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.

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.

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.

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.

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.