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?

Use the masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) produced to calculate the moles of each. Molar mass of \(\mathrm{CO}_{2} = 12.01 + (2 \times 16.00) = 44.01 \: \mathrm{g}/\mathrm{mol}\) Molar mass of \(\mathrm{H}_{2}\mathrm{O} = (2 \times 1.01) + 16.00 = 18.02 \: \mathrm{g}/\mathrm{mol}\) Moles of \(\mathrm{CO}_{2} = \frac{16.01 \: \mathrm{mg}}{44.01 \: \mathrm{g/mol}} = 0.3635 \: \mathrm{mol}\) Moles of \(\mathrm{H}_{2}\mathrm{O} = \frac{4.37 \: \mathrm{mg}}{18.02 \: \mathrm{g/mol}} = 0.2426 \: \mathrm{mol}\)

Use the amount of moles of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) to find the moles of carbon (C) and hydrogen (H) in the compound. Moles of C: 1 mole of \(\mathrm{CO}_{2}\) has 1 mole of C, so moles of C = 0.3635 \(\mathrm{mol}\) Moles of H: 1 mole of \(\mathrm{H}_{2}\mathrm{O}\) has 2 moles of H, so moles of H = 0.2426 \(\mathrm{mol} \times 2 = 0.4852 \: \mathrm{mol}\)

Use the moles of C and H found previously to calculate their respective masses. Mass of C = 0.3635 \(\mathrm{mol} \times 12.01 \: \mathrm{g}/\mathrm{mol} = 4.364 \: \mathrm{mg}\) Mass of H = 0.4852 \(\mathrm{mol} \times 1.01 \: \mathrm{g}/\mathrm{mol} = 0.490 \: \mathrm{mg}\)

Use the initial mass of the compound and subtract the masses of C and H to find the mass of oxygen (O). Mass of O = Total mass - Mass of C - Mass of H = 10.68 \: \mathrm{mg} - 4.364 \: \mathrm{mg} - 0.490 \: \mathrm{mg} = 5.826 \: \mathrm{mg}$

Use the mass of O to determine the moles of O in the compound. Moles of O = \(\frac{5.826 \: \mathrm{mg}}{16.00 \: \mathrm{g}/\mathrm{mol}} = 0.3641 \: \mathrm{mol}\)

Determine the ratio of moles of C, H, and O in the compound by dividing each by the smallest mole value and rounding to the nearest whole number. Mole ratio = \(\frac{0.3635 \: \mathrm{mol} \: C}{0.3635 \: \mathrm{mol}} : \frac{0.4852 \: \mathrm{mol} \: H}{0.3635 \: \mathrm{mol}} : \frac{0.3641 \: \mathrm{mol} \: O}{0.3635 \: \mathrm{mol}} = 1 : 1.33 : 1\) We round this to the nearest whole number ratio, which gives: Empirical Formula = \(\mathrm{CH}_{2}\mathrm{O}\)

Use the molar mass of the empirical formula and the given molar mass of the compound to determine the molecular formula. Molar mass of empirical formula, \(\mathrm{CH}_{2}\mathrm{O} = 12.01 + (2 \times 1.01) + 16.00 = 30.03 \: \mathrm{g}/\mathrm{mol}\) N = \(\frac{Molar \: mass \: of \: molecular \: formula}{Molar \: mass \: of \: empirical \: formula} = \frac{176.1 \: \mathrm{g}/\mathrm{mol}}{30.03 \: \mathrm{g}/\mathrm{mol}} = 5.86\) We round this to the nearest whole number, which is 6. Multiply the empirical formula by 6 to get the molecular formula: \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\) The empirical formula of the compound is \(\mathrm{CH}_{2}\mathrm{O}\), and the molecular formula is \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\).