Question

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157g of the compound produced 0.213g of CO2 and 0.0310g of H2O. In another experiment, it is found that 0.103g of the compound produces 0.0230g of NH3. What is the empirical formula of the compound?

Answer 1

Answer:

$C_{7} H_{5}N_{3}O_{6}$

Explanation:

First reaction gives you the number of moles or the mass from Carbon and hydrogen

for carbon:

$0,213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} =0.005molC$

$0.213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} .\frac{12gC}{1molC} = 0.058gC$

Analogously for hydrogen:

0.0310g$H_{2}O$ 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.

$0.023gNH_{3} .\frac{1molNH_{3}}{17gNH_{3}} .\frac{1molN}{1molNH_{3}} .\frac{14gN}{1molN} \frac{100}{0.103gsample} =18.4%N$

now

$\frac{18.4gN}{100gsample} .0.157gsample=0.0289gN in the first reaction$

this is equivalet to 0.002mol of N

with this information you can find the mass of oxygen by matter conservation.

$gO=total mass-(gN+gC+gH)=0.157-(0.0289+0.058+0.0034)=0.0666gO$

this is equivalent to 0.004molO

finally you divide all moles obtained between the smaller number of mole (this is mol of H)

$C\frac{0.0048}{0.0034} H\frac{0.0034}{0.0034} N\frac{0.002}{0.0034} O\frac{0.004}{0.0034} =C_{1.4} HN_{0.6} O_{1.2}$

and you can multiply by  5   to obtain: $C_{7} H_{5}N_{3}O_{6}$