2021-Nov-Chemistry_paper_2__TZ0_SL Detailed Solution

Solution:

We can calculate the empirical formula of the compound by using the masses of carbon, hydrogen, and oxygen in the sample. To do this, we first need to calculate the number of moles of carbon and hydrogen in the sample using the masses of $\mathrm{CO_2}$ and $\mathrm{H_2O}$ produced, as we did earlier:

$$n_{\mathrm{C}} = \frac{m_{\mathrm{CO_2}}}{M_{\mathrm{CO_2}}} = \frac{8.802~\mathrm{g}}{44.01~\mathrm{g/mol}} \approx 0.1998~\mathrm{mol}$$

$$n_{\mathrm{H}} = 2\times\frac{m_{\mathrm{H_2O}}}{M_{\mathrm{H_2O}}} = 2\times\frac{3.604~\mathrm{g}}{18.015~\mathrm{g/mol}} = 0.3998~\mathrm{mol}$$

Next, we can calculate the masses of carbon and hydrogen in the sample using their respective molar masses:

$$m_{\mathrm{C}} = n_{\mathrm{C}}\times M_{\mathrm{C}} = 0.1998~\mathrm{mol} \times 12.01~\mathrm{g/mol} = 2.401~\mathrm{g}$$

$$m_{\mathrm{H}} = n_{\mathrm{H}}\times M_{\mathrm{H}} = 0.3998~\mathrm{mol} \times 1.008~\mathrm{g/mol} = 0.402~\mathrm{g}$$

Finally, we can calculate the mass of oxygen in the sample by subtracting the masses of carbon and hydrogen from the total mass of the sample:

$$m_{\mathrm{O}} = m_{\mathrm{sample}} – m_{\mathrm{C}} – m_{\mathrm{H}} = 4.406~\mathrm{g} – 2.401~\mathrm{g} – 0.402~\mathrm{g} = 1.603~\mathrm{g}$$

Now we have the masses of carbon, hydrogen, and oxygen in the sample, and we can calculate the empirical formula of the compound by dividing each mass by its respective molar mass and then dividing by the smallest resulting value:

$$\begin{array}{c|ccc} \mathrm{Element} & \mathrm{C} & \mathrm{H} & \mathrm{O} \\ \hline \mathrm{Mass~(g)} & 2.401 & 0.402 & 1.603 \\ \mathrm{Molar~mass~(g/mol)} & 12.01 & 1.008 & 16.00 \\ \mathrm{Number~of~moles} & 0.1998 & 0.3998 & 0.1002 \\ \mathrm{Mole~ratio}\approx  & 2 & 4 & 1  \end{array}$$

So the empirical formula of the compound is also $\mathrm{C_2H_4O}$.