Chemistry ideal gas law question..

A mixture of gases collected over water at 14°C has a total pressure of 1.198 atm and occupies
72 mL. How many grams of water escaped into the vapor phase?

I used the ideal gas equation and got 6.45 x 10^-4 g H20 (escaped into vapor phase)

Did I do this right?

EDITED ANSWER, BASED ON EXTRA DETAILS:

OK, I see you've updated your post with some more information!

Your answer isn't correct (still), but I'll show you the correct solution and you can figure out where/why you made your mistake. I'll ignore the background stuff (like going over the ideal gas law) that I had in my original answer, since you can just look down.

PV = nRT
n = P*V/(R*T)

Here:
P (water) = 12.0 torr
V = 0.072 L
T = 287.15 K
R (note: I'm using an R that has Torr as the unit of pressure here, just because--you could easily use one with atm as the unit of pressure by simply converting torr to atm) = 62.36367 L*Torr/(K*mol)

We have to assume that all the gaseous water atoms are ones that escaped into the vapor phase from the water. Based on the problem, that assumption isn't really stretching things much at all.

Plug in numbers and...
n = 4.8247*10^(-5) mol

Convert mols to g using molar mass of water and we get:
8.69 * 10^(-4) grams of water

This is approximately 30% off as per the percent error formula than our original (with some MAJOR assumptions) answer, in case you're curious.

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ORIGINAL ANSWER, NOT BASED ON EXTRA DETAILS:

No, you didn't do it right, but it may be a calculation error somewhere... Here's a complete solution, so you can find out what you did wrong and learn from it (we DO have to make some considerable assumptions to solve what you gave, so if those aren't true based on the problem statement then this answer is not correct):

We know: PV = nRT