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

First, we need to convert the mass percentages of each element into moles. We will do this by dividing the mass percentages by their respective atomic masses: C: \(\frac{14.52\%}{12.01~\text{g/mol}} = 1.21~\text{mol}\) H: \(\frac{1.83\%}{1.01~\text{g/mol}} = 1.81~\text{mol}\) Cl: \(\frac{64.3\%}{35.45~\text{g/mol}} = 1.81~\text{mol}\) O: \(\frac{13.35\%}{16.00~\text{g/mol}} = 0.83~\text{mol}\) Now, find the simplest whole number ratio by dividing each mole amount by the smallest value amongst them: C: \(\frac{1.21}{0.83} = 1.46 \approx 1\) H: \(\frac{1.81}{0.83} = 2.18 \approx 2\) Cl: \(\frac{1.81}{0.83} = 2.18 \approx 2\) O: \(\frac{0.83}{0.83} = 1.00 \approx 1\) This gives us the empirical formula: \(C_{1}H_{2}Cl_{2}O_{1}\), or \(CH_2Cl_2O\).

Next, we will find the molecular formula by comparing the empirical formula mass with the given molar mass of the compound: Empirical formula mass: \(C + 2H + 2Cl + O = 12.01 + 2(1.01) + 2(35.45) + 16.00 = 165.38~\text{g/mol}\) The given molar mass of the compound is \(165.4~\text{g/mol}\). Since the empirical formula mass and the given molar mass are approximately equal, the molecular formula is the same as the empirical formula: Molecular formula: \(CH_2Cl_2O\)

We will now draw the Lewis structure of the compound, keeping in mind the bonding information provided: The Cl atoms are bonded to a single C atom; There is a C-C bond between two carbon atoms; There are two C=O bonds. 1. Place the least electronegative element in the center: C 2. Attach the other C atom (C-C bond) and two Cl atoms to the central C. 3. Arrange the two O atoms in a double bond with the second C atom. 4. Distribute the remaining electrons to complete the octets for the surrounding atoms. Cl \ C-C=O / Cl The Lewis structure for chloral hydrate (\(CH_2Cl_2O\)) is now complete.