty good to me. Try to remember it diagrammatically (AAS particularly) rather than in a series of rote-learned dot points. It will help you understand it better. Perhaps you already have, and you're just translating it into words for me to see, which is great!
woohoo, finally. and yeah, i know what the setup looks like and haven't memorised any dotpoints, i just wrote them out here because i can't use pictures...
anyway, second round of questions:
1. How does quantitative analysis with IR work? (p95 of heinemann, there's about a paragraph on it). My understanding: Choose a significant peak on the spectrum of "unknown" analyte; figure out what it is (let's say it's CH3); plot a series of samples with known CH3 concentration; read the CH3 concentration of unknown off the graph. Is that about right? We never did it at school...
2. NMR is "the only technique that can be used to determine exact 3D structure of biological molecules that cannot be crystallised"....what does this even mean? What's crystalisation, and why does it help you analyse things?
3. Do we need to know/understand the relationship C=λv?
4. For mass spec, how much do we need to know about the process of fragmentation and the formation of free radicals? Because basically all I know is that there's an electron beam, it shoots electrons off atoms (somehow) to make them into cations, electrons are now unpaired so they split off into two parts, one part is a cation, one is a free radical...the radical gets vaccuumed away and the cation is detected and plotted on the graph? And this can happen in a variety of different combinations, hence the different peaks on the spectrum?
Post by: Mao on March 26, 2008, 06:24:02 pm
1. How does quantitative analysis with IR work? (p95 of heinemann, there's about a paragraph on it). My understanding: Choose a significant peak on the spectrum of "unknown" analyte; figure out what it is (let's say it's CH3); plot a series of samples with known CH3 concentration; read the CH3 concentration of unknown off the graph. Is that about right? We never did it at school...that is roughly right
it is exactly the same principle as other spectroscopy, where known concentrations are used to make a calibration curve. It is very similar to UV-Vis, but where that graph is concentration vs absorbance at a particular wavelength, this is concentration vs transmittence at a particular wavenumber.
The initial IR reading serves to identify which wavenumber should be used (a particular peak), much like how the UV Vis machine is first optimized to a particular wavelenth to allow maximum absorbance.
3. Do we need to know/understand the relationship C=λv?from my understanding, we only need to understand that higher wavelength means lower frequency and lower energy, vice versa. We wouldnt be needed to use the actual formula.
4. For mass spec, how much do we need to know about the process of fragmentation and the formation of free radicals? Because basically all I know is that there's an electron beam, it shoots electrons off atoms (somehow) to make them into cations, electrons are now unpaired so they split off into two parts, one part is a cation, one is a free radical...the radical gets vaccuumed away and the cation is detected and plotted on the graph? And this can happen in a variety of different combinations, hence the different peaks on the spectrum?that is already far more than required. What we need to know is that the compound is ionised in a flame and becomes a cation (dont need to know how), and in the process it may be broken down into smaller parts. Then they are sent down a curved tube with magnetic plates on both sides. The cations curves according to the mass/charge ratio (z), which means they move at different angles, and reach the detector at different positions (or hit the wall, in which case the magnetic plates are adjusted and the detector recalibrated accordingly), and this gives a graph of relative percentage abundance vs mass/charge. For the purpose of this course, we also need to know how to identify the peaks (guessing/from quesion/data from other analytical techniques, etc)
Post by: bec on March 26, 2008, 06:53:15 pm
so with mass spec, we probably wouldn't need to understand what this is on about? (page 113 from the book)
The relative intensities of the ions depend on:
- the energy of the bombarding electrons
- the stability of the ion fragments formed
- the ease with which ions can lose atoms.
...because i don't. And I've pretty much given up being curious about chem now and just want to learn what I absolutely have to...which is a bit sad but pretty much necessary...
Post by: Mao on March 26, 2008, 07:30:57 pm
I have never heard of any of those things, and I dont think you'll need them
the study design for Chem is very vague, but this is what it says:
Quote from: VCAA n00bishness
principles and applications of spectroscopic techniques and interpretation of qualitative and quantitative data from...
my interpretation is the general principle of how it works, what it works with, what are the results, and how to read the results...
that applies to all the techniques
Post by: bec on March 26, 2008, 07:44:36 pm
i'm starting to think i complicate things by about 1000%....
Post by: Collin Li on March 26, 2008, 09:29:28 pm
Quote
2. NMR is "the only technique that can be used to determine exact 3D structure of biological molecules that cannot be crystallised"....what does this even mean? What's crystalisation, and why does it help you analyse things?
You don't need to worry about this: it's just a contextual statement that is demonstrating the power of NMR. To crystallise something basically means you have the ability to create a purified sample of the product. If you can do that, usually you can identify what it is - because of what type of reagents you need, etc. NMR doesn't require this - it feels the 3d shape through magnetism, rather than through chemical reaction.
Post by: bec on March 26, 2008, 09:33:36 pm
Post by: Collin Li on March 26, 2008, 09:39:01 pm
Quote
4. For mass spec, how much do we need to know about the process of fragmentation and the formation of free radicals? Because basically all I know is that there's an electron beam, it shoots electrons off atoms (somehow) to make them into cations, electrons are now unpaired so they split off into two parts, one part is a cation, one is a free radical...the radical gets vaccuumed away and the cation is detected and plotted on the graph? And this can happen in a variety of different combinations, hence the different peaks on the spectrum?
You should know everything that you said in this paragraph. About the free radical being "vacuumed away," I'm not too sure about that. The point is that the free radical is not a charged particle, and hence it will not be deflected by the magnetic field of the mass spectrometer. This means it will just fly a straight path, instead of being deflected towards the detector.
Post by: bec on March 29, 2008, 02:37:04 pm
Atomic absorption spectroscopy (AAS) and colorimetry both involve absorption of light. Both can be used to determine the amount of copper in a solution.
a What species absorbs the light when copper nitrate is analysed by:
i colorimetry?
ii AAS?
b Which technique would be simplest for the analysis of 0.5 M copper nitrate solution? Explain your answer.
A:
a i The copper(II) ion, Cu2+.
ii Cu atoms.
b Colorimetry would be the easiest technique to determine the concentration of a 0.5 M copper nitrate solution; AAS would require several dilutions in order to bring the concentration of copper into the linear range of the calibration curve.
What I don't understand:
1. How do we know that only the Cu2+ ions absorb light in colorimetry? What about the nitrate, how do we know that doesn't?
2. For part (b), I'm confused: won't colorimetry also require a calibration curve? If we're selective about the standard solutions we use for AAS, wouldn't the sample solution be in the linear range of that as it is?
Post by: Collin Li on March 29, 2008, 04:01:02 pm
1. How do we know that only the Cu2+ ions absorb light in colorimetry? What about the nitrate, how do we know that doesn't?
2. For part (b), I'm confused: won't colorimetry also require a calibration curve? If we're selective about the standard solutions we use for AAS, wouldn't the sample solution be in the linear range of that as it is?
1. As I've explained before, metallic atoms are much more vulnerable to spectroscopic techniques. This is because their electrons are more easily excited because the nucleus has a weaker grip on the electrons than non-metallic atoms. The nitrate group is made of non-metallic atoms (N and O), so it is not very spectroscopically active.
2. This is probably outside of the scope of the course. What the question is basically implying is that AAS is very sensitive, which is why you'd need several dilutions. The "linear range" is an important concept. Basically, when we make calibration curves, we assume the relationship between concentration and absorption is linear. However, this is only true for a particular range (called the linear range). The linear relationship breaks down when the concentrations are too high. You can imagine that if you continually increased the concentration of some solution, the absorption would eventually plateau (like an asymptote). This is why there is a linear range, and it depends on how sensitive an instrument is.
The part which is outside the scope of the course is knowing, quantitatively, how sensitive particular instruments are. You should know AAS is more sensitive than colorimetry, but you shouldn't be expected to know whether a 0.5M solution works better in colorimetry or AAS.
Post by: Mao on March 29, 2008, 04:33:03 pm
I have uploaded some notes on this particular topic, dl it if you want to
http://notes.vcenotes.com/?step=downloader&download=35
also this thread...
http://vcenotes.com/forum/index.php/topic,2116.0.html
Post by: bec on March 29, 2008, 04:53:45 pm
hahaha after reading your notes and the thread you linked me to i was still thinking that my original understanding had been right, so i came back to my initial question and realised that, yet again , the textbook was wrong: the graph they gave in the question had different integration traces to the graph in the answers (which i copied here)
so i'll delete my post since it makes no sense!
Post by: bec on March 29, 2008, 05:14:55 pm
I was right with (a) but part b i got wrong. According to the solutions, it's:
Propane: 7 Propan-1-ol: 1 Tartaric acid: 4
Before I go any further, is this another mistake in the textbook? If the book is right then there's obviously something i'm missing...
Post by: Mao on March 29, 2008, 05:52:02 pm
for propane, it is next to 6 CH bonds, so following the n+1 rule, the peak for the * should be split into 7
for propan-1-ol thought, the *(H) is next to 5 CH bonds. from my knowledge, the adjacent CO bond should not shield anymore than the H in OH, so there should be 6 peaks
as for tartaric acid, the only unshielded H nearby is the lone CH below it, so there should be 2 peaks
Post by: bec on March 29, 2008, 06:07:02 pm
with the tartaric acid one though, wouldn't it just be a single peak? I thought peak splitting was all about the adjacent atoms - the ones they were actually linked to?
Post by: droodles on March 29, 2008, 06:14:24 pm
Post by: Mao on March 29, 2008, 06:37:19 pm
yeah i got the same for the first twofor tartaric acid, you have essentially a H*-C-C-H, or CH*CH (the OH coming off we dont care about, because they are all "shielded" and dont affect this)
with the tartaric acid one though, wouldn't it just be a single peak? I thought peak splitting was all about the adjacent atoms - the ones they were actually linked to?
so for our H*, there is 1 hydrogen next to it
and following the n+1 rule, you'll have two peaks
Post by: bec on April 01, 2008, 01:37:30 pm
I said Cu2+ but the answer is "The copper hexaaqua ion, Cu(H2O)62+". wha?
Also, in question asking me compare samples analysed using UV-vis and IR, the answer was this:
IR SPEC
A very wide range of organic and molecules including solids, liquids and gases
UV-VIS SPEC
Relatively low molecular weight organic molecules which are coloured or have conjugated double or triple bonds eg caffeine in soft drink, sunscreen in a skin cream
1) do we need to know this much detail?
2) what's a conjugated bond?
3) why is low molecular weight necessary?
thanks!
Post by: Collin Li on April 01, 2008, 02:17:59 pm
What species absorbs the light when copper nitrate is analysed by UV-visible spec?
I said Cu2+ but the answer is "The copper hexaaqua ion, Cu(H2O)62+". wha?
That's true, but I'm not actually sure if that is expected knowledge. I'm not sure whether transition metal complexes are in the course any more (I think they show up in Unit 1), but the blue copper sulfate solution is actually from the transition metal complex - the copper hexaaqua ion. The colour of the copper ion fades (to white) when you dehydrate it with sulfuric acid (a dehydrating agent).
If you remember from your transition metal studies (that is, if you did any), the complexes that they form are colourful. Copper is a transition metal, and that is why it produces a beautiful blue colour when in complex with the H2O ligands.
Post by: Collin Li on April 01, 2008, 02:22:44 pm
A conjugated double bond is a pattern of bonds, alternating between single and double. You don't need to know this, and you don't need to know this is suitable for UV-visible (don't need to know why either - I suspect it's due to the double bonds that can interact with UV-light).
Post by: bec on April 01, 2008, 02:48:17 pm
If you remember from your transition metal studies (that is, if you did any), the complexes that they form are colourful. Copper is a transition metal, and that is why it produces a beautiful blue colour when in complex with the H2O ligands.
nope, never did any. and i've never heard of ligands. has anyone else learnt this stuff?
if not i'm assuming it's not on the course and the book just put it in there for "fun"...
Post by: Mao on April 01, 2008, 03:18:26 pm
something to do with subshells and other things (which also links to donuts somehow, but i forget)
Post by: bec on April 02, 2008, 11:51:00 am
a) Why is the test solution held at 50C for 15min before proceeding with the analysis?
(This is question 2 on a past exam but I had a quick look and it's not on the 2002-2007 exams, so the assessment report isn't on the net. could be the 2001 exam 1?)
Post by: cara.mel on April 02, 2008, 12:14:13 pm
Post by: bec on April 02, 2008, 12:30:54 pm
so with the same question, it says that the concentration of dichromate in the test solution is 0.0028M.
Then,
"Calculate the amount, in mole, of ethanol that reacted with dichromate in the original test solution."
How do I do that? I know i need to subtract n(C2H5OH reacted) from n(total C2H5OH) but i don't know how to get the no. mols reacted.
i thought of using n=cV, but i only "c" so that would be impossible.
UNLESS... i assume that there is still 100mL of solution there, even though it's been heated for so long and there would have been evaporation.
is that what i need to do? if not, how do you work it out?
Post by: Mao on April 02, 2008, 06:20:30 pm
if not, you should check the assessment report...
Post by: bec on April 08, 2008, 07:59:12 pm
I'm trying to write a list of two possible sources of error for all the instrumental spec techniques, can anyone help me fill it up?
AAS: incorrecty positioned line of best fit to creat callibration curve; standard solutions made up incorrectly
UV-VIS: same poss. errors as AAS
Colorimetry: Fingerprints on cuvette, standard solutions made up incorrectly
IR: Incorrect interpretation of graph + ???
NMR: ???
thanks
Post by: bec on April 09, 2008, 07:46:29 pm
a) Detecting the presence of a performance enhancing drug in a urine sample
b) Determining the concentration of a sodium chloride solution
The answer is (a) - and in the worked solutions the reason given for (b) not being correct is that "AAS can be used to determine the concentration of sodium in a solution".
But does that mean that you COULDN'T use UV-vis to analyse that?
2. Which of the following correctly matches a technique with the required analysis?
a) mass spec to suggest the presence of a CH3 group in an organic molecule
b) UV-vis spec to determine the magnesium content of a fertiliser
Answer is (a), and it says that "magnesium content would be determined using AAS" which i KNOW would be better but does that mean that a UV-vis spectrometer couldn't perform the analysis of magnesium content?
3. Name the spectroscopic technique that would be used to perform the following analyses:
a) Identification of a particular drug in a blood sample (I said AAS, correct A is UV-vis)
b) Detection of the C=O group present in an organic compound (I said NMR, correct is IR)
c) Measurement of the concentration of Mn2+ ions in a waste water sample (I said UV but should be AAS)
Can anyone help clear these up? thanks
Post by: Mao on April 09, 2008, 08:24:54 pm
metallic ions (and ions in general) a full outershell, making it very stable and have little absorbing potential. metals in AAS however goes through the burner and are reduced to ground state, hence why AAS is very useful for detecting metallic ions.
I'm presuming that UV spectrum are not absorbed by ions. however, transition metals such as copper form colourful complexes, and the bondings in these complexes absorb UV, so *some* metallic ions can be analysed by UV
for q3:
a) since drugs are not metallic ions, AAS would not be used. but as for identification, i thought IR would be more suitable, as it can be used for identification, UV-vis tells the concentration, not molecular structure
b) C13 NMR could be used, but IR is easier, more cost effective
c) metallic ion = AAS. if it was permanganate (or other compounds), UV-vis would be suitable.
Post by: bucket on April 10, 2008, 09:55:02 pm
I seem to have forgotten everything to do with gasses...
i don't know how to do
"calculate the concentration of SO2 in the polluted air in g m-1"
I have worked out;
m(SO2)=0.185g
and
v(air)=10.0m3
If you need any other numbers let me know... because i don't know what measurements are required to do such a question.
Post by: Collin Li on April 11, 2008, 07:40:45 am
It cannot be
A concentration of
Post by: bec on April 21, 2008, 06:07:38 pm
For example, 2-methylhexane is written as CH3CH(CH3)CH2CH2CH2CH3
i was assuming the functional groups were in brackets, but then i saw that 4-nonanol is written as CH3CH2CH2CHOHCH2CH2CH2CH2CH3
is this just one of those things that you can write either way, or is there a reason for it?
Post by: Mao on April 21, 2008, 07:38:56 pm
CH3CH(CH3)CH2CH2CH2CH3
the CH3 in brackets is the methyl side chain attached to the second side-chain, hence 2-methyl hexane
if that wasnt written in that way, it'll be very difficult to figure out what it is, as it can be read as heptene (even though it has two extra hydrogens
as for the 4-nonanol, the OH doesnt need to be in brackets, as it is implied that it is attached to the carbon from the side, not actually being part of the backbone chain itself.
it that made sense at all =)
Post by: bec on April 21, 2008, 07:53:01 pm