Chapter 5, Problem 72E

Interpretation Introduction

Interpretation:Empirical formula and molecular formula of organic compound are to be determined.

Concept introduction: The formula to calculate volume as per ideal gas law is as follows:

  $M=\frac{mRT}{PV}$

Where,

• $R$ is gas constant.
• $V$ denotes the volume.
• $n$ denotes number of moles.
• $T$ denotes temperature.
• $P$ denotes pressure.

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

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

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

  $1\text{ atm}=\text{760 torr}$

Answer to Problem 72E

Empirical formula and molecular formula of compound are ${\text{C}}_{12}{\text{H}}_{21}\text{NO}$ and ${\text{C}}_{24}{\text{H}}_{42}{\text{N}}_2{\text{O}}_2$ respectively.

Explanation of Solution

For ${\text{CO}}_{\text{2}}$ since $1{\text{ mol CO}}_2$ has $1\text{ mol C}$ thus, moles of $\text{C}$ from amount of ${\text{CO}}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}\text{Amount of C}\left(\text{mol}\right)=\left(\text{0}\text{.2766 }{\text{g CO}}_2\right)\left(\frac{{\text{1 mol CO}}_2}{\text{44}{\text{.01 g CO}}_2}\right)\left(\frac{1\text{ mol C}}{1{\text{ mol CO}}_2}\right)\\ =0.00628\text{ mol C}\end{array}$

The formula to convert moles to mass in grams is as follows:

  $\text{Mass of C}=\left(\text{Moles of C}\right)\left(\text{molar mass of C}\right)$

Moles of $\text{C}$ is $0.00628\text{ mol}$ .

Molar mass of $\text{C}$ is $\text{12}\text{.01 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Mass of C}=\left(\text{Moles of C}\right)\left(\text{molar mass of C}\right)\\ =\left(0.00628\text{ mol}\right)\left(\text{12}\text{.011 g/mol}\right)\\ =0.075429\text{ g}\end{array}$

For ${\text{H}}_2\text{O}$ since $1{\text{ mol H}}_2\text{O}$ has $\text{2 mol H}$ thus, moles of $\text{H}$ from amount of $\text{H}$ is calculated as follows:

  $\begin{array}{c}\text{Amount of H}\left(\text{mol}\right)=\left(0.0991{\text{g H}}_2\text{O}\right)\left(\frac{{\text{1 mol H}}_2\text{O }}{\text{18}{\text{.01 g H}}_2\text{O}}\right)\left(\frac{\text{2 mol H}}{1{\text{ mol H}}_2\text{O}}\right)\\ =0.011004\text{ mol H}\end{array}$

The formula to convert moles to mass in grams is as follows:

  $\text{Mass of H}=\left(\text{Moles of H}\right)\left(\text{molar mass of H}\right)$

Moles of $\text{H}$ is $0.011004\text{ mol}$ .

Molar mass of $\text{H}$ is $\text{1}\text{.008 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Mass of H}=\left(\text{Moles of H}\right)\left(\text{molar mass of H}\right)\\ =\left(0.011004\text{ mol}\right)\left(\text{1}\text{.008 g/mol}\right)\\ =0.01109\text{ g}\end{array}$

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

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

The temperature in Celsius is $\text{0 }°\text{C}$ .

Substitute the value in the above formula.

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}\\ =0°\text{C}+273\text{ K}\\ =273\text{ K}\end{array}$

Total pressure as per the Dalton’s law of partial pressure is calculated as follows:

  $P_{\text{total}}=P_{{\text{N}}_{\text{2}}}+P_{{\text{H}}_2\text{O}}$

The value of $P_{{\text{H}}_2\text{O}}$ is $\text{23}\text{.8 torr}$ .

The value of $P_{\text{total}}$ is $\text{760 torr}$ .

Substitute the value in above formula.

  $\begin{array}{c}{P}_{\text{total}}=P_{{\text{N}}_{\text{2}}}+P_{{\text{H}}_2\text{O}}\\ \text{760 torr}=P_{{\text{N}}_{\text{2}}}+\text{23}\text{.8 torr}\end{array}$

Rearrange to evaluate $P_{{\text{N}}_{\text{2}}}$ .

  $\begin{array}{c}{P}_{{\text{N}}_{\text{2}}}=\text{760 torr}-\text{23}\text{.8 torr}\\ =736.2\text{ torr}\end{array}$

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

  $1\text{ atm}=\text{760 torr}$

Thus, $736.2\text{ torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(736.2\text{ torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =0.9686\text{ atm}\end{array}$

The conversion factor to convert liters to milliliters is as follows:

  $\text{1 L}=\text{1000 mL}$

Hence convert $\text{27}\text{.6 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{27}\text{.6 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =0.0276\text{ L}\end{array}$

The formula to calculate moles as per ideal gas law is as follows:

  $n=\frac{PV}{RT}$

The value of $P$ is $0.9686\text{ atm}$ .

The value of $V$ is $0.0276\text{ L}$ .

The value of $T$ is $295\text{ K}$ .

The value of $R$ is $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ .

Substitute the values in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(0.9686\text{ atm}\right)\left(0.0276\text{ L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(\text{273 K}\right)}\\ =0.001193\text{ mol}\end{array}$

Since $1{\text{ mol N}}_2$ gives $\text{2 mol N}$ so moles of $\text{N}$ formed is calculated as follows:

  $\begin{array}{c}\text{Moles of N}=\left(\text{2}\right)\left(0.001193\text{ mol}\right)\\ =0.002386\text{ mol}\end{array}$

The formula to calculate mass from moles is given as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$

For $\text{N}$

Number of moles is $0.002386\text{ mol}$ .

Molar mass is $\text{14 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Mass}=\left(0.002386\text{ mol}\right)\left(\text{14 g/mol}\right)\\ =0.033404\text{ g}\end{array}$

Since $0.033404\text{ g N}$ is obtained from $\text{0}\text{.4831 g}$ of compound thus mass of $\text{N}$ from $0.1023\text{ g}$ of compound is calculated as follows:

  $\begin{array}{c}\mathrm{Mass}ofN=\left(0.1023\text{ g}\right)\left(\frac{0.033404\text{ g N}}{\text{0}\text{.4831 g}}\right)\\ =0.0070735\text{ g}\end{array}$

Since mass of $\text{1 mol N}$ is $\text{14 g}$ so moles of $\text{N}$ are calculated as follows:

  $\begin{array}{c}\text{Moles}ofN=\left(0.007080\text{ g}\right)\left(\frac{1\text{ mol N}}{\text{14 g N}}\right)\\ =0.0005057\text{ mol}\end{array}$

The formula to calculate mass of $\text{O}$ is as follows:

  $\text{Mass of O}=\text{Mass of sample}-\left(\text{mass of C}+\text{mass of H}+\text{mass of N}\right)$

Mass of sample is $0.1023\text{ g}$ .

Mass of $\text{C}$ is $0.075429\text{ g}$ .

Mass of $\text{H}$ is $0.01109\text{ g}$ .

Mass of $\text{N}$ is $0.0070735\text{ g}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Mass of O}=\text{Mass of sample}-\left(\text{mass of C}+\text{mass of H}+\text{mass of N}\right)\\ =0.1023\text{ g}-\left(0.075429\text{ g}+0.01109\text{ g}+0.0070735\text{ g}\right)\\ =0.0087075\text{ g}\end{array}$

Since mass of $\text{1 mol O}$ is $\text{16 g}$ so moles of $\text{O}$ in the mixture are calculated as follows:

  $\begin{array}{c}\text{Moles}of\text{O}=\left(0.0087075\text{ g}\right)\left(\frac{1\text{ mol O}}{\text{16 g O}}\right)\\ =0.0005442\text{ mol O}\end{array}$

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

  ${\text{C}}_{0.00628}{\text{H}}_{0.011004}{\text{N}}_{0.0005070}{\text{O}}_{0.0005442}$

The smallest subscript is 0.0045819. So, divide each subscript by 0.0045819 as follows:

  ${\text{C}}_{\frac{0.00628}{0.0005070}}{\text{H}}_{\frac{0.011004}{0.0005070}}{\text{N}}_{\frac{0.0005070}{0.0005070}}{\text{O}}_{\frac{0.0005442}{0.0005070}}\to {\text{C}}_{12}{\text{H}}_{21}{\text{N}}_1{\text{O}}_1$

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

  $1\text{ atm}=\text{760 torr}$

Thus, $\text{750 torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{750 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =0.9868\text{ atm}\end{array}$

The conversion factor to convert liters to milliliters is as follows:

  $\text{1 L}=\text{1000 mL}$

Hence convert $\text{256 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{256 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =0.256\text{ L}\end{array}$

The formula to calculate volume as per ideal gas law is as follows:

  $M=\frac{mRT}{PV}$

The value of $P$ is $10\text{ atm}$ .

The value of $m$ is $\text{0}\text{.800 g}$ .

The value of $V$ is $0.256\text{ L}$ .

The value of $T$ is $373\text{ K}$ .

The value of $R$ is $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ .

Substitute the values in above equation to calculate $M$ .

  $\begin{array}{c}M=\frac{mRT}{PV}\\ =\frac{\left(\text{0}\text{.800 g}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(373\text{ K}\right)}{\left(10\text{ atm}\right)\left(0.256\text{ L}\right)}\\ =9.56\text{ g/mol}\end{array}$

Formula to calculate empirical formula mass is as follows:

  ${\text{Empirical formula mass of C}}_{12}{\text{H}}_{21}\text{NO}=\left[\begin{array}{l}\left(12\right)\left(\text{M of C}\right)+\left(21\right)\left(\text{M of H}\right)\\ +\left(1\right)\left(\text{M of N}\right)+\left(1\right)\left(\text{M of O}\right)\end{array}\right]$

  $\text{M}$ of $\text{C}$ is $48.471\text{ g/mol}$ .

  $\text{M}$ of $\text{H}$ is $1.008\text{ g/mol}$ .

  $\text{M}$ of $\text{N}$ is $\text{14 g/mol}$ .

  $\text{M}$ of $\text{O}$ is $\text{16 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}{\text{Empirical formula mass of C}}_{12}{\text{H}}_{21}\text{NO}=\left[\begin{array}{l}\left(12\right)\left(\text{M of C}\right)+\left(21\right)\left(\text{M of H}\right)\\ +\left(1\right)\left(\text{M of N}\right)+\left(1\right)\left(\text{M of O}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(12\right)\left(12.01\text{ g/mol}\right)+\left(21\right)\left(\text{1}\text{.008 g/mol}\right)\\ +\left(1\right)\left(\text{14 g/mol}\right)+\left(1\right)\left(\text{16 g/mol}\right)\end{array}\right]\\ =195.288\text{ g/mol}\end{array}$

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

  $1\text{ atm}=\text{760 torr}$

Thus, $\text{256 torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{256 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =0.3368\text{ atm}\end{array}$

The formula to calculate molecular weight from density is as follows:

  $M=\frac{dRT}{P}$

The value of $P$ at STP is $\text{0}\text{.3368 atm}$ .

The value of $d$ is $\text{4}\text{.02 g/mol}$ .

The value of $T$ is $373\text{ K}$ .

The value of $R$ is $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ .

Substitute the values in above equation.

  $\begin{array}{c}M=\frac{dRT}{P}\\ =\frac{\left(\text{4}\text{.02 g/mol}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(\text{400 K}\right)}{\left(\text{0}\text{.3368 atm}\right)}\\ =391.782\text{ g/mol}\end{array}$

The formula to compute whole number multiple is as follows:

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

Molar mass is $391.782\text{ g/mol}$ .

Empirical formula mass is $195.288\text{ g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Whole}-\text{number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}\\ =\frac{391.782\text{ g/mol}}{195.288\text{ g/mol}}\\ \approx 2\end{array}$

Multiply the subscripts in ${\text{C}}_{12}{\text{H}}_{21}\text{NO}$ by 2 to obtain molecular formula.

  $\text{Molecular formula}={\text{C}}_{2\left(12\right)}{\text{H}}_{2\left(21\right)}{\text{N}}_{2\left(1\right)}{\text{O}}_{2\left(1\right)}\to {\text{C}}_{24}{\text{H}}_{42}{\text{N}}_2{\text{O}}_2$

Conclusion

Empirical formula and molecular formula of compound are ${\text{C}}_{12}{\text{H}}_{21}\text{NO}$ and ${\text{C}}_{24}{\text{H}}_{42}{\text{N}}_2{\text{O}}_2$ respectively.