Question

Maleic acid is prepared by the catalytic oxidation of benzene. It is a dicarboxylic acid; that is, it has two carboxylic acid groups. (a) Combustion of 0.125 g of the acid gives \(0.190 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.0388 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Calculate the empirical formula of the acid. (b) A 0.261-g sample of the acid requires \(34.60 \mathrm{mL}\) of \(0.130 \mathrm{M} \mathrm{NaOH}\) for complete titration (so that the H ions from both carboxylic acid groups are used). What is the molecular formula of the acid? (c) Draw a Lewis structure for the acid. (d) Describe the hybridization used by the C atoms (e) What are the bond angles around each C atom?

Short answer

(a) Empirical formula is CHO. (b) Molecular formula is C4H4O4. (c) Lewis structure shows double bonds and carboxylic groups. (d) Carbonds are sp2 hybridized. (e) Bond angles are ~120°.

Step-by-step solution

Calculate the Moles of CO2 and H2O

Normalize the mass of products to find moles. Convert the mass of \( \mathrm{CO}_2 \) and \( \mathrm{H}_2 \mathrm{O} \) to moles: \[ \text{Moles of } \mathrm{CO}_2 = \frac{0.190}{44.01} = 0.00432 \text{ mol} \]\[ \text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{0.0388}{18.015} = 0.00215 \text{ mol} \]

Determine Moles of Carbon and Hydrogen

From the moles of \( \mathrm{CO}_2 \), find moles of Carbon and from the moles of \( \mathrm{H}_2 \mathrm{O} \), find moles of Hydrogen:\[ \text{Moles of C} = 0.00432 \text{ mol} \]\[ \text{Moles of H} = 2 \times 0.00215 = 0.00430 \text{ mol} \]

Calculate Mass of Oxygen in the Acid

Find the mass of carbon and hydrogen in the sample:\[ \text{Mass of C} = 0.00432 \times 12.01 = 0.05187 \text{ g} \]\[ \text{Mass of H} = 0.00430 \times 1.008 = 0.00433 \text{ g} \]The mass of oxygen is:\[ \text{Mass of O} = 0.125 - 0.05187 - 0.00433 = 0.0688 \text{ g} \]

Calculate Moles of Oxygen

Convert the mass of oxygen to moles:\[ \text{Moles of O} = \frac{0.0688}{16.00} = 0.00430 \text{ mol} \]

Determine Empirical Formula

Find the ratio of moles:- C: 0.00432 mol- H: 0.00430 mol- O: 0.00430 molSince all are approximately equal, the empirical formula is \( \mathrm{CHO} \).

Find Moles of NaOH Used

Using the volume and molarity of the \( \mathrm{NaOH} \) solution:\[ \text{Moles of } \mathrm{NaOH} = 0.03460 \times 0.130 = 0.004498 \text{ mol} \]

Determine Molecular Formula

The acid has two carboxylic groups, so each mole of acid reacts with 2 moles of \( \mathrm{NaOH} \). Therefore, 0.002249 moles of the acid are present in 0.261 g of sample.The molar mass \( M \) of the acid is:\[ M = \frac{0.261}{0.002249} = 116.0 \text{ g/mol} \]Since the empirical formula \( \mathrm{CHO} \) has a molar mass of about 29 g/mol, the molecular formula is \( \mathrm{C_4H_4O_4} \).

Draw the Lewis Structure

The Lewis structure of maleic acid, \( \mathrm{C_4H_4O_4} \), shows four carbon atoms in a chain, two of them with carboxylic acid groups attached: [O=C-C=C-C=O]-OH with an OH group and a =O double bonded to the second and fourth carbon.

Hybridization of Carbon Atoms

In maleic acid, the four carbons exhibit \( \mathrm{sp^2} \) hybridization due to their double bonds.

Determine Bond Angles

In \( \mathrm{sp^2} \) hybridization, the bond angles around each carbon are approximately 120 degrees.

Key concepts

Empirical Formula

An empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. To determine the empirical formula, you start by converting the amount of each element present into moles. This involves using the mass of the element and its molar mass — for example, for carbon, the molar mass is 12.01 g/mol.
In the given exercise, maleic acid's empirical determination was based on the combustion products, which are carbon dioxide (\( \mathrm{CO}_2 \)) and water (\( \mathrm{H}_2 \mathrm{O} \)). By analyzing these products, the moles of carbon and hydrogen were calculated and further used to find the moles of oxygen remaining in the compound.
The ratios calculated were found to be approximately equal, leading to the empirical formula \( \mathrm{CHO} \). This means that in the simplest form of this compound, carbon, hydrogen, and oxygen are present in a 1:1:1 ratio.

Molecular Formula

The molecular formula provides the actual number of atoms of each element in a molecule of the compound. Calculating this involves determining the compound’s molar mass and comparing it to the molar mass of its empirical formula.
In this exercise, the molecular formula of maleic acid was found using a titration method. The amount of NaOH used in the titration revealed the number of moles of the acid. Comparing this to the mass involved gave a molar mass of 116.0 g/mol.
Since the empirical formula \( \mathrm{CHO} \) has a molar mass of 29 g/mol, the molecular formula was determined to be \( \mathrm{C_4H_4O_4} \). This means that the actual molecule is four times the size of the simplest \( \mathrm{CHO} \) unit.

Hybridization

Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds. In organic molecules, this concept helps understand shapes and bond angles.
In maleic acid, hybridization of the carbon atoms is crucial. Each carbon involved with a double bond is \( \mathrm{sp^2} \) hybridized, meaning one s and two p orbitals mix together. This generally results in a trigonal planar structure around these carbons, allowing for the formation of double bonds.
The remaining orbitals not involved in hybridization represent the double bond functionalities. This allows for overlap between p orbitals of adjacent atoms, forming \( \pi \) bonds — essential in a molecule like maleic acid.

Bond Angles

Bond angles in a molecule are affected by the types of hybridization and the shapes adopted by atoms. In a system like maleic acid with \( \mathrm{sp^2} \) hybridization, the geometry around each carbon atom typically leads to bond angles of about 120 degrees.
These angles correspond with the trigonal planar structure of the molecule's double-bonded carbon atoms. Since the p orbitals are perpendicular to the plane of the molecule, the distinct arrangement around each carbon optimizes bond overlap and molecular stability.
Understanding these angles is crucial as it helps visualize the spatial arrangement of atoms, which has implications for reactivity and interaction with other molecules.