A compound contains 47.08\(\%\) carbon, 6.59\(\%\) hydrogen, and 46.33\(\%\) chlorine by mass; the molar mass of the compound is 153 g/mol. What are the empirical and molecular formulas of the compound?

Assume there are 100 grams of the compound. The mass of each element is given as a percentage of the total mass. Therefore, - Carbon: 47.08 g - Hydrogen: 6.59 g - Chlorine: 46.33 g Now, we will convert the mass of each element to moles using their respective molar masses: - Carbon: 12.01 g/mol (C) - Hydrogen: 1.01 g/mol (H) - Chlorine: 35.45 g/mol (Cl) Moles of each element: - Carbon: \( \frac{47.08}{12.01} \approx 3.92 \) moles - Hydrogen: \( \frac{6.59}{1.01} \approx 6.52 \) moles - Chlorine: \( \frac{46.33}{35.45} \approx 1.31 \) moles

To find the empirical formula, we need to determine the simplest whole number ratio of the moles of each element: - Carbon: \( \frac{3.92}{1.31} \approx 3 \) - Hydrogen: \( \frac{6.52}{1.31} \approx 5 \) - Chlorine: \( \frac{1.31}{1.31} \approx 1 \) Now, we can see that the empirical formula is C3H5Cl.

Using the molar mass of the empirical formula, we can find the molecular formula. The molar mass of the empirical formula is as follows: - Carbon: 3 x 12.01 g/mol (C) = 36.03 g/mol - Hydrogen: 5 x 1.01 g/mol (H) = 5.05 g/mol - Chlorine: 1 x 35.45 g/mol (Cl) = 35.45 g/mol Total molar mass of the empirical formula: \(36.03 + 5.05 + 35.45 \approx 76.53 \: g/mol \) Now, we will find the ratio of the molar mass of the compound to the molar mass of the empirical formula: Ratio: \( \frac{153}{76.53} \approx 2 \) Since the ratio is approximately 2, this means that the molecular formula is twice the empirical formula. Therefore: Molecular formula: (C3H5Cl)2 = C6H10Cl2 The empirical formula of the compound is C3H5Cl, and the molecular formula is C6H10Cl2.