Calculate the m1m peak for the hypothetical compound...

Transcript

00:01 Hello, this question it is related to spectroscopy.

00:07 So here it is given a compound formula that is compound is c5h11 cl and this is nothing but one chloropentane.

00:29 So here if we see the normal structure of this we have one carbon 2, 3, 4, 5 and to this chlorine atom it is attached there.

00:39 So m plus peak it should be, see this m plus it is the highest intensity peak in cluster of peaks at this highest m by z.

00:54 For example in this graph if we say that that is the highest peak it is shown by this line.

01:00 So this this is called as this m plus and to this line the shortest one it will be that is the this one will be considered as m plus 1 and m plus 1 and m plus 2 it shows the presence of it shows the presence of chlorine or bromine and here as we have in this compound that is chlorine that means chlorine will be there and m plus 1 it is it represents then and we can say it is caused by carbon 13 and isotope which is present in the molecule and this m plus it is the highest peak so therefore here this m plus p should be at m by z it is 106 it is 106 because 106 it is the molar mass of this one chloropentane so that will be the m plus one peak and m plus one peak it will be here in the next in the next case it is this this will be the m plus one so therefore this will become one zero seven and hence m plus two it is the next one that is one zero eight plus two plus two to m plus this the this is the peak and this is the molar mass of this is the molar mass of the compound which is given here that is one chloropentine and to this plus one it will be one zero seven plus two it is one zero eight now as we know that that is the if we need to find out the percentage of carbon atom, percentage of number of carbon atom then we have a relation that is relative intensity of m plus 1 peak divided by as the carbon 13 abundance it is 1 .1 so therefore we can say that it is 0 .011 multiply by relative intensity of mp by this formula we can find out these values but here first let us see this if here it is given n plus it is given here that is 27 if m plus is the 27 then see this m plus 2 it shows that it is a peak about chlorine atom which is present in the molecule and is having two isotopes that is chlorine with mass 35 and chlorine with mass 37 and if we see their abundance that is 75 % is the abundance and 25 % of this chlorine 37 so therefore here ratio will be 3 ratio 1 therefore the peak due to chlorine atom it is 3 raised to 1 so here given it is m plus 1 that is 27 so if here we can say that is m plus ratio to m plus two it will be equal to three ratio one and hence it is 27 it is given m plus two it is equal to three ratio one and hence we can find out this value that is m plus two it is equal to three nine times 27 so therefore n plus two it is nine...

Question

Calculate the M+1/M+ peak for the hypothetical compound C15H25Cl2 if the M+ peak is 85.0% in height or relative abundance. Give your answer to 2 places after the decimal.

Question

Calculate the M+1/M+ peak for the hypothetical compound C15H25Cl2 if the M+ peak is 85.0% in height or relative abundance. Give your answer to 2 places after the decimal.

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