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Thread: NMR course

  1. #1 NMR course 
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    so I have quite the conundrum. I have to take this course in NMR to complete my degree. the long and short of it is, that I have to take it as an external credit that is equivalent to the course at my home institution.

    I have no prior experience in NMR, and I have attended all the lectures and have gone to the prof. a few times, for help. I have to admit i feel for the guy as I'm sure he is annoyed that they allowed me to take this course with no prereq.

    So I have my final assignment and I would love to kill it as it is worth 25% of my final mark. So i wanted to see if there is anyone out there that would be willing to provide guidance. I am in the process of doing the assignment and I'm just looking for feedback. and i am at the point now that I am willing to pay for somebodies time.

    Anyways if anybody is willing to help please let me know.

    Thanks


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  3. #2  
    Moderator Moderator AlexP's Avatar
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    There's no need to pay anyone here. We're happy to help as long as you put some serious effort into it, which it sounds like you're willing to. I don't know who has how much knowledge of NMR, but feel free to post what you need guidance in and we'll see what we can do, but no promises.


    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  4. #3  
    Forum Cosmic Wizard i_feel_tiredsleepy's Avatar
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    Ug I haven't dealt with NMR in 5 years, and I wasn't very good at it back then when it was just a small component of a chem class.

    Hopefully, there is somebody around here that knows a bit more about it.

    Is it one of those things where you have to look at the graph and determine what the molecule is, or is it more complicated?
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  5. #4  
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    so here are two questions:

    1.An NMR spectrum was recorded in a quantiative way by a professional spectroscopist with the utmost care. The spectroscopist looked at the data after 3 hours of data accumulation. She measured the signal to noise ratio as 11.2 but allowed the data to continue to accumulate. When the data accumulation was complete the signal to noise ratio was 203.8

    a) what was the total time for the data accumulations?
    b) The weight of the sample in the NMR tube was 4.6mg. How many mg of sample would have to be dissolved in the NMR tube to get a singal to noise ratio of 203.8 in only 3 hours.

    2. A student placed a sample in an NMR magnet, set the spectrometer up for 31P using a simple one-pulse sequence, tuned the probe and was about to collect the data when the phone rang....the student quickly set up his paramters as follows:

    Spectral Width =30,000Hz
    Acquistion time = 500msec
    Recycle delay = 500msec
    Pulse width = 3usec (the 90 degree pulse was 9usec)
    Number of scans = 7200

    Before he left to play squash he examined the data after the first scana nd saw two peaks in the spectrum with intensity ratio 1:1. This was entirley consistent with the 1:1 molar mixture he was expecting. He returned after playing squash to find his data acquisition had finished. He was disappointed to find that the intensity ratio of the peaks was now approx. 3:1. Assume that the sample was stable over time and that the peaks really should have been in a 1:1 intensity ratio, why was the spectrum from the first scan quantitative but the spectrum from all 7200 scans not quantiative? How could this problem be avoided?
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  6. #5  
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    so question number 1a) seems pretty straight forward to me. let me know what you think

    S/N = 11.2
    Time = 3 hours
    final S/N = 203.8
    final time = X

    11.2/3 = 203.8/X
    X = 54.59 hours

    but as far as b) im not sure where to start. I mean i know that the S/N is proportional to the sqrt of the number of scans and directly proportional to the concentration. but even with that information i dont know if it is helpful for b.
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  7. #6  
    gc
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    Here's a hint for 1a:

    What is the relationship between the time that the data is accumulated for and the S/N ratio?
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  8. #7  
    gc
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    Hint for question 1b:

    What happens to the S/N ratio as you increase the amount of sample? For example, if you doubled the amount of sample from 4.6 mg to 9.2 mg, what would the S/N ratio be after 3 hours?
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  9. #8  
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    for hint 1b) i would think that as the concentration is doubling so is the S/N ratio so 22.4?

    so if the s/n = 203.8/11.2 = 18.196
    then 18.196 x 4.6 = 83.mg

    am i on the right track?
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  10. #9  
    gc
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    Quote Originally Posted by rumberg
    for hint 1b) i would think that as the concentration is doubling so is the S/N ratio so 22.4?

    so if the s/n = 203.8/11.2 = 18.196
    then 18.196 x 4.6 = 83.mg

    am i on the right track?
    That looks right to me. Good job!
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  11. #10  
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    Quote Originally Posted by gc
    Here's a hint for 1a:

    What is the relationship between the time that the data is accumulated for and the S/N ratio?
    gc,

    this is the only way i think that makes sense to get this answer

    S/N = 11.2
    Time = 3 hours
    final S/N = 203.8
    final time = X

    11.2/3 = 203.8/X
    X = 54.59 hours


    maybe im just not getting your clue..

    btw thanks a lot for your help
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  12. #11  
    gc
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    Quote Originally Posted by rumberg
    gc,

    this is the only way i think that makes sense to get this answer

    S/N = 11.2
    Time = 3 hours
    final S/N = 203.8
    final time = X

    11.2/3 = 203.8/X
    X = 54.59 hours


    maybe im just not getting your clue..

    btw thanks a lot for your help
    No problem

    You're close, but your answer is assuming that the S/N ratio increases linearly with data accumulation time. Think about how the S/N ratio increases as you accumulate more scans.
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  13. #12  
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    Quote Originally Posted by gc

    No problem

    You're close, but your answer is assuming that the S/N ratio increases linearly with data accumulation time. Think about how the S/N ratio increases as you accumulate more scans.
    by george i think i got it...

    so 203.8/11.2 = 18.2 so the S/N increases by a factor of 18.2

    It takes 331.24 times as many transients (18.2^2)

    since 3 hours has already passed it would be 3 x 331.24 =993.72 hours

    how is that?

    edited : trying to figure out something, because that amount a time just cannot be right.
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  14. #13  
    gc
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    Quote Originally Posted by rumberg
    by george i think i got it...

    so 203.8/11.2 = 18.2 so the S/N increases by a factor of 18.2

    It takes 331.24 times as many transients (18.2^2)

    since 3 hours has already passed it would be 3 x 331.24 =993.72 hours

    how is that?

    edited : trying to figure out something, because that amount a time just cannot be right.
    Excellent!

    Yeah, it does sound like a ridiculous amount of time, but that's the answer I get.
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  15. #14  
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    GC i wanted to know if I posted or emailed you my assignment with my answers if you would mind taking a look at it for me. It is due tomorrow and I just want to ensure i get the best possible mark.

    I'm sorry I know that it is asking way to much.
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  16. #15  
    gc
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    Quote Originally Posted by rumberg
    GC i wanted to know if I posted or emailed you my assignment with my answers if you would mind taking a look at it for me. It is due tomorrow and I just want to ensure i get the best possible mark.

    I'm sorry I know that it is asking way to much.
    Feel free to PM me, or you could just post it here if you want.
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