Here's the problem:

A compound contains only C, H, N, and O. Combustion of 0.157 g of the compound produces 0.213 g CO2 and 0.0310 g H2O.  In another experiment, it is found that 0.103 g of the compound produces 0.0230 g NH3.  What is the empirical formula of the compound?  Hint: Cobustion involves reacting with excess O2.  Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O.  Also assume that all the nitrogen ends up in the NH3 in the second experiment.
From Zumdahl Chemistry 5e, Chapter 3, #121

I started by writing the chemical equation:


Is this correct?  Can I add NH3 to the equation?  It has to go somewhere.

Second, I used a proportion to find the amound of NH3 produced if 0.157 g of compound combusted.  I got 0.0351g NH3.  I then found the amount of O2 that was used in combustion by subtracting reactant masses from product masses and got 0.122 g O2 combusted.  Next, I found how many grams of each element was used up in each product by using molar masses.  I found that there was 0.0581g C, 0.00970g H, 0.0289g N, and 0.061g O used in the compound.  I then used % composition in the compound and found 37.0%C, 6.18%H, 18.4% N, and 38.9% O by mass.  I then divided those numbers by the molar mass of each element to get 6.13 mol H, 3.08 mol C, 1.31 mol N, and 2.43 mol O.  Even allowing for err in rounding, these numbers are nowhere near integers nor do they have a small GCF to multiply by to get integers.  Book answer is C7H5N3O6.  This is an AP Chem summer assignment so it is not mission critical that I arrive at the correct answer, but help would be greatly appreciated.  

this is wrong, since combustion of the compound results in NO2, not NH3.

--> experiment 2 must be used separately from experiment 1 as done in the calculation

this is my answer:

1: we have 0,157 g and let that react with O2 to form 0,213 g CO2 and 0,0310 g H2O

--> this results into the following calculation:

--> 0,213 g CO2 contains:
0,213 / 44,01 = 0,0048398091 mol C
2 * 0,0048398091 = 0,0096796183 mol O

--> 0,0310 g H2O contains:
0,0310 / 18,016 = 0,0017206927 mol O
2 * 0,0017206927 = 0,0034413854 mol H

So the compound of 0,157 g Contains:
0,0048398091 mol C
0,0034413854 mol H

--> so the ratio C : H = 7 : 5

the mass% C and H are 37m% and 2,21% respectively

so 0,103 g compound contains 0,03811 g C and 0,0022763 g H
--> that's 0,0022582341 mol H and 0,003173189 mol C

the compuond contained 0,0230 / 17,034 = 0,0013502407 mol N

so the mass is now 0,0593031721 g so 0,0436968279 g is O
--> that's 0,0027310517 mol O
--> N : O = 1 : 2

now let's complete the whole story:

N : C = 1 : 2,35, so that's 3 : 7

so the book is totally correct, since N : O = 1:2 we get indeed the simplest formula of:

C7H5N3O6

corrected %mass composition:

carbon: 37%
hydrogen: 2.2%
nitrogen: 18.4%
oxygen: 42.4%

I also arrive at the same answer: C7H5N3O6

Just curious.
How does this balance?

C7H5N3O6 + O2 --> CO2 + H2O + NH3

Adding coefficients to the NH3 product to add more N product increases the needed coefficient for the CHNO reactant to balance the H, and increasing the CHNO coefficient to balance the H increases the NH3 coefficient to increase the N supply.

combustion will give NO2 instead of NH3.

Not necesarilly. In most cases of organic compunds combustion nitrogen evolves as N₂.

This works but I had to use a negative stoichimetric coefficient, so its not to be trusted at all. But here you go.

C7H5N3O6 + (3)O2 --> (7)CO2 + (-2)H2O + (3)NH3

Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O.  Also assume that all the nitrogen ends up in the NH3 in the second experiment.

although the question might be theoretically wrong, we were already told to assume that all nitrogen ends up as ammonia.