Chapter 7, Problem 35PE

(a)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{63}\text{.6\%}\text{N}\text{\&36}\text{.4\%}\text{O}$ 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 35PE

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

Explanation of Solution

Given,

The percentage composition of nitrogen is $\text{63}\text{.6\%}\text{.}$

The percentage composition of oxygen is $\text{36}\text{.4\%}\text{.}$

The atomic mass of nitrogen is $14.01\text{g/mol}\text{.}$

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 nitrogen $\text{=}\left(63.6\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{14.01\text{g}}\text{=}4.54\text{mol}$

  The moles of oxygen $\text{=}\left(36.4\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=2}\text{.28}\text{mol}$

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.

  $\text{N=}\frac{4.54\text{mol}}{2.28\text{mol}}\text{=2}$

  $\text{O=}\frac{2.28\text{mol}}{2.28\text{mol}}\text{=1}$

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

(b)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{46}\text{.71\%}\text{N}\&53.3\text{\%}\text{O}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 35PE

The empirical formula of the compound is $\text{NO}\text{.}$

Explanation of Solution

Given,

The percentage composition of nitrogen is $\text{46}\text{.71\%}\text{.}$

The percentage composition of oxygen is $\text{53}\text{.3\%}\text{.}$

The atomic mass of nitrogen is $14.01\text{g/mol}\text{.}$

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 nitrogen $\text{=}\left(46.7\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{14.01\text{g}}\text{=}3.33\text{mol}$

  The moles of oxygen $\text{=}\left(53.3\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=3}\text{.33}\text{mol}$

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.

  $\text{N=}\frac{3.33\text{mol}}{3.33\text{mol}}\text{=1}$

  $\text{O=}\frac{3.33\text{mol}}{3.33\text{mol}}\text{=1}$

The empirical formula of the compound is $\text{NO}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{25}\text{.9\%}\text{N}\&74.1\text{\%}\text{O}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 35PE

The empirical formula is ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\text{.}$

Explanation of Solution

Given,

The percentage composition of nitrogen is $\text{25}\text{.9\%}\text{.}$

The percentage composition of oxygen is $\text{74}\text{.1\%}\text{.}$

The atomic mass of nitrogen is $14.01\text{g/mol}\text{.}$

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 nitrogen $\text{=}\left(25.9\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{14.01\text{g}}\text{=}1.85\text{mol}$

  The moles of oxygen $\text{=}\left(74.1\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=4}\text{.63}\text{mol}$

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.

  $\text{N=}\frac{1.85\text{mol}}{1.85\text{mol}}\text{=1}$

  $\text{O=}\frac{4.63\text{mol}}{1.85\text{mol}}\text{=2}\text{.5}$

Since the value is not a whole number multiply each by two.  The empirical formula is ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{43}\text{.4\%}\text{Na,}\text{11}\text{.3\%C}\text{\&45}\text{.3\%}\text{O}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 35PE

The empirical formula is ${\text{Na}}_{\text{2}}{\text{CO}}_3\text{.}$

Explanation of Solution

Given,

The percentage composition of sodium is $\text{43}\text{.4\%}\text{.}$

The percentage composition of oxygen is $\text{45}\text{.3\%}\text{.}$

The percentage composition of carbon is $\text{11}\text{.3\%}\text{.}$

The atomic mass of sodium is $22.99\text{g/mol}\text{.}$

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

The atomic mass of carbon is $\text{12}\text{.01}\text{g/mol}\text{.}$

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 sodium $\text{=}\left(43.4\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{22.99\text{g}}\text{=}1.89\text{mol}$

  The moles of oxygen $\text{=}\left(45.3\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=2}\text{.83}\text{mol}$

  The moles of carbon $\text{=}\left(11.3\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{12.01\text{g}}\text{=}0.941\text{mol}$

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.

  $\text{Na=}\frac{1.89\text{mol}}{0.941\text{mol}}\text{=2}\text{.0}$

  $\text{C=}\frac{0.941\text{mol}}{0.941\text{mol}}\text{=1}\text{.0}$

  $\text{O=}\frac{2.83\text{mol}}{0.941\text{mol}}\text{=3}\text{.0}$

The empirical formula is ${\text{Na}}_{\text{2}}{\text{CO}}_3\text{.}$

(e)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{18}\text{.8\%}\text{Na,}29.0\text{\%Cl}\text{\&52}\text{.3\%}\text{O}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 35PE

The empirical formula is ${\text{NaClO}}_4\text{.}$

Explanation of Solution

Given,

The percentage composition of sodium is $\text{18}\text{.8\%}\text{.}$

The percentage composition of oxygen is $\text{53}\text{.3\%}\text{.}$

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

The atomic mass of sodium is $22.99\text{g/mol}\text{.}$

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

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 sodium $\text{=}\left(18.8\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{22.99\text{g}}\text{=}0.818\text{mol}$

  The moles of oxygen $\text{=}\left(52.3\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=3}\text{.27}\text{mol}$

  The moles of chlorine $\text{=}\left(29\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{35.45\text{g}}\text{=}3.27\text{mol}$

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.

  $\text{Na=}\frac{0.818\text{mol}}{0.818\text{mol}}\text{=1}\text{.0}$

  $\text{Cl=}\frac{0.818\text{mol}}{0.818\text{mol}}\text{=1}\text{.0}$

  $\text{O=}\frac{3.27\text{mol}}{0.818\text{mol}}\text{=4}\text{.0}$

The empirical formula is ${\text{NaClO}}_4\text{.}$

(f)

Interpretation Introduction

Interpretation:

The empirical formula the compound with percentage composition of $\text{72}\text{.02\%}\text{Mn}\text{\&27}\text{.98}\text{\%}\text{O}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 35PE

The empirical formula is ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{.}$

Explanation of Solution

Given,

The percentage composition of manganese is $\text{72}\text{.02\%}\text{.}$

The percentage composition of oxygen is $\text{27}\text{.98\%}\text{.}$

The atomic mass of manganese is $54.94\text{g/mol}\text{.}$

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 manganese $\text{=}\left(72.02\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{54.94\text{g}}\text{=}1.311\text{mol}$

  The moles of oxygen $\text{=}\left(27.98\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=1}\text{.749}\text{mol}$

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.

  $\text{Mn=}\frac{1.311\text{mol}}{1.311\text{mol}}\text{=1}\text{.0}$

  $\text{O=}\frac{1.749\text{mol}}{1.311\text{mol}}\text{=1}\text{.334}$

Since the value is not a whole number multiply each by three.  The empirical formula is ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{.}$