91Ó°ÊÓ

• Combustion analysis is a method used in both organic and analytical chemistry to determine the elemental composition (more precisely, empirical formula) of a pure organic compound by combusting the sample under conditions that allow the combustion products to be quantitatively analysed.
• After determining the number of moles of each combustion product, the empirical formula or a partial empirical formula of the original compound can be calculated.

Here we determine H:C ratio.

$$\begin{gathered} \frac{\mathrm{n}\left(\mathrm{H}\right)}{\mathrm{n}\left(\mathrm{C}\right)}=\frac{{\displaystyle \frac{\mathrm{m}}{\mathrm{M}}}\left(\mathrm{H}\right)}{{\displaystyle \frac{\mathrm{m}}{\mathrm{M}}}\left(\mathrm{C}\right)} \\ \frac{6.76×0.12}{1.00794}=\frac{6.707Â\pm 0.119}{5.926Â\pm 0.0341} \\ \frac{71.17Â\pm 0.41}{12.0107} \\ \frac{6.707Â\pm 0.119}{5.926Â\pm 0.0341}=\frac{6.707Â\pm 0.119}{5.926Â\pm 0.0341} \\ \frac{6.707Â\pm 0.119}{5.926Â\pm 0.0341}=1.132Â\pm 0.021 \end{gathered}$$

Next, if the stoichiometry of C is 8 as written in the task, then we will write the stoichiometry coefficient of H as$8\left(1.132Â\pm 0.021\right)=9.06Â\pm 0.17$

Here we determine N:C ratio.

$$\begin{gathered} \frac{\mathrm{n}\left(\mathrm{N}\right)}{\mathrm{n}\left(\mathrm{C}\right)}=\frac{{\displaystyle \frac{\mathrm{m}}{\mathrm{M}}}\left(\mathrm{N}\right)}{{\displaystyle \frac{\mathrm{m}}{\mathrm{M}}}\left(\mathrm{C}\right)} \\ \frac{{\displaystyle \frac{10.34Â\pm 0.08}{14.00674}}}{71.17Â\pm 0.41}=\frac{0.7832Â\pm 0.0057}{5.926Â\pm 0.0341} \\ 12.0107 \end{gathered}$$

$$\begin{gathered} \frac{0.7832Â\pm 0.0057}{5.926Â\pm 0.0341}=\frac{0.7382Â\pm 0.774\%}{5.925Â\pm 0.576\%} \\ \frac{0.7832Â\pm 0.0057}{5.926Â\pm 0.0341}=0.1246Â\pm 0.0012 \end{gathered}$$

Next, if the stoichiometry of C is 8 as written in the task, then we will write the stoichiometry coefficient of N as $8\left(0.1246Â\pm 0.0012\right)=0.9968Â\pm 0.0096$.

This would give us the empirical formula of ${\mathrm{C}}_8{\mathrm{H}}_{Â\pm Â\pm \mathrm{x}}{\mathrm{N}}_{\mathrm{n}Â\pm \mathrm{y}}$to be ${\mathrm{C}}_8{\mathrm{H}}_{9.06}Â\pm 0.17{\mathrm{N}}_{0.997Â\pm 0.010}$

Therefore the stoichiometry coefficient of H is $8\left(1.132Â\pm 0.021\right)=9.06Â\pm 0.17$and the stoichiometry coefficient of N is $8\left(0.1246Â\pm 0.012\right)=0.9968Â\pm 0.0096$