Chapter 7, Problem 36PE

(a)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{64}\text{.1\%}\text{Cu,}\text{35}\text{.9\%}\text{Cl}$ 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.

Answer to Problem 36PE

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

Explanation of Solution

Given,

The percentage composition of copper is $\text{64}\text{.1\%}\text{.}$

The percentage composition of chlorine is $\text{35}\text{.9\%}\text{.}$

The atomic mass of copper is $63.55\text{g/mol}\text{.}$

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

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

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

  The moles of copper $\text{=}\left(64.1\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{63.55\text{g}}\text{=}1.0\text{mol}$

  The moles of chlorine $\text{=}\left(35.9\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{35.45\text{g}}\text{=1}\text{.0}\text{mol}$

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.

  $\text{Cu=}\frac{1.0\text{mol}}{1.0\text{mol}}\text{=1}$

  $\text{Cl=}\frac{1.0\text{mol}}{1.0\text{mol}}\text{=1}$

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

(b)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{47}\text{.2\%}\text{Cu,}52.8\text{\%}\text{Cl}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 36PE

The empirical formula of the compound is ${\text{CuCl}}_2\text{.}$

Explanation of Solution

Given,

The percentage composition of copper is $\text{47}\text{.2\%}\text{.}$

The percentage composition of chlorine is $\text{52}\text{.8\%}\text{.}$

The atomic mass of copper is $63.55\text{g/mol}\text{.}$

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

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

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

  The moles of copper $\text{=}\left(47.2\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{63.55\text{g}}\text{=}\text{0}\text{.743mol}$

  The moles of chlorine $\text{=}\left(52.8\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{35.45\text{g}}\text{=1}\text{.49mol}$

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.

  $\text{Cu=}\frac{0.743\text{mol}}{0.743\text{mol}}\text{=1}$

  $\text{Cl=}\frac{1.49\text{mol}}{0.743\text{mol}}\text{=2}$

The empirical formula of the compound is ${\text{CuCl}}_2\text{.}$

(c)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{51}\text{.9\%}\text{Cr,}48.1\text{\%}\text{S}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 36PE

The empirical formula of the compound is ${\text{Cr}}_2{\text{S}}_3\text{.}$

Explanation of Solution

Given,

The percentage composition of chromium is $\text{51}\text{.9\%}\text{.}$

The percentage composition of sulfur is $\text{48}\text{.1\%}\text{.}$

The atomic mass of chromium is $52\text{g/mol}\text{.}$

The atomic mass of sulfur is $32\text{g/mol}\text{.}$

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

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

  The moles of chromium $\text{=}\left(51.9\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{52\text{g}}\text{=}\text{0}\text{.998mol}$

  The moles of sulfur $\text{=}\left(48.1\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{32\text{g}}\text{=1}\text{.50mol}$

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.

  $\text{Cr=}\frac{0.998\text{mol}}{0.998\text{mol}}\text{=1}$

  $\text{S=}\frac{1.50\text{mol}}{0.998\text{mol}}\text{=1}\text{.50}$

Since the value if fractional multiply both the values by two.  The empirical formula of the compound is ${\text{Cr}}_2{\text{S}}_3\text{.}$

(d)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{55}\text{.3\%}\text{K,}\text{14}\text{.6\%}\text{P\&30}\text{.1\%}\text{O}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 36PE

The empirical formula of the compound is ${\text{K}}_3{\text{PO}}_4\text{.}$

Explanation of Solution

Given,

The percentage composition of potassium is $\text{55}\text{.3\%}\text{.}$

The percentage composition of phosphorus is $\text{14}\text{.6\%}\text{.}$

The percentage composition of oxygen is $30.1\%.$

The atomic mass of potassium is $39.10\text{g/mol}\text{.}$

The atomic mass of phosphorus is $30.97\text{g/mol}\text{.}$

The atomic mass of oxygen is $16\text{g/mol}\text{.}$

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

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

  The moles of potassium $\text{=}\left(55.3\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{39.10\text{g}}\text{=}1.41\text{mol}$

  The moles of phosphorus $\text{=}\left(14.6\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{30.97\text{g}}\text{=0}\text{.471mol}$

  The moles of oxygen $\text{=}\left(30.1\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=1}\text{.88}\text{mol}$

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.

  $\text{K=}\frac{1.41\text{mol}}{0.471\text{mol}}\text{=3}$

  $\text{P=}\frac{0.471\text{mol}}{0.471\text{mol}}\text{=1}$

  $\text{O=}\frac{1.88\text{mol}}{0.471\text{mol}}\text{=4}$

The empirical formula of the compound is ${\text{K}}_3{\text{PO}}_4\text{.}$

(e)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{38}\text{.9\%}\text{Ba,}\text{29}\text{.4\%}\text{Cr\&31}\text{.7\%}\text{O}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 36PE

The empirical formula of the compound is ${\text{BaCr}}_2{\text{O}}_7\text{.}$

Explanation of Solution

Given,

The percentage composition of barium is $\text{38}\text{.9\%}\text{.}$

The percentage composition of chromium is $\text{29}\text{.4\%}\text{.}$

The percentage composition of oxygen is $31.7\%.$

The atomic mass of barium is $137.3\text{g/mol}\text{.}$

The atomic mass of chromium is $52\text{g/mol}\text{.}$

The atomic mass of oxygen is $16\text{g/mol}\text{.}$

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

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

  The moles of barium $\text{=}\left(38.9\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{137.3\text{g}}\text{=}0.283\text{mol}$

  The moles of chromium $\text{=}\left(29.4\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{52\text{g}}\text{=0}\text{.565mol}$

  The moles of oxygen $\text{=}\left(31.7\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=1}\text{.98}\text{mol}$

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.

  $\text{Ba=}\frac{\text{0}\text{.283}\text{mol}}{0.283\text{mol}}\text{=1}$

  $\text{Cr=}\frac{0.565\text{mol}}{0.283\text{mol}}\text{=2}$

  $\text{O=}\frac{1.98\text{mol}}{0.283\text{mol}}\text{=7}$

The empirical formula of the compound is ${\text{BaCr}}_2{\text{O}}_7\text{.}$

(f)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{3}\text{.99\%}\text{P,}\text{82}\text{.3\%}\text{Br}\text{\&}\text{13}\text{.7\%}\text{Cl}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 36PE

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

Explanation of Solution

Given,

The percentage composition of bromine is $\text{82}\text{.3\%}\text{.}$

The percentage composition of phosphorus is $\text{3}\text{.99\%}\text{.}$

The percentage composition of chlorine is $13.7\%.$

The atomic mass of bromine is $79.90\text{g/mol}\text{.}$

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

The atomic mass of phosphorus is $30.97\text{g/mol}\text{.}$

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

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

  The moles of phosphorus $\text{=}\left(3.99\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{30.97\text{g}}\text{=}0.129\text{mol}$

  The moles of bromine $\text{=}\left(82.3\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{79.90\text{g}}\text{=1}\text{.03mol}$

  The moles of chlorine $\text{=}\left(13.7\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{35.45\text{g}}\text{=0}\text{.386}\text{mol}$

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.

  $\text{P=}\frac{\text{0}\text{.129}\text{mol}}{0.129\text{mol}}\text{=1}$

  $\text{Br=}\frac{1.03\text{mol}}{0.129\text{mol}}\text{=8}$

  $\text{Cl=}\frac{0.386\text{mol}}{0.129\text{mol}}\text{=3}$

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