Chapter 11, Problem 11.94QP

Interpretation Introduction

Interpretation: An unknown compound with antibiotic properties is given to be dissolved in water to make a $\text{100}\text{.0}\text{mL}$ solution. The solution is given to be non-conducting with an osmotic pressure of $\text{9}\text{.94}\text{torr}$ at $23.6°\text{C}$. The molecular formula of the compound is to be calculated.

Concept introduction: When a compartment is separated by a semipermeable membrane, solvent always flows from lower concentration to higher concentration. This process of solvent flow is known as osmosis. Now if another side external pressure is applied then this process reversed and this applied reverse pressure is known as osmotic pressure.

Empirical formula of any compound represents simplest relative whole number ratio of atoms of each element.

Molecular formula of any compound represents actual number of atoms of each element present in one molecule.

To determine: The molecular formula of the compound.

Answer to Problem 11.94QP

Solution

The molecular formula of the compound ${\text{C}}_{21}{\text{H}}_{\text{35}}{\text{N}}_{\text{14}}$.

Explanation of Solution

Explanation

Given

The given mass of compound is $27.40\text{mg}$.

The given volume of solvent is $100\text{mL}$.

The given osmotic pressure of solution is $9.94\text{torr}$.

The given temperature is $23.6°\text{C}$.

The given mass percent of the element carbon is $42.34\%$.

The given mass percent of the element hydrogen is $5.92\%$.

The given mass percent of the element nitrogen is $32.93\%$.

The given mass in terms of gram (g) unit is given as,

$\begin{array}{c}1\text{Gram}\left(\text{g}\right)=1000\text{mg}\\ \text{1}\text{mg}=\frac{1}{1000}\text{g}\\ \text{27}\text{.40}\text{mg}=27.40\times \frac{1}{1000}\text{g}\\ =27.40\times {10}^{-3}\text{g}\end{array}$

The given volume in terms of liter (L) unit is given as,

$\begin{array}{c}1\text{Liter}\left(\text{L}\right)=1000\text{mL}\\ \text{1}\text{mL}=\frac{1}{1000}\text{L}\\ \text{100}\text{mL}=100\times \frac{1}{1000}\text{L}\\ \text{100}\text{mL}=0.1\text{L}\end{array}$

The given pressure in terms of atmosphere (atm) unit is given as,

$\begin{array}{c}1\text{Torr}=\frac{1}{760}\text{atm}\\ 9.94\text{Torr}=9.94\times \frac{1}{760}\text{atm}\\ 9.94\text{Torr}=0.01308\text{atm}\end{array}$

The given temperature in terms of Kelvin (K) unit is given as,

$\begin{array}{c}\text{Temperature}=\left(1°\text{C}+273\right)\text{K}\\ 23.6°\text{C}\text{Temperature}=\left(23.6°\text{C}+273\right)\text{K}\\ =296.6\text{K}\end{array}$

To determine empirical formula of the compound certain rules have been followed which are given as,

1. i. First find the relative atomic mass by dividing the percent amount by the atomic mass of the respective element.
2. ii. Then, to find the simplest ratio of the elements find the lowest relative atomic number and divide all the numbers with it.
3. iii. Now, multiply all the ratios with an integer to make whole number ratio.

With the help of given percent of elements it is clear that the compound is comprises of carbon, hydrogen and nitrogen. Therefore, the empirical formula of the compound is calculated as,

$\begin{array}{cccccc}\text{Element}& \text{Percent}& \text{Atomic}\text{mass}& \begin{array}{l}\text{Relative}\text{atomic}\\ \text{number}\end{array}& \text{Simplest}\text{ratio}& \begin{array}{l}\text{Simplest}\text{whole}\\ \text{no}\text{ratio}\end{array}\\ \text{Carbon}\left(\text{C}\right)& 42.34& 12& \frac{42.34}{12}=3.53& \frac{3.53}{2.35}=1.5& \begin{array}{l}1.5\times 2\\ =3.0\end{array}\\ \text{Hydrogen}\left(\text{H}\right)& 5.92& 1& \frac{5.92}{1}=5.92& \frac{5.92}{2.35}=2.5& \begin{array}{l}2.5\times 2\\ =5\end{array}\\ \text{Nytrogen}\left(\text{N}\right)& 32.93& 14& \frac{32.93}{14}=2.35& \frac{2.35}{2.35}=1& \begin{array}{l}1\times 2\\ =2\end{array}\end{array}$

From the above table the empirical formula of compound is given as,

${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{N}}_{\text{2}}$

Now the empirical formula weight is calculated as,

$\begin{array}{c}{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{N}}_{\text{2}}=\left(12\times 3\right)\text{g/mol}+\left(5\times 1\right)\text{g/mol}+\left(14\times 2\right)\text{g/mol}\\ =\left(36\right)\text{g/mol}+\left(5\right)\text{g/mol}+\left(28\right)\text{g/mol}\\ =69\text{g/mol}\end{array}$

The molecular formula is n times of empirical formula and it is represented as,

$\text{Molecular formula}={\left(\text{Emperical}\text{formula}\right)}_n$ (1)

The value of n is calculated as,

$n=\frac{\text{Molecular}\text{mass}}{\text{Emprical}\text{formula}\text{mass}}$ (2)

The molecular mass (molar mass) is to be calculated with the help of molarity and given value of osmotic pressure.

With the help of the given value of osmotic pressure, the concentration of solvent in terms of molarity is calculated by the formula,

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

Where,

• $\pi$ is the osmotic pressure.
• $M$ is the molarity.
• $\text{R}$ is the universal gas constant.
• $T$ is the temperature in Kelvin.

Substitute the values in the above formula.

$\begin{array}{l}M=\frac{\pi }{\text{R}T}\\ M=\frac{0.01308\text{atm}}{\left(0.0821\text{L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(296.6\text{K}\right)}\\ M=5.37\times {10}^{-4}\text{mol}/\text{L}\end{array}$

Molarity of a solution is defined as the number of moles solute present in per liter of the solvent.

With the help of molarity the number of moles is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\text{Molarity}\times \text{Volume}$

Substitute the values in the above formula,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Molarity}\times \text{Volume}\\ =5.37\times {10}^{-4}\text{mol}/\text{L}\times 0.1\text{L}\\ =5.37\times {10}^{-5}\text{mol}\end{array}$

The molar mass of the solute is calculated by the formula,

$\text{Molar mass}=\frac{\text{Given}\text{mass}}{\text{Number}\text{of}\text{moles}}$

Substitute the values in the above formula,

$\begin{array}{c}\text{Molar mass}=\frac{\text{Given}\text{mass}}{\text{Number}\text{of}\text{moles}}\\ \text{Molar mass}=\frac{27.40\times {10}^{-3}\text{g}}{5.37\times {10}^{-5}\text{mol}}\\ =510.24\text{g}/\text{mol}\end{array}$

Now, substitute the values of molar mass and empirical formula mass in equation (2).

$\begin{array}{c}n=\frac{\text{Molecular}\text{mass}}{\text{Emprical}\text{formula}\text{mass}}\\ n=\frac{510.24\text{g}/\text{mol}}{\text{69}\text{g/mol}}\\ n\approx 7\end{array}$

Substitute the value of n in equation (1).

$\begin{array}{c}\text{Molecular formula}={\left(\text{Emperical}\text{formula}\right)}_n\\ ={\left({\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{N}}_{\text{2}}\right)}_7\\ ={\text{C}}_{21}{\text{H}}_{\text{35}}{\text{N}}_{\text{14}}\end{array}$

Conclusion

The molecular formula of the compound ${\text{C}}_{21}{\text{H}}_{\text{35}}{\text{N}}_{\text{14}}$.