Chapter 7, Problem 75AE

(a)

Interpretation Introduction

Interpretation:

The empirical formula and molecular formula of a compound with percent composition $\text{7}\text{.79}\text{\%}\text{C}\text{\&}\text{92}\text{.91}\text{Cl}$ has to be given.

Concept Introduction:

Empirical Formula:

The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound.  It can be the same as the compound’s molecular formula but not always.  An empirical formula can be calculated from information about the mass of each element in a compound or from the percentage composition.

The steps for determining the empirical formula of a compound as follows:

• Obtain the mass of each element present in grams.
• Determine the number of moles of each atom present.
• Divide the number of moles of each element by the smallest number of moles.
• Convert the numbers to whole numbers.  The set of whole numbers are the subscripts in the empirical formula.

Molecular formula:

The molecular formula is the expression of the number of atoms of each element in one molecule of a compound if the molar mass value is known the molecular formula is calculated by the empirical formula.

  $\text{n}\text{=}\frac{\text{Molar}\text{mass}}{\text{Mass}\text{of}\text{the}\text{empirical}\text{formula}}$

Answer to Problem 75AE

The empirical formula of the compound is ${\text{CCl}}_4\text{.}$

The molecular formula of compound is ${\text{CCl}}_4\text{.}$

Explanation of Solution

Given,

The molar mass of the compound is $153.8\text{g}\text{.}$

The percentage of carbon is $\text{7}\text{.79\%}\text{.}$

The percentage of chlorine is $\text{92}\text{.91\%}\text{.}$

Assuming that $\text{100}\text{g}$ of material, the percent of each element equals the grams of each element.

The atomic mass of carbon is $12\text{g/mol}\text{.}$

The atomic mass of chlorine is $35.45\text{g/mol}\text{.}$

The grams of each element has to be converted to moles as,

  The moles of carbon $\text{=}\left(7.79\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{\text{12}\text{g}}\text{=0}\text{.6491}\text{mol}$

  The moles of chlorine $\text{=}\left(92.91\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{35.45\text{g}}\text{=2}\text{.601}\text{mol}$

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  $\text{C=}\frac{0.649\text{mol}}{0.649\text{mol}}\text{=1}\text{.0}$

  $\text{Cl=}\frac{2.601\text{mol}}{0.649\text{mol}}\text{=4}\text{.0}$

The empirical formula of the compound is ${\text{CCl}}_4\text{.}$

The molecular formula of the compound can be calculated as,

Mass of the empirical formula $\text{=153}\text{.8}\text{g}$

  $\text{n}\text{=}\frac{\text{Molar}\text{mass}}{\text{Mass}\text{of}\text{the}\text{empirical}\text{formula}}$

  $\text{n}\text{=}\frac{153.88\text{g}}{153.88\text{g}}$

  $\text{n}\text{=1}$

The molecular formula of compound is ${\text{CCl}}_4\text{.}$

(b)

Interpretation Introduction

Interpretation:

The empirical formula and molecular formula of a compound with percent composition $10.13\text{\%}\text{C}\text{\&}89.87\text{Cl}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 75AE

The empirical formula of the compound is ${\text{CCl}}_3\text{.}$

The molecular formula of compound is ${\text{C}}_2{\text{Cl}}_6\text{.}$

Explanation of Solution

Given,

The molar mass of the compound is $236.7\text{g}\text{.}$

The percentage of carbon is $\text{10}\text{.13\%}\text{.}$

The percentage of chlorine is $\text{89}\text{.87\%}\text{.}$

Assuming that $\text{100}\text{g}$ of material, the percent of each element equals the grams of each element.

The atomic mass of carbon is $12\text{g/mol}\text{.}$

The atomic mass of chlorine is $35.45\text{g/mol}\text{.}$

The grams of each element has to be converted to moles as,

  The moles of carbon $\text{=}\left(10.13\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{\text{12}\text{g}}\text{=0}\text{.8435}\text{mol}$

  The moles of chlorine $\text{=}\left(89.87\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{35.45\text{g}}\text{=2}\text{.535}\text{mol}$

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  $\text{C=}\frac{0.8435\text{mol}}{0.8435\text{mol}}\text{=1}\text{.0}$

  $\text{Cl=}\frac{2.535\text{mol}}{0.649\text{mol}}\text{=3}\text{.0}$

The empirical formula of the compound is ${\text{CCl}}_3\text{.}$

The molecular formula of the compound can be calculated as,

Mass of the empirical formula $\text{=118}\text{.4}\text{g}$

  $\text{n}\text{=}\frac{\text{Molar}\text{mass}}{\text{Mass}\text{of}\text{the}\text{empirical}\text{formula}}$

  $\text{n}\text{=}\frac{236.7\text{g}}{118.4\text{g}}$

  $\text{n}\text{=2}$

The molecular formula of compound is ${\text{C}}_2{\text{Cl}}_6\text{.}$

(c)

Interpretation Introduction

Interpretation:

The empirical formula and molecular formula of a compound with percent composition $25.26\text{\%}\text{C}\text{\&}74.74\text{Cl}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 75AE

The empirical formula of the compound is $\text{CCl}\text{.}$

The molecular formula of compound is ${\text{C}}_6{\text{Cl}}_6\text{.}$

Explanation of Solution

Given,

The molar mass of the compound is $284.8\text{g}\text{.}$

The percentage of carbon is $\text{25}\text{.26\%}\text{.}$

The percentage of chlorine is $\text{74}\text{.74\%}\text{.}$

Assuming that $\text{100}\text{g}$ of material, the percent of each element equals the grams of each element.

The atomic mass of carbon is $12\text{g/mol}\text{.}$

The atomic mass of chlorine is $35.45\text{g/mol}\text{.}$

The grams of each element has to be converted to moles as,

  The moles of carbon $\text{=}\left(25.26\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{\text{12}\text{g}}\text{=2}\text{.103}\text{mol}$

  The moles of chlorine $\text{=}\left(74.74\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{35.45\text{g}}\text{=2}\text{.108}\text{mol}$

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  $\text{C=}\frac{2.103\text{mol}}{2.103\text{mol}}\text{=1}\text{.0}$

  $\text{Cl=}\frac{2.103\text{mol}}{2.108\text{mol}}\text{=1}\text{.0}$

The empirical formula of the compound is $\text{CCl}\text{.}$

The molecular formula of the compound can be calculated as,

Mass of the empirical formula $\text{=47}\text{.46}\text{g}$

  $\text{n}\text{=}\frac{\text{Molar}\text{mass}}{\text{Mass}\text{of}\text{the}\text{empirical}\text{formula}}$

  $\text{n}\text{=}\frac{248.8\text{g}}{47.46\text{g}}$

  $\text{n}\text{=6}$

The molecular formula of compound is ${\text{C}}_6{\text{Cl}}_6\text{.}$

(d)

Interpretation Introduction

Interpretation:

The empirical formula and molecular formula of a compound with percent composition $11.25\text{\%}\text{C}\text{\&}88.75\text{Cl}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 75AE

The empirical formula of the compound is ${\text{C}}_3{\text{Cl}}_8\text{.}$

The molecular formula of compound is ${\text{C}}_3{\text{Cl}}_8\text{.}$

Explanation of Solution

Given,

The molar mass of the compound is $319.6\text{g}\text{.}$

The percentage of carbon is $\text{11}\text{.25\%}\text{.}$

The percentage of chlorine is $\text{88}\text{.75\%}\text{.}$

Assuming that $\text{100}\text{g}$ of material, the percent of each element equals the grams of each element.

The atomic mass of carbon is $12\text{g/mol}\text{.}$

The atomic mass of chlorine is $35.45\text{g/mol}\text{.}$

The grams of each element has to be converted to moles as,

  The moles of carbon $\text{=}\left(11.25\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{\text{12}\text{g}}\text{=0}\text{.9367}\text{mol}$

  The moles of chlorine $\text{=}\left(88.75\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{35.45\text{g}}\text{=2}\text{.504}\text{mol}$

The empirical formula can be calculated as,

The number of moles can be converted to moles to whole numbers by dividing by the small number.

  $\text{C=}\frac{0.9397\text{mol}}{0.9397\text{mol}}\text{=1}\text{.0}$

  $\text{Cl=}\frac{2.504\text{mol}}{0.9367\text{mol}}\text{=2}\text{.6}$

Since the value is fractional multiply each value by 3.  The empirical formula of the compound is ${\text{C}}_3{\text{Cl}}_8\text{.}$

The molecular formula of the compound can be calculated as,

Mass of the empirical formula $\text{=319}\text{.6}\text{g}$

  $\text{n}\text{=}\frac{\text{Molar}\text{mass}}{\text{Mass}\text{of}\text{the}\text{empirical}\text{formula}}$

  $\text{n}\text{=}\frac{319.6\text{g}}{319.6\text{g}}$

  $\text{n}\text{=1}$

The molecular formula of compound is ${\text{C}}_3{\text{Cl}}_8\text{.}$