In the course material you have been introduced to multiple charging for electrospray spectra. Each peak in the spectrum below can be used to provide a measure of the mass of this protein, you can work out the charge associated with each peak, and measures can be averaged to find the overall mass. Use simultaneous equations to find the charge associated with each peak in the spectra below, which will allow you to calculate the average mass of each species. Include an estimate of the error associated with this measure.Check samples also need this type calculation x30510-0200600659.8714.7700779.5800857.4900952.610001071.51100120013001400+M5, 0.1-1.3min 1(7-29) 4. B/S [1509← 2023 MSC CHROMATOGRAPHY AN... Q L :687.5620-651,346007001:39800883 68900951.56958.911030.74 1124.4310001158 23 1281 67 1373.99 1463.06 1570.061100 1200 1300 1400 1500 1600Each of the peaks in this spectrum correspond to the intact protein (no fragmentation!) but withdifferent levels of charge. This is an electrospray mass spectrum, hence you get multiple charging ofthe analyte. So To find molecular weight of this protein you need to multiply the m/z of each peak bythe number of charges (and then remember to subtract the number of charges, cos in electrospraythe charges are protons and have a mass of 1 each, and you don't want to include this in your sum.You can do this process for quite a few peaks in the spectrum and average the calculated masses togive you a more accurate answer, and this will also give an error associated with the measurement.Firs, however, you need to work out what the charge is for each peak. To do this you usesimultaneous equations.773.362-727.942=727.94Choose two adjacent peaks in the series (e.g. 773.36 & 727.94). Use the following equations:773.36 = M/Z (where M is the mass of the protein and Z is the number of charges that gives rise tothe peak at 773.36727.94 M/(Z+1) (M is still the same, but to give the next peak along in the series the charge hasgone up by 1!Rearrange these equations and solve for Z:773.362 = M and 727.94Z+727.94 M therefore 773.362 = 727.94Z+727.94O1729.74 1840 28 1932 071274m1700 1800 1900 2000Z=16.03. Since the charge is a proton, it needs to be an integer, hence Z=16.Z(773.36-727.94) = 727.94 therefore Z= 727.94 / (773.36-727.94)Lets work out the mass: M = 773.36*16= 12,373.76 take off 16 (the proton charge) = 12,357.76And 727.94 * 17 12,374.98 take off 17 12,357.98And 824.85 15 = 12,372.75 take off 15 = 12,357.75And 687.56* 18 = 12,376.08 take off 18 = 12,358.08 etcPAGE 6 OF 6O

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The two adjacent peaks are 714.7 and 779.5M/Z = 779.5 and M/(Z+1) = 714.7779.5∗Z=714.7(Z+1) => Z…

659.8 4588 714.7 779.5 857.4 24