An unknown substance is found to contain only carbon and hydrogen. It is a liquid that boils at \(49^{\circ} \mathrm{C}\) at 1 atm pressure. Upon analysis it is found to contain 85.7\(\%\) carbon and 14.3\(\%\) hydrogen by mass. At \(100^{\circ} \mathrm{C}\) and 735 torr, the vapor of this unknown has a density of 2.21 \(\mathrm{g} / \mathrm{L} .\) When it is dissolved in hexane solution and bromine water is added, no reaction occurs. What is the identity of the unknown compound?

First, we need to find the empirical formula of the unknown compound using the given mass percentages of carbon and hydrogen. We will assume 100g of the substance, which gives us 85.7g of carbon and 14.3g of hydrogen. Then, convert the mass to moles by dividing by their respective atomic masses: Carbon: \(\frac{85.7\,\text{g}}{12.01\,\text{g/mol}} = 7.14\,\text{mol}\) Hydrogen: \(\frac{14.3\,\text{g}}{1.008\,\text{g/mol}} = 14.2\,\text{mol}\) Now, we will find the mole ratio by dividing each number of moles by the smallest mole value among them: Carbon: \(\frac{7.14}{7.14}=1\) Hydrogen: \(\frac{14.2}{7.14}=1.99 \approx 2\) Hence, the empirical formula of the unknown compound is CH2.

The density of the vapor of the unknown compound is given at 100°C and 735 torr: 2.21 g/L. We can determine the molar mass (MM) of the compound using the Ideal Gas Law: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. First, we need to convert the temperature and pressure into Kelvin and atm: Temperature: \(100^{\circ}\text{C} + 273.15 = 373.15\,\text{K}\) Pressure: \(\frac{735\,\text{torr}}{760\,\text{torr/atm}} = 0.967\,\text{atm}\) Now we can substitute the values and solve for n: \(0.967\,\text{atm} \times 1\,\text{L} = n \times 0.0821\,\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}} \times 373.15\,\text{K}\) \(n = \frac{0.967 \times 1}{0.0821 \times 373.15} = 0.0315\,\text{mol}\) Since we have the vapor density and number of moles, we can now calculate the molar mass: Molar Mass = \(\frac{2.21\,\text{g}}{0.0315\,\text{mol}} = 70.16\,\text{g/mol}\)

Using the molar mass of the compound (70.16 g/mol) and the empirical formula (CH2), we can determine the molecular formula: Empirical formula mass of CH2: 12.01 g/mol (C) + 2 x 1.008 g/mol (H) = 14.03 g/mol Divide molar mass by empirical formula mass to find the number of empirical units per molecule: \(\frac{70.16\,\text{g/mol}}{14.03\,\text{g/mol}} = 5\) Since there are 5 empirical formula units per molecule, the molecular formula of the unknown compound is: C5H10

Since the compound does not react with bromine water, it does not have any carbon-carbon double bonds or triple bonds, meaning it must be an alkane. The molecular formula of the unknown compound is C5H10, which does not correspond to an alkane. Therefore, it must be a cyclic hydrocarbon. Thus, the identity of the unknown compound is cyclopentane (C5H10) as it perfectly matches the molecular formula, boiling point, and reaction information provided in the exercise.