Answer:
[tex]C_{7} H_{5}N_{3}O_{6}[/tex]
Explanation:
First reaction gives you the number of moles or the mass from Carbon and hydrogen
for carbon:
[tex]0,213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} =0.005molC[/tex]
[tex]0.213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} .\frac{12gC}{1molC} = 0.058gC\\[/tex]
Analogously for hydrogen:
0.0310g[tex]H_{2}O[/tex] have 0.0034gH or 0.0034mol of H
In the second reaction you can obtain the amount of nitrogen as a percentage and find the mass of N in the first sample.
[tex]0.023gNH_{3} .\frac{1molNH_{3}}{17gNH_{3}} .\frac{1molN}{1molNH_{3}} .\frac{14gN}{1molN} \frac{100}{0.103gsample} =18.4%N[/tex]
now
[tex]\frac{18.4gN}{100gsample} .0.157gsample=0.0289gN in the first reaction[/tex]
this is equivalet to 0.002mol of N
with this information you can find the mass of oxygen by matter conservation.
[tex]gO=total mass-(gN+gC+gH)=0.157-(0.0289+0.058+0.0034)=0.0666gO[/tex]
this is equivalent to 0.004molO
finally you divide all moles obtained between the smaller number of mole (this is mol of H)
[tex]C\frac{0.0048}{0.0034} H\frac{0.0034}{0.0034} N\frac{0.002}{0.0034} O\frac{0.004}{0.0034} =C_{1.4} HN_{0.6} O_{1.2}[/tex]
and you can multiply by 5 to obtain: [tex]C_{7} H_{5}N_{3}O_{6}[/tex]
