Chapter 13, Problem 86QP

Interpretation Introduction

Interpretation:

The molecular formula of the compound is to be determined.

Concept introduction:

Molar mass is the mass of one mole of a substance in grams. Molar mass of a compound is equal to the sum of the atomic masses of all elements present in that compound.

The relation between molarity and osmotic pressure is mathematically expressed as follows:

$\pi =M\text{R}T$

Here, $\pi$ is the osmotic pressure, R is the solution constant, $T$ is the temperature, and $M$ is the molarity of the solution.

The number of moles of solute are calculated as follows:

$\text{Number of moles of solute}=\text{Molarity }\left(m\right)\times \text{Mass of solvent}$

The molar mass of unknown solute is calculated as follows:

$\text{Molar mass of unknown solute}=\frac{\text{Mass of solute}}{\text{Moles of solute}}$

The molecular formula is the formula that gives the exact number of atoms present in that molecule.

Answer to Problem 86QP

Solution: ${\text{C}}_{15}{\text{H}}_{20}{\text{O}}_{10}{\text{N}}_5$

Explanation of Solution

Given information: $\text{41}\text{.8\%}$ carbon, $\text{4}\text{.7\%}$ hydrogen, $\text{37}\text{.3\%}$ oxygen, and $\text{16}\text{.3\%}$ nitrogen. Mass of solute is $0.65\text{ g}$. Mass of solvent (benzene) is $\text{7}\text{.480 g}$. The volume of the solution is $300\text{ mL}$. Osmotic pressure is $1.43\text{atm}$ and the temperature is $27°\text{C or 300 K}$.

The molarity using osmotic pressure is mathematically calculated as follows:

$\pi =M\text{R}T$

Here, $\pi$ is the osmotic pressure, R is the solution constant, $T$ is the temperature, and $M$ is the molarity of the solution.

Molarity can be determined by substituting the values in the above equation as follows:

$\begin{array}{c}M=\frac{\pi }{\text{R}T}\\ =\frac{1.43\cancel{\text{atm}}}{\left(0.0821\text{L}\text{.}\cancel{\text{atm}}\text{/}\cancel{\text{K}}\text{.mol}\right)\left(300\cancel{\text{K}}\right)}\\ =0.0581\text{mol/L}\end{array}$

The volume of the solution is $0.3000\text{ L}$. So, the number of moles of solute are calculated as follows:

$\begin{array}{c}\text{Number of moles of solute}=\text{Molarity }\left(m\right)\times \text{Mass of solvent}\\ =\left(\frac{0.0581\text{ mol}}{1\cancel{\text{L}}}\times 0.3000\text{}\cancel{\text{L}}\right)\\ =0.0174\text{mol}\end{array}$

The molar mass of unknown solute is calculated as follows:

$\begin{array}{c}\text{Molar mass of unknown solute}=\frac{\text{Mass of solute}}{\text{Moles of solute}}\\ =\left(\frac{\text{7}\text{.480 g}}{0.0174\text{ mol}}\right)\\ =430\text{ g/mol}\end{array}$

Now, calculate the number of moles of various constituents(assuming mass in $100\text{ g}$ of substance).

Moles of carbon are calculated as follows:

$\begin{array}{c}\text{Moles C}=\frac{\text{Given mass}}{\text{Atomic mass}}\\ =\left(\frac{41.8\text{mol}}{12.01\text{ g}}\right)\\ =3.48\text{ mol}\text{C}\end{array}$

Moles of hydrogen are calculated as follows:

$\begin{array}{c}\text{Moles H}=\frac{\text{Given mass}}{\text{Atomic mass}}\\ =\left(\frac{4.7\text{mol}}{1.008\text{ g}}\right)\\ =4.7\text{ mol}\text{H}\end{array}$

Moles of oxygen are calculated as follows:

$\begin{array}{c}\text{Moles O}=\frac{\text{Given mass}}{\text{Atomic mass}}\\ =\left(\frac{37.3\text{ mol}}{16\text{ g}}\right)\\ =2.33\text{ mol}\text{}\text{ O}\end{array}$

Moles of nitrogen are calculated as follows:

$\begin{array}{c}\text{Moles N}=\frac{\text{Given mass}}{\text{Atomic mass}}\\ =\frac{16.3\text{mol}}{14.01\text{g}}\\ =1.16\text{ mol}\text{}\text{ N}\end{array}$

The molar ratio of carbon, hydrogen, oxygen, and nitrogen is $3.48:4.7:2.33:1.16$.

Therefore, the empirical formula of the compound is ${\text{C}}_{3.48}{\text{H}}_{4.7}{\text{O}}_{2.33}{\text{N}}_{1.16}$.

The empirical formula mass is calculated as follows:

$\begin{array}{c}m=(3\times \text{at}\text{.mass of C+4}\times \text{at}\text{.mass of H+2}\times \text{at}\text{.mass of O+at}\text{.mass of N)}\\ =(3\times 12+4\times 1+2\times 16+1\times 14)\\ =86\text{ g}\end{array}$

Divide the molar mass by the empirical formula mass to get the value of $n$ as follows:

$\begin{array}{c}n=\frac{\text{Molar mass}}{\text{Empirical formula mass}}\\ =\left(\frac{429.44}{86}\right)\\ =5\end{array}$

Now, the molecular formula can be obtained using the expression as follows:

$\begin{array}{c}\text{Molecular formula}=\text{Empirical formula}\times n\\ ={\text{C}}_3{\text{H}}_4{\text{O}}_2\text{N}\times \text{4}\\ {\text{=C}}_{15}{\text{H}}_{20}{\text{O}}_{10}{\text{N}}_5\end{array}$

Conclusion

The molecular formula of the compound is ${\text{C}}_{15}{\text{H}}_{20}{\text{O}}_{10}{\text{N}}_5$.