Please Help-Chem Questions.

[bec]

k thanks

[Collin Li]

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.

[bec]

Q:
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?

[Collin Li]

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.

[Mao]

look at the numbers just on top of the peaks, they are integration traces, basically area under the curve. It usually is provided with the graph.

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

[bec]

thanks for that mao!
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!

[bec]

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...

[Mao]

mmm i think so

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

[bec]

yeah i got the same for the first two

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?

[droodles]

you're COOH mao

[Mao]

yeah i got the same for the first two

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?

for 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)

so for our H*, there is 1 hydrogen next to it

and following the n+1 rule, you'll have two peaks

[bec]

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?


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!

[Collin Li]

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.

[Collin Li]

UV-visible can be used for metallic ions too, so I am not sure why it says "organic molecules." I am not aware of the low molecular weight requirement, and I don't think you need to know it (you do need to know of this requirement for GC, however).

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).

[bec]

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"...