Question

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

Short answer

The empirical formula is$C_7{H}_5{N}_3{O}_6$.

Step-by-step solution

Calculating the mass of C and H in a compound

To calculate the mass,

$\begin{array}{rcl}m\left(C\right)& =& 0.213gC{O}_2·\frac{12.0.11gC}{44.009gC{O}_2}\\ & =& 0.058gC\\ m\left(H\right)& =& 0.0310g{H}_{2}O·\frac{2.016gH}{18.015g{H}_{2}O}\\ & =& 0.0035gH\end{array}$

Calculating the mass in a compound

To calculate the mass of ammonia,

$\begin{array}{rcl}m\left(N{H}_3\right)& =& 0.157gcompound\frac{0.023gN{H}_3}{0.103gCompound}\\ & =& 0.035gN{H}_3\\ m\left(H\right)& =& 0.035gN{H}_3·\frac{14.007gN}{17.031gN{H}_3}\\ & =& 0.029gNin0.157gofcompound\end{array}$

Calculating the mass of O

To calculate the mass of oxygen,

$\begin{array}{rcl}m\left(O\right)& =& 0.157g-0.058g-0.0035g-0.029g-0.0665g\\ n\left(C\right)& =& 0.058gH·\frac{1mol}{12.011gC}\\ & =& 0.0048mol\\ n\left(H\right)& =& 0.00347mol\\ n\left(N\right)& =& 0.002mol\\ n\left(O\right)& =& 0.00415mol\end{array}$

Then,

$$\begin{gathered} n\left(C\right):n\left(H\right):n\left(N\right):n\left(O\right)=0.0048:0.00347:0.002:0.00415:0.002 \\ n\left(C\right):n\left(H\right):n\left(N\right):n\left(O\right)=7.2:5.2:3:6.2 \end{gathered}$$

The empirical formula is$C_7{H}_5{N}_3{O}_6$