I've been working on spectroscopy practice problems for my organic chemistry course and am stuck on these two. The practice problems call for the molecular formula and skeletal structures for the compound. Please help! Thanks!

Hi
Your questions are very simple
mass spectra gives the molar mass and IR informs about the main functional group
13C and 1H NMR gives the skeleton of the molecule
1) Now the odd molar mass and IR spectra shows the presence of NH2 (~3300cm^_1)and aromatic ring(~3000cm^-1)
NMR clearly indicates the presence of aromatic ring( signals from 7ppm -8ppm) and the ring has 5 H atoms on it.Two H atoms are on N atom which is present on the ring
13C also confirms the presence of 6 aromatic carbon atoms
so the formula goes like this
C6H5NH2 with molar mass of 93
2) Similarly in this IR shows the presence of aromatic and alkane part also (Pl check the IR table to identify the signals )
~2900 cm^-1 is the C-H sp3 stretch and C-H sp2 stretch is ~3000cm^-1
There is one strectch of C-O near 1000cm^-1
NMR confirms the aromatic ring with 5 hydrogen atoms and one singlet of three H atoms near 4ppm shows the methoxy group on the ring.13C also confirms 6 carbon which are in the aromatic ring and one carbon which is aliphatic with no unsaturation.
so the formula is - C6H5OCH3 with molar mass of 108