Problem 106 - 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 \(19.35 \% \mathrm{O}\) 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 \(\mathrm{C}\) 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 CCl₂H₂O, and the molecular formula is C₂Cl₄H₄O₂. The Lewis structure of the chloral hydrate molecule is: O = C - C - Cl || | O Cl | H | H

Step 1: Convert mass percentages to moles

First, assume a 100 g sample of the compound, which makes the given percentages equal to grams. Now, convert the grams to moles using the atomic masses: C = 14.52 g / \(12.01 \mathrm{~g/mol}\) = 1.21 mol H = 1.83 g / \(1.01 \mathrm{~g/mol}\) = 1.81 mol Cl = 64.30 g / \(35.45 \mathrm{~g/mol}\) = 1.82 mol O = 19.35 g / \(16.00 \mathrm{~g/mol}\) = 1.21 mol

Step 2: Find the simplest whole-number ratio of moles

Divide each of the moles by the smallest value among them: C = \(1.21 / 1.21 = 1\) H = \(1.81 / 1.21 ≈ 1.50\) Cl = \(1.82 / 1.21 ≈ 1.50\) O = \(1.21 / 1.21 = 1\) Now, since we have \(1.50\) as a ratio for two elements, we round these values to the nearest simple whole numbers: C = 1 H = 2 Cl = 2 O = 1 The empirical formula is CCl₂H₂O.

Step 3: Find the molecular formula

To find the molecular formula, determine the whole-number factor between the molar mass of the empirical formula and the given molar mass: Molar mass of the empirical formula: 12.01 + 2 * 35.45 + 2 * 1.01 + 16.00 = 101.94 g/mol Whole-number factor (approximation): \(165.4 \mathrm{~g/mol} / 101.94 \mathrm{~g/mol} ≈ 1.62\), which we round to the nearest whole number, 2. Multiply each subscript in the empirical formula by the whole-number factor to get the molecular formula: C(1 * 2)Cl(2 * 2)H(2 * 2)O(1 * 2) = C₂Cl₄H₄O₂ The molecular formula is C₂Cl₄H₄O₂.

Step 4: Draw the Lewis structure

Start by counting the total valence electrons in the molecule: C₂Cl₄H₄O₂: (2 * 4) + (4 * 7) + (4 * 1) + (2 * 6) = 40 valence electrons Now, for drawing the Lewis structure, follow the given assumptions: a) Cl atoms bond to a single C atom b) There is a C-C bond c) There are two C-O bonds 1. Connect the two carbon atoms with a single bond. 2. Connect each chlorine atom to one of the carbon atoms with a single bond. 3. Connect each oxygen atom to the other carbon atom with a double bond. 4. Add remaining hydrogen atoms to the carbon that has Cl connected to it. 5. Fill in the remaining octets with lone pairs. O = C - C - Cl || | O Cl | H | H The Lewis structure of the chloral hydrate molecule is as above.

Key Concepts

Mass Percentage Composition

Understanding mass percentage composition is vital for determining the composition of elements within a compound. When the percentage of an element is given, it represents how much of that element is present in a compound relative to its total mass.
To make it easier to work with, chemists often assume a 100-gram sample of the compound, allowing the percentage to be treated as grams. For example, in the compound chloral hydrate, 14.52% of carbon translates to 14.52 grams when assuming a 100 gram sample.
This conversion is the first step in determining how many moles of each element are present. You compare this to the atomic masses (found on the periodic table):

• Carbon (C) has an atomic mass of 12.01 g/mol.
• Hydrogen (H) has an atomic mass of 1.01 g/mol.
• Chlorine (Cl) has an atomic mass of 35.45 g/mol.
• Oxygen (O) has an atomic mass of 16.00 g/mol.

By converting each element's mass to moles, you can determine the empirical formula of the compound, reflecting the simplest whole-number ratio of atoms within the compound.

Atomic Mass Conversion

Atomic mass conversion is a crucial step for finding empirical and molecular formulas. It involves turning the mass percentage of elements into moles using their respective atomic masses.
This process allows chemists to determine how many atoms of each element exist in a particular mass of a compound. The mole is a standard unit in chemistry that represents Avogadro's number, which is roughly 6.022 x 10²³ atoms.
In our exercise with chloral hydrate, using the percentages given:

• Carbon's percentage is divided by its atomic mass: 14.52 g / 12.01 g/mol = 1.21 mol
• Hydrogen: 1.83 g / 1.01 g/mol = 1.81 mol
• Chlorine: 64.30 g / 35.45 g/mol = 1.82 mol
• Oxygen: 19.35 g / 16.00 g/mol = 1.21 mol

The smallest value of moles here (1.21 mol) becomes the divisor, allowing for the calculation of the simplest whole-number ratio. These ratios are then adjusted to nearest whole numbers if necessary, resulting in an empirical formula like CCl₂H₂O.

Lewis Structure Drawing

Drawing Lewis structures is essential for visualizing the arrangement of atoms in a molecule. It highlights how atoms bond and any lone pairs of electrons.
For chloral hydrate, the Lewis structure involves several steps using specific bonding rules provided in the exercise:

• Begin by counting total valence electrons in the molecule. Here, it is calculated as (2 * 4) + (4 * 7) + (4 * 1) + (2 * 6) = 40 valence electrons.
• Follow these guidelines for bond formation:
  • Chlorine atoms bond to a single carbon atom.
  • There is a C-C bond between the carbon atoms.
  • Two C-O bonds exist, implying a double bond for each.

Starting with connecting two carbon atoms, assign a single bond. Attach each chloral atom to a carbon with single bonds, and each oxygen to the remaining carbon atom through double bonds. Add hydrogen atoms to the carbon connected to the chlorine atoms, ensuring all octets are complete.
This procedure ensures a clear, structured depiction of the chloral hydrate molecule, satisfying its valency and bonding requirements as provided and ensuring no valence electron is left unaccounted for.