Question

Question: Complete combustion of $\mathbf{2}\mathbf{.}\mathbf{40}\mathit{g}$of a compound of carbon, hydrogen, and oxygen yielded$\mathbf{5}\mathbf{.}\mathbf{46}\mathbf{}{\mathbf{\text{gCO}}}_{\mathbf{2}}$ and role="math" localid="1663418772271" $\mathbf{2}\mathbf{.}\mathbf{23}\mathbf{}{\mathbf{\text{gH}}}_{\mathbf{2}}\mathbf{\text{O}}$. Whenrole="math" localid="1663418784865" $\mathbf{8}\mathbf{.}\mathbf{69}\mathit{g}$ of the compound was dissolved in $\mathbf{281}\mathit{g}$of water, the freezing point of the solution was found to be $\mathbf{-}\mathbf{0}\mathbf{.}{\mathbf{97}}^{\mathbf{o}}\mathit{C}$. What is the molecular formula of the compound?

Short answer

The molecular formula of the compound is ${\text{C}}_3{\text{H}}_6\text{O}$.

Step-by-step solution

Given data

When $8.69g$of the compound was dissolved in $281g$of water, the freezing point of the solution was found to be -0.97 °C.

Definition of Solution

In chemistry, a solution is a special type of homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.

Calculation of the molality

The difference in temperature is given as:

$$\begin{gathered} \Delta T_f=T_{\text{solvent}}-T_{\text{solution}} \\ \Delta T_f=0.{00}^{o}C-(-0.{97}^{o}C) \\ \Delta T_f=0.97{}^{o}C \\ \Delta T_f=0.97K \end{gathered}$$

The molality is calculated as:

$$\begin{gathered} \Delta T_f=K_{f}m \\ 0.97\text{K}=1.86{\text{Kkgmol}}^{-1}\times m \\ m=\frac{0.97\text{K}}{1.86{\text{Kkgmol}}^{-1}} \\ m=0.52\text{mol}/\text{kg} \end{gathered}$$

Calculation of the molar mass

Number of moles can be calculated as:

$$\begin{gathered} {\mathbf{n}}_{\text{solute}}=molality\times massofsolvent \\ {n}_{\text{solute}}=0.52\frac{\text{mol}}{\text{kg}}\times 281\text{g}\times \frac{\text{kg}}{1000\text{g}} \\ {n}_{\text{solute}}=0.146\text{mol} \end{gathered}$$

Therefore,$$\begin{gathered} \text{Molar mass}=\frac{8.69\text{g}}{0.146\text{mol}} \\ \text{Molar mass}=59.5\text{g}/\text{mol} \end{gathered}$$

Calculation of the number of moles

Number of moles, and mass of $C$.

$$\begin{gathered} 5.46{\text{gCO}}_2\times \frac{1{\text{molCO}}_2}{44.01{\text{gCO}}_2}\times \frac{1\text{molC}}{1\text{mole}{\sigma }_2}=0.124\text{molC} \\ 0.124\text{mol}C\times \frac{12.0107\text{gC}}{1\text{molC}}=1.49\text{g} \end{gathered}$$

Number of moles, and mass of $H$.

$$\begin{gathered} 2.23{\text{gH}}_2\text{O}\times \frac{1{\text{molH}}_2\text{O}}{18.02{\text{gH}}_2\text{O}}\times \frac{2\text{molH}}{1{\text{molH}}_2\text{O}}=0.248\text{molH} \\ 0.248\text{molH}\times \frac{1.008\text{gH}}{1\text{molH}}=0.250\text{g} \end{gathered}$$

Number of moles, and mass of $O$.

$$\begin{gathered} 2.40\text{g}-(1.49\text{g}+0.250\text{g})=0.66\text{g} \\ \frac{0.66\text{g}}{15.9994\text{g}}=0.0412\text{mol} \end{gathered}$$

Calculation for the molecular formula

The empirical formula of the compound is ${\text{C}}_{0.124}{\text{H}}_{0.248}{\text{O}}_{0.0412}$.

Divide with the smallest number.

$C_{\frac{0.124}{0.0412}}{H}_{\frac{0.248}{0.0412}}{O}_{\frac{0.0412}{0.0412}}={\text{C}}_3{\text{H}}_6{\text{O}}_1^0$

$$\begin{gathered} n=\frac{\text{Molar mass of compound}}{\text{Empirical mass of the compound}} \\ n=\frac{59.5\text{g}/\text{mol}}{58.08\text{g}/\text{mol}} \\ n=1.02 \\ n\approx 1 \end{gathered}$$

Thus, the molecular formula of the compound is ${\text{C}}_3{\text{H}}_6\text{O}$.