Problem 123 - Chemistry Principles and Practice

Question

A compound is analyzed and found to contain \(54.53 \%\) carbon, \(9.15 \%\) hydrogen, and \(36.32 \%\) oxygen by mass. A mass spectrometry experiment shows that the molar mass is \(44 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula? There are two reasonable ways to draw noncyclic skeleton structures of this molecule. Draw the Lewis structure for each, indicating the bond angles and hybridization of each central atom.

Step-by-Step Solution

Answer

The molecular formula is \(\text{C}_2\text{H}_4\text{O}\), with structures: acetaldehyde and ethylene oxide.

Step 1: Convert Percentage to Grams

Since the percentages are given, assume you have a 100 g sample of the compound. This will make the math simpler by turning the percentages directly into grams: 54.53 g of Carbon (C), 9.15 g of Hydrogen (H), and 36.32 g of Oxygen (O).

Step 2: Convert Grams to Moles

Use the atomic masses from the periodic table to convert the grams of each element to moles. Carbon has an atomic mass of approximately 12.01 g/mol, Hydrogen 1.008 g/mol, and Oxygen 16.00 g/mol. - Moles of C = \( \frac{54.53}{12.01} \approx 4.54 \text{ moles} \) - Moles of H = \( \frac{9.15}{1.008} \approx 9.07 \text{ moles} \) - Moles of O = \( \frac{36.32}{16.00} \approx 2.27 \text{ moles} \)

Step 3: Determine Empirical Formula

To find the empirical formula, divide each element's mole quantity by the smallest number of moles calculated in the previous step, which is for oxygen (2.27 moles).- Ratio of C: \( \frac{4.54}{2.27} \approx 2 \) - Ratio of H: \( \frac{9.07}{2.27} \approx 4 \) - Ratio of O: \( \frac{2.27}{2.27} = 1 \) This gives the empirical formula: \( \text{C}_2\text{H}_4\text{O} \).

Step 4: Determine the Molecular Formula

The molar mass of the empirical formula \( \text{C}_2\text{H}_4\text{O} \) is calculated as follows:\( \text{C}: 2 \times 12.01 + \text{H}: 4 \times 1.008 + \text{O}: 1 \times 16.00 = 44.05 \text{ g/mol} \)Since this matches the given molar mass of 44 g/mol, the molecular formula is the same as the empirical formula: \( \text{C}_2\text{H}_4\text{O} \).

Step 5: Draw Lewis Structures

There are two possible non-cyclic structures for \( \text{C}_2\text{H}_4\text{O} \):1. Acetaldehyde (ethanal): O atom is bonded to one of the Cs, the structure is H-C(H)=C=O.- Bond angles: Approx. 120° around the C=C bond (sp\(^2\) hybridization).2. Ethylene oxide: O atom is bonded to two Cs, forming an epoxide ring. - Bond angles: Approx. 60° in the three-membered ring, C-O bonds (sp\(^3\) hybridization on Cs and sp\(^3\) on O).

Key Concepts

Empirical Formula Calculation

The empirical formula of a compound is the simplest integer ratio of the atoms involved. In the exercise, we start by analyzing the elemental composition given in percentages. Imagine you have a sample weighing 100 grams because it simplifies the conversion of percentages directly into grams.
Next, convert these grams to moles. This is done using the atomic masses from the periodic table:

• Carbon (C) has an atomic mass of approximately 12.01 g/mol.
• Hydrogen (H) is about 1.008 g/mol.
• Oxygen (O) is 16.00 g/mol.

Divide the mass of each element in grams by its atomic mass to get moles:- Moles of C: 54.53 / 12.01- Moles of H: 9.15 / 1.008- Moles of O: 36.32 / 16.00
The next step is to find the simplest ratio of these moles. To do this, divide the number of moles of each element by the smallest number of moles obtained. This ratio gives us the empirical formula, in this case, \( \text{C}_2\text{H}_4\text{O} \). Remember, the empirical formula may or may not represent the actual molecule, but it is a step closer to finding the molecular formula.

Mass Spectrometry Analysis

Mass spectrometry is a powerful tool used for identifying the molar mass of a compound, which helps in determining the molecular formula from the empirical formula.
In our example, the mass spectrometer indicated that the molar mass is 44 g/mol. When you calculate the molar mass of the empirical formula \( \text{C}_2\text{H}_4\text{O} \), it also amounts to approximately 44 g/mol:

• Carbon: 2 × 12.01 g/mol = 24.02 g/mol
• Hydrogen: 4 × 1.008 g/mol = 4.032 g/mol
• Oxygen: 16.00 g/mol = 16.00 g/mol

Adding these together confirms that the molar mass aligns perfectly with the empirical formula, meaning the empirical and molecular formulas are identical here. Realize that mass spectrometry not only identifies molar mass but also provides insight into the arrangement and forms of atoms within a molecule.

Lewis Structures

Lewis structures are visual representations of the bonds between atoms in a molecule and the lone pairs of electrons that may exist in the molecule. For the molecule \( \text{C}_2\text{H}_4\text{O} \), two alternate non-cyclic skeleton structures possible are acetaldehyde (ethanal) and ethylene oxide.
For acetaldehyde:

• The Lewis structure involves an aldehyde group, where the oxygen is double-bonded to one carbon. The carbon atom in the aldehyde group has a hybridization of sp\(^2\) with bond angles of approximately 120°.

For ethylene oxide:

• The structure forms a three-membered ring involving two carbons and an oxygen. The ring imparts a significant amount of angle strain, reducing the bond angles to around 60°. Here, the oxygen and carbons are sp\(^3\) hybridized.

Understanding and drawing Lewis structures require acknowledging how electrons are shared between atoms and how molecular geometry affects properties.

Hybridization and Bond Angles

Hybridization is a key concept in molecular chemistry. It explains the mixing of atomic orbitals to form new hybrid orbitals, suited to pair with other atoms in a molecule.
In the exercise, acetaldehyde features carbons involved in sp\(^2\) hybridization which accommodates planar bonding shapes with bond angles close to 120°. These angles are typical due to the trigonal planar arrangement around the central carbon atom connected to the carbonyl group (C=O).
In contrast, the ethylene oxide structure displays a three-membered ring. The carbons and oxygen are sp\(^3\) hybridized, normally ideal for 109.5° angles. However, due to the ring's tension and size, bond angles narrow dramatically to about 60°.
Knowing the type of hybridization helps predict one of the key geometrical properties—bond angles—crucial for visualizing molecular structure and predicting reactivity and physical properties.