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?

First, let's convert the mass of CO2 and H2O to moles. To do this, we need the molar masses of CO2 and H2O. The molar mass of CO2 is 12.01 g/mol (for C) + 2 * 16.00 g/mol (for O) = 44.01 g/mol. The molar mass of H2O is 2 * 1.01 g/mol (for H) + 16.00 g/mol (for O) = 18.02 g/mol. Now, let's convert the mass of CO2 and H2O to moles: - Moles of CO2 = (16.01 mg CO2) * (1 g / 1000 mg) * (1 mol CO2 / 44.01 g CO2) = \(3.636 \times10^{-4} mol\) - Moles of H2O = (4.37 mg H2O) * (1 g / 1000 mg) * (1 mol H2O / 18.02 g H2O) = \(2.427 \times10^{-4} mol\) Finally, let's determine the moles of carbon and hydrogen in the compound: - Moles of C = \(3.636 \times10^{-4} mol\) - Moles of H =\(2 \times 2.427 \times10^{-4} mol\) = \(4.854 \times10^{-4} mol\)

We can find the moles of oxygen by using the initial moles of the compound. First, we need to convert the mass of compound to moles: - Moles of compound = (10.68 mg compound) * (1 g / 1000 mg) * (1 mol compound / 176.1 g compound) = \(6.063 \times 10^{-5} mol\) Since the combustion reaction is as follows: Compound + O₂ → CO₂ + H₂O Total moles of O atoms in the compound = (total moles of O atoms in CO₂) + (total moles of O atoms in H₂O) Let's find the total moles of O atoms in CO₂ and H₂O: - Moles of O in CO₂ = \(2 \times 3.636 \times 10^{-4} mol\) = \(7.273 \times 10^{-4} mol\) - Moles of O in H₂O = \(2.427 \times 10^{-4} mol\) Now, let's calculate the moles of O in the compound: - Moles of O = \(7.273 \times 10^{-4} mol + 2.427\times 10^{-4} mol\) = \(9.7 \times 10^{-4} mol\)

Divide each number of moles by the smallest value to obtain the simplest whole-number ratio: \(C : \frac{3.636 \times10^{-4}}{6.063 \times 10^{-5}} = 6 ; H : \frac{4.854 \times10^{-4}}{6.063\times 10^{-5}} = 8 ; O : \frac{9.7 \times10^{-4}}{6.063\times 10^{-5}} = 16\) So, the empirical formula is C₆H₈O₁₆.

First, calculate the empirical formula mass: C₆H₈O₁₆: \(6(12.01)+8(1.01)+16(16.00) \approx 352.20\, g/mol \) Since the molar mass of the compound is about half of the empirical formula mass, the molecular formula should be half the empirical formula. Therefore, the molecular formula is C₃H₄O₈.