Chapter 7, Problem 37PE

(a)

Interpretation Introduction

Interpretation:

The empirical formula of the compound containing $\text{26}\text{.08}\text{Zn,4}\text{.79}\text{g}\text{C}\text{\&19}\text{.14}\text{g}\text{O}$ 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 37PE

The empirical formula of the compound is ${\text{ZnCO}}_{\text{3}}\text{.}$

Explanation of Solution

Given,

The grams of zinc is $\text{26}\text{.08}\text{g}\text{.}$

The grams of carbon is $\text{4}\text{.79}\text{g}\text{.}$

The grams of oxygen is $\text{19}\text{.14}\text{g}\text{.}$

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

The atomic mass of zinc is $65.39\text{g/mol}\text{.}$

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

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

  The moles of carbon $\text{=}\left(4.76\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{\text{12}\text{g}}\text{=0}\text{.399}\text{mol}$

  The moles of Zinc $\text{=}\left(26.08\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{65.39\text{g}}\text{=0}\text{.3988}\text{mol}$

  The moles of oxygen $\text{=}\left(19.14\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{\text{16}\text{g}}\text{=1}\text{.196}\text{mol}$

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.

  $\text{C=}\frac{0.3988\text{mol}}{0.3988\text{mol}}\text{=1}\text{.0}$

  $\text{Zn=}\frac{0.3988\text{mol}}{0.3988\text{mol}}\text{=1}\text{.0}$

  $\text{O=}\frac{1.196\text{mol}}{0.3988\text{mol}}\text{=3}\text{.0}$

The empirical formula of the compound is ${\text{ZnCO}}_{\text{3}}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The empirical formula of the compound containing $\text{57}\text{.66}\text{g}\text{C}\text{\&7}\text{.26}\text{g}\text{H}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 37PE

The empirical formula of the compound is ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}\text{.}$

Explanation of Solution

Given,

The grams of the compound is $150\text{g}\text{.}$

The grams of carbon is $5.76\text{g}\text{.}$

The grams of hydrogen is $7.26\text{g}\text{.}$

The grams of chlorine is $\text{150}\text{g-}\left(\text{57}\text{.66}\text{g+7}\text{.26}\text{g}\right)\text{=8}\text{.51}\text{g}\text{Cl}$

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

The atomic mass of hydrogen is $1.008\text{g/mol}\text{.}$

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 $\text{=}\left(57.66\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{\text{12}\text{g}}\text{=4}\text{.801}\text{mol}$

  The moles of hydrogen $\text{=}\left(7.26\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{1.008\text{g}}\text{=7}\text{.20}\text{mol}$

  The moles of Chlorine $\text{=}\left(85.1\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{35.45\text{g}}\text{=2}\text{.40}\text{mol}$

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.

  $\text{C=}\frac{4.801\text{mol}}{2.40\text{mol}}\text{=2}\text{.0}$

  $\text{Cl=}\frac{2.40\text{mol}}{2.40\text{mol}}\text{=1}\text{.0}$

  $\text{H=}\frac{7.20\text{mol}}{2.40\text{mol}}\text{=3}\text{.0}$

The empirical formula of the compound is ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The empirical formula of the compound containing $\text{42}\text{.0}\text{g}\text{V}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 37PE

The empirical formula of the compound is ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\text{.}$

Explanation of Solution

Given,

The grams of the compound is $75\text{g}\text{.}$

The grams of vanadium is $42.0\text{g}\text{.}$

The grams of oxygen is $\text{75}\text{g-}\text{42}\text{.0}\text{g=33}\text{g}\text{O}$

The atomic mass of vanadium is $50.94\text{g/mol}\text{.}$

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

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

  The moles of vanadium $\text{=}\left(42.0\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{50.94\text{g}}\text{=0}\text{.824}\text{mol}$

  The moles of oxygen $\text{=}\left(33\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=2}\text{.06}\text{mol}$

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.

  $\text{V=}\frac{0.824\text{mol}}{0.824\text{mol}}\text{=1}\text{.0}$

  $\text{O=}\frac{2.06\text{mol}}{2.06\text{mol}}\text{=2}\text{.50}$

Since the value is fractional multiply the value by two.  The empirical formula of the compound is ${\text{V}}_{\text{2}}{\text{O}}_{\text{5}}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The empirical formula of the compound containing $\text{67}\text{.35}\text{g}\text{Ni}\text{,43}\text{.46}\text{g}\text{O}$ and $\text{23}\text{.69}\text{g}\text{P}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 37PE

The empirical formula of the compound is ${\text{Ni}}_{\text{3}}{\text{O}}_{\text{8}}{\text{P}}_{\text{2}}\text{.}$

Explanation of Solution

Given,

The grams of nickel is $\text{67}\text{.35}\text{g}\text{.}$

The grams of oxygen is $\text{43}\text{.46}\text{g}\text{.}$

The grams of phosphorus is $\text{23}\text{.69}\text{g}\text{.}$

The atomic mass of nickel is $58.69\text{g/mol}\text{.}$

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

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

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

  The moles of nickel $\text{=}\left(67.35\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{58.69\text{g}}\text{=1}\text{.148}\text{mol}$

  The moles of oxygen $\text{=}\left(48.96\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{16\text{g}}\text{=3}\text{.0}\text{mol}$

  The moles of phosphorus $\text{=}\left(23.96\text{g}\right)\text{×}\frac{\text{1}\text{mol}}{30.97\text{g}}\text{=0}\text{.7649}\text{mol}$

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.

  $\text{Ni=}\frac{1.148\text{mol}}{0.7649\text{mol}}\text{=1}\text{.5}$

  $\text{O=}\frac{3.060\text{mol}}{0.7649\text{mol}}\text{=4}\text{.0}$

  $\text{P=}\frac{0.7649\text{mol}}{0.7649\text{mol}}\text{=1}\text{.0}$

Since the value is fractional multiply the value by two.  The empirical formula of the compound is ${\text{Ni}}_{\text{3}}{\text{O}}_{\text{8}}{\text{P}}_{\text{2}}\text{.}$