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Hi Kumodgurav, I am going to make some assumptions about the information that you gave me. You are reporting a retention time that suggests that your data is from an LC/MS run. Is it possible that you used an electrospray ionization (ESI) detector in the negative ion mode? ESI frequently produces adduct ions that are a combination of the analyte molecule plus some an ion present in the buffer. The mass difference of 45 amu fits nicely with the idea that the observed parent ion is the neutral drug molecule plus one formate ion [M + HCO2]- See: www.acdlabs.com/blog/common-adduct-and-fragment-ions-in-mass-spectrometry/

That's a good question and you need to ask it whenever someone reports a value for a formal potential, because people do different things. However, most often people work at conditions other than unit activity for all of the reactants and products for the half reaction of interest. Rarely do people work at pH 0 (unit H+ ion activity), for example. The pH is probably one of the most influential parameters that varies among experiments. Temperature is also frequently varied. Excellent experimentalists are careful to report temperature, ionic strength, pH, reactant concentrations for any measurements that are not at standard state.

Think of a molecule with the formula that you have given containing atoms of only the most abundant isotope of each type. (If you don't have a table of isotopes for elements common to organic molecules, you can find one on-line.) For most non-metallic elements, the most abundant isotope tends to be the lightest. Do this for carbon, hydrogen and chlorine separately. Multiply those isotope masses by the number of atoms of that element that appears in each molecule and sum these numbers for all the elements in the molecule.

@@garymabbott4064 A big thank you for your answer Gary! This really helps me clarify this concept. I had another question: We are interested in the potential and current flowing from the working electrode for obvious reasons. The counter electrode however is mostly there to get rid of the electrons our working electrode produces. I was wondering how the potential of this counter electrode is regulated? dependent on which reaction happens there, it seems to me that a different voltage would be needed at the working electrode to perform both the half reactions? If a reaction at the counter electrode happens that needs a high driving force voltage wise, it would seem to me that the working electrode would need to work harder (have a higher potential) to make sure the reaction happens? Or is there some sort of mechanism in the technical part that senses how many volts are needed at the counter electrode for that half reaction to happen to make up for it? I have also been struggling a bit with the concept of an electrolyte. To me, it seems that as long a reduction reactions happens at the cathode, and an oxidation reaction happens at the anode, current is flowing. Does the electrolytes facilitate the transition of the electrons from the electrodes to the species? why does the 'current' wants to go through the solvent instead of taking the electrodes as starting and ending point of a circuit?

There are these situations when the molecular ion is not the tallest peak in the high mass cluster: When the sample is contaminated and more than one compound is being ionized at the same time. We can usually rule this out, if our separation and/or clean up steps are rigorous. However, when the signal is weak background peaks (trace level contaminants and electronic noise) can complicate interpretation. When the compound of interest breaks apart so easily that no molecular ion survives to give a signal. This happens with long alkyl chains, for example. That is a tough puzzle, but one can get a lot of information by applying the same strategy to the cluster of ions at the highest mass with the caveat that this cluster is due to a fragment of the molecular ion is quite possibly an even mass ion. (See Fred McLafferty, Interpretation of Mass Spectra for an in-depth discussion.) When the molecule has multiple chlorine and/or bromine atoms. In those cases the probability of having a molecule with at least one heavy halogen isotope is greater than a molecule with all light isotopes. But these cases are easy to spot, since the tall peaks are spaced by two mass units from each other. Remember that everything heavier than the molecular ion with all light isotopes must be the result of ions with a combination heavy isotopes (and the same chemical structure, namely, the molecule minus one electron). Other ionization methods that we were not considering here also give a different signal for the molecular ion. Electrospray ionization (ESI) is very important method in LC/MS and in desorption ESI (or DESI). These methods lead to much less fragmentation, but form “adducts” (combinations of the molecule plus a hydrogen ion or sometimes a sodium ion). Here the signals in the high mass cluster must be explainable by combinations of isotopes for the structure of the whole molecule plus a hydrogen, that is, (M+H).

Thanks for the explaination. So, sir in the nutshell, do we have to consider only those elements whose M+1 intensity percentage in the table is less than the M+1 intensity given in question? Because if (5.1% of 75.2 =3.8) < 4.1 is consider here, we might not be able to put Carbon in the formula.

Thanks sir.

@@emaanimtiaz4371 Electrochemistry can be confusing, in part because of all of the jargon. The term “polarizable electrode” means that voltage (the energy for an electron) between the surface of the electrode and the solution can be changed by imposing energy (from a battery or electrical power source) from outside the cell. We need at least one other electrode in the cell in order to do that experiment. A reference electrode, such as a silver/silver chloride electrode, can be used for that purpose. A good reference electrode will not move from its “rest potential” (its value that can be calculated from the Nernst Equation for the reference reaction, such as AgCl + e  Ag + Cl-). We say that the reference electrode is non-polarizable. By selecting a non-polarizable reference electrode for a voltammetry experiment, virtually all of the energy (voltage) that we apply from the outside circuit ends up polarizing the working electrode. That is, all of the energy applied to the cell from the outside changes the voltage at the interface between the working electrode and the solution. If that voltage is big enough, it can force an electron exchange process there. A platinum wire or carbon electrodes are good working electrodes since they are polarizable and will exchange electrons readily with species in solution. One point of possible confusion with electrochemical measurements is that there are two common types of experiments. Voltammetry is the type I was referring to in the paragraph above. Voltammetry is based on applying an external voltage to the system which forces an electron exchange at the working electrode. The current is measured and its magnitude is proportional to the concentration of the electroactive species in solution. Voltammetry uses a polarizable working electrode. The other common experiment is potentiometry. The rest voltage of the working electrode is determined by the solution conditions and it is this quantity that we measure in a potentiometric experiment. In that case the working electrode or sensor should not be polarizable. We try to prevent significant amount of current going through the cell, since it could distort the voltage that appears at the working electrode. That is, allowing current in the system would push the working electrode away from the voltage that naturally develops at its surface because of the concentration of the species that it senses. I hope that helps.

So, the bottom line is that a polarizable electrode CAN exchange electrons with something in solution, if we push it to the right voltage.

Your question goes to the heart of the field of chromatography. Basically, when two compounds elute a the same time, you need to change the separation conditions (either the stationary phase or the mobile phase or both) in order to retain one more strongly than the other.

Hi Dawson, The term “slit width” refers to the physical width of the opening used to pass a narrow portion of the spectrum of light to be used in the experiment. The light from the lamp is dispersed by a grating into a rainbow across a wall inside the box that houses the monochromator. The opening in the wall (called the slit) allows only a narrow band (a small range of wavelengths) to pass out of the monochromator and on to the detector. In some contexts, slit width may also refer to the range of wavelengths (usually in nanometers for visible light) that pass through the monochromator (that is, through the physical slit). More often this wavelength range is called the "spectral bandpass". The narrower the slit width the better job the monochromator does in isolating the wavelength for the atomic absorption transition. However, beyond a certain point decreasing the slit width is detrimental because it weakens the signal intensity and, consequently, poorer sensitivity.