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 water is produced when hydrogen in the cumene reacts with oxygen, we can calculate the number of moles of hydrogen in the sample by using the mass of water produced. We know that the mass of water produced is \(42.8 \mathrm{mg}\), so we can convert this mass to moles: \[\text{moles of H}_2\text{O} = \frac{42.8\, \mathrm{mg}}{18.015\, \mathrm{g/mol}} \times \frac{1\, \mathrm{g}}{1000\, \mathrm{mg}} = 0.002375\, \mathrm{mol}.\] Since one water molecule is formed from two hydrogen atoms, the moles of hydrogen in cumene is twice the moles of water: \[\text{moles of H} = 2 \times \text{moles of H}_2\text{O} = 2 \times 0.002375\, \mathrm{mol} = 0.004750\, \mathrm{mol}.\]

Carbon dioxide is produced when carbon in the cumene reacts with oxygen. We can calculate the mass of CO2 produced by subtracting the mass of water from the initial mass of cumene: \[\text{mass of CO}_2 = 47.6\, \mathrm{mg} - 42.8\, \mathrm{mg} = 4.8\, \mathrm{mg}.\] Now, we can calculate the moles of carbon in the cumene by converting the mass of CO2 to moles of carbon: \[\text{moles of CO}_2 = \frac{4.8\, \mathrm{mg}}{44.01\, \mathrm{g/mol}} \times \frac{1\, \mathrm{g}}{1000\, \mathrm{mg}} = 0.000109\, \mathrm{mol}.\] Since there is one mole of carbon for every mole of CO2, the moles of carbon in cumene are the same as the moles of CO2 produced (0.000109 mol).

To find the empirical formula, we need to find the whole number ratio of moles of carbon and hydrogen. Divide both the moles of H and C by the smallest value to find the ratio: \[\frac{\text{moles of C}}{\text{smallest value}} = \frac{0.000109}{0.000109} = 1\] \[\frac{\text{moles of H}}{\text{smallest value}} = \frac{0.004750}{0.000109} = 43.58 \approx 44\] The empirical formula for cumene is \(\text{C}_1\text{H}_{44}\).

We are given that the molar mass of cumene is between \(\text{115 to 125 g/mol}\) and the empirical formula mass can be calculated as follows: Empirical formula mass = \(1 \times 12.01\, \mathrm{g/mol} + 44 \times 1.008\, \mathrm{g/mol} = 56.35\, \mathrm{g/mol}\) Now, we can determine the molecular formula by finding the ratio of the molecular mass to the empirical formula mass, and multiplying the empirical formula by this ratio. n = (molecular mass)/(empirical formula mass) Since the molecular mass must be between \(\text{115 and 125 g/mol}\), we can estimate the lowest and highest possible values of n: \[\text{Lowest value of n} = \frac{115\, \mathrm{g/mol}}{56.35\, \mathrm{g/mol}} = 2.04 \approx 2\] \[\text{Highest value of n} = \frac{125\, \mathrm{g/mol}}{56.35\, \mathrm{g/mol}} = 2.22 \approx 2\] Since both values of n are approximately equal to 2, the molecular formula of cumene is \(\text{C}_2\text{H}_{88}\).