Chapter 23, Problem 121SCQ

(a)

Interpretation Introduction

Interpretation:

The empirical formula of the given acid has to be determined.

Concept introduction:

Maleic acid is prepared by the catalytic oxidation of benzene.It is a dicarboxylic acid ,that is, it has two carboxylic acid groups.

It is cis– isomer of butenedioic acid.Molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$.

The structure of maleic acid is as shown below.

Empirical formula of a compound is the smallest integer ratio of numbers of each element presented in that compound.

Molecular formula of a compound is integer multiple of empirical formula, the integer is depend upon the mass of empirical formula and the molecular mass of the compound.

The number of moles of any substance can be determined using the equation,                     

              $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Answer to Problem 121SCQ

Empirical formula of the acid is $\text{CHO}$

Explanation of Solution

Assuming the molecular formula of the compound is ${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}{\text{O}}_{\text{z}}$.

It dissociates into carbondioxide and water. The equation is as follows.

${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}{\text{O}}_{\text{z}}\to {\text{CO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$

Let’s calculate the moles of C in ${\text{CO}}_{\text{2}}$:

$\begin{array}{c}\text{Moles of C =}\frac{{\text{Given weight of CO}}_{\text{2}}}{\begin{array}{l}{\text{Molecular weight of CO}}_{\text{2}}\\ \text{= }\frac{\text{1×0}\text{.190}}{\text{44}}\end{array}}\\ \text{= 4}{\text{.32×10}}^{\text{-3}}\end{array}$

$\begin{array}{c}\text{Weigh of Carbon = 4}{\text{.32 × 10}}^{\text{-3}}\text{ × 12}\\ \text{= 0}\text{.0518 g}\end{array}$

Let’s calculate the moles of $\text{H}$ in ${\text{H}}_{\text{2}}\text{O}$.

$\begin{array}{c}\text{Moles of H =}\frac{{\text{Given weight of H}}_{\text{2}}\text{O}}{\begin{array}{l}{\text{Molecular weight of H}}_{\text{2}}\text{O}\\ \text{= }\frac{\text{2×0}\text{.0388}}{\text{18}}\end{array}}\\ \text{= 4}{\text{.31×10}}^{\text{-3}}\text{ mol}\end{array}$

Weight of Hydrogen = $\text{1 × 4}{\text{.31×10}}^{\text{-3}}\text{g}$

Let’s calculate the weight of Oxygen:

$\begin{array}{c}\text{Weight of Oxygen =Weight of acid - (Weight of C + Weight of H)}\\ \text{= 0}\text{.125 - (0}\text{.0518 + 4}{\text{.3×10}}^{\text{-3}}\text{)}\\ \text{= 0}\text{.0689 g}\end{array}$

Let’s calculate the mole of Oxygen:

$\begin{array}{c}\text{Moles of }\text{}\text{O =}\frac{\text{0}\text{.0689}}{\text{16}}\\ \text{= 4}{\text{.3×10}}^{\text{-3}}\text{ mol}\end{array}$

Let’s calculate the mole ratio of each element:

$\begin{array}{l}\text{Mole of C:H:O = 4}{\text{.3×10}}^{\text{-3}}\text{ : 4}{\text{.3×10}}^{\text{-3}}\text{ : 4}{\text{.3×10}}^{\text{-3}}\\ \text{ = 1:1:1}\end{array}$

Therefore, the empirical formula of the given compound is ${\text{C}}_{\text{1}}{\text{H}}_{\text{1}}{\text{O}}_{\text{1}}\approx \text{ CHO}$.

(b)

Interpretation Introduction

Interpretation:

The molecular formula of the given acid has to be determined.

Concept introduction:

Maleic acid is prepared by the catalytic oxidation of benzene.It is a dicarboxylic acid ,that is, it has two carboxylic acid groups.

It is cis– isomer of butenedioic acid.Molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$.

The structure of maleic acid is as shown below.

Empirical formula of a compound is the smallest integer ratio of numbers of each element presented in that compound.

Molecular formula of a compound is integer multiple of empirical formula, the integer is depend upon the mass of empirical formula and the molecular mass of the compound.

The number of moles of any substance can be determined using the equation,                    

              $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Answer to Problem 121SCQ

Molecular formula of the acid is ${\text{C}}_4{\text{H}}_4{\text{O}}_{\text{4}}$

Explanation of Solution

According to the law of gram equivalents, equivalent acid is equal to the equivalents of base.

$\begin{array}{c}\text{Equivalent of acid = Equivalent of base}\\ \frac{\text{0}\text{.261}}{\left(\frac{\text{Molecular weight}}{\text{2}}\right)}\text{= 34}{\text{.6×10}}^{\text{-3}}\text{ × 0}\text{.13}\\ \text{Molecular weight = 116}\text{.05 g/mol}\end{array}$

Let’s calculate the molecular formula:

$\begin{array}{c}\text{n = }\frac{\text{Molecular weigth}}{\text{Empirical formula weight}}\\ \text{=}\frac{\text{116}\text{.05}}{\text{29}}\\ \text{= 4}\end{array}$

Substitute the ‘n’ value we get molecular formula of acid.

${\text{C}}_{\text{x}}{\text{H}}_{\text{y}}{\text{O}}_{\text{z}}\simeq {\text{ C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of the given acid has to be drawn.

Concept introduction:

Maleic acid is prepared by the catalytic oxidation of benzene.It is a dicarboxylic acid ,that is, it has two carboxylic acid groups.

It is cis– isomer of butenedioic acid.Molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$.

The structure of maleic acid is as shown below.

Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.

It is also known as Lewis dot structures which represent the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.

Explanation of Solution

The Lewis structure indicates the all unpaired electrons present in the atom in the molecules.

The Lewis structure of Maleic is as follows.

(d)

Interpretation Introduction

Interpretation:

The hybridization used by the carbon atom in the given acid compound has to be described.

Concept introduction:

Maleic acid is prepared by the catalytic oxidation of benzene.It is a dicarboxylic acid ,that is, it has two carboxylic acid groups.

It is cis– isomer of butenedioic acid.Molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$.

The structure of maleic acid is as shown below.

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

The Lewis structure indicates the all unpaired electrons present in the atom in the molecules.

The Lewis structure of Maleic is as follows.

The geometry around the entire carbon atoms in the molecule is trigonal planar, which means these carbon atoms used $s{p}^2$ hybridized orbitals for the formation of the compound.

(e)

Interpretation Introduction

Interpretation:

The bond angles around each C-atom in the given molecule has to be identified.

Concept introduction:

Maleic acid is prepared by the catalytic oxidation of benzene.It is a dicarboxylic acid ,that is, it has two carboxylic acid groups.

It is cis– isomer of butenedioic acid.Molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_{\text{4}}$.

The structure of maleic acid is as shown below.

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& Geometry& Bondangle\\ A{X}_2& sp& Linear& {180}^{°}\\ A{X}_3,A{X}_{2}B& s{p}^2& Trigonalplanar& {120}^{°}\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& Tetrahedral& {109.5}^{°}\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& Trigonalbipyramidal& {120}^{°},{90}^{°}\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& Octahedral& {90}^{°}\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

The Lewis structure of Maleic is as follows.

The geometry around the entire carbon atoms in the molecule is trigonal planar, which means these carbon atoms used $s{p}^2$ hybridized orbitals for the formation of the compound.

The geometry around each carbon atom in the given compound is trigonal planar and so the bond angle is ${120}^o$