Acompound contains only c, h, and n. combustion of 35.0 mg of the compound produces 33.5 mg of co2 and 41.1 mg of h2o. what is the empirical formula of the compound.

I'll tell you how to solve it- 
all the C ends up as CO2 
mg of C = (12.011/44.011) x 33.5 mg CO2 = 9.142 mg C 
all of the H ends up as H2) 
mg H = (2.016/18.016) x 41.1 mg H2O = 4.599 mg H 
mg N = 35.0 - 9.14 - 4.60 = 21.26 
CxHyNz represents the empirical formula 
where x = moles of C atoms, y = moles of H atoms, and z = moles of N atoms in (1 mole of) the molecule 
Calculate the moles of C, H, and N that you found. 
Determine the ratio of the moles x : y : z 
Do that and then look at my calculations- 
9.14 mg C / 12.011 mg/mmol = 0.76 millimol C 
4.60 mg H / 1.008 mg/mmol = 4.60 millinol H 
21.26 mg N / 14.006 mg/mmol = 1.52 millimol N 
divide by the smallest number (0.76) 
C = 1 
H = 6 
N = 2