Sorry for the delay, this is what I've managed to do so far:

The M+2/M intensity is about 1/3 so Cl is present.

The M peak is odd so it contains N too (teacher said that if there's N it's only one for these exercises).

Then I divided 171 by 13 and I got C_{13}H_{15} formula, then I needed to substract the Cl and the N so it is a CH_{2} for the N and C_{2}H_{11} for the Cl.

This gives C_{10}H_{2} remaining, so there should be more H than the initial guess. My guess was 6 or a multiple of 6 but I don't see how to "guess" it now. Do I just try 12, if it doesn't work 18 and so on? Or just add 6H to that formula? Maybe my initial guess is wrong but adding the H in the NMR spectra is 6 no matter what you do.

I'm stuck with the formula, I think if I manage to get it the remaining part should be easier.

Never, ever do that that division by 13 to find the C count. Instead, measure the ratio of the intensity of the molecular ion at [M+1] to the molecular ion [M].

Deduce approximate C atom count from this ratio.

You have a MW = 171, so N =odd count (1 atom) and Cl =1 atom.

171 - 14 N - 35Cl = 122; no evidence of S or P or other halogen.

Set up a table as shown below,

C H O = 122

10 2 0

9 14 0

8 26 0

8 10 1

Etc until you get a combination of numbers of C, H and O atoms that fits your given data.

Did you look up the "Number of rings & double bond" equivalents?

If not, you should.

Regards,

Motoball