Chapter 3, Problem 3B.37E

(a)

Interpretation Introduction

Interpretation:

Molar mass of the gaseous fluorinated methane with density $8\text{g}\cdot {\text{L}}^{-1}$ has to be determined.

Concept Introduction:

Expression that relates density, molar mass, pressure, temperature, and gas constant for an ideal gas is as follows:

  $d=\frac{MP}{RT}$

Here,

$d$ is density of object.

$M$ is molar mass.

$P$ is pressure.

$R$ is gas constant.

$T$ is the temperature.

Answer to Problem 3B.37E

Molar mass of the gaseous fluorinated methane with density $8\text{g}\cdot {\text{L}}^{-1}$ is $70.08466\text{g}\cdot {\text{mol}}^{-1}$.

Explanation of Solution

The formula to calculate density is as follows:

  $d=\frac{MP}{RT}$        (1)

Rearrange equation (1) to calculate $M$.

  $M=\frac{dRT}{P}$        (2)

Substitute $300\text{K}$ for $T$, $8\text{g}\cdot {\text{L}}^{-1}$ for $d$, $0.08205746\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $2.81\text{atm}$ for $P$ in equation (2).

  $\begin{array}{c}M=\frac{\left(8\text{g}\cdot {\text{L}}^{-1}\right)\left(0.08205746\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}{\left(2.81\text{atm}\right)}\\ =70.08466\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Molar mass of gaseous fluorinated methane is $70.08466\text{g}\cdot {\text{mol}}^{-1}$.

(b)

Interpretation Introduction

Interpretation:

Formula of the gaseous fluorinated methane compound has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.37E

Formula of the gaseous fluorinated methane compound is $0.6964\text{g}\cdot {\text{L}}^{-1}$.

Explanation of Solution

Formula of the gaseous fluorinated methane compound contains $\text{C, H, F}$.

Consider the number of $\text{H}$ be $x$ and $4-x$ be the number of $\text{F}$ as the formula for methane is ${\text{CH}}_4$.

Formula to calculate molar mass of the compound is as follows:

  $\left[\begin{array}{l}\left\{\begin{array}{l}\left(\text{number of C}\right)\\ \left(\text{atomic mass of C}\right)\end{array}\right\}+\\ \left\{\begin{array}{l}\left(\text{number of H}\right)\\ \left(\text{atomic mass of H}\right)\end{array}\right\}+\\ \left\{\begin{array}{l}\left(\text{number of F}\right)\\ \left(\text{atomic mass of F}\right)\end{array}\right\}\end{array}\right]=\left(\text{molar mass of compound}\right)$        (3)

Substitute 1for number of $\text{C}$, $x$ for number of $\text{H}$, $4-x$ for number of $\text{F}$, $1.00784\text{g/mol}$ for atomic mass of $\text{H}$, $12.0107\text{g/mol}$ for atomic mass of $\text{C}$, $18.998403\text{g/mol}$ for atomic mass of $\text{F}$ and $70.08466\text{g}\cdot {\text{mol}}^{-1}$ for molar mass of compound in equation (3).

  $\begin{array}{c}\left[\begin{array}{l}\left\{\begin{array}{l}\left(\text{1}\right)\\ \left(12.0107\text{g/mol}\right)\end{array}\right\}+\left\{\begin{array}{l}\left(x\right)\\ \left(1.00784\text{g/mol}\right)\end{array}\right\}+\\ \left\{\begin{array}{l}\left(\text{4}-x\right)\\ \left(18.998403\text{g/mol}\right)\end{array}\right\}\end{array}\right]=\left(70.08466\text{g}\cdot {\text{mol}}^{-1}\right)\\ \left[\begin{array}{l}\left(12.0107\text{g/mol}\right)+\left(x\right)\left(1.00784\text{g/mol}\right)+\\ \left(75.9936\text{g/mol}\right)-\left(x\right)\left(18.998403\text{g/mol}\right)\end{array}\right]=\left(70.08466\text{g}\cdot {\text{mol}}^{-1}\right)\\ \left[\begin{array}{l}\left(88.0043\text{g/mol}\right)-\left(x\right)\left(17.990563\text{g/mol}\right)+\\ \end{array}\right]=\left(70.08466\text{g}\cdot {\text{mol}}^{-1}\right)\\ x=0.9960577\\ \approx 1\end{array}$

Since number of $\text{H}$ is $x$ and $4-x$ is number of $\text{F}$ so number of $\text{H}$ is $1$ and $3$ is number of $\text{F}$.

Hence the formula for compound will be ${\text{CHF}}_3$

(c)

Interpretation Introduction

Interpretation:

Density of fluorinated gaseous methane at $1\text{atm}$ and $298\text{K}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.37E

Density of fluorinated gaseous methane at $1\text{atm}$ and $298\text{K}$ is $2.8660\text{g}\cdot {\text{L}}^{-1}$.

Explanation of Solution

Substitute $298\text{K}$ for $T$, $70.08466\text{g}\cdot {\text{mol}}^{-1}$ for $M$, $0.08205746\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $1\text{atm}$ for $P$ in equation (1).

  $\begin{array}{c}d=\frac{\left(70.08466\text{g}\cdot {\text{mol}}^{-1}\right)\left(1\text{atm}\right)}{\left(0.08205746\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}\\ =2.8660\text{g}\cdot {\text{L}}^{-1}\end{array}$