Combustion analysis of an organic compound gave the composition $\mathbf{71}\mathbf{.}\mathbf{17}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{41}\mathbf{wt}\mathbf{\%}\mathbf{C}\mathbf{,}\mathbf{6}\mathbf{.}\mathbf{76}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{wt}\mathbf{\%}\mathbf{H}$, and$\mathbf{10}\mathbf{.}\mathbf{34}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{08}\mathbf{wt}\mathbf{\%}\mathbf{N}$ . Find the coefficients and and their uncertainties and in the formula ${\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{\pm }\mathbf{\pm }\mathbf{x}}{\mathbf{N}}_{\mathbf{n}\mathbf{\pm }\mathbf{y}}$.

• 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.

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

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

Here we determine N:C ratio.

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

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

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

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

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