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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.

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 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.