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 \(\mathrm{g} / \mathrm{mol}\) . (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 \(C-C\) bond and two \(C-O\) bonds in the compound.

Short answer

The empirical formula of the compound is CH\(_2\)Cl\(_2\)O, the molecular formula is C\(_2\)H\(_4\)Cl\(_4\)O\(_2\), and the Lewis structure is: Cl Cl | | H - C - C - O - O | | H H

Step-by-step solution

Moles Calculation

Assuming we have a 100 g sample of the compound, the mass of each element present can be calculated as follows: Carbon: \(14.52\% \times 100 \mathrm{g} = 14.52 \mathrm{g}\) Hydrogen: \(1.83\% \times 100 \mathrm{g} = 1.83 \mathrm{g}\) Chlorine: \(64.30\% \times 100 \mathrm{g} = 64.30 \mathrm{g}\) Oxygen: \(13.35\% \times 100 \mathrm{g} = 13.35 \mathrm{g}\) Now, we'll convert these masses to moles using the molar mass of each element: Carbon: \(\frac{14.52 \mathrm{g}}{12.01 \mathrm{g/mol}} = 1.21\) moles Hydrogen: \(\frac{1.83 \mathrm{g}}{1.008 \mathrm{g/mol}} = 1.81\) moles Chlorine: \(\frac{64.30 \mathrm{g}}{35.45 \mathrm{g/mol}} = 1.81\) moles Oxygen: \(\frac{13.35 \mathrm{g}}{16.00 \mathrm{g/mol}} = 0.834\) moles Step 2: Determine the empirical formula.

Empirical Formula Calculation

To find the empirical formula, we need to divide the moles of each element by the smallest number of moles: The smallest moles value is 0.834 (Oxygen). Now, divide each element's moles by 0.834: Carbon: \(\frac{1.21}{0.834} = 1.45 \approx 1\) Hydrogen: \(\frac{1.81}{0.834} = 2.17 \approx 2\) Chlorine: \(\frac{1.81}{0.834} = 2.17 \approx 2\) Oxygen: \(\frac{0.834}{0.834} = 1\) So, the empirical formula is CH\(_2\)Cl\(_2\)O. Step 3: Calculate the molecular formula.

Molecular Formula Calculation

First, find the empirical formula mass by adding the molar mass of all elements present in the empirical formula: Empirical formula mass: \( C + 2H + 2Cl + O = 12.01 + 2(1.008) + 2(35.45) + 16.00 = 83.48 \mathrm{g/mol}\) Now, divide the given molar mass (165.4 g/mol) by the empirical formula mass (83.48 g/mol): Number of units: \(\frac{165.4 \mathrm{g/mol}}{83.48 \mathrm{g/mol}} = 1.98 \approx 2\) Since the number of units is approximately two, we can determine the molecular formula by multiplying the empirical formula by two: Molecular formula: \(2(\text{CH}_2\text{Cl}_2\text{O}) = \text{C}_2\text{H}_4\text{Cl}_4\text{O}_2\) Step 4: Draw the Lewis structure.

Lewis Structure Drawing

According to the problem's assumptions, the chlorine atoms are bonded to a single carbon atom, and there is a carbon-carbon bond and two carbon-oxygen bonds. The Lewis structure will look like this: Cl Cl | | H - C - C - O - O | | H H To conclude: a) The empirical formula is CH\(_2\)Cl\(_2\)O. b) The molecular formula is C\(_2\)H\(_4\)Cl\(_4\)O\(_2\). c) The Lewis structure is shown above.

Key concepts

Elemental Analysis

Elemental analysis involves determining the percentage composition of each element in a compound. This step is crucial when discovering the compound's empirical formula, which shows the simplest ratio of each type of atom in the compound. For chloral hydrate, we begin by assuming we have 100 grams of the compound. This simplification makes it easy to directly translate the percentages of each element to grams:

• Carbon: 14.52% translates to 14.52 grams.
• Hydrogen: 1.83% translates to 1.83 grams.
• Chlorine: 64.30% translates to 64.30 grams.
• Oxygen: 13.35% translates to 13.35 grams.

By using the atomic mass of each element, we then convert these masses to moles, allowing us to compare the number of atoms of each element present. This comparison is key to determining the empirical formula, as it relies on the identification of the simplest integer ratio between the elements.

Lewis Structures

Lewis structures are diagrams that depict the bonds between atoms and the lone pairs of electrons in a molecule. They are essential for visualizing the arrangement of atoms within a compound. In the case of chloral hydrate, we assume that the chlorine atoms are each bonded to a single carbon atom, and there is a single bond between the two carbon atoms. A Lewis structure shows how these atoms are connected and how electrons are distributed among them. When drawing a Lewis structure:

• Place the element with the most available bonding spots, typically a less electronegative atom, in the center.
• Use lines to represent covalent bonds and dots to represent lone pairs of electrons. Each line corresponds to a pair of shared electrons, while dots stand for electrons not involved in bonding.
• Ensure every atom meets its respective stable valence shell, often following the octet rule, where atoms typically need 8 electrons to be considered stable (except for hydrogen, which requires only 2).

In chloral hydrate, the arrangement involves carbon, hydrogen, chlorine, and oxygen atoms, with certain atoms forming double or single bonds to fulfill octet configurations.

Moles Calculation

Calculating moles is a fundamental concept in chemistry that involves determining the amount of a substance in terms of the number of atoms or molecules. This process uses the molar mass of elements, defined as the mass of one mole of atoms of an element, expressed in grams per mole. For chloral hydrate, given a 100-gram sample, we convert the mass of each element into moles:

• For Carbon: Divide its mass (14.52 g) by its molar mass (12.01 g/mol) to get 1.21 moles.
• For Hydrogen: Divide its mass (1.83 g) by its molar mass (1.008 g/mol) to get 1.81 moles.
• For Chlorine: Divide its mass (64.30 g) by its molar mass (35.45 g/mol) to obtain 1.81 moles.
• For Oxygen: Divide its mass (13.35 g) by its molar mass (16.00 g/mol) to reach 0.834 moles.

These mole calculations allow us to identify the simplest ratios between the atoms in a compound and determine the empirical formula. Always remember that these ratios give the simplification of atoms that helps in defining the chemical identity of the compound.