Question

An unknown organic compound is found on elemental analysis to contain \(68.1 \%\) carbon, \(13.7 \%\) hydrogen, and \(18.2 \%\) oxygen by mass. It is slightly soluble in water. Upon careful oxidation it is converted into a compound that behaves chemically like a ketone and contains \(69.7 \%\) carbon, \(11.7 \%\) hydrogen, and \(18.6 \%\) oxygen by mass. Indicate two or more reasonable structures for the unknown.

Short answer

The unknown organic compound with the empirical formula C5H12O is likely soluble in water due to the presence of a polar functional group such as a hydroxyl group (OH). Two reasonable structures for the compound are 1-pentanol (CH3-CH2-CH2-CH2-CH2-OH) and 2-pentanol (CH3-CH2-CH2-CH(CH3)-OH). These structures allow for hydrogen bonding with water, making them soluble, and can be converted into ketones upon oxidation, consistent with the given information about the second compound.

Step-by-step solution

Calculate the empirical formulas of both compounds

To find the empirical formula, we will consider a 100 g sample of each compound. We will use the percentage composition given to calculate the moles of each element in the sample and then find the simplest whole number ratio. For Compound 1: Mass of Carbon = 68.1 g Mass of Hydrogen = 13.7 g Mass of Oxygen = 18.2 g Moles of Carbon = Mass / Molar mass = \( 68.1 g / 12.01 g/mol = 5.67 \mathrm{mol} \) Moles of Hydrogen = \( 13.7 g / 1.01 g/mol = 13.6 \mathrm{mol} \) Moles of Oxygen = \( 18.2 g / 16.0 g/mol = 1.14 \mathrm{mol} \) To find the simplest whole number ratio, we divide the moles of each element by the smallest value. In this case, the smallest value is 1.14. Carbon ratio = \( 5.67 / 1.14 = 5 \) Hydrogen ratio = \( 13.6 / 1.14 = 12 \) Oxygen ratio = \( 1.14 / 1.14 = 1 \) So, the empirical formula for Compound 1: C5H12O Likewise, we follow the same steps for Compound 2. Moles of Carbon = \( 69.7/12.01 = 5.8 \mathrm{mol} \) Moles of Hydrogen = \( 11.7/1.01 = 11.1 \mathrm{mol} \) Moles of Oxygen = \( 18.6/16.0 = 1.16\mathrm{mol} \) Using the smallest value 1.16: Carbon ratio = \( 5.8 / 1.16 = 5 \) Hydrogen ratio = \( 11.1 / 1.16 = 9.5 \) (approximately 10 for simplicity) Oxygen ratio = \( 1.16/ 1.16 = 1\) So, the empirical formula for Compound 2: C5H10O

Generate reasonable structures for the unknown compound

Now, we need to propose structures for the first compound that make it soluble in water. We know that the first compound has the empirical formula C5H12O. We need to create a compound containing 5 carbon atoms, 12 hydrogen atoms, and 1 oxygen atom that has the potential to be soluble in water. In order to be soluble in water, the compound must have a polar functional group such as a hydroxyl group (OH). Here are two reasonable possibilities: 1. 1-pentanol (CH3-CH2-CH2-CH2-CH2-OH) 2. 2-pentanol (CH3-CH2-CH2-CH(CH3)-OH) These structures provide a polar functional group (hydroxyl group) that can form hydrogen bonds with water, making them soluble in water. Furthermore, upon oxidation, these alcohols can be converted into ketones, which is consistent with the information given in the problem that the second compound behaves like a ketone. In conclusion, two reasonable structures for the unknown organic compound with the empirical formula C5H12O are 1-pentanol and 2-pentanol.

Key concepts

Understanding Percentage Composition in Chemistry

Percentage composition is a critical concept in chemistry that represents the percent by mass of each element within a compound. To understand percentage composition, imagine having a 100-gram sample of a substance; the percentage then directly corresponds to the mass of each present element. For example, if a compound has a carbon content of 68.1%, this means that there are 68.1 grams of carbon in a 100-gram sample of the compound.

Calculating the percentage composition is straightforward: we divide the mass of the specific element by the total mass of the compound, and then multiply by 100. This calculation is valuable for determining the empirical formula of a compound, which speaks to the simplest whole-number ratio of atoms within it. The empirical formula is an essential stepping stone to deducing molecular structures and understanding the stoichiometry of reactions.

Mole Calculations: The Foundation of Chemical Quantities

In chemistry, the concept of the mole is used as a fundamental counting unit. It represents Avogadro's number (\(6.022 \times 10^{23}\) entities) of atoms, molecules, ions, or other particles. When we discuss mole calculations, we're dealing with the conversion between the mass of a substance and the amount of substance in moles. To perform these calculations, we use the molar mass of an element, which is the mass of one mole of that element.

As seen in the exercise, by dividing the mass of each element by its molar mass, we determine the number of moles. This calculation allows us to find the empirical formula. One must be careful to always express the amounts in whole number ratios, simplifying if necessary. This whole number ratio gives us insights into the ratios of atoms as they actually occur in a substance, guiding us towards an understanding of the compound's formula and its reactivity.

Structural Isomers: Same Molecular Formula, Different Arrangement

Structural isomers are molecules that have the same molecular formula but differ in the arrangement of their atoms. This difference in arrangement can lead to compounds with vastly different chemical and physical properties. For the initial compound in our exercise with the empirical formula C5H12O, there could be numerous isomers since the carbons can be arranged in various chain lengths, and the functional groups (such as hydroxyl groups) can be located at different positions.

For example, the difference between 1-pentanol and 2-pentanol lies in the position of the hydroxyl group. In 1-pentanol, the OH group is attached to the first carbon in the chain, while in 2-pentanol, it is attached to the second carbon. These subtleties can significantly influence solubility, boiling point, and reactivity. Recognizing structural isomers is important in organic chemistry because it opens a world of possibilities for the same empirical formula, serving as a fascinating example of chemistry's complexity and diversity.