Q13-123CP - Chemistry: Molecular Nature Of Matter And Change

Question

A chemist is studying small organic compounds for their potential use as an antifreeze. When 0.243 g of a compound is dissolved in 25.0 mL of water, the freezing point of the solution is $\mathbf{-}\mathbf{0}\mathbf{.}{\mathbf{201}}^{\mathbf{o}}\mathit{C}\mathbf{.}$

1. Calculate the molar mass of the compound (d of water 1.00 g/mL).
2. Analysis shows that the compound is 53.31 mass % C and 11.18 mass % H, the remainder being O. Calculate the empirical and molecular formulas of the compound.
3. Draw a Lewis structure for a compound with this formula that forms H bonds and another for one that does not.

 

Step-by-Step Solution

Answer

1. The compound's molar mass is 90 g/mol.
2. Empirical formula and molecular formula of the compound arE$C_2{H}_{5.05}O and C_4{H}_{10}{O}_2$
3. Lewis structures are given below.

 

Step 1: Molar mass

To calculate the molar mass of compound first calculate the molality of the solution.

Calculate the molality by using the following formula;

 $\mathit{\Delta }{\mathit{T}}_{\mathbf{f}}\mathbf{=}{\mathit{K}}_{\mathbf{f}}\mathbf{\times }\mathit{m}$

Where $\Delta T_f$ is the point depression, K_f is freezing point depression $(1.{86}^{\circ }C/m)$ constant and m is molality.

Freezing point depression is calculated as

 $\begin{array}{l}{T}_f=T_f-{T_f}^{\circ }\\ =0^{\circ }C-(-{201}^{\circ }\mathrm{C})\\ ={201}^{\circ }C\\ \end{array}$

Therefore, 

 $$\begin{gathered} \Delta T_f=K_f\times m \\ {201}^{\circ }C=(1.{86}^{\circ }C/m)\times m \\ m=0.108m \end{gathered}$$

 

Mass of water can be calculated by multiplying its volume with density.

$$\begin{gathered} 25m\times 1g/mL=25g\times \frac{1\mathrm{kg}}{1000\mathrm{g}} \\ =0.0250kg \end{gathered}$$

Now, moles of compound can be calculated by multiplying its molality with mass of water.

$$\begin{gathered} n=0.108mol/kg \times 0.0250kg \\ =0.00270mol \end{gathered}$$

Finally, molar of compound is calculated by dividing its mass by its moles. Mass of compound is given 0.243g.

$$\begin{gathered} M=\frac{0.243g}{0.00270mol} \\ =90g/mol \end{gathered}$$

Thus, the compound's molar mass is 90g/mol.

 

Step 2: Empirical and molecular formulas

It is given that compound has 53.31 mass % C and 11.18 mass % H, the remainder being O.

Remaining O content

$$\begin{gathered} O\%=100-\left(53.31+11.18\right) \\ =35.51 \end{gathered}$$

Assume 100g compound and calculate the moles of C, H and O by dividing their % content by their molecular mass as follow;

   $moles\hspace{0.33em}of\hspace{0.33em}carbon=53.31g\times \frac{1mol}{12g}=4.44mol$

$moles\hspace{0.33em}of\hspace{0.33em}hydrogen=11.18g\times \frac{1mol}{1g}=11.18mol$

$moles\hspace{0.33em}of\hspace{0.33em}oxygen=35.51g\times \frac{1mol}{16g}=2.21mol$

2.21 mol is the smallest content present in compound. Thus, empirical formula of compound can be calculated as

$\begin{array}{l}C=\frac{4.44mol}{2.21mol}\\ =2\end{array}$

$$\begin{gathered} H=\frac{11.18mol}{2.21mol} \\ =5.05 \end{gathered}$$

$\begin{array}{l}O=\frac{2.21mol}{2.21mol}\\ =1\end{array}$

Thus, $C_2{H}_{5.05}O$is the empirical formula for compound.

To calculate the molecular formula of the compound first we have to find out the mass of empirical formula which can be calculated as

$C_2{H}_5.05O=2\times 12+5.05\times 1+1\times \times 16=45.05g$

$molecular formula=empirical formula\times \frac{molar mass}{empirical formula of mass}$

$$\begin{gathered} =C_2{H}_{5.05}O\times \left(\frac{90}{45.5}\right) \\ =C_2{H}_{5.05}O\times 1.972 \\ =C_4{H}_{10}{O}_2 \end{gathered}$$

Step 3: Lewis structure of the compound

The compound having molecular formula C₄H₁₀O₂ can be cis1,2-butanediol and trans2,3-butanediol. In cis1,2-butanediol form it forms hydrogen bonding but not in trans2,3-butanediol.

The Lewis structures are given below