Problem 103 - Chemistry: The Central Science in SI Units, Global Edition

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 \(13.35 \% \mathrm{O}\) by mass, and has a molar mass of \(165.4 \mathrm{~g} / \mathrm{mol} .(\mathbf{a})\) 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 Cl atoms bond to a single \(C\) atom and that there are a \(\mathrm{C}-\mathrm{C}\) bond and two \(\mathrm{C}-\mathrm{O}\) bonds in the compound.

Step-by-Step Solution

Answer

The empirical formula of the compound is \(CH_2Cl_2O\), and since the empirical formula mass is approximately equal to the given molar mass, the molecular formula is the same: \(CH_2Cl_2O\). The Lewis structure for chloral hydrate is: Cl \ C-C=O / Cl

Step 1: Find the mole ratio of the elements

First, we need to convert the mass percentages of each element into moles. We will do this by dividing the mass percentages by their respective atomic masses: C: \(\frac{14.52\%}{12.01~\text{g/mol}} = 1.21~\text{mol}\) H: \(\frac{1.83\%}{1.01~\text{g/mol}} = 1.81~\text{mol}\) Cl: \(\frac{64.3\%}{35.45~\text{g/mol}} = 1.81~\text{mol}\) O: \(\frac{13.35\%}{16.00~\text{g/mol}} = 0.83~\text{mol}\) Now, find the simplest whole number ratio by dividing each mole amount by the smallest value amongst them: C: \(\frac{1.21}{0.83} = 1.46 \approx 1\) H: \(\frac{1.81}{0.83} = 2.18 \approx 2\) Cl: \(\frac{1.81}{0.83} = 2.18 \approx 2\) O: \(\frac{0.83}{0.83} = 1.00 \approx 1\) This gives us the empirical formula: \(C_{1}H_{2}Cl_{2}O_{1}\), or \(CH_2Cl_2O\).

Step 2: Find the molecular formula

Next, we will find the molecular formula by comparing the empirical formula mass with the given molar mass of the compound: Empirical formula mass: \(C + 2H + 2Cl + O = 12.01 + 2(1.01) + 2(35.45) + 16.00 = 165.38~\text{g/mol}\) The given molar mass of the compound is \(165.4~\text{g/mol}\). Since the empirical formula mass and the given molar mass are approximately equal, the molecular formula is the same as the empirical formula: Molecular formula: \(CH_2Cl_2O\)

Step 3: Draw the Lewis structure

We will now draw the Lewis structure of the compound, keeping in mind the bonding information provided: The Cl atoms are bonded to a single C atom; There is a C-C bond between two carbon atoms; There are two C=O bonds. 1. Place the least electronegative element in the center: C 2. Attach the other C atom (C-C bond) and two Cl atoms to the central C. 3. Arrange the two O atoms in a double bond with the second C atom. 4. Distribute the remaining electrons to complete the octets for the surrounding atoms. Cl \ C-C=O / Cl The Lewis structure for chloral hydrate (\(CH_2Cl_2O\)) is now complete.

Key Concepts

Molecular Formula

Understanding the molecular formula of a compound is like unraveling the genetic code of molecules. It tells us the exact number of each type of atom in a molecule. This is different from an empirical formula, which only gives a simplified ratio among the elements.

When deciphering the molecular formula, it's important to first find the empirical formula—the simplest version of the compound that maintains the same relative proportions of elements. In the example of chloral hydrate, the empirical formula matches the molecular formula, indicating that the simplest form is also the true makeup of the compound.

To confirm the molecular formula, we compare the empirical formula mass to the given molar mass. If they match, or are very close, it means the empirical formula is indeed the molecular formula. For chloral hydrate, both calculated and given molar masses were nearly identical, reinforcing that the molecular formula is indeed \(CH_2Cl_2O\).

• Empirical Formula Mass: Calculate from the atomic weights based on the empirical formula.
• Molecular Mass: Provided by experimental data, crucial to verify against calculated empirical mass.

In cases where they differ, you'd need to determine how many times the empirical formula's mass fits into the molecular mass, multiplying subscripts in the empirical formula by this number to obtain the molecular formula.

Lewis Structure

Imagine assembling a puzzle where not just the pieces but how they connect matters. That's akin to drawing Lewis structures, which let us visualize a molecule's geometry, showing the connections between atoms and any lone pairs of electrons. For a molecule like chloral hydrate, this isn't just about numbers; it's about seeing how atoms like carbon (C), hydrogen (H), chlorine (Cl), and oxygen (O) link up.

To create a Lewis structure, remember a few essential steps:

• Least Electronegative Atom: Begin by placing the least electronegative atom in the center, usually carbon.
• Bonding: Connect according to bond rules—in chloral hydrate, Cl atoms attach to one C, both C's connect via a single bond, and O bonds to the second C with double bonds.
• Electron Distribution: Distribute electrons to fulfill octet rules, ensuring all atoms satisfy their valency needs.

For chloral hydrate, this means visualizing two Cl atoms bonded to one C atom, a second C connected by a single C-C bond, and double bonds to O atoms, resulting in a balanced assembly that follows the given connectivity rules. This clarity helps in understanding how molecules might interact in reactions and their potential behavior or state of matter.

Mole Ratio

The concept of mole ratio serves as the backbone of determining chemical formulas. It helps in translating percentage masses into a concrete visualization of each atom's presence in a compound. For chloral hydrate, mastering this concept was key to finding its empirical and molecular formulas.

Achieving the correct mole ratio involves several steps:

• Convert Mass to Moles: Use each element's atomic mass to turn mass percentages into moles. For example, divide the percentage of carbon by its atomic mass (12.01 g/mol) to get moles of carbon.
• Normalize Ratios: Divide all mole values by the smallest mole value obtained among the elements.


• Simplify to Whole Numbers: Adjust ratios to the nearest whole number to simplify into an empirical formula (e.g., approximate C: 1, H: 2, Cl: 2, O: 1).

In this case, the process confirmed the empirical formula as \(CH_2Cl_2O\), showing the relative amounts of each element. Understanding mole ratio helps in creating a bridge between theoretical chemistry and real-world chemical behavior, making it easier to predict how substances will react and form new bonds.