A compound used as an insecticide that contains only \(C, H\), and \(C l\) is subjected to combustion analysis, yielding \(24.78 \% \mathrm{C}\) and \(2.08 \% \mathrm{H}\). (a) What is the empirical formula for this compound? (b) What is the molecular formula if its actual formula is four times the mass of its empirical formula?

First, we need to assume a sample mass of the compound, say 100 grams. This way, the given percentages can convert to the mass of each element in grams. Then, we will convert the mass to moles using the atomic masses of carbon, hydrogen, and chlorine. - Mass of Carbon (C) = 24.78 g - Mass of Hydrogen (H) = 2.08 g - Mass of Chlorine (Cl) = 100 g - (24.78 g + 2.08 g) = 73.14 g Now, we will convert the masses to moles by dividing the mass by the respective atomic masses of the elements. - Moles of C = 24.78 g / 12.01 g/mol ≈ 2.065 moles - Moles of H = 2.08 g / 1.008 g/mol ≈ 2.063 moles - Moles of Cl = 73.14 g / 35.45 g/mol ≈ 2.063 moles

To find the empirical formula, we need to express the ratio of the moles of the elements in the compound as the smallest possible whole numbers. Divide the moles of each element by the smallest number of moles, then round it to the nearest whole number. - Ratio of C atoms: 2.065 moles / 2.063 ≈ 1 - Ratio of H atoms: 2.063 moles / 2.063 ≈ 1 - Ratio of Cl atoms: 2.063 moles / 2.063 ≈ 1 Thus, the empirical formula of the compound is CHCl.

The problem states that the molecular formula has a mass four times larger than the empirical formula. To find the molecular formula, first, calculate the molar mass of the empirical formula. - Molar mass of CHCl = 12.01 g/mol (C) + 1.008 g/mol (H) + 35.45 g/mol (Cl) ≈ 48.47 g/mol Now, determine the molar mass of the molecular formula, which is four times the mass of the empirical formula. - Molar mass of molecular formula = 4 × 48.47 g/mol ≈ 193.88 g/mol Since the molecular formula has a mass four times larger than the empirical formula, the number of atoms of each element in the molecular formula will be four times that in the empirical formula. - Molecular formula = C₄H₄Cl₄ So, the molecular formula of the compound is C₄H₄Cl₄.