Chapter 11, Problem 11.94QP

Interpretation Introduction

Interpretation: A is given to be dissolved in $1.00\text{g}$ of chloroform. The boiling point solution is given to be increasing by $2.45°\text{C}$. The molar mass and the molecular formula of the caffeine compound are to be calculated.

Concept introduction: When a system 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 molar mass and the molecular formula of the eugenol compound.

Answer to Problem 11.94QP

Solution

The molar mass of the caffeine compound is $\underset{\_}{194g/mol}$.

The molecular formula of the caffeine compound is ${\text{C}}_8{\text{H}}_{10}{\text{N}}_4{\text{O}}_2$.

Explanation of Solution

Explanation

Given

The given mass of caffeine compound is $150\text{mg}$.

The given mass of camphor is $10.0\text{g}$.

The given freezing point depression constant $\left(K_{\text{f}}\right)$ is $39.7°\text{C/m}$.

The given decreased freezing point $\left(\Delta T_{\text{f}}\right)$ of camphor is $3.07°\text{C}$.

The given mass percent of the element carbon is $49.49\%$.

The given mass percent of the element hydrogen is $5.15\%$.

The given mass percent of the element nitrogen is $28.87\%$.

The given mass percent of the element oxygen is the remainder of all these masses and it is given as,

$\begin{array}{c}\text{Maas}\text{\%}\text{of}\text{O}=100-\left(49.49+5.15+28.87\right)\\ =100-83.51\\ =16.49\end{array}$

The compound caffeine is solute and the compound camphor is solvent here.

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

$\begin{array}{c}1\text{Gram}\left(\text{g}\right)=1000\text{mg}\\ \text{1}\text{mg}=\frac{1}{1000}\text{g}\\ 150\text{mg}=150\times \frac{1}{1000}\text{g}\\ =0.150\text{g}\end{array}$

The given mass of solvent camphor in terms of kilogram (kg) unit is given as,

$\begin{array}{c}1\text{Kilogram}\left(\text{kg}\right)=1000\text{g}\\ \text{1}\text{g}=\frac{1}{1000}\text{kg}\\ 10.0\text{g}=10.0\times \frac{1}{1000}\text{kg}\\ =10.0\times {10}^{-3}\text{kg}\end{array}$

Molar mass of solute is also calculated with the help of freezing point depression $\left(\Delta T_{\text{f}}\right)$ and is given by the formula,

$\Delta T_{\text{f}}=K_{\text{f}}m$

Where,

• $m$ is the molality of the solution.
• $K_{\text{f}}$ is the freezing point depression constant.
• $\Delta T_{\text{f}}$ is the freezing point depression.

With the help of above formula the molality of solution is calculated as,

$\begin{array}{l}\Delta T_{\text{f}}=K_{\text{f}}m\\ m=\frac{\Delta T_{\text{f}}}{K_{\text{f}}}\end{array}$

Substitute the given values in above formula.

$\begin{array}{l}m=\frac{\Delta T_{\text{f}}}{K_{\text{f}}}\\ m=\frac{3.07°\text{C}}{39.7°\text{C/m}}\\ m=0.0773\text{m}\end{array}$

Molality of a solution defined as the number of moles solute present in per kilogram of the solvent. Therefore, according to the above definition the molality of the solution is expressed as,

$\text{Molality}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent}\text{in}\text{kg}}$

Hence, the calculated amount of molality reveals that $0.0773\text{moles}$ of solute are dissolved per kilogram of the solvent.

According to the given amount of solute and solvent, $0.150\text{g}$ of caffeine is present in $0.01\text{kg}$ of camphor. Therefore, for $1\text{kg}$ solvent camphor, $15.0\text{g}$ of caffeine is present.

From the above calculated values it is clear that $0.0773\text{moles}$ have the mass of $15.0\text{g}$. It means mass of one mole of solute is calculated as,

$\begin{array}{c}0.0773\text{moles}=15.0\text{g}\\ 1\text{mol}=\frac{15.0\text{g}}{0.0773}\\ =194\text{g}\end{array}$

The molar mass of caffeine is $194\text{g/mol}$.

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 oxygen. Therefore, the empirical formula of the compound is calculated as,

$\begin{array}{cccccc}\text{Element}& \text{Percent}& \text{Atomic}\text{mass}& \begin{array}{l}\text{Relative}\text{number}\\ \text{of}\text{atoms}\end{array}& \text{Simplest}\text{ratio}& \begin{array}{l}\text{Simplest}\text{whole}\\ \text{no}\text{ratio}\end{array}\\ \text{Carbon}\left(\text{C}\right)& 49.49& 12& \frac{49.49}{12}=4.1& \frac{4.1}{1}\simeq 4.0& 4\\ \text{Hydrogen}\left(\text{H}\right)& 5.15& 1& \frac{5.15}{1}=5.1& \frac{5.1}{1}\simeq 5.0& 5\\ \text{nitrogen}\left(\text{N}\right)& 28.87& 14& \frac{28.87}{14}=2.0& \frac{2.0}{1}=2& 2\\ \text{Oxygen}\left(\text{O}\right)& 16.49& 16& \frac{16.49}{16}=1.0& \frac{1.0}{1.0}=1& 1\end{array}$

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

${\text{C}}_4{\text{H}}_5{\text{N}}_2{\text{O}}_{\text{1}}$

Now the empirical formula weight is calculated as,

$\begin{array}{c}{\text{C}}_4{\text{H}}_5{\text{N}}_2{\text{O}}_{\text{1}}=\left(12\times 4\right)\text{g/mol}+\left(5\times 1\right)\text{g/mol}+\left(14\times 2\right)\text{g/mol}+\left(16\times 1\right)\text{g/mol}\\ =\left(48\right)\text{g/mol}+\left(5\right)\text{g/mol}+\left(28\right)\text{g/mol}+\left(16\right)\text{g/mol}\\ =97\text{g/mol}\end{array}$

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)

Substitute the values in the above formula.

$\begin{array}{l}n=\frac{\text{Molecular}\text{mass}}{\text{Emprical}\text{formula}\text{mass}}\\ n=\frac{194\text{g/mol}}{97\text{g/mol}}\\ n=2\end{array}$

Substitute the value of n in equation (1).

$\begin{array}{c}\text{Molecular formula}={\left(\text{Emperical}\text{formula}\right)}_n\\ ={\left({\text{C}}_4{\text{H}}_5{\text{N}}_2{\text{O}}_{\text{1}}\right)}_2\\ ={\text{C}}_8{\text{H}}_{10}{\text{N}}_4{\text{O}}_2\end{array}$

Conclusion

The molar mass of the caffeine compound is $\underset{\_}{194g/mol}$.

The molecular formula of the caffeine compound is ${\text{C}}_8{\text{H}}_{10}{\text{N}}_4{\text{O}}_2$.