Question

An organic compound is analyzed and found to contain \(66.7 \%\) carbon, \(11.2 \%\) hydrogen, and \(22.1 \%\) oxygen by mass. The compound boils at \(79.6^{\circ} \mathrm{C}\). At \(100^{\circ} \mathrm{C}\) and \(0.970 \mathrm{~atm}\), the vapor has a density of \(2.28 \mathrm{~g} / \mathrm{L}\). The compound has a carbonyl group and cannot be oxidized to a carboxylic acid. Suggest a structure for the compound.

Short answer

The structure of the compound is most likely propanone (CH3-CO-CH2-CH3), also commonly known as acetone, based on the calculated molecular formula of C3H6O, the presence of a carbonyl group as a ketone and the condition that it cannot be oxidized to a carboxylic acid.

Step-by-step solution

Find the Empirical Formula

Using the mass percentages, imagine having 100 g of the compound. Then the amount (in g) of each element would be: - Carbon: 66.7 g - Hydrogen: 11.2 g - Oxygen: 22.1 g Now, calculate the amount (in moles) of each element: - Carbon: \(\frac{66.7}{12.01}\) mol - Hydrogen: \(\frac{11.2}{1.008}\) mol - Oxygen: \(\frac{22.1}{16.00}\) mol Then, divide each mole value by the smallest one to find the mole ratio of each element in the compound: - Carbon: \(\frac{66.7/12.01}{22.1/16.00}\) \(\approx 3\) - Hydrogen: \(\frac{11.2/1.008}{22.1/16.00}\) \(\approx 6\) - Oxygen: \(\frac{22.1/16.00}{22.1/16.00}\) \(\approx 1\) Hence, the empirical formula is C3H6O.

Determine the Molecular Formula

To find the molecular formula, we need to first calculate the molar mass of the empirical formula, which is: Molar mass of C3H6O = 3(12.01) + 6(1.008) + 16.00 = 58.08 g/mol Next, use the equation for finding the molar mass of a gas using the density, temperature, and pressure information: Molar mass = \(\frac{density \times R \times T}{P}\) Where R is the ideal gas constant (0.0821 L.atm/mol.K), T is the temperature in Kelvin (100 + 273.15 = 373.15 K), and P is the pressure in atm (0.970 atm). Molar mass = \(\frac{2.28 \times 0.0821 \times 373.15}{0.970}\) = 58.00 g/mol Since the molar mass of the molecular formula is nearly equal to the molar mass of the empirical formula, the molecular formula is also C3H6O.

Identify Functional Groups

The given information states that the compound has a carbonyl group and cannot be oxidized to a carboxylic acid. This indicates that the carbonyl group is not present as an aldehyde (which can be oxidized to carboxylic acids); hence, it is present as a ketone.

Propose the Structure

The molecular formula obtained is C3H6O. Given the presence of a carbonyl group as a ketone, the structure of the organic compound can be proposed as: CH3-CO-CH2-CH3 Hence, the structure of the compound is most likely propanone, or commonly known as acetone.

Key concepts

Empirical Formula Calculation

Understanding the empirical formula of a compound is the foundation of organic chemistry. An empirical formula represents the simplest whole-number ratio of the elements within a compound. To calculate the empirical formula, you must convert the mass percentages of each component to moles by dividing them by the respective atomic or molar masses. After that, to arrive at the simplest ratio, you should divide each mole value by the smallest of the calculated mole values.

In the given exercise, after converting the mass percentages to moles for carbon, hydrogen, and oxygen, and then finding the simplest whole-number mole ratio, the empirical formula is determined to be C3H6O. This tells us that for every three carbon atoms, there are six hydrogen atoms and one oxygen atom in the simplest form of the compound.

Molecular Formula Determination

Once the empirical formula is identified, the next step is determining the molecular formula, which represents the actual number of atoms of each element in a molecule. To do this, compare the empirical formula's molar mass to the actual molar mass of the compound. This comparison can reveal whether the molecular formula is the same as the empirical formula or a multiple of it.

The exercise shows that the molar mass calculated from the compound's density, temperature, and pressure closely matches that of the empirical formula. Hence, no multiplication is required, and the molecular formula is concluded to be the same as the empirical formula: C3H6O. It's important to understand that while the empirical formula is useful, the molecular formula gives a complete picture of the compound's actual makeup.

Functional Groups in Organic Chemistry

Functional groups are specific groups of atoms within molecules that determine the characteristic chemical reactions of those molecules. They are critical for identifying the structure and behavior of organic compounds. In the exercise, the presence of a carbonyl group, which consists of a carbon atom double-bonded to an oxygen atom, is identified, but the information provided rules out the possibility of it being an aldehyde.

Recognizing that the compound cannot be oxidized to a carboxylic acid points to the carbonyl group being part of a ketone rather than an aldehyde. This functional group identification is a vital clue for proposing the correct structure of the organic compound.

Organic Compound Boiling Point

The boiling point of an organic compound is not only important for practical laboratory work but also provides insights into the compound's structure. It is influenced by factors such as molecular size, the shape of the molecule, and the type of intermolecular forces present. Compounds with stronger intermolecular forces generally have higher boiling points.

In our exercise, though, the boiling point alone does not give us a definitive structure; it does suggest a relatively small and less complex molecule due to the relatively low boiling point of 79.6°C. These hints, combined with the functional group information, lead us to conclude that the compound is likely to be a small ketone, such as propanone (acetone), which aligns with the physical properties and structure suggested in the solution.