[bobbym46]

   My answer is that this molecule is the s configuration and I just wanted to double check if I was right ? can anyone affirm that it is S or R?

here is my reasoning...

I drew the molecule but switched the groups so the methyl was in the back, and figured out the molecule was R, so if u switch them back the original must be S right ?

[Babcock_Hall]

Your answer looks correct to me.  Your way of solving it is not my favorite method, but works if applied carefully.  I prefer to hold one substituent in the plane of the screen constant (either the propyl group or the carboxylic acid) and spin the other three substituents, placing the lowest priority group in the back.

[bobbym46]

Hmmm thanks for the help, I understand what you mean, but still dont understand if it is r or s.. I tried to make it in this 3d molecule viewer to better understand and came up with an S configuration

[Babcock_Hall]

As you are showing it, the F atom is in back, is it not?  If you can rotate your model to put the methyl group in back, that should allow you to assign it.  Here is a ChemDraw rendition of what I described.  This method does not change the absolute configuration of the carbon atom.