Post by: Water on April 01, 2011, 09:53:20 pm
So I said NMR, HPLC, MASS, Infrared.
Why can't I use HPLC though?
D)
Book said only GC. But, why can't I used HPLC?
The book said volumetric, AAS , Gravimetric
But why can't I use UV Vis?
Post by: vea on April 01, 2011, 10:13:47 pm
Post by: Water on April 01, 2011, 10:16:25 pm
14B) UV Visible is only for coloured or organic compounds.
vea, the book said
UV Vis can be used for identifying the presence of metal ions; even if the metal ion itself is not colored, it may be possible to analyse it by converting it into a colored compound.
Also UV Vis occurs for such electron transitions of atoms, ions or molecules
>;
Could the book be wrong for this answer? For question 14B that is.
Post by: azngirl456 on April 01, 2011, 10:18:43 pm
Post by: Greatness on April 01, 2011, 10:18:55 pm
Post by: Water on April 01, 2011, 10:26:04 pm
Post by: azngirl456 on April 01, 2011, 10:36:00 pm
Azngirl, in HLPC, isn't the sample, liquid throughout the process?
Yes that's correct, which means that you can't use a gas in HPLC. You can use liquid alkanes as the sample for HPLC, but not gas ones.
Because it just says mixture, you're not sure of the states, so I would just go with GLC...because it's possible to make a liquid into a gas
Post by: Water on April 01, 2011, 10:39:05 pm
Azngirl, in HLPC, isn't the sample, liquid throughout the process?
Yes that's correct, which means that you can't use a gas in HPLC. You can use liquid alkanes as the sample for HPLC, but not gas ones.
Because it just says mixture, you're not sure of the states, so I would just go with GLC...because it's possible to make a liquid into a gas
Not sure about the UV Vis one though, we analyzed K2MNO4 with colorimetry, so how is it considered for only organic chemistry as well? ;O
Post by: luken93 on April 01, 2011, 10:52:26 pm
I think you may be reading it wrong Water...14B) UV Visible is only for coloured or organic compounds.
vea, the book said
UV Vis can be used for identifying the presence of metal ions; even if the metal ion itself is not colored, it may be possible to analyse it by converting it into a colored compound.
Also UV Vis occurs for such electron transitions of atoms, ions or molecules
>;
Could the book be wrong for this answer? For question 14B that is.
"it may be possible to analyse it by converting it into a colored compound." - Silver Nitrate isn't coloured, however you could analyse if Silver was present and/or it's concentration by making it into a coloured solution (can't think of any). But for Silver Nitrate specifically, it isn't coloured, so you can't analyse SILVER NITRATE with UV Vis...
Post by: Water on April 01, 2011, 10:56:57 pm
And for Silver Nitrate, you would need to precipitate it.
ARGH, this is so confusing. Heineman Book page 120.
UV VIS : Typical Analyte : Low Molecular mass organic molecules
Typical Sample: Liquid and Gas Samples
But on page 85: UV VIS has wide range of applications for virtually everything.
So Frustrating! *slits wrist*
Post by: luken93 on April 01, 2011, 11:27:59 pm
AAS: Will give you the concentration of metal (Ag) ions by putting it through the AAS instruments
Gravimetric: Add a substance that will form an Ag precip, from there you can determine the concentration
In UV however, you actually have to change the substance. In the others, another substance is added to allow it to become equi-molar, or to precipitate, but in UV you can't actually use AgNO3 if you get what I mean?
It makes sense to me saying it like that, ultimately you could determine the concentration by creating a coloured compound with the Ag in it, but you can't actually determine the concentration without creating a different substance?
I think I'm going around in circles, and as you say you also need a substance for Grav and Vol as well...
Post by: Water on April 02, 2011, 03:43:50 pm
Post by: luken93 on April 02, 2011, 07:05:15 pm
luken93, I went to my maths tutor today, he's a chemical engineer and he said you can find the concentration of AgNO3 via UV Vis ! No Reason why should not be able to, as I suspected :tickedoff:Hahaha well then the book is indeed wrong, I was just trying to justify the book, I'm sorry :P
Post by: zibb3r on April 02, 2011, 11:41:33 pm
Post by: Mao on April 03, 2011, 01:09:25 am
In UV however, you actually have to change the substance. In the others, another substance is added to allow it to become equi-molar, or to precipitate, but in UV you can't actually use AgNO3 if you get what I mean?
Yes, there are many dye bases to which we can react with the analyte selectively to find their concentration with photospectrometry. In the typical case, we are probably looking at a dye that will react with NO3- only (We add the dye base in excess, and all the nitrate will form a coloured dye). It is correct that we cannot measure some things directly.
Post by: luken93 on April 03, 2011, 08:36:59 am
So then what actually is the right answer? Or are we going a lot deeper than the book requires?
In UV however, you actually have to change the substance. In the others, another substance is added to allow it to become equi-molar, or to precipitate, but in UV you can't actually use AgNO3 if you get what I mean?
Yes, there are many dye bases to which we can react with the analyte selectively to find their concentration with photospectrometry. In the typical case, we are probably looking at a dye that will react with NO3- only (We add the dye base in excess, and all the nitrate will form a coloured dye). It is correct that we cannot measure some things directly.
Post by: Water on April 03, 2011, 08:44:00 am
So then what actually is the right answer? Or are we going a lot deeper than the book requires?
In UV however, you actually have to change the substance. In the others, another substance is added to allow it to become equi-molar, or to precipitate, but in UV you can't actually use AgNO3 if you get what I mean?
Yes, there are many dye bases to which we can react with the analyte selectively to find their concentration with photospectrometry. In the typical case, we are probably looking at a dye that will react with NO3- only (We add the dye base in excess, and all the nitrate will form a coloured dye). It is correct that we cannot measure some things directly.
I suspect that, what hes saying is, AgNO3, you make all of the NO3 in AgNO3 react with the dye. And through there, we can find the concentration/mol of AgNO3, similar to gravimetric analysis with excess precipitating agent. Though we can have the chem man himself say it/ or correct me as "always" :)
Post by: luken93 on April 03, 2011, 09:05:15 am
Post by: Mao on April 03, 2011, 12:45:47 pm
So then what actually is the right answer? Or are we going a lot deeper than the book requires?
In UV however, you actually have to change the substance. In the others, another substance is added to allow it to become equi-molar, or to precipitate, but in UV you can't actually use AgNO3 if you get what I mean?
Yes, there are many dye bases to which we can react with the analyte selectively to find their concentration with photospectrometry. In the typical case, we are probably looking at a dye that will react with NO3- only (We add the dye base in excess, and all the nitrate will form a coloured dye). It is correct that we cannot measure some things directly.
I suspect that, what hes saying is, AgNO3, you make all of the NO3 in AgNO3 react with the dye. And through there, we can find the concentration/mol of AgNO3, similar to gravimetric analysis with excess precipitating agent. Though we can have the chem man himself say it/ or correct me as "always" :)
You are correct.
What I'm trying to say is if you just put a 'neat' solution of AgNO3 into the UV-vis spectrometer, most probably nothing will happen. However, you can react your sample to form other stuff which will have absorption in the UV-vis spectrum.
The book is trying to target the knowledge that UV-vis is generally suitable for coloured compounds, AAS is generally suitable for metal ion solutions.
Post by: Water on April 06, 2011, 05:52:10 pm
Does 2- methyl But-4-ene exist? And if no, why not? Because Heineman only gave me 4 solutions, but it doesn't nclude this one?
Post by: Greatness on April 06, 2011, 06:15:13 pm
Post by: Water on April 06, 2011, 06:25:06 pm
Post by: jasoN- on April 06, 2011, 06:30:28 pm
This is 2-methylbutene
CH3
\ | | /
-C-C-C=C
/ | \
(http://www.lookchem.com/UserFilesUpload/26760-64-5.gif)
i don't see why it won't exist if that's what you're asking
what are the four solutions you're given?
Post by: Water on April 06, 2011, 06:47:43 pm
The problem is, you could draw it out
CH3 CH(CH3)CHCH2
Your counting from your nearing side groups first before you count double bonds I"m assuming. So from this...
I'd get 2 MethylBent-4-ene.
Post by: jasoN- on April 06, 2011, 07:00:28 pm
But you count from double bonds before side groups
Post by: Water on April 06, 2011, 07:03:38 pm
you count from double bonds before side groups
But it was asking for isomers of Pentene
CH3 CH(CH3)CHCH2
So It'd be 3-methylbent-1ene ;)
Post by: Water on April 09, 2011, 12:19:24 pm
MnO4- or Cr2O7
CH3CH2OH (aq) ---------------------> CH3COOH
H+
What is the H+ meant to be symbolic of? What is its purpose?
Post by: andy456 on April 09, 2011, 12:22:00 pm
This is probably wrong but: I think it has something to do with the electrochemical series. As in the MnO4 won't react unless in acidic solution.
Post by: luken93 on April 09, 2011, 01:05:21 pm
It means the solution must be acidic but I can't remember why....Yeah that's pretty much correct, you pronounce it as "Acidified Permangante/Dichromate"
This is probably wrong but: I think it has something to do with the electrochemical series. As in the MnO4 won't react unless in acidic solution.
Post by: Mao on April 09, 2011, 02:04:10 pm
Post by: Water on April 12, 2011, 01:05:41 am
b) When a stick of chalk is placed upright into a beaker that contains a small quantity of black ink, various colours are observed to move up the chalk
My Explaination: The stick of chalk acts as a stationary phase, where the black ink will act as both moving phase and solvent. The Different and adsorption rates will separate the components within the black ink.
Is this appropriate for an answer, or is it flawed?.
Post by: Greatness on April 12, 2011, 11:10:28 am
The Different and adsorption rates will separate the components within the black ink.Hmm your answer seems ok, but maybe include a bit more detail. Is this an exam quesiton or textbook question?
Perhaps change that into something like: the components of the ink continually undergo adsorption onto the stationary phase followed by desorption back into the mobile phase. The degree to which this occurs causes the different components of the ink to separate.
Post by: Water on April 12, 2011, 04:07:32 pm
Post by: Water on April 21, 2011, 08:19:30 pm
a) Write balanced chemical equations for the reaction of both As4S6 and As4S4 with oxygen.
b) Calculate the mass of both As4S6 and As4S4 in the original Mixture.
So for my answers, I had m(As4S6) = 0.531g and for m(As4S4) = 0.469g....
I would like to consult with VN, to see if I"m correct. Thankyou :)
Post by: luken93 on April 21, 2011, 09:00:35 pm
(The solid is Arsenic Hexoxide)
b)
Hmm, still getting over a gram. Whether they aren't using VCAA molar masses, I'm not really sure...
Post by: Water on April 21, 2011, 09:08:00 pm
My method was abit tedious, wasn't sure, if how it was set out was right. I generally lose marks for setting out, cause my teacher is a dick like that.
n(As4O6) = 0.905g / 395.68gmol
= 0.002287
m(As) in (As4O6) = 0.002287 x 4 x 74.92gmol-1
=0.6854g
m(S) in mixture = 1.00g - 0.6854g
= 0.3146g
m(As) in (As4S6) = 1/2 x 0.6854g
= 0.3427g
m(S) in As4S6 = 6/10 x 0.3146g
= 0.18876gram
m(As4S6) = 0.3427g + 0.18876g
= 0.531gram
m(As) in (As4S4) = 1/2 x 0.6854g
= 0.3427gram
m(S) in As4S4 = 4/10 x 0.3146g
= 0.1258gram
m(As4S4) = 0.3427g + 0.1258gram
= 0.469gram
m(As4S4) + m(as4S6) = 0.531g + 0.469gram = 1.00gram
Post by: luken93 on April 21, 2011, 09:24:38 pm
Although, couldn't you just work out the ratios of the molar masses of the two?
Mr(As4S6) = 492.2
Mr(As4S4) = 428.0
Ratio As4S6 : As4S4
= 1.15 : 1
= 0.53488 : 0.46511
In the end, that is what you are trying to find out, the mole ratios are the same?
Post by: Water on April 21, 2011, 09:29:07 pm
Ratio As4S6 : As4S4
= 1.15 : 1
= 0.53488 : 0.46511
How'd you get from 1.15, to 0.53... and 1 to 0.46511, mathematically
I Knew there was a much more efficient way than mines. Mine was just pathetic if it was examination period Lol.
And, would mine be considered wrong? Cause answers look different to yours, and I used more calculations, hence more room for error.
Post by: luken93 on April 21, 2011, 09:36:59 pm
Ummm, I'm not really sure who's wrong. I think it may be best if I head back and do mine exactly again. Stay posted
Post by: Water on April 24, 2011, 04:20:38 pm
Post by: Greatness on April 24, 2011, 04:36:26 pm
Post by: luken93 on April 24, 2011, 06:14:25 pm
Post by: Water on May 19, 2011, 09:31:52 pm
Calculate the amount, in mole, of phosphorus in the 20.00ml volume of solution.
So,
My method of calculation is, find mol of Mg2P2O7.
Then n(MgNH4PO4) = n(Mg2P2O7)
n (P) = n (MgNH4PO4)
Am I correct in my method?
Post by: Water on May 19, 2011, 10:42:31 pm
Post by: Mao on May 20, 2011, 12:18:58 am
n(MgNH4PO4) = 2 n(Mg2P2O7) ?
Anyways, a more direct method would be
n(P) = 2 n(Mg2P2O7)
Post by: Water on May 20, 2011, 12:39:23 am
The precipitate was filtered, washed and then converted by heating into Mg2P2O7
I wasn't so sure, about this whole process. But, if its simply converted, wouldn't the mols remain the same? I also used
MgNH... as it was in a solution...whilst the latter Mg2P2O7m is dried and stuff, hence perhaps isn't compatible with the question?
): I've got no solutions Haha atm haha xD.
Post by: Mao on May 20, 2011, 12:50:49 am
The ionic equation is
Thus n(P) = n(MgNH4PO4) = 2 n(Mg2P2O7)
:)
Post by: Water on May 20, 2011, 12:56:11 am