Chapter 13, Problem 13.75P

Interpretation Introduction

Interpretation:

The molecular formula of Lactic acid has to be determined.

Concept Introduction:

The expression to determine molecular weight as per ideal gas equation is given as follows:

  $M=\frac{mRT}{PV}$

Here,

$\text{R}$ denotes gas constant.

$V$ denotes the volume.

$n$ denotes the temperature.

$P$ denotes the pressure.

$m$ denotes mass.

$M$ denotes molecular mass.

In order to obtain the molecular formula of the compound, Multiply the whole number with the subscript of each element present in the empirical formula.

Answer to Problem 13.75P

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

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Substitute $\text{150 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=\text{150 }°\text{C}+273\text{ K}\\ =423\text{ K}\end{array}$

The conversion factor to convert $\text{mL}$ to $\text{L}$ is as follows:

  $1\text{ mL}={10}^{-3}\text{ L}$

Thus $\text{300 mL}$ is converted to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{300 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{\text{1 mL}}\right)\\ =0.3\text{ L}\end{array}$

The conversion factor to convert $\text{atm}$ to $\text{torr}$ is as follows:

  $1\text{ atm}=\text{760 torr}$

Thus, $\text{326 torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{326 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =0.42894\text{ atm}\end{array}$

The expression to determine molecular weight as per ideal gas equation is as follows:

  $M=\frac{mRT}{PV}$        (2)

Substitute $0.3338\text{ g}$ for $m$, $0.3\text{ L}$ for $V$, $0.42894\text{ atm}$ for $P$, and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $423\text{ K}$ for $T$ in equation (2).

  $\begin{array}{c}M=\frac{\left(0.3338\text{ g}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(423\text{ K}\right)}{\left(0.42894\text{ atm}\right)\left(0.3\text{ L}\right)}\\ =90.04\text{ g/mol}\end{array}$

Molecular mass is $90.04\text{ g/mol}$.

The formula to convert mass to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$        (3)

Substitute $\text{0}\text{.3999 g}$ for mass and $\text{12}\text{.01 g/mol}$ for molar mass of $\text{C}$ in equation (3) to calculate moles of $\text{C}$.

  $\begin{array}{c}\text{Moles of C}=\frac{\text{0}\text{.3999 g}}{\text{12}\text{.01 g/mol}}\\ =0.033297\text{ mol}\end{array}$

Substitute $\text{0}\text{.0673 g}$ for mass and $\text{1}\text{.008 g/mol}$ for molar mass of $\text{H}$ in equation (3) to calculate the moles of $\text{H}$.

  $\begin{array}{c}\text{Moles of H}=\frac{\text{0}\text{.0673 g}}{\text{1}\text{.008 g/mol}}\\ =0.0667\text{ mol}\end{array}$

Substitute $\text{0}\text{.5328 g}$ for mass and $16.00\text{ g/mol}$ for molar mass of $\text{O}$ in equation (3) to calculate moles of $\text{O}$.

  $\begin{array}{c}\text{Moles of O}=\frac{\text{0}\text{.0673 g}}{\text{1}\text{.008 g/mol}}\\ =0.0333\text{ mol}\end{array}$

With moles of $\text{C}$, $\text{H}$ and $\text{O}$ as subscripts and write a preliminary formula as follows:

  ${\text{C}}_{0.033297}{\text{H}}_{0.0667}{\text{O}}_{0.0333}$

The smallest subscript is 0.033297. So, divide each subscript by 0.033297 as follows:

  ${\text{C}}_{\frac{0.033297}{0.033297}}{\text{H}}_{\frac{0.0667}{0.033297}}{\text{O}}_{\frac{0.0333}{0.033297}}\to {\text{C}}_{\text{1}}{\text{H}}_{\text{2}}{\text{O}}_{\text{1}}$

The expression to calculate the empirical formula mass is as follows:

  $\text{Empirical formula mass of }=\left(1\right)\left(\text{M of C}\right)+\left(2\right)\left(\text{M of H}\right)+\left(1\right)\left(\text{M of O}\right)$        (4)

Substitute $12.01\text{ g/mol}$ for $\text{M}$ of $\text{C}$, $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$ , and $16.00\text{ g/mol}$ for $\text{M}$ of $\text{O}$ in equation (4).

  $\begin{array}{c}\text{Empirical formula mass}=\left(1\right)\left(12.01\text{ g/mol}\right)+\left(2\right)\left(1.008\text{ g/mol}\right)+\left(1\right)\left(16.00\text{ g/mol}\right)\\ =30.026\text{ g/mol}\end{array}$

The formula to calculate the whole number multiple is as follows:

  $\text{Whole-number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$        (5)

Substitute $90.04\text{ g/mol}$ for the molar mass and $30.026\text{ g/mol}$ for the empirical formula mass in equation (5) to compute whole number multiple.

  $\begin{array}{c}\text{Whole}-\text{number multiple}=\frac{90.04\text{ g/mol}}{30.026\text{ g/mol}}\\ =2.99\\ \approx 3\end{array}$

Multiply the subscripts in ${\text{C}}_{\text{1}}{\text{H}}_{\text{2}}{\text{O}}_{\text{1}}$ by 3 to obtain molecular formula.

  $\text{Molecular formula}\to \left({\text{C}}_{\text{1}\left(3\right)}{\text{H}}_{\text{2}\left(3\right)}{\text{O}}_{\text{1}\left(3\right)}\right)\to {\text{C}}_3{\text{H}}_6{\text{O}}_3$

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