Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of 47.6 \(\mathrm{mg}\) cumene produces some \(\mathrm{CO}_{2}\) and 42.8 \(\mathrm{mg}\) water. The molar mass of cumene is between 115 and 125 $\mathrm{g} / \mathrm{mol}$ . Determine the empirical and molecular formulas.

Since cumene contains only carbon and hydrogen, the hydrogen atoms come from the water produced in the combustion reaction. To find the moles of hydrogen atoms, we will first find the moles of water, remembering that 1 mole of water is produced from 2 moles of hydrogen atoms. Calculate the moles of water produced: Moles of water = (Mass of water) / (Molar mass of water) Moles of water = (42.8 mg) / (18.015 g/mol) Note that we have to convert mg to g: Moles of water = (42.8 x 10^-3 g) / (18.015 g/mol) = 0.00238 mol of water Now, determine moles of hydrogen atoms: Moles of hydrogen atoms = 2 x (Moles of water) = 2 × 0.00238 mol = 0.00476 mol

All carbon in cumene must have been converted to CO2 during combustion. Thus, we can find the mass of carbon in cumene by finding the mass of CO2 with the following relationship: 1 mole of CO2 = 1 mole of carbon atom Mass of CO2 = Mass of cumene - Mass of water Mass of CO2 = 47.6 mg - 42.8 mg = 4.8 mg Now, calculate moles of carbon atoms: Moles of Carbon atoms = (Mass of CO2) / (Molar mass of CO2) Moles of Carbon atoms = (4.8 x 10^-3 g) / (44.01 g/mol) = 0.000109 mol of carbon

Determine the smallest whole-number ratio of moles of carbon to moles of hydrogen: Moles C : Moles H = 0.000109 : 0.00476 To find the smallest whole-number ratio, divide both numbers by the smaller one: Moles C : Moles H = 0.000109/0.000109 : 0.00476/0.000109 = 1 : 43.7 Since we have to round to whole numbers, the empirical formula is C1H44.

The molecular formula is a whole-number multiple of the empirical formula. First, find the molar mass of the empirical formula: C1H44 -> 1 carbon atom + 44 hydrogen atoms = (12.01 g/mol) + 44 × (1.008 g/mol) = 56.37 g/mol Now, use the given molar mass range of cumene (between 115 and 125 g/mol) to find the whole-number multiple "n": n × (Empirical formula's molar mass) = Molecular formula's molar mass n = (Molecular formula's molar mass) / (Empirical formula's molar mass) Divide the lower limit of the molar mass range by the molar mass of the empirical formula: n = 115 g/mol / 56.37 g/mol = 2.04 Since n must be a whole number and we obtained a value close to 2, the molecular formula is twice the empirical formula: Molecular formula = C2H88 So, the empirical formula of cumene is C1H44, and the molecular formula is C2H88.