The following data is available from the question:

MS: M =   114 g/mol (even, no isotope pattern for Cl or Br)

EA: Calculates empirical formula = C₃H₅O (IR confirms O is present).
Which with MW from MS gives molecular formula = C₆H₁₀O₂ .  From here we get the IHD = 2

IR: There are significant absorptions at 1720cm^-1 due to a C=O and at 1640cm^-1 due to a C=C, C-O possible due to bands between 1200 and 1100cm^-1. There is no OH (about 3500cm^-1).

13C nmr: The proton decoupled spectrum shows a total of 6 peaks indicating 6 types of C. By analysis of the chemical shifts, we have a C=O (probably an acid derivative) at 167 ppm, 2 types in the region 125-140 ppm is C=C, 61 ppm is typical of a C-O and those at 18 and 14 ppm are most likely from hydrocarbon portions.

1H nmr: First of all we have 5 types of H showing up. After this, it's a good idea to tabulate the information to make sure you get it all correctly matched up:
 

The most significant structural information in the H nmr data is that we have an disubstituted alkene (note that the MW, MF, number of ArH all rule out the it being an benzene system), an ethyl group : -CH₂-CH₃ most likely as an ethoxy group: -O-CH₂-CH₃ and an isolated methyl group : -CH₃

Summary....
The MS indicated MW =  114 g/mol.
The IR showed the presence of  C=O, C=C and C-O bonds.
13C peak at 167 ppm suggests that the C=O is a carboxylic acid derivative
H nmr also gives a disubstituted alkene, a -O-CH₂-CH₃ and an uncoupled  -CH₃.
Use this to check the molecular formula : C₆H₁₀O₂ = 6 x 12 + 10 x 1 + 2 x 16 = 114 g/mol
So with all this information we have the following pieces: C=O, -O-CH₂-CH₃, -CH₃ and H-C=C-H or H₂C=C.

Altogether...
 

With the pieces we have :  C=O, -O-CH2-CH3, -CH3 and H-C=C-H or H2C=C. IR and 13C suggest an ester so we have : CO2CH2-CH3 This means the two substituents on the alkene C=C are the ester group and the -CH3 There are three possible arrangements: 1,1- , cis-1,2- or trans-1,2- If it were 1,2- then the methyl group would have to have an H neighbour and therefore appear as a doublet (i.e. in a CH3-CH= system). Therefore, it must be the 1,1-isomer where is has no adjacent H.....

ethyl methacrylate or ethyl 2-methylpropenoate

The final step should always be to check what you have drawn. The easiest thing to check is usually the coupling patterns you would expect to see, and the chemical shifts of each unit.
You should be asking yourself : "Does my answer give me what the H-nmr shows ?"