91Ó°ÊÓ

Since the combustion of cumene produces water, and water contains only hydrogen and oxygen, the mass of hydrogen in the cumene can be determined by the mass of hydrogen in the water produced. In every water molecule, there are two hydrogen atoms and one oxygen atom. We know that the molar mass of hydrogen (\(H\)) is approximately 1 g/mol, and the molar mass of oxygen (\(O\)) is approximately 16 g/mol. To find the mass of hydrogen in cumene, we first need to find the molar amount of water produced: \(42.8 \: \mathrm{mg} \: H_2O \times \frac{1 \: \mathrm{mol} \: H_2O}{18.016 \: \mathrm{g} \: H_2O} \times \frac{1000 \: \mathrm{mg}}{1 \: \mathrm{g}} = 2.376 \times 10^{-3} \: \mathrm{mol} \: H_2O\) Next, we convert the moles of water to moles of hydrogen: \(2.376 \times 10^{-3} \: \mathrm{mol} \: H_2O \times \frac{2 \: \mathrm{mol} \: H}{1 \: \mathrm{mol} \: H_2O} = 4.752 \times 10^{-3} \: \mathrm{mol} \: H\) Now, we can find the mass of hydrogen in cumene: \(4.752 \times 10^{-3} \: \mathrm{mol} \: H \times \frac{1 \: \mathrm{g}}{\mathrm{mol} \: H} \times \frac{1000 \: \mathrm{mg}}{1 \: \mathrm{g}} = 4.75 \: \mathrm{mg} \: H\)

Since cumene contains only carbon and hydrogen, the mass of carbon in cumene can be determined by subtracting the mass of hydrogen found in Step 1 from the original mass of cumene: \(47.6 \: \mathrm{mg} \: \text{cumene} - 4.75 \: \mathrm{mg} \: H = 42.85 \: \mathrm{mg} \: C\)

Now that we have the mass of carbon and hydrogen, we will convert these values into moles: Moles of carbon: \(42.85 \: \mathrm{mg} \: C \times \frac{1 \: \mathrm{mol} \: C}{12.01 \: \mathrm{g} \: C} \times \frac{1000 \: \mathrm{mg}}{1 \: \mathrm{g}} = 3.57 \times 10^{-3} \: \mathrm{mol} \: C\) Moles of hydrogen: \(4.75 \: \mathrm{mg} \: H \times \frac{1 \: \mathrm{mol} \: H}{1 \: \mathrm{g} \: H} \times \frac{1000 \: \mathrm{mg}}{1 \: \mathrm{g}} = 4.75 \times 10^{-3} \: \mathrm{mol} \: H\)

To find the empirical formula, we need to find the ratio of moles of carbon to hydrogen. Carbon-to-hydrogen mole ratio: \(\frac{3.57 \times 10^{-3} \: \mathrm{mol} \: C}{4.75 \times 10^{-3} \: \mathrm{mol} \: H} \approx \frac{1}{1.33} \approx \frac{3}{4}\) So, the empirical formula is \(C_3H_4\).

To find the molecular formula, we will compare the empirical formula molar mass to the given molar mass range. The empirical formula molar mass is: \(3 \times 12.01 \: \mathrm{g/mol} \: C + 4 \times 1 \: \mathrm{g/mol} \: H = 40.03 \: \mathrm{g/mol}\) Now, we can find the multiple of the empirical formula molar mass that is within the given molar mass range: \(115 \: \mathrm{g/mol} \le 40.03 \: \mathrm{g/mol} \times n \le 125 \: \mathrm{g/mol}\) Solving for \(n\), we find that \(n \approx 3\). Therefore, the molecular formula is: \(C_3H_4 \times 3 = C_9H_{12}\) So, the empirical formula of cumene is \(C_3H_4\) and the molecular formula is \(C_9H_{12}\).