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Belowmasr spucinuman unknow comocundIdentity Ihe molecular mass of this compoundb) Provide Molecular lormula Irom this mass ionPerfom hydrogen deficiency ndex (HD...

Question

Belowmasr spucinuman unknow comocundIdentity Ihe molecular mass of this compoundb) Provide Molecular lormula Irom this mass ionPerfom hydrogen deficiency ndex (HDI) analysis this fonula and detemmine Ihe numbor double bond equivalents are conlained molocule [please show working]Provide structures for the fragments lost provide mass ions at 71, 58 and 43 respeclively?

Below masr spucinum an unknow comocund Identity Ihe molecular mass of this compound b) Provide Molecular lormula Irom this mass ion Perfom hydrogen deficiency ndex (HDI) analysis this fonula and detemmine Ihe numbor double bond equivalents are conlained molocule [please show working] Provide structures for the fragments lost provide mass ions at 71, 58 and 43 respeclively?



Answers

What is the mass of the molecular ion formed from compounds having each molecular formula:
(a) $C_3H_6O$; (b) $C_{10}H_{20}$; (c) $C_8H_8O_2$; (d) methamphetamine $(C_{10}H_{15}N)$?
How to use the mass of the molecular ion to propose molecular formulas for an unknown is shown
in Sample Problem 13.2. In this process, keep in mind the following useful fact. Hydrocarbons
like methane $(CH_4)$ and hexane $(C_6H_{14})$, as well as compounds that contain only C, H, and 0 atoms, always have a molecular ion with an $even$ mass. An odd molecular ion generally indicates
that a compound contains nitrogen.

Here we are looking to determine the molecular masses of different substances. That's what we need to do is add the molecular mass of each of the elements within the compound together. So for the first one, M c A S four is equal to the mass of calcium, add the mass of sulfa, ad The four lots of the massive oxygen. Because we've got four oxygen atoms, This is equal to 136.14 g per mole. And then we do the exact same thing for the remaining examples. So for the mass of C three H eight Hydrocarbon, this is 44 .0956 g per mole. Next door, we have the C six H four, S 02, And H22 Is equal to 172.20 grams per mole. The final example we have here, The mass of UO two three P 042 is 1000 Went 03 g per more.

This question wants us to analyze the mass spec data for false moral chloride. Um so first, the cat on fragment with a mass to charge ratio of 85 corresponds to P o F. Two, whereas the cat ion fragment with a mass to charge ratio of 69 corresponds to pf two. Um, next question is asking which Pete, which to Pete in the mass spec provide evidence that the oxygen atom is connected to the phosphorus Adam and not connected to any of the foreign. Adams and the Peaks, which um support the fact that oxygen is connected to foster, um, are the peaks at 88 at 47 which correspond to the cat ions, PF three and P O, respectively. Eso The peak at 88 is the most conclusive evidence that the oxygen is bound to the phosphorus atom as the difference between it and the parent alien peak is equal to the mass. A charge ratio of oxygen. The P get 47 corresponds to the mass of phosphor er and oxygen being bound together without the presence of flooring Adams. So it means that knocking off the floor Eames did not remove the oxygen, which can only be explained, Um, if false for and oxygen are bound together

We are taking a look at the empirical formula. So that is the simplest formula with the whole number ratio of atoms. So what we have to do is take the mass of each element and divided by the molar mass to get our value of moles. So we have not point 270 so platinum, not 0.542 for nitrogen, 1.62 moles for hydrogen, 1.8 moles for chlorine. And so what we do is divide them by the lowest number of moles, not 0.270 And what we get is following ratio 1 to 2 to 6 to 4. So from this I know that my empirical formula is Petey and to age six cl four. And so what we can do is deduced some complexes. And so we have the following, we are pt in the center and H three and H three. Then we have full chlorine as instructed by the formula at the top. Alternatively, what we can have is two of the irons, the chloride and irons free. If we want to pt and H 32 cl two, I was too plus charge and we have cl minus where we've got to lots of them. That gives us a square plane are. Instead we have NH three NH three. Again, we can have the system the trans of these, where we have a two plus charge and then we have the two chloride and irons.

This is the answer to Chapter thirteen. Problem number two from the Smith Organic Chemistry Textbook on this problem. Eyes giving us three ions from three mass spectrometry experiments on DH asking us to rights impossible formulas for each ion. And so, essentially, what you want to do is just remember thie atomic weight of each different element on DH. See what kind of combinations you come up with. So for Ah, for an ion of seventy two. Easiest answer is C five h twelve on I've drawn the math out down here. Carbon weighs twelve. There's five of them. Hydrogen weighs one. There's twelve of them. S O sixty plus twelve is going to give you seventy two. Alternatively, you could include an oxygen. Remember, oxygen weighs sixteen. So four carbons in twelve apiece. Eight hydrogen is at one apiece on one oxygen at sixteen on. That also adds up to seventy two. So the answer to the next the next I on one hundred is pretty much, you know, exactly the same. You just find combinations that will add up to one hundred. And so again, the simplest one just carbon and hydrogen is going to be eight carbons. A twelve apiece and four Hodgins that want a piece again? If you wanted to include an oxygen, Um, you know, just as it's drawn out here, see six h twelve o. If you do the math that will add up to one hundred for you on DH, then I do just want oh, focus on the third part of this problem. Just a little bit eso We're asked to find formulas. Basically, that add up to seventy three. And what I was always taught was when you're looking at mass spectrometry problems, if you haven't a Nayan with an odd number. So in this case seventy three, you should immediately think that nitrogen is present. And so the two answers that I've drawn here both contain nice regions. S O nitrogen weighs fourteen, but what it's actually doing is forcing the hydrogen is to assume odd numbers. And that's how you get your odd number. And that's how you get to your seventy three. Because remember carbons or twelve oxygen's or sixteen nitrogen czar fourteen so you can never add to even numbers together or however many even numbers together and yet not number. But when you include a nitrogen. It's going tio force the hydrogen Sze to be an odd number. And that's how you get your ID number. And so again, I've drawn the math for each of these out in green. Uh, and that is the answer to Chapter thirteen. Problem number two.


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