Chapter 13, Problem 128QRT

(a)

Interpretation Introduction

Interpretation:

The empirical formula of a compound ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{Cr}$ has to be calculated.

Concept introduction:

Molecular formula of a compound is derived from empirical formula mass and molar mass of the compound as,

    $\frac{molarmass}{empiricalmolarmass}$

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

  $\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Explanation of Solution

Given information as: The compound contains $\text{73}\text{.94\%}$ of Carbon and $\text{8}\text{.27\%}$ of Hydrogen and the remaining is Chromium.

Here, the total amount is considered as $\text{100}\text{g}$; from which the amount of Chromium is calculated as $\text{100}\text{g}\text{-}\text{73}\text{.94}\left(\text{of}\text{carbon}\right)\text{-}\text{8}\text{.27}\left(\text{g}\text{of}\text{Hydrogen}\right)\text{=}\text{17}\text{.79}\text{g}\text{Cr}\text{.}$

Calculate the moles of Carbon:

  $\text{73}\text{.94}\text{g}\text{C}\text{×}\frac{\text{1}\text{mol}\text{C}}{\text{12}\text{.0107}\text{g}\text{C}}\text{=}\text{6}\text{.156}\text{mol}\text{C}$

Calculate the moles of Hydrogen:

  $8.27\text{g}\text{H}\text{×}\frac{\text{1}\text{mol}H}{\text{1}\text{.0079}\text{g}\text{H}}\text{=}\text{8}\text{.21}\text{mol}\text{H}$

Calculate the moles of Chromium:

  $\text{17}\text{.79}\text{g}\text{Cr}\text{×}\frac{\text{1}\text{mol}\text{Cr}}{\text{51}\text{.996}\text{g}\text{Cr}}\text{=}\text{0}\text{.342}\text{mol}\text{Cr}$

From the above, the mole ratio of Carbon, Hydrogen and Chromium is,

  $\begin{array}{l}\text{Mole ratio = }\text{Carbon : Hydrogen : Chromium}\\ \\ \text{=}\text{6}\text{.156}\text{mol}\text{ : 8}\text{.21 mol : 0}\text{.342 mol}\end{array}$

Thus, the simplified ratio becomes, $\text{18}\text{C}\text{:}\text{24H}\text{:}\text{1}\text{Cr}$

Therefore, the empirical formula is $C_{18}{H}_{24}Cr.$

(b)

Interpretation Introduction

Interpretation:

The molecular formula of a compound ${\left({\text{C}}_{\text{18}}{\text{H}}_{\text{24}}\text{Cr}\right)}_{\text{n}}$ has to be calculated.

Concept introduction:

Molecular formula of a compound is derived from empirical formula mass and molar mass of the compound as,

    $\frac{molarmass}{empiricalmolarmass}$

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

  $\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Explanation of Solution

Using the formula of osmotic pressure as shown below,

  Temperature, $\text{T}{\text{= 25}}^{\text{o}}\text{C + 273}\text{.15}\text{= 298 K}\text{.}$

Osmotic pressure $\text{(π) = MRT}$

  $\begin{array}{l}\text{Π = cRT}\\ \\ \text{c}\text{=}\frac{\text{Π}}{\text{RT}}\text{=}\frac{\left(\text{3}\text{.17}\text{mmHg}\right)\left(\frac{\text{1atm}}{\text{760}\text{mmHg}}\right)}{\left(\text{0}\text{.08206}\frac{\text{L}\text{.atm}}{\text{mol}\text{.K}}\right)\left(\text{298}\text{K}\right)}\\ \\ \text{=}\text{1}\text{.71}\times {\text{10}}^{-4}\text{ mol/L}\text{.}\end{array}$

The addition of solute does not change the volume of the solution; the volume of the solvent is equal to the volume of the solution. Hence,

  $\begin{array}{l}\left(\text{100}\text{mL}\text{Chloroform}\right)\left(\frac{\text{1 L}\text{Chloroform}}{\text{1000 mL}\text{Chloroform}}\right)\left(\frac{\text{1}\text{L}\text{solution}}{\text{1L}\text{Chloroform}}\right)\left(\frac{\text{1}\text{.71}{\text{×10}}^{\text{-4}}\text{ mol}\text{solute}}{\text{1}\text{L solution}}\right)\\ \text{=}\text{1}\text{.71}\times {\text{10}}^{-5}\text{ mol}\text{solute}\end{array}$

  $\text{Molar}\text{mass of compound = }\frac{\text{5}\text{.0 mg}}{\text{1}\text{.71}{\text{×10}}^{\text{-5}}\text{ mol}}\text{×}\frac{\text{1}\text{g}}{\text{1000}\text{mg}}\text{=}\text{292}\text{g/mol}\text{.}$

Thus, molecular mass of the compound is $292g/mol.$

As known, the molecular formula is a multiple of the empirical formula: ${\left({\text{C}}_{\text{18}}{\text{H}}_{\text{24}}\text{Cr}\right)}_{\text{n}}.$

The molar mass of the empirical formula $\left({\text{C}}_{\text{18}}{\text{H}}_{\text{24}}\text{Cr}\right)$ is $\text{292}\text{.37}\text{g/mol}\text{.}$

Therefore, the molecular formula is $\left(C_{18}{H}_{24}Cr\right).$