Chapter 3, Problem 3D.8E

Interpretation Introduction

Interpretation:

Molecular formula of compound used to make PVC and take $7.73\min$ to effuse has to be determined.

Concept Introduction:

Subscript of empirical formula is multiplied by value of $n$ to give molecular formula. The value of $n$ can be calculated through division of molecular mass by empirical formula mass as follows:

  $n=\frac{\text{molecular mass}}{\text{empirical formula mass}}$

Effusion is explained as movement of the gas molecule through small hole.

Diffusion can be explained as when one gas molecule mix with another gas molecule by random motion.

Graham’s law of effusion gives relation of rate of effusion with molar mass. Rate of effusion and square root of molar mass are inversely related. The mathematical expression is as follows:

  $\frac{\text{Effusion rate of A}}{\text{Effusion rate of B}}=\sqrt{\frac{M_{\text{B}}}{M_{\text{A}}}}$

Here,

$M_{\text{B}}$ is molar mass of gas $\text{B}$.

$M_{\text{A}}$ is molar mass of gas $\text{A}$.

Answer to Problem 3D.8E

The molecular formula for compound used to make PVC and take $7.73\min$ to effuse is ${\text{C}}_2{\text{H}}_3\text{Cl}$.

Explanation of Solution

Compound PVC has $38.4\text{ \% C}$, $4.82\text{ \% H}$, and $\text{56}\text{.8 \% Cl}$. Consider mass of compound be $100\text{ g}$. Thus $100\text{ g}$ of PVC compound would contain $38.4\text{ g C}$, $4.82\text{ g H}$, and $\text{56}\text{.8 g Cl}$.

The expression to determine moles is as follows:

  $\text{Moles}\left(\text{mol}\right)=\frac{\text{Given mass}\left(\text{g}\right)}{\text{Molar mass}\left(\text{g/mol}\right)}$        (1)

Substitute $38.4\text{ g}$ for given mass and $12.011\text{ g/mol}$ for molar mass in equation (1) to calculated moles of $\text{C}$.

  $\begin{array}{c}\text{Moles}\left(\text{mol}\right)=\frac{38.4\text{ g}}{12.011\text{ g/mol}}\\ =3.20\text{ mol}\end{array}$

Substitute $4.82\text{ g}$ for given mass and $1.00794\text{ g/mol}$ for molar mass in equation (1) to calculated moles of $\text{H}$.

  $\begin{array}{c}\text{Moles}\left(\text{mol}\right)=\frac{4.82\text{ g}}{1.00794\text{ g/mol}}\\ =4.78\text{ mol}\end{array}$

Substitute $\text{56}\text{.8 g}$ for given mass and $35.45\text{ g/mol}$ for molar mass in equation (1) to calculated moles of $\text{Cl}$.

  $\begin{array}{c}\text{Moles}\left(\text{mol}\right)=\frac{\text{56}\text{.8 g}}{35.45\text{ g/mol}}\\ =1.60\text{ mol}\end{array}$

Since each mole value should be divided by smallest value thus divide all mole values by 1.60 to determined subscripts of respective elements.

The subscripts for $\text{C}$ can be calculated as follows:

  $\begin{array}{c}\text{Subscript for C}=\frac{3.20\text{ mol C}}{1.60}\\ =2\text{ mol C}\end{array}$

The subscripts for $\text{H}$ can be calculated as follows:

  $\begin{array}{c}\text{Subscript for H}=\frac{4.78\text{ mol H}}{1.60}\\ \simeq 3\text{ mol H}\end{array}$

The integer or mixed fraction for $\text{Cl}$ can be calculated as follows:

  $\begin{array}{c}\text{Subscript for Cl}=\frac{1.60\text{ mol Cl}}{1.60}\\ =1\text{ mol Cl}\end{array}$

The final empirical formula for PVC is ${\text{C}}_2{\text{H}}_3\text{Cl}$.

Let molecular formula of compound is ${\text{C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n$.

The expression for molar masses of ${\text{C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n$ and $\text{Ar}$ gases and their time for effusion is as follows:

  $\frac{{\text{Time for C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n\text{ to effuse}}{\text{Time for Ar to effuse}}=\sqrt{\frac{M_{{\text{C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n}}{M_{\text{Ar}}}}$        (2)

Rearrange equation (2) to calculate value of $M_{{\text{C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n}$.

  $M_{{\text{C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n}=\left(M_{\text{Ar}}\right){\left(\frac{{\text{Time for C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n\text{ to effuse}}{\text{Time for Ar to effuse}}\right)}^2$        (3)

Substitute $39.95\text{ g/mol}$ for $M_{\text{Ar}}$, $7.73\text{ min}$ for time for ${\text{C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n$ to effuse, and $6.18\text{ min}$ for time for $\text{Ar}$ to effuse in equation (2).

  $\begin{array}{c}{M}_{{\text{C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n}=\left(39.95\text{ g/mol}\right){\left(\frac{7.73\text{ min}}{6.18\text{ min}}\right)}^2\\ =62.50\text{ g/mol}\end{array}$

Thus molar mass of ${\text{C}}_{2n}{\text{H}}_{3n}{\text{Cl}}_n$ is $62.50\text{ g/mol}$.

The empirical formula mass can be calculated as follows:

  $\text{Mass}\left({\text{C}}_2{\text{H}}_3\text{Cl}\right)=\left[\begin{array}{l}2\left(\text{atomic mass of C}\right)+3\left(\text{atomic mass of H}\right)+\\ 1\left(\text{atomic mass of Cl}\right)\end{array}\right]$        (3)

Substitute $12.011\text{ g}$ for atomic mass of $\text{C}$, $1.00794\text{ g}$ for atomic mass of $\text{H}$, and $35.45\text{ g}$ for atomic mass of $\text{H}$ in equation (3).

  $\begin{array}{c}\text{Mass}\left({\text{C}}_2{\text{H}}_3\right)=\left[2\left(12.011\text{ g}\right)+3\left(1.00794\text{ g}\right)+1\left(35.45\text{ g}\right)\right]\\ =62.50\text{ g}\end{array}$

The value of $n$ can be calculated through division of molecular mass by empirical formula mass as follows:

  $n=\frac{\text{molecular mass}}{\text{empirical formula mass}}$        (4)

Substitute $62.50\text{ g}$ for molecule mass and $62.50\text{ g}$ for empirical formula mass in equation (4).

  $\begin{array}{c}n=\frac{62.50\text{ g}}{62.50\text{ g}}\\ \approx 1\end{array}$

Hence, molecular formula for compound is ${\text{C}}_2{\text{H}}_3\text{Cl}$.