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All possible arrangements of $\ce{Br2}$ molecule: $\displaystyle 79 + 79 = 158$ $\displaystyle \color{red}{79 + 81} = 160$ $\displaystyle \color{red}{81 + 79} = 160$ $\displaystyle 81 + 81 = 162$ The amount of $\ce{^{79}Br}$ and $\ce{^{81}Br}$ in nature is roughly the same, thus each permutation is equally probable. There are two arrangements that lead to $... 14 I'm not sure I completely understand what exactly your problem is, but: You want to resolve to signals at approx. m/z = 1060.2 and m/z = 1061.2. So$\Delta m = 1$The required resolution is:$\frac{m}{\Delta m} = \frac{1061}{1} \approx 1060$: The spectrometer in (a) has only 1000 which is too low. The spectrometer in (b) has 5000 which is quite more than ... 14 Without knowing more details, it is hard to guess, but at this$m/z$, it seems likely that the peak is the result of one$^{12}\ce{C}$being substituted by one$^{13}\ce{C}$. It is more useful to assume a uniform mass of 12 for carbon when analyzing such a spectrum. Note that with more carbons in a larger molecule, it becomes more likely that at least one ... 12 In mass spectrometry, ions with the same nominal mass (but possibly differing exact masses) are called isobaric. 11 I am not an expert in ICP-MS, but an ICP-MS should be considered an ICP ion source coupled with a mass spectrometer. There are some key issues raised by the coupling between both instruments due to the high temperature of the ICP source as well as due to the fact that one wants to analyse atomic ions and remove any interfering molecular ions (mostly oxides, ... 11 The peak at$m/z = 59$with lower intensity in respect to the one at$m / z =58$(the molecular ion) is not overseen. Mass spectroscopy is capable to deliver information about the isotopic composition of your sample, too. Since -- accounting over all carbon atoms of your sample -- about 1.1% are of the non-radioactive isotope of$\ce{^{13}C}$, you may use ... 11 Here is the "periodic table for biomolecules" (leaving out hydrogen, please ignore silicon): The pie chart shows the isotope distribution. Of the elements C, N O, P, S and H, sulfur has the highest abundance minor isotope (around 4%). On the other end of the spectrum,phosphorous is a mono-isotopic element, with just one stable isotope. There is an ... 10 The MSMS part is the detection of the LC in this case (nb. you don't need to do LC-MSMS, you can do direct infusion quite happily if you don't need to separate anything), so the actual experiment wouldn't need to take much longer than any other LC (as CHM says). However, calibration + set-up of these things, and extensive cleaning afterwards, can make it a ... 10 Yes it is possible, but is very expensive and would be orders of magnitude more costly than what people are willing to pay for recycled materials. Let me give two data points to explain why it is expensive. Some commercial material is manufactured by mass spectrometry. The starting material isn't garbage, but reasonably chemically purified metals of ... 10 Being an NMR fan myself I would inspect that NMR spectrum: The integrals suggest you have 11$\ce{^1H}$or a multiple thereof (the number under each peak is the normalized integral, which is proportional to the number of protons represented by the multiplet). That leaves you with$\pu{87 Da -11 Da}=\pu{76 Da}$to explain. If you throw in an oxygen you ... 10 I would probably also use the method Buck has suggested, but let’s say the NMR broke down or somebody is measuring a$\ce{^13C}$of$\pu{2.5mg}$meaning it will be blocked until tomorrow; in this case, we can still extract more information from the mass spectrum. In addition to the molecule peak at 122, you have: a chlorine-containing fragment$m/z=93$a ... 9 The abbreviation$m/z$is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number. It has long been called the mass-to-charge ratio although$m$is not the ionic mass nor is$z$a multiple or the elementary (electronic) charge,$e$. The abbreviation$m/e$is, therefore, not recommended. Thus, for example, ... 8 First of all, you should indicate if you consider the fragmentation in electron ionization or in other ionization conditions. Considering it is the electron ionization conditions and that you refer to this spectrum, one can postulate for a reasonable mechanism, which can be derived from what is known on the fragmentation of benzoic acid and esters. I am not ... 8 Question 1: What’s the difference between an electric field and a magnetic field? And why does the magnetic field bend the ions while the electron field does not. Is this just a consequence of the shape of the apparatus? Could you use a magnetic field to accelerate a particle and a electric field to do the bending? I’m just really confused as to the ... 8 A way to understand this that may be familiar is that of the Punnett square from biology, since the two isotopes have nearly 50/50 split in nature. \begin{array}{c|cc} & \ce{^{79}Br} & \ce{^{81}Br} \\\hline \ce{^{79}Br} & \ce{^{158}Br} & \ce{^{160}Br} \\ \ce{^{81}Br} & \ce{^{160}Br} & \ce{^{162}Br} \\ \... 7 Is this even how GC/MS results work? As cbeleites said the method you described is a proper technique but not likely to be appropriate given the information you cited. In GC/MS you should have two sets of information. The first is the GC Total Ion Chromatograph (TIC) which will have time as the x-axis and response (abundance) as the y-axis. For each ... 7 In$\ce{Br2}$, the possible combinations of isotopes are:$\ce{^{79}Br-^{79}Br} : \mathrm{m/z}~158\ce{^{79}Br-^{81}Br} : \mathrm{m/z}~160\ce{^{81}Br-^{79}Br} : \mathrm{m/z}~160\ce{^{81}Br-^{81}Br} : \mathrm{m/z}~162$Bromine's isotopic distribution is essentially a$50:50$ratio of mass$79$and mass$81$. That means that there's a$25\%$chance ... 7 Reverse engineering, or deformulation, is a common industry practice. This is a very common type of request for commercial analytical laboratories, and is generally not cheap. You can discover just how widespread it is, and how many commercial providers there are with a quick Google search. To get yourself a reasonable starting point, you should request a ... 7 2.3:10, 23:100 and (23/123):(100/123) are all correct values for the ratio of the relative abundance of B-10 to B-11. However, only the last value, (23/123):(100/123) or 18.7:81.3 shows the normalized ratio of relative abundance of these isotopes. It is this normalized value that adds up to 100. 7 The answer probably has to do with how the mass spectrum was obtained. Most mass spectra that are provided online, including the ones on the NIST WebBook, are electron ionisation mass spectra. That means that you put the sample in a chamber, accelerate electrons to a really high speed (giving them approximately$70~\mathrm{eV}$of energy), and then bombard ... 7 Generally speaking: knowing the history of your sample (e.g. earlier steps of synthesis allowing / excluding elements that could numerically fit into the list of suggested fragments) and additional spectroscopic characterisation (UV-Vis, IR, Raman, NMR, etc.) may be helpful. The problem you describe may be attenuated by using a mass spectrometer of higher ... 7 I like both answers provided before me where one has used exclusive use of internet to suggest structure by NMR spectrum, and the other has used thorough analysis of mass spectrum. Although these two are valuable techniques, I feel OP needs to analyse step by step analysis of spectra given to predict the structure since he/her seemingly in a graduate course, ... 6 The mass spectrometer observes the mass to charge ratio (m/z). Most of the time, small organic molecules like butane will be singly charged by the ionizer, which means that the mass can be read off directly since z =1. If butane was somehow ionized to the +2 state, then the mass spec would observe 58/2=29, which is the mass of butane (58 g/mol) divided by ... 6 The major isotopes of chlorine are 35 and 37. If 4 chlorines are present in a molecule, then the following isotopic combinations of chlorine are possible 35, 35, 35, 35 35, 35, 35, 37 35, 35, 37, 37 35, 37, 37, 37 37, 37, 37, 37 So there will be 5 molecular ion peaks. You could calculate their relative heights by factoring in the relative isotopic ... 6 You have already got an answer by Dr. Karsten, but let me add that modern mass spectra have become sensitive beyond imagination. Secondly, the resolution of mass is no longer a problem. Fourier transform ion-cyclotron resonance mass spectrometer (a mouthful) is extremely sensitive to mass differences. Yes, if you have a low concentration, your main analyte ... 6 I wanted to back up my comment and make sure it's justified, so I decided to flip through Gross'Mass spectrometry: a textbook [1] and it looks like there are indeed basically two possible variations, including distance between$m/z$values, as you suggested: {Distance | difference | mass difference} between {peaks | signals | peak tips} {Difference in$m/z$... 5 This type of ionization is known as electron ionization (there are many types of ionization methods used for MS). One possible reaction for EI is: $$\ce{M +e- ->M^{•+} +2e-}$$ source If the incoming electron has the correct energy to interact and passes close enough to the molecule, it's electric field can impart enough energy to one of the molecule's ... 5 Okay, let's tackle this. First of all you worked out correctly that there are four different combinations, resulting in four different peaks. You have also given the probability for finding a$\ce{{}^{35}_{17}Cl}$is$a=33.3\%\approx\frac13$and the probability for finding a$\ce{{}^{37}_{17}Cl}$is$b=66.7\%\approx\frac23$. Since you have three positions to ... 5 I found it quite hard to follow both the question and the answer, so I am going to state a few relevant things and paraphrase the answers of @LDC3 to add some clarity. Chemical shift (ppm) and peak area (integral) These are independent quantities. One tells you about the environment (shift) the protons are in, and the other tells you how many protons are ... 5 Absolute mass can be determined electrochemically, once the coulomb is defined in terms of numbers of electrons, as through the Millikan oil-drop experiment, to be$6.24×10^{18}\$ electrons per coulomb. See Determination of Avogadro's Number and Absolute Determination of the Electrochemical Equivalent and the Atomic Weight of Zinc. One thing that ...