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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

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.

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.

Start by counting the total valence electrons in the molecule: Câ‚‚Clâ‚„Hâ‚„Oâ‚‚: (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.