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 chlorine-containing fragment $m/z=63$
- a chlorine-free fragment $m/z=73$
- and some more stuff at lower masses that doesn’t analyse itself quite as easily.
Those three peaks should derive from fragmentation processes, so we can get an idea of what groups we have from looking at the mass difference (i.e. the bit that fragmented away).
For the peak at 93: $122-93=29$. The chlorine atom is retained. The most common groups to leave are methyl, ethyl etc., and 29 happens to be exactly the mass of $\ce{CH3CH2}$ (methyl is 15, methylene is 14; you internalise those numbers pretty quickly). Therefore, your molecule should have an ethyl group somewhere.
The chlorine-free peak at 73: $122-73=49$. That’s an uncommon number at first sight but remember we lost a chlorine. Taking that into account, $49-35=14$ and we find another methylene group. So we can say that there is a terminal $\ce{CH2Cl-{}}$ group.
Finally, the peak at 63: $122-63=59$. That again doesn’t say everything at once, but what happens if we remove the 29 we found earlier for the ethyl group? $59-29=30$. 30 could mean two methyl groups but that’s somewhat unlikely given how few fragments we have overall—especially since we would also expect a peak with a mass difference of 15 for a single methyl group. But $30=14+16$, meaning there could be a $\ce{-CH2-O-{}}$ group attached to the ethyl group. Other possibilities for 16 might be $\ce{NH2}$, but the molecule disobeys the odd-nitrogen rule and two amino groups doesn’t look good with the data.
If we take what we have, we have a strong suspicion that there might be a $\ce{CH3-CH2-CH2-O-{}}$ group and a $\ce{CH2Cl-{}}$ group. Adding these together gives an intermediate mass of $59+49=108$ or 14 missing from the complete molecule. It seems reasonable to assume a $\ce{CH2}$ group connecting the fragments.
We should consider this all a working hypothesis until we can then take a look at the NMR spectrum which beautifully confirms that molecular structure.