Chapter 8, Problem 38QRT

Interpretation Introduction

Interpretation:

Molecular formula of the given hydrocarbon has to be determined using the given data.

Concept Introduction:

Ideal gas Equation:

  $\text{PV = nRT}$

Where,

$P$ is the Pressure

$\text{V}$ is the Volume

$\text{n }$ is the number of moles

$\text{R}$ is the Gas constant

$\text{T}$ is the Temperature.

Number of moles (n):

  $\text{n}\text{ =}\frac{\text{m}}{\text{M}}$

M is the molar mass and m is the mass of the given substance.

Pressure can be expressed in different units.  But it is normally reported in

$\text{mm}\text{Hg}\left(\text{millimeters}\text{of}\text{mercury}\right)$

General unit conversions:

  $\begin{array}{l}\text{1}\text{Pa}\text{=}{\text{1Nm}}^{-2}{\text{ = 1 kgm}}^{-1}{s}^{-2};{\text{ 1 kPa = 10}}^{\text{3}}\text{ Pa}\\ \\ \text{1}\text{bar}\text{=}{\text{10}}^{\text{5}}{\text{Pa = 10}}^{\text{3}}{\text{ kPa; mbar = 10}}^{\text{2}}\text{ Pa}\\ \\ \text{1}\text{atm}\text{=}1.01325\times {10}^5\text{ Pa = 101}\text{.325 kPa}\\ \\ 1\text{ atm = 760 Torr = 760}\text{mm}\text{Hg}\\ \\ 1\text{ atm = 14}{\text{.7 lb/in}}^{\text{2}}\left(psi\right)\text{ = 1}\text{.01325 bar}\end{array}$

Answer to Problem 38QRT

The molecular formula is $C_6{H}_6.$

Explanation of Solution

Given data is shown below,

  $\begin{array}{c}\text{M of }C\text{ = }12.0107\text{ g/mol}\\ \\ \text{M of }H\text{ = 1}\text{.0079 g/mol}\\ \\ {\text{m}}_{C_x{H}_y}\text{ = 0}\text{.293 g }\\ \\ \text{P = }374\text{ mmHg}\times \frac{1\text{ atm}}{760\text{ mmHg}}\\ \\ =\text{ 0}\text{.492 atm}\\ \\ \text{T = 23}°\text{C = }\left(22+273\right)K=\text{296 K}\\ \\ V\text{ = 185 mL = 0}\text{.185 L}\end{array}$

Sample of $C_x{H}_y$ contains $92.26\%\text{ C}$ and hence, the amount of $H$ in $\text{100 g }{C}_x{H}_y$ is determined as follows,

  $\begin{array}{c}{m}_H\text{ = }100\text{ g }{C}_x{H}_y-\text{ 92}\text{.26 g C}\\ \\ \text{= 7}\text{.74 g H}\end{array}$

Amount of $H$ is $\text{7}\text{.74 g}\text{.}$

Number of moles of $C$ and $H$ is determined as follows,

  $\begin{array}{c}{\text{n}}_C\text{ = }\frac{{\text{m}}_C}{{\text{M}}_C}\\ \\ =\frac{\text{92}\text{.29 g}}{\text{12}\text{.0107 g/mol}}\\ \\ =7.684\text{ mol C}\text{.}\\ \\ {\text{n}}_H\text{ = }\frac{{\text{m}}_H}{{\text{M}}_H}\\ \\ =\frac{\text{7}\text{.74 g}}{\text{1}\text{.0079 g/mol}}\\ \\ =7.68\text{ mol H}\text{.}\end{array}$

Empirical formula can be obtained by simplifying the number of moles of $C$ and $H$ as shown below,

  $\begin{array}{ccc}{n}_C& :& \begin{array}{l}{n}_H\\ \end{array}\\ \text{7}\text{.684 mol }& :& \begin{array}{l}\text{7}\text{.68 mol}\\ \end{array}\\ 1& :& 1\end{array}$

Therefore, empirical formula is $CH.$  The molar mass of the empirical formula is $\text{13}\text{.02 g/mol}\text{.}$  Here,

Number of moles of $C_x{H}_y$ can be calculated using Ideal gas equation as shown below,

  $\begin{array}{c}PV\text{ = nRT}\\ \\ \left(0.492\text{ atm}\right)\left(\text{0}\text{.185 L}\right)\text{= n}\left(\text{0}{\text{.0821 L atm mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\right)\left(\text{296 K}\right)\\ \\ n=\frac{\left(0.492\text{ atm}\right)\left(\text{0}\text{.185 L}\right)}{\left(\text{0}{\text{.0821 L atm mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\right)\left(\text{296 K}\right)}\\ \\ =\text{ 3}\text{.75}\times {\text{10}}^{-3}{\text{ mol C}}_x{H}_y\end{array}$

Number of moles of $C_x{H}_y$ is $\text{3}\text{.75}\times {\text{10}}^{-3}\text{ mol}.$

Therefore,

Molar mass is determined as follows,

  $\begin{array}{c}\text{n = }\frac{\text{m}}{\text{M}}\\ \\ \text{3}\text{.75}\times {\text{10}}^{-3}\text{ mol}=\frac{\text{0}\text{.293 g }}{\text{M}}\\ \\ =78.2{\text{ g C}}_x{H}_y\end{array}$

Molar mass is $78.2\text{ g}\text{. }$

The value of x and y can be determined by taking ratio of molecular molar mass and molar mass of empirical molar mass where $x=y.$

  $\begin{array}{c}x\text{ = y =}\frac{\text{78}\text{.2 g/mol }}{\text{13}\text{.02 g/mol}}\\ \\ =\text{ 6}\text{.01}\end{array}$

Therefore,

The molecular formula is $C_6{H}_6.$