Chapter 8, Problem 116QRT

(a)

Interpretation Introduction

Interpretation:

Molar mass of the compound has to be determined.

Concept Introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\text{PV = nRT}$

Here,

  n is the moles of gas

    P is the Pressure

  V is the Volume

    T is the Temperature

    R is the gas constant

Molar mass can be determined using the given equation,

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

Here, n is the number of moles.

  M is the Molar mass.

    m is the Mass.

Answer to Problem 116QRT

Molar mass of the compound is $\text{64}\text{.1 g/mol}\text{.}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}P=\text{ 750 mmHg}\\ \\ \text{= 750 mmHg}\times \frac{1\text{ atm}}{\text{760 mmHg}}\text{ = 0}\text{.987 atm}\\ \\ \text{T = 50}°\text{C = }\left(50+273\right)K=\text{323 K}\\ \\ \text{V = 125 mL = 0}\text{.125 L}\end{array}$

Number of moles of the compound is determined using Ideal gas equation as given,

  $\begin{array}{c}PV=\text{ nRT}\\ \\ \left(\text{0}\text{.987 atm}\right)\times \left(\text{0}\text{.125 L}\right)\text{ = n}\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(\text{323 K}\right)\\ \\ n\text{= }\frac{\left(\text{0}\text{.987 atm}\right)\times \left(\text{0}\text{.125 L}\right)}{\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(\text{323 K}\right)}\\ \\ =4.65\times {10}^{-3}\text{ mol}\end{array}$

Molar mass of the compound is determined as follows,

  $\begin{array}{c}\text{n}\text{ =}\frac{Mass}{\text{Molar mass}}\\ \\ 4.65\times {10}^{-3}\text{ mol}=\frac{\text{0}\text{.298 g}}{\text{Molar mass}}\\ \\ \text{Molar mass = }\frac{\text{0}\text{.298 g}}{4.65\times {10}^{-3}\text{ mol}}\\ \\ =\text{ 64}\text{.1 g/mol}\end{array}$

Molar mass of the compound is $64.1g/mol.$

(b)

Interpretation Introduction

Interpretation:

Empirical formula and molecular formula for the compound has to be given.

Concept Introduction:

Empirical Formula:

Empirical formula of a compound is the simplest formula which provides the lowest positive whole number ratio of atoms that exists in the compound. The molecular formula of a compound can either be same as the empirical formula or a multiple of it.

Percent composition:

Percent composition is nothing but providing the mass percent of each element present in the compound.

Answer to Problem 116QRT

Empirical formula of the compound is $\text{CHF}\text{.}$

Molecular formula for the compound is ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{F}}_{\text{2}}.$

Explanation of Solution

Given data:

The percent composition of the ingredient in the compound is given below:

  $\text{37}\text{.5\% C, 3}\text{.15\% H, 59}\text{.3\% F}$

Empirical Formula:

Let us assume that the mass of the compound is $100\text{ g}$

From the percent composition given, calculate the number of moles of each element in the ingredient.

Mole calculation for carbon

  $\begin{array}{c}\text{Mole of C = 37}\text{.5 g C}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g C}}\\ \\ =\text{3}\text{.12 mol C}\end{array}$

Mole calculation for hydrogen,

  $\begin{array}{c}\text{Mole of H = 3}\text{.15 g}\text{H}\times \frac{\text{1 mol H}}{\text{1}\text{.01 g H}}\\ \\ =\text{3}\text{.13 mol H}\end{array}$

Mole calculation of fluorine,

  $\begin{array}{c}\text{Mole of F = 59}\text{.3 g F}\times \frac{\text{1 mol F}}{\text{19 g F}}\\ \\ =\text{3}\text{.12 mol F}\end{array}$

On dividing all the three numbers by the smallest integer $\left(\text{3}\text{.12 mol}\right)$, the mole ratio of $\text{C:H:O}$ is calculated as follows,

Empirical formula of the compound was determined.

  $\begin{array}{ccccc}C& :& H& :& F\\ \frac{3.12}{3.12}& :& \frac{3.13}{3.12}& :& \frac{3.12}{3.12}\\ \text{1}& :& 1& :& 1\\ & & & & \end{array}$

Hence, the empirical formula will be $CHF.$

Molecular formula can also be written as ${\left(\text{CHF}\right)}_{\text{n}}.$

Empirical mass is calculated as follows,

$\begin{array}{c}Empirical\text{ mass = Mass of C + Mass of H + Mass of F}\\ \\ \text{= }\left(\text{12}\text{.01 g/mol}\right)+\left(1.01\text{ g/mol}\right)+\left(\text{19 g/mol}\right)\\ \\ =\text{ 32}\text{.02 g/mol}\end{array}$

Molar mass of the compound is $\text{64}\text{.1 g/mol}$

Molecular formula for the compound can be determined using the given formula,

  $\begin{array}{c}n=\frac{Molarmass}{Empiricalmass}\\ \\ =\frac{\text{64}\text{.1 g/mol}}{\text{32}\text{.02 g/mol}}\\ \\ =\text{ 2}\\ \\ Molecular\text{ formula = 2}\left(Empiricalformula\right)\\ \\ ={\text{ C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{F}}_{\text{2}}\end{array}$

Molecular formula for the compound is $C_2{H}_2{F}_2.$

(c)

Interpretation Introduction

Interpretation:

Lewis structure of each of the isomer of compound has to be determined.

Concept-Introduction:

Lewis structure

Electron dot structure also known as Lewis dot structure represents the number of valence electrons of an atom or constituent atoms bonded in a molecule.  Each dot corresponds to one electron.

Explanation of Solution

Molecular formula for the compound is ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{F}}_{\text{2}}.$

Three isomers (two cis and one trans) are possible for the compound having this molecular formula.

The Lewis electron dot structure for given molecules are determined by first drawing the skeletal structure for the given molecules, then the total number of valence electrons for all atoms present in the molecules are determined.

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed considering each atom contains eight electrons in its valence shell.

Draw Lewis structure of the compound:

Outer valence electrons of Carbon Oxygen and Fluorine are four, six and seven respectively.

  $\begin{array}{l}C\text{ = 2}\times 4\text{ = 8}\\ \text{H = 2}\times \text{1 = 2}\\ \text{F = 2}\times \text{7 = 14}\\ Total{\text{ e}}^{-}\text{ = 8+ 2 + 14 = 24}\\ \text{5 Bonds = }\left(5\times 2\right)=10\\ {\text{Remaining e}}^{-}\text{ = 24}-10=14\end{array}$

Here, one double bond is required to complete the complete the octets of all the atoms.

After the distribution of electrons, both fluorine atoms get two lone pair of electrons.

The Lewis structure of each of the isomer of compound follows as,