Chapter 13, Problem 13.30P

Interpretation Introduction

Interpretation:

The molecular mass and formula of unknown liquid has to be determined.

Concept Introduction:

The expression to determine molecular weight as per ideal gas equation is given as follows:

  $M=\frac{mRT}{PV}$

Here,

$\text{R}$ denotes gas constant.

$V$ denotes the volume.

$n$ denotes the temperature.

$P$ denotes the pressure.

$m$ denotes mass.

$M$ denotes molecular mass.

Answer to Problem 13.30P

The molecular mass and formula of unknown liquid is $60.0553\text{ g/mol}$ and ${\text{C}}_3{\text{H}}_8{\text{O}}_1$  respectively.

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Substitute $\text{10}\text{.8 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=\text{10}\text{.8 }°\text{C}+273\text{ K}\\ =283.8\text{ K}\end{array}$

The formula to convert ${\text{cm}}^3$ to $\text{L}$ is as follows:

  $1{\text{ cm}}^3={10}^{-3}\text{ L}$

Thus, ${\text{250 cm}}^3$ is converted to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left({\text{250 cm}}^3\right)\left(\frac{{10}^{-3}\text{ L}}{{\text{1 cm}}^3}\right)\\ =0.25\text{ L}\end{array}$

The formula to convert $\text{mg}$ to $\text{g}$ is as follows:

  $1\text{ mg}={10}^{-3}\text{ g}$

Thus, $635\text{ mg}$ is converted to $\text{g}$ as follows:

  $\begin{array}{c}\text{Mass}=\left(635\text{ mg}\right)\left(\frac{{10}^{-3}\text{ g}}{1\text{ kg}}\right)\\ =0.635\text{ g}\end{array}$

The expression to determine molecular weight as per ideal gas equation is given as follows:

  $M=\frac{mRT}{PV}$        (2)

Substitute $0.635\text{ g}$ for $m$, $0.25\text{ L}$ for $V$, $\text{0}\text{.998 bar}$ for $P$, and $0.083145\text{ L}\cdot \text{bar/mol}\cdot \text{K}$ for $R$, $283.8\text{ K}$ for $T$ in equation (2).

  $\begin{array}{c}M=\frac{\left(0.635\text{ g}\right)\left(0.083145\text{ L}\cdot \text{bar/mol}\cdot \text{K}\right)\left(283.8\text{ K}\right)}{\left(\text{0}\text{.998 bar}\right)\left(0.25\text{ L}\right)}\\ =60.0553\text{ g/mol}\end{array}$

Molecular mass of the liquid is $60.0553\text{ g/mol}$.

The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$        (3)

Substitute $\text{2}\text{.637 g}$ for mass and $\text{44}\text{.01 g/mol}$ for molar mass of ${\text{CO}}_2$ in equation (3) to calculate moles of ${\text{CO}}_2$.

  $\begin{array}{c}{\text{Moles of CO}}_2=\frac{\text{2}\text{.637 g}}{\text{44}\text{.01 g/mol}}\\ =0.059918\text{ mol}\end{array}$

Substitute $\text{1}\text{.439 g}$ for mass and $\text{18}\text{.0152 g/mol}$ for molar mass of ${\text{H}}_2\text{O}$  in equation (3) to calculate the moles of ${\text{H}}_2\text{O}$.

  $\begin{array}{c}{\text{Moles of H}}_2\text{O}=\frac{\text{1}\text{.439 g}}{\text{18}\text{.0152 g/mol}}\\ =0.07987\text{ mol}\end{array}$

Substitute $\text{1}\text{.200 g}$ for mass of liquid and $60.0553\text{ g/mol}$ for molar mass of ${\text{H}}_2\text{O}$  in equation (3) to calculate the moles of ${\text{H}}_2\text{O}$.

  $\begin{array}{c}\text{Moles of unknown liquid}=\frac{\text{1}\text{.200 g}}{60.0553\text{ g/mol}}\\ =0.01998\text{ mol}\end{array}$

Moles of $\text{C}$ in ${\text{CO}}_2$ are calculated as follows:

  $\begin{array}{c}\text{Amount of C}\left(\text{mol}\right)=\left(0.059918\text{ mol}\right)\left(\frac{1\text{ mol C}}{1{\text{ mol CO}}_2}\right)\\ =0.059918\text{ mol C}\end{array}$

Moles of $\text{H}$ in ${\text{H}}_2\text{O}$ are calculated as follows:

  $\begin{array}{c}\text{Amount of H}\left(\text{mol}\right)=\left(0.07987\text{ mol}\right)\left(\frac{\text{2 mol H}}{1{\text{ mol H}}_2\text{O}}\right)\\ =0.15974\text{ mol H}\end{array}$

The formula to calculate the mass of carbon in the sample is as follows:

  $\text{Mass of C}=\left(\text{Moles of C}\right)\left(\frac{12.01\text{ g C}}{1\text{ mol C}}\right)$        (4)

The formula to calculate the mass of hydrogen in the sample is as follows:

  $\text{Mass of H}=\left(\text{Moles of H}\right)\left(\frac{1.008\text{ g H}}{1\text{ mol H}}\right)$        (5)

Substitute $0.059918\text{ mol}$ for the amount of $\text{C}$ in the equation (4).

  $\begin{array}{c}\text{Mass of C}=\left(0.059918\text{ mol C}\right)\left(\frac{12.01\text{ g C}}{1\text{ mol C}}\right)\\ =0.719615\text{ g C}\end{array}$

Substitute $0.15974\text{ mol}$ for the moles of $\text{C}$ in the equation (5).

  $\begin{array}{c}\text{Mass of H}=\left(0.15974\text{ mol H}\right)\left(\frac{1.008\text{ g H}}{1\text{ mol H}}\right)\\ =0.1610179\text{ g H}\end{array}$

The formula to calculate mass of $\text{O}$ is as follows:

  $\text{Mass of O}\left(\text{g}\right)=\text{Mass of sample}\left(\text{g}\right)-\left(\text{Mass of C}+\text{Mass of H}\right)$        (6)

Substitute $1.200\text{ g}$ for the mass of the sample, $0.719615\text{ g}$ for massof $\text{C}$, and $0.1610179\text{ g}$ in equation (6).

  $\begin{array}{c}\text{Mass of O}\left(\text{g}\right)=1.200\text{ g}-\left(0.719615\text{ g}+0.1610179\text{ g}\right)\\ =0.31936\text{ g}\end{array}$

Substitute $0.31936\text{ g}$ for mass and $16.00\text{ g/mol}$ for molar mass of $\text{O}$ in equation (3) to calculate moles of $\text{O}$.

  $\begin{array}{c}\text{Moles of O}=\frac{0.31936\text{ g}}{16.00\text{ g/mol}}\\ =0.019960\text{ mol}\end{array}$

Write moles of $\text{C}$$\text{H}$ and $\text{O}$ as subscripts to obtain a preliminary formula as follows:

  ${\text{C}}_{0.059918}{\text{H}}_{0.15974\text{ mol}}{\text{O}}_{0.0199604}$

The smallest subscript is 0.0199604. So, divide each subscript by 0.019960 as follows:

  ${\text{C}}_{\frac{0.059918}{0.0199604}}{\text{H}}_{\frac{0.15974}{0.0199604}}{\text{O}}_{\frac{0.0199604}{0.0199604}}\to {\text{C}}_3{\text{H}}_8{\text{O}}_1$

The expression to calculate the empirical formula mass of ${\text{C}}_3{\text{H}}_8{\text{O}}_1$ is as follows:

  $\text{Empirical formula mass}=\left[\begin{array}{l}\left(3\right)\left(\text{M of C}\right)+\left(8\right)\left(\text{M of H}\right)\\ +\left(1\right)\left(\text{M of O}\right)\end{array}\right]$        (7)

Substitute $12.01\text{ g/mol}$ for $\text{M}$ of $\text{C}$, $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$ , $16\text{ g/mol}$ for $\text{M}$ of $\text{O}$ in equation (7).

  $\begin{array}{c}\text{Empirical formula mass}=\left(3\right)\left(12.01\text{ g/mol}\right)+\left(8\right)\left(1.008\text{ g/mol}\right)+\left(1\right)\left(\text{16 g/mol}\right)\\ =60.094\text{ g/mol}\end{array}$

The formula to calculate the whole number multiple is as follows:

  $\text{Whole-number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$        (8)

Substitute $60.0553\text{ g/mol}$ for the molar mass of the compound and $60.094\text{ g/mol}$ for the empirical formula mass in equation (8).

  $\begin{array}{c}\text{Whole}-\text{number multiple}=\frac{60.0553\text{ g/mol}}{60.094\text{ g/mol}}\\ =0.999\\ \approx 1\end{array}$

The emprical

Multiply the subscripts in ${\text{C}}_3{\text{H}}_8{\text{O}}_1$ by 1 to obtain molecular formula.

  $\text{Molecular formula}=\left({\text{C}}_{3\left(1\right)}{\text{H}}_{8\left(1\right)}{\text{O}}_{1\left(1\right)}\right)\to {\text{C}}_3{\text{H}}_8{\text{O}}_1$

The compound is benzene and has a molecular formula of ${\text{C}}_3{\text{H}}_8{\text{O}}_1$.