Problem 103 - Chemistry The Central Science In Si Units

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? \((\mathbf{b})\) 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 \(\mathrm{C}\) 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

(a) Empirical: \( \text{C}_3\text{H}_4\text{Cl}_4\text{O}_2 \). (b) Molecular: \( \text{C}_3\text{H}_4\text{Cl}_4\text{O}_2 \). (c) Lewis structure: Cl and O bonds to a central C, with C–C bonds as described.

Step 1: Calculate Moles of Each Element

Given mass percentages: \( \text{C: } 14.52\% \), \( \text{H: } 1.83\% \), \( \text{Cl: } 64.30\% \), \( \text{O: } 13.35\% \). Assume 100 g of compound for simplicity.\\(\text{Moles of C} = \frac{14.52}{12.01} = 1.209 \) mol \\(\text{Moles of H} = \frac{1.83}{1.008} = 1.815 \) mol \\(\text{Moles of Cl} = \frac{64.30}{35.45} = 1.815 \) mol \\(\text{Moles of O} = \frac{13.35}{16.00} = 0.834 \) mol.

Step 2: Determine Simplest Whole Number Ratio

Divide each mole value by the smallest number, which is \(0.834\):\\( \frac{1.209}{0.834} \approx 1.45 \) for C, \\( \frac{1.815}{0.834} \approx 2.18 \) for H and Cl, \\( \frac{0.834}{0.834} = 1 \) for O. \Multiply by 2 to get whole numbers: 3C, 4H, 4Cl, 2O.

Step 3: Write Empirical Formula

The empirical formula is based on the simplest whole-number ratio: \( \text{C}_3\text{H}_4\text{Cl}_4\text{O}_2 \).

Step 4: Confirm Empirical Formula Mass & Calculate Molecular Formula

Calculate empirical formula mass: \\(3 \times 12.01 + 4 \times 1.008 + 4 \times 35.45 + 2 \times 16.00 = 165.4 \text{ g/mol}\). \The empirical formula mass matches the given molar mass, so the molecular formula is \(\text{C}_3\text{H}_4\text{Cl}_4\text{O}_2\).

Step 5: Draw the Lewis Structure

With \(\text{C}_3\text{H}_4\text{Cl}_4\text{O}_2\), arrange the atoms:- Central C attached to Cl atoms.- C–C bond between central carbons.- Each C connects to Cl and O.- Include C–O bonds and hydrogen distribution according to typical valency, forming two \(\text{C--O} \) bonds.

Key Concepts

Molecular Formula

When determining the molecular formula of a compound like chloral hydrate, we need to know the ratio of atoms in a molecule. The molecular formula tells us the exact number of each type of atom present in a molecule. To find the molecular formula, we start with the empirical formula, which represents the simplest whole-number ratio of elements in the compound.

The empirical formula we found is \( \text{C}_3\text{H}_4\text{Cl}_4\text{O}_2 \). This means in the empirical unit, there are 3 carbon atoms, 4 hydrogen atoms, 4 chlorine atoms, and 2 oxygen atoms. We calculated the molar mass of the empirical formula to be 165.4 g/mol, which matches the given molar mass of the compound. This tells us that the empirical formula is, in fact, the molecular formula, as they have the same mass.

Knowing the molecular formula is crucial because it provides information about the actual chemical makeup and the proportions of each element in the compound, which is important for understanding its chemical behavior and properties.

Lewis Structure

The Lewis structure is a diagram that shows the bonds between atoms in a molecule and any unshared electron pairs. Drawing the Lewis structure for chloral hydrate helps visualize the arrangement of atoms and the bonding pattern.

To draw the Lewis structure of chloral hydrate with the molecular formula \( \text{C}_3\text{H}_4\text{Cl}_4\text{O}_2 \), we must consider the specific bonding instructions provided:

• There is a C–C bond, meaning two carbon atoms are directly bonded.
• Two chlorine atoms bond to the central carbon atom.
• Two carbon-oxygen (C–O) bonds are present in the structure.

Begin your structure with a central carbon, assuming it connects to two other carbons on the sides. Attach four chlorine atoms to the carbon atoms, ensuring they satisfy the typical valency of one bond per chlorine. Then, connect the oxygen atoms such that each is in a C--O bond, ensuring that they have two pairs of lone electrons.The Lewis structure not only illustrates the atoms present in the molecule but also determines the three-dimensional shape, which can affect how the molecule interacts with others. It informs us about the molecule’s geometry and helps understand its chemical properties and reactivity.

Chemical Composition Analysis

Chemical composition analysis means determining the percentage of each element present in a compound. In our exercise, this analysis involved determining the mass percentages of carbon, hydrogen, chlorine, and oxygen in chloral hydrate.

Assuming 100 grams of the compound, the mass percentages provided allowed us to convert these directly into masses of each element in grams.

• Carbon: 14.52% means 14.52 g of carbon.
• Hydrogen: 1.83% means 1.83 g of hydrogen.
• Chlorine: 64.30% means 64.30 g of chlorine.
• Oxygen: 13.35% means 13.35 g of oxygen.

To find the amount of each element present in moles, divide each element's mass by its atomic mass:

• \( \text{Moles of C} = \frac{14.52}{12.01} \approx 1.209 \text{ mol} \)
• \( \text{Moles of H} = \frac{1.83}{1.008} \approx 1.815 \text{ mol} \)
• \( \text{Moles of Cl} = \frac{64.30}{35.45} \approx 1.815 \text{ mol} \)
• \( \text{Moles of O} = \frac{13.35}{16.00} \approx 0.834 \text{ mol} \)

By calculating the moles, you effectively translate the mass data into a format useful for determining the empirical formula, which helps deduce the composition of molecules and relate it to molecular weight.