Chapter 2, Problem 129GQ

Empirical and molecular formulas.

(a) Fluorocarbonyl hypofluorite is composed of 14.6% C, 30.0% O, and 46.3% F. The molar mass of the compound is 82 g/mol. Determine the empirical and molecular formulas of the compound,

(b) Azulene, a beautiful blue hydrocarbon, is 93.71% C and has a molar mass of 128.16 g/mol. What are the empirical and molecular formulas of azulene?

Interpretation Introduction

Interpretation: The empirical and molecular formulas for the given fluorocarbonyl hypofluorite should be determined.

Concept introduction:

• Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
• Equation for finding Molecular formula from the empirical formula,

    $\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula mass}}\times \text{ Empirical formula}$

• Equation for number moles from mass and molar mass,

  $Numberofmoles=\frac{Massingrams}{Molarmass}$

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

Answer to Problem 129GQ

Empirical formula and the molecular formula of the given compound is ${\text{CF}}_{\text{2}}{\text{O}}_{\text{2}}$

Explanation of Solution

Given: Mass percent of carbon, oxygen and Florine in the given compound fluorocarbonyl hypofluorite are $14.6\%,39.0\%and46.3\%$ respectively.

Mass percent of an element means, 100g of compound contains that mass percent of an element in grams. Therefore, mass of  $14.6g$ carbon $39.0g$ of oxygen and $\text{46}\text{.3 }g$ of F are present respectively in 100g of fluorocarbonyl hypofluorite.

Equation for number moles from mass and molar mass is,

  $Numberofmoles=\frac{Massingrams}{Molarmass}$

Therefore, the number of moles of C is,

    $Numberofmoles=\frac{14.6g}{12.01g/mol}=1.215mol$

The number of moles of oxygen is,

    $Numberofmoles=\frac{39g}{16g/mol}=2.437mol$

The number of moles of F is,

    $Numberofmoles=\frac{46.3g}{19g/mol}=2.437mol$

So, the mole ratio between elements in fluorocarbonyl hypofluorite is,

    $C:F:O=1.215:2.437:2.437$

Dividing the every element’s number of moles by the smallest number of mole.

    $\begin{array}{l}\text{C:F:O}\text{=}\frac{\text{1}\text{.215}}{\text{1}\text{.215}}\text{:}\frac{\text{2}\text{.437}}{\text{1}\text{.215}}\text{:}\frac{\text{2}\text{.437}}{\text{1}\text{.215}}\\ \text{=}\text{1}\text{:}\text{2:2}\\ \text{=}\text{1}\text{C}\text{:}\text{2F:}\text{2O}\end{array}$

Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.

Therefore empirical formula of the given compound is ${\text{CF}}_{\text{2}}{\text{O}}_{\text{2}}$.

Equation for finding Molecular formula from the empirical formula,

    $\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula mass}}\times \text{ Empirical formula}$

Empirical formula mass and molar mass of the compound is $82g$ substituting this in the above equation,

    $\begin{array}{l}\text{Molecular}\text{formula}\text{of}\text{the}\text{compound}\text{=}\frac{\text{82}}{\text{82}}{\text{×CF}}_{\text{2}}{\text{O}}_{\text{2}}\\ \\ \text{=}{\text{CF}}_{\text{2}}{\text{O}}_{\text{2}}\end{array}$

Interpretation Introduction

Interpretation: The empirical and molecular formulas for the given azulene should be determined.

Concept introduction:

• Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
• Equation for finding Molecular formula from the empirical formula,

    $\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula mass}}\times \text{ Empirical formula}$

• Equation for number moles from mass and molar mass,

  $Numberofmoles=\frac{Massingrams}{Molarmass}$

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

Answer to Problem 129GQ

Empirical formula is $C_5{H}_4$ and the molecular formula of the compound is $C_{10}{H}_8$

Explanation of Solution

Given

Mass percent of carbon in the given compound azulene is $93.71\%$ the remaining part is the percent of hydrogen in the compound and it is $6.29\%$

Mass percent of an element means, 100g of compound contains that mass percent of an element in grams. Therefore, mass of  $93.71g$ carbon and $6.29g$ of H are present respectively in 100g of the given compound azulene.

Equation for number moles from mass and molar mass is,

  $Numberofmoles=\frac{Massingrams}{Molarmass}$

Therefore, the number of moles of C is,

    $Numberofmoles=\frac{93.71g}{12.01g/mol}=7.802mol$

The number of moles of hydrogen is,

    $Numberofmoles=\frac{6.29g}{1g/mol}=6.29mol$

So, the mole ratio between elements in the given compound azulene is,

    $C:H=7.802:6.29$

Dividing the every element’s number of moles by the smallest number of mole.

        $\begin{array}{l}\text{C:}\text{H}\text{=}\frac{7.802}{6.29}:\frac{6.29}{6.29}\\ =\text{1}\text{.25:}1\\ \end{array}$

Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.

To get the whole number ratio, multiplying the above ratio with 4.

$\begin{array}{l}\text{C:H}\text{=}\text{4×}\left(\text{1}\text{.25}\text{:}\text{1}\right)\\ \text{=}\text{5:4}\\ \text{=}\text{5C:}\text{4H}\end{array}$

Therefore empirical formula of the given compound is ${\text{C}}_{\text{5}}{\text{H}}_{\text{4}}$.

Equation for finding Molecular formula from the empirical formula,

    $\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula mass}}\times \text{ Empirical formula}$

Empirical formula mass of the compound is $64.05g$ and molar mass of the compound is $128.16g/mol$ substituting this in the above equation,

    $\begin{array}{l}\text{Molecular}\text{formula}\text{of}\text{the}\text{compound}\text{=}\frac{\text{64}\text{.05}}{\text{128}\text{.16}}{\text{×C}}_{\text{5}}{\text{H}}_{\text{4}}\\ \\ \text{=}{\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\end{array}$