I adore your content and I have recommended your channel and lessons for use in my chemistry class on AS Access. You really are an exceptional teacher :D
THANK YOU SO MUCH! I have been looking for something for nearly a week now (i'm bad at research it seems) and my teacher has been ill so i couldn't ask her. Normally the internet is the biggest hep but i just couldn't find my answer. Then i stumble across this and my week long struggle and stress has finally come to an end. THANK YOU FOR THIS THANKS THANKS!!!
6:22 why is ch2oh a radical????....... and would it also be possible if e.g. we fragment the CH3CH2 off, for the OH to get a positive charge instead. do they get the positive charge randomly????
It depends on what type of positive fragment you see. When you split the molecular ion into its fragments, the one with the m/z value relative to the peak, is the one which takes the positive ion and the rest of the molecular then becomes a radical. So, it doesn't really matter as long as the spectrometer picks up the positive peak.
Sorry to bother you, but I have another question: A student mixes 100 cm3 of 0.200 mol dm-3 NaCl(aq) with 100 cm3 of 0.200 mol dm-3 Na2CO3(aq). This is on my sample paper for ocr. I thought that there are 3 atoms of Na therefore on the product side there would be 3 Na ions. Next, I Calculated the mols of NaOH (or either since they have the same conc/volume) and got 0.02 mols. Then I did 0.02 x 3 (because there is 3 atoms of Na+ ions on the product side) Lastly, I did 0.06=c x 200/1000 (200 as the overall solution has 200 cm3) and rearranged to get c= 0.3 mol dm-3. The thing that's bothering me is whether I was correct in thinking there is 3 atoms of Na+ ions. what if I balanced the equation and it gave me a number in front of the Na product? Thanks, and again sorry to bother you
Thank you very much for the video. How do you determine which fragment will be the positive ion and which will be the radical? Do you not do both as it could be either? Thanks
like wouldn't it be random fragmentation. And also why couldn't the CH3 become the radical instead of the CH2OH couldnt it be the other way round (5:33
BAMitsFRESHPINK I think it could be either, he might've just given the CH3 the charge because its a really common fragment. Also I think that the exam question will have a mass spectra which only detects positive charges, so you don't have to worry about which radicals are being formed
Remember that the molecule loses one electron in the ionisation chamber of the mass spectrometer giving it an initial 1+ charge. If a further bond is broken (homolytically) to produce fragments, one part will carry the 1+ charge and be a radical cation (this is detected by the spectrometer) the other will not carry the charge and be a radical (not detected). There are various ways that fragmentation can occur, involving different types of bond fission. I teach A level Chemistry which does not require such an in depth look at bond breaking so I keep it as simple as possible.
the m+1 peak represents the naturally occurring isotope of carbon which is c-13 but you'd normally ignore the m+1 peak and focus on the molecular ion peak for these questions. hope that helped and good luck if you have chem unit 2 today :)
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Very helpful, didn't quite get it in class but this has really cleared up the stuff I didn't understand
I adore your content and I have recommended your channel and lessons for use in my chemistry class on AS Access.
You really are an exceptional teacher :D
THANK YOU SO MUCH! I have been looking for something for nearly a week now (i'm bad at research it seems) and my teacher has been ill so i couldn't ask her. Normally the internet is the biggest hep but i just couldn't find my answer. Then i stumble across this and my week long struggle and stress has finally come to an end. THANK YOU FOR THIS THANKS THANKS!!!
You're welcome! Glad you found it :)
Best vid on ive found on mass spec. cheers x
6:22 why is ch2oh a radical????....... and would it also be possible if e.g. we fragment the CH3CH2 off, for the OH to get a positive charge instead. do they get the positive charge randomly????
Thank you for your video :) Also how do you know which one is the radical?
Cheers sir really clear explanation 👌
Glad you liked it
thanks a lot for these videos very helpful
Thanks for the comment Patrick. Good luck with your exams!
cheers
how do you know what height to make the peak?
Much helpful explanation, Cheers man.
Great Explanation. Thanks, Sir.
Wouldn't the OH+ be a charge too? Why are you only splitting the molecules vertically?
This was very helpful. Thank you.
thanks for the video, can u help me analyze my new compound?
how do you know which fragment is positive and which is the radical
It depends on what type of positive fragment you see. When you split the molecular ion into its fragments, the one with the m/z value relative to the peak, is the one which takes the positive ion and the rest of the molecular then becomes a radical.
So, it doesn't really matter as long as the spectrometer picks up the positive peak.
can you do a video for constructing B6H6 mass spectra?
So when you use the electron gun to knock off an electron, will you always knock 1 off otherwise the mass would be half if 2 were knocked off?
Correct and sometimes that happens and you see peaks with half the mass of the others
Sorry to bother you, but I have another question: A student mixes 100 cm3 of 0.200 mol dm-3 NaCl(aq) with 100 cm3 of 0.200 mol dm-3 Na2CO3(aq). This is on my sample paper for ocr. I thought that there are 3 atoms of Na therefore on the product side there would be 3 Na ions. Next, I Calculated the mols of NaOH (or either since they have the same conc/volume) and got 0.02 mols. Then I did 0.02 x 3 (because there is 3 atoms of Na+ ions on the product side) Lastly, I did 0.06=c x 200/1000 (200 as the overall solution has 200 cm3) and rearranged to get c= 0.3 mol dm-3. The thing that's bothering me is whether I was correct in thinking there is 3 atoms of Na+ ions. what if I balanced the equation and it gave me a number in front of the Na product?
Thanks, and again sorry to bother you
Sir please tell me which compound used as a reference compound in mass spectra please sir tommorow is my paper of organic please
wouldn't the CH3 fragment be a radical?
Does the height of a peak ie relative abundance % matter? How to distinguish which peak is higher?
Thank you!!
Lidia Stralys Peak height is linked to the amount of the fragment ion that forms relative to the others. This isn’t tested in the exam
@@MaChemGuy great! Many thanks!
Thank you very much for the video. How do you determine which fragment will be the positive ion and which will be the radical? Do you not do both as it could be either? Thanks
It could very well be either. At A level you don’t need to know
@@MaChemGuy Ok thank you!
How do you know where the molecule fragmentation would take place though?
like wouldn't it be random fragmentation.
And also why couldn't the CH3 become the radical instead of the CH2OH couldnt it be the other way round (5:33
BAMitsFRESHPINK I think it could be either, he might've just given the CH3 the charge because its a really common fragment. Also I think that the exam question will have a mass spectra which only detects positive charges, so you don't have to worry about which radicals are being formed
i was soo happy to see ur video related to mass spectrometry but after 2:00 mins the video is not focused :'(
cutiee pie Strange, looks fine to me
MaChemGuy :-[
cutiee pie maybe your computer is buffering
MaChemGuy no
cutiee pie Mystery then. Video been up for 2 years or so and no mention of picture quality
why is the ch3 a positive ion and not the other molecule
Both fragments can actually carry the positive charge but some are more able than others. The neutral fragment is a radical btw
So does the molecule fragment hetrolytically?
Remember that the molecule loses one electron in the ionisation chamber of the mass spectrometer giving it an initial 1+ charge. If a further bond is broken (homolytically) to produce fragments, one part will carry the 1+ charge and be a radical cation (this is detected by the spectrometer) the other will not carry the charge and be a radical (not detected). There are various ways that fragmentation can occur, involving different types of bond fission. I teach A level Chemistry which does not require such an in depth look at bond breaking so I keep it as simple as possible.
does the m+1 peak represent all carbons in the fragment ion or only one
the m+1 peak represents the naturally occurring isotope of carbon which is c-13 but you'd normally ignore the m+1 peak and focus on the molecular ion peak for these questions. hope that helped and good luck if you have chem unit 2 today :)
No way is this the real MaChemGuy???
Cannot believe it
???
MaChemGuy I once heard of a legendary chemistry tutor, who’s lessons were the stuff of legend. I just didn’t believe it until now!
Benjamin Kidane Haha! That’s got to be up there as one of my best comments 👍
Great video thank you
not good
11th AS Thanks 🙏🏻