Question

The compound chloral hydrate, known in detective stories as knockout drops, is composed of \(14.52 \% \mathrm{C}\), \(1.83 \% \mathrm{H}, 64.30 \% \mathrm{Cl}\), and \(19.35 \%\) O by mass and has a molarmass of \(165.4 \mathrm{~g} / \mathrm{mol} .\) (a) What is the empirical formula of this substance? (b) What is the molecular formula of this substance? (c) Draw the Lewis structure of the molecule, assuming that the \(\mathrm{Cl}\) atoms bond to a single \(C\) atom and that there are a \(C-C\) bond and two C \(-\) O bonds in the compound.

Short answer

The empirical formula of the compound is \(\mathrm{CH_2Cl_2O}\), and the molecular formula is \(\mathrm{C_2H_4Cl_4O_2}\). The Lewis structure has a single C-C bond, four Cl atoms bonded to one of the carbon atoms, and two C-O bonds connected to the other carbon atom. The structure has complete octets, with the chlorine atoms forming single bonds to the carbon, and the oxygen atoms having a single bond to the carbon and one lone pair each.

Step-by-step solution

Calculate moles of each element

First, we will assume that we have 100 grams of this compound, which will simplify calculations. Then, we will convert the mass percentages to moles using the molar masses of each element. For \(\mathrm{C}\): \(\frac{14.52 \mathrm{~g}}{12.01 \mathrm{~g/mol}} = 1.21 \mathrm{~moles}\) For \(\mathrm{H}\): \(\frac{1.83 \mathrm{~g}}{1.01 \mathrm{~g~mol^{-1}}} = 1.81 \mathrm{~moles}\) For \(\mathrm{Cl}\): \(\frac{64.30 \mathrm{~g}}{35.45 \mathrm{~g/mol}} = 1.81 \mathrm{~moles}\) For \(\mathrm{O}\): \(\frac{19.35 \mathrm{~g}}{16.00 \mathrm{~g/mol}} = 1.21 \mathrm{~moles}\)

Determine the empirical formula

To get the empirical formula, we have to divide each number of moles by the smallest number of moles, which in our case is 1.21. For \(\mathrm{C}\), \(\mathrm{H}\), \(\mathrm{Cl}\), and \(\mathrm{O}\), the ratios are approximately: C: \(\frac{1.21}{1.21} = 1\) H: \(\frac{1.81}{1.21} = 1.5 \approx 2\) Cl: \(\frac{1.81}{1.21} = 1.5 \approx 2\) O: \(\frac{1.21}{1.21} = 1\) Thus, the empirical formula is \(\mathrm{CH_2Cl_2O}\).

Determine the molecular formula

Now, we calculate the molar mass of the empirical formula: \(1(12.01) + 2(1.01) + 2(35.45) + 1(16.00) = 98.93 \mathrm{~g/mol}\) Next, we will find the ratio between the molecular molar mass and the empirical molar mass: \(\frac{165.4}{98.93} \approx 1.67 \approx 2\) Since this ratio is approximately 2, the molecular formula is twice the empirical formula: \(\mathrm{C_2H_4Cl_4O_2}\)

Draw the Lewis structure

Lastly, we need to draw the Lewis structure using the information provided about the bonding. First, we can calculate the total number of valence electrons in the molecule: \(2(4) + 4(1) + 4(7) + 2(6) = 40\) The Lewis structure should have: - A C-C bond - 4 Cl atoms bonded to a single C atom - 2 C-O bonds Let's arrange the atoms as mentioned: - Place two carbon atoms next to each other (C1 and C2) - Bond the four chlorine atoms to C1 - Bond the two oxygen atoms to C2 Now, complete the octets for all atoms by adding lone pairs: - Cl atoms already have complete octets. - Add one lone pair to the two oxygen atoms. - Add two lone pairs to C2 atom. Thus, we have obtained the Lewis structure for the molecule \(\mathrm{C_2H_4Cl_4O_2}\).

Key concepts

Chemical Composition Analysis

Understanding the chemical composition of substances is fundamental in chemistry. Chemical composition analysis involves determining the percentage by mass of each element within a compound. It's like a recipe that describes what and how much of each ingredient is in a mixture. In the exercise, the percentages of carbon (C), hydrogen (H), chlorine (Cl), and oxygen (O) in chloral hydrate were given. To calculate the empirical formula, which is the simplest whole-number ratio of atoms in a compound, percentages were converted to moles since moles provide a consistent method to compare amounts of substances.

To perform the conversion, you divide the mass percentage of each element by its respective molar mass. This step ensures that you're working with the right stoichiometric proportions. With the mole quantities calculated, the next step is to derive the ratios to find the smallest whole number ratio of the atoms, thus leading to the empirical formula. This aspect of chemical composition analysis is crucial for students to grasp as it provides the building blocks for further calculations and understanding of a compound's structure.

Stoichiometry

Stoichiometry is akin to mathematical relationships within a recipe; it's the part of chemistry that deals with the quantitative aspects of chemical reactions. In the context of the empirical and molecular formula determination, stoichiometry helps us to relate the moles of each element within a compound to its molar mass.

By assuming a sample mass of 100 grams in our exercise, the stoichiometric calculations become more straightforward. After finding the moles of each element, the smallest number of moles is used to normalize the mole ratios. This normalized ratio translates directly into the subscripts for the empirical formula. However, stoichiometry doesn't stop there; the molar mass of the empirical formula is then compared to the molar mass of the compound to find the molecular formula, further illustrating the stoichiometric ratios at play. This compound's molar mass, which must account for all the atoms in the molecule, led us to discover that the empirical formula needed to be multiplied by approximately two to achieve the molecular formula.

Lewis Structures

Lewis structures are diagrams that show the distribution of valence electrons among the atoms in a molecule. They're like blueprints for molecules, illustrating how atoms are bonded together and how electrons are distributed. In the exercise, the Lewis structure for chloral hydrate needed to incorporate certain bonding requirements, such as a C-C bond and two C-O bonds.

To sketch the Lewis structure, it is essential to count the total number of valence electrons that need to be distributed. These electrons come from the valence shell of each atom and are involved in forming bonds or as non-bonding pairs. In chloral hydrate, with all the valence electrons accounted for, we arrange the atoms accordingly, ensuring each chlorine and oxygen atom achieves its octet — a stable arrangement of eight electrons typically associated with the noble gas configuration. It's like solving a puzzle where the pieces are the atoms and electrons, and the picture you're completing is the molecule itself. This visual representation helps to predict the molecular geometry as well as the reactivity and physical properties of the molecule.