Problem 102 - Chemistry: The Central Science

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 molar mass 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 \(\mathrm{C}-\mathrm{O}\) bonds in the compound.

Step-by-Step Solution

Answer

The molecular formula of the compound chloral hydrate is C₂H₃Cl₃O₂. The Lewis structure, based on the given assumptions, can be represented as: Cl | Cl - C - Cl | C - O | O | H

Step 1: Convert mass percentages to moles

We'll use the given mass percentage of each element and assume we have 100 grams of the compound. This means that we have: - \(14.52 \ grams\) of carbon (C) - \(1.83 \ grams\) of hydrogen (H) - \(64.3 \ grams\) of chlorine (Cl) - \(19.35 \ grams\) of oxygen (O) Now, we'll convert these masses into moles using the molar mass of each element: - Carbon: \(14.52 \mathrm{\: g} \times \frac{1 \: \mathrm{mol}}{12.01\mathrm{\:g}} = 1.21 \: \mathrm{mol}\) - Hydrogen: \(1.83 \mathrm{\: g} \times \frac{1 \: \mathrm{mol}}{1.01\mathrm{\:g}} = 1.81 \: \mathrm{mol}\) - Chlorine: \(64.3 \mathrm{\: g} \times \frac{1 \: \mathrm{mol}}{35.45\mathrm{\:g}} = 1.81 \: \mathrm{mol}\) - Oxygen: \(19.35 \mathrm{\: g} \times \frac{1 \: \mathrm{mol}}{16.00\mathrm{\:g}} = 1.21 \: \mathrm{mol}\)

Step 2: Determine the mole ratio

To find the mole ratio, we'll divide each number of moles by the lowest value. In this case, the lowest value is 1.21. So we get: - Carbon: \(1.21/1.21 = 1\) - Hydrogen: \(1.81/1.21\approx 1.5\) - Chlorine: \(1.81/1.21\approx 1.5\) - Oxygen: \(1.21/1.21 = 1\) However, we can't have half of an atom in the empirical formula, so we need an even ratio. Therefore, we'll multiply each value by 2, giving us the following mole ratio: - Carbon: \(1 \times 2 = 2\) - Hydrogen: \(1.5 \times 2 = 3\) - Chlorine: \(1.5 \times 2 = 3\) - Oxygen: \(1 \times 2 = 2\)

Step 3: Find the empirical formula

Now that we have the mole ratio, we can write the empirical formula as: C₂H₃Cl₃O₂

Step 4: Find the molecular formula

We know that the molar mass of this compound is \(165.4 \mathrm{~g} / \mathrm{mol}\). Let's calculate the molar mass of the empirical formula: - Carbon: \( 2 \times 12.01 = 24.02 \: \mathrm{g/mol}\) - Hydrogen: \( 3 \times 1.01 = 3.03 \: \mathrm{g/mol}\) - Chlorine: \( 3 \times 35.45 = 106.35 \: \mathrm{g/mol}\) - Oxygen: \( 2 \times 16.00 = 32.00 \: \mathrm{g/mol}\) Total molar mass of the empirical formula = \(24.02 + 3.03 + 106.35 + 32.00 = 165.40 \mathrm{g/mol}\) As the molar mass of the empirical formula is the same as the given molar mass of the compound, we can conclude that the empirical and molecular formulas are the same. Thus, the molecular formula is also C₂H₃Cl₃O₂.

Step 5: Draw the Lewis structure

We will now draw the Lewis structure for this compound, following the given assumptions: 1. The chlorine atoms bond to a single carbon atom. 2. There is a carbon-carbon bond. 3. There are two carbon-oxygen bonds. The central carbon atom is bonded to the three chlorine atoms and the other carbon atom. The other carbon atom is bonded to the two oxygen atoms and a hydrogen atom. The remaining two hydrogen atoms are bonded to the chlorine atoms (one on each). The unshared electron pair on each oxygen atom forms a double bond with the carbon atom. Upon drawing the structure, we get: Cl | Cl - C - Cl | C - O | O | H So, the molecular formula of the compound is C₂H₃Cl₃O₂, and the Lewis structure is as shown above using the given assumptions.

Key Concepts

Stoichiometry

Stoichiometry is the field of chemistry that entails the calculation of the quantities of reactants and products involved in a chemical reaction. It is a crucial concept for understanding the principles of balanced equations, mole ratios, and conservation of mass. A stoichiometry problem often begins by identifying the ratio in which elements combine to form compounds, which is vital in determining the empirical formula of a substance.

For example, when solving for the empirical formula of chloral hydrate, we use stoichiometry to convert the mass percentages of each element to moles, which then allows us to find their mole ratios. These ratios indicate the simplest whole-number ratios of the atoms in the compound. Understanding stoichiometry can further help students comprehend how to scale these mole ratios up or down in order to find the molecular formula, which represents the actual number of atoms of each element in a molecule of the compound. As such, grasping stoichiometry fundamentally underpins tasks like finding empirical and molecular formulas.

Mole-to-Mass Conversion

Mole-to-mass conversion is a technique commonly used in stoichiometry to determine how much material is involved in a chemical reaction. It is based on the concept of the mole, which is a basic unit in chemistry representing a specific quantity of particles, typically Avogadro's number (\(6.022 \times 10^{23}\) particles per mole).

Mole-to-mass conversion involves using the molar mass, which is the mass in grams of one mole of a substance. During the conversion process in our example of chloral hydrate, we calculate the amount of each element in moles by taking the given mass of each element and dividing it by its molar mass. This step is crucial because it allows us to compare different elements on an equal basis. The molar mass acts as a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure in the laboratory. It's important for students to become comfortable with this conversion as it is used in a wide variety of problems in chemistry, including those related to empirical and molecular formulas.

Lewis Structures

Lewis structures are visual representations of molecules showing how the atoms are bonded together and the arrangement of valence electrons. These diagrams help predict the shapes of molecules, how they might interact with other molecules, and the types of bonds that hold the atoms together – single, double, or triple bonds.

When drawing a Lewis structure for a molecule like chloral hydrate, one has to consider specific rules and assumptions, such as the duet and octet rule, which dictate the numbers of bonds that atoms typically form to have a stable electron configuration. For chloral hydrate, we ensured that chlorine atoms formed single bonds with a central carbon atom, that there was a carbon-carbon bond, and there were two carbon-oxygen bonds. Drawing the Lewis structure requires an understanding of the shared electron pairs (bonds) and unshared electron pairs (lone pairs) around an atom.

Representation of Electrons

In the Lewis structure, dots or lines represent bonding and non-bonding valence electrons. Mastery of Lewis structures gives a strong foundation for predicting molecular geometry and reactivity in further chemistry studies.