Chemistry: Moles Question

Question:
A compound was known to contain C, H, N, and O. When 0.0513 g sample was burned and the products were 0.0842 g of CO2, 0.0207 g of H20, and 0.0107 g of N2. What was the empirical formula of the compound? The answer is C10 H12 N4 O5

I got carbon, hydrogen, and nitrogen right, its just that im having problem with oxygen. and the thing is all similar problems that have to do with oxygen, i always get stuck on them, why?

I got 0.001914 mol for Carbon, 0.002298 mol for Hydrogen, 0.00076392 mol for Nitrogen, and my oxygen is wrong.

I got this ratio so far

C:H:N:O
2.5 : 3 : 1 : x

Helps plox, 10 points if you can explain this stuff >.> got test on this garbage tomorrow :p

(0.0842 g CO2) / (44.00964 g CO2/mol) x (1/1) x (12.01078 g C/mol) = 0.022979 g C
(0.0207 g H2O) / (18.01532 g H2O/mol) x (2/1) x (1.007947 g H/mol) = 0.0023163 g H
(0.0107 g N2) / (28.01344 g N2/mol) x (2/1) x (14.00672 g N/mol) = 0.010700 g N

(0.0513 g) - (0.022979 g C) - (0.0023163 g H) - (0.010700 g N) = 0.015305 g O

(0.022979 g C) / (12.01078 g C/mol) = 0.0019132 mol C
(0.0023163 g H) / (1.007947 g H/mol) = 0.0022980 mol H
(0.010700 g N) / (14.00672 g N/mol) = 0.00076392 mol N
(0.015305 g O) / (15.99943 g O/mol) = 0.00095660 mol O

Divide by the smallest number of moles:
(0.0019132 mol C) / 0.00095660 = 2.00
(0.0022980 mol H) / 0.00095660 = 2.40
(0.00076392 mol N) / 0.00095660 = 0.800
(0.00095660 mol O) / 0.00095660 = 1.00

To achieve mole ratios near integers, multiply the above by 5, then round to the nearest whole number:
C10H12N4O5