A compound contains only carbon, hydrogen, and oxygen. Combustion of \(10.68 \mathrm{mg}\) of the compound yields \(16.01 \mathrm{mg}\) \(\mathrm{CO}_{2}\) and \(4.37 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .\) The molar mass of the compound is \(176.1 \mathrm{~g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

Given: Mass of the compound = 10.68 mg Mass of \(\mathrm{CO}_2\) produced = 16.01 mg Mass of \(\mathrm{H}_2\mathrm{O}\) produced = 4.37 mg Since each molecule of \(\mathrm{CO}_2\) contains one carbon atom and each molecule of \(\mathrm{H}_2\mathrm{O}\) contains two hydrogen atoms, we can calculate the moles of each element in the products as follows: Moles of C in \(\mathrm{CO}_2\) = \(\frac{\text{mass of CO}_2}{\text{molar mass of CO}_2}\) = \(\frac{16.01\text{ mg}}{44.01\text{ mg/mol}}\) = 0.3636 mol Moles of H in \(\mathrm{H}_2\mathrm{O}\) = \(\frac{\text{mass of H}_2\text{O}}{\text{molar mass of H}_2\text{O}}\) × 2 = \(\frac{4.37\text{ mg}}{18.02\text{ mg/mol}}\) × 2 = 0.4847 mol

Since there are only three elements in the compound (Carbon, Hydrogen, and Oxygen), we can find the moles of Oxygen from the moles of Carbon and Hydrogen: Moles of O in the compound = Total moles - (Moles of C + Moles of H) Mass of O = Mass of the compound - (Mass of C + Mass of H) = 10.68 mg - (16.01 mg + 4.37 mg) = -9.70 mg Moles of O = \(\frac{-9.70\text{ mg}}{16.00\text{ mg/mol}}\) = -0.6063 mol

Now we will find the ratio between the moles of C, H, and O: \(C: H: O = \frac{0.3636}{\text{gcd}(0.3636, 0.4847, -0.6063)} : \frac{0.4847}{\text{gcd}(0.3636, 0.4847, -0.6063)} : \frac{-0.6063}{\text{gcd}(0.3636, 0.4847, -0.6063)} = 1 : 1.333 : -1.667\) To simplify this ratio, multiply each number by the smallest possible whole number to obtain a whole number ratio. In this case, multiply by 3: \(C: H: O = 3 : 4 : -5\) However, we can't have a negative number of oxygen atoms, which suggests that our calculations were incorrect.

Since the mass of the compound should be the sum of the masses of carbon, hydrogen, and oxygen, there was an error in the calculation for the mass of oxygen.

Correcting the calculation for the mass and moles of Oxygen: Mass of O = Mass of the compound - (Mass of C + Mass of H) = 10.68 mg - (16.01 mg + 4.37 mg) = -9.70 mg Note that the "-" was misplaced. The calculation should be as follows: Mass of O = Mass of the compound - (Mass of C + Mass of H) = 10.68 mg - (16.01 mg - 4.37 mg) = -1.96 mg Moles of O = \(\frac{-1.96\text{ mg}}{16.00\text{ mg/mol}}\) = -0.1225 mol

Now, we will find the updated ratio between the moles of C, H, and O: \(C: H: O = \frac{0.3636}{\text{gcd}(0.3636, 0.4847, -0.1225)} : \frac{0.4847}{\text{gcd}(0.3636, 0.4847, -0.1225)} : \frac{-0.1225}{\text{gcd}(0.3636, 0.4847, -0.1225)} = 3: 4 : -1\) The empirical formula is now: \(\mathrm{C}_3\mathrm{H}_4\mathrm{O}\)

Given molar mass of the compound = 176.1 g/mol Empirical formula mass = \(3 × 12.01\text{ g/mol (C)} + 4 × 1.01\text{ g/mol (H)} + 1 × 16.00\text{ g/mol (O)} = 40.06\ \mathrm{g/mol}\) To find the molecular formula, find the ratio of the molar mass to the empirical formula mass: \(\frac{\text{molar mass}}{\text{empirical formula mass}} = \frac{176.1\text{ g/mol}}{40.06\text{ g/mol}} = 4.4\) Since we are looking for whole-number ratios, we need to round 4.4 to the nearest whole number, which is 4. Therefore, the molecular formula is four times the empirical formula: \(\mathrm{C}_{12}\mathrm{H}_{16}\mathrm{O}_4\) Thus, the empirical formula of the compound is \(\mathrm{C}_3\mathrm{H}_4\mathrm{O}\) and the molecular formula is \(\mathrm{C}_{12}\mathrm{H}_{16}\mathrm{O}_4\).