Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 \(\mathrm{K}\). Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 \(\mathrm{K}\). This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon

Given the density of the mixture (1.391 g/L) and the fact that the volume of the mixture is four times that of the hydrocarbon, we can calculate the mass of the mixture: mass = density × volume We know that the volume of hydrocarbon = V (not given), so the volume of mixture = 4V mass = 1.391 g/L × 4V = 5.564 g To find the moles of the mixture, we use the ideal gas law equation: \(PV = nRT\), where P is the pressure, V is the volume, n is the moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin. We are given the pressure and temperature of the mixture, so we can plug those values into the equation: 1.51 atm × (4V) = n × 0.0821 L atm/mol K × 375 K Now, we can solve for n (moles of the mixture): n = (1.51 atm × 4V) / (0.0821 L atm/mol K × 375 K)

Since CO2 and H2O are the products of the combustion reaction, their partial pressures add up to give the total pressure of the mixture. We are given the total pressure, so we can write: Partial pressure of CO2 + partial pressure of H2O = 1.51 atm

We can use the ideal gas law equation again for CO2 and H2O separately: \(PV_{CO2} = n_{CO2}RT\) and \(PV_{H2O} = n_{H2O}RT\) Divide the first equation by the second equation to get: \(\frac{n_{CO2}}{n_{H2O}} = \frac{PV_{CO2}}{PV_{H2O}} = \frac{P_{CO2}}{P_{H2O}}\) Now, we can solve for the moles of CO2 and H2O using the partial pressures obtained in Step 2.

Using the stoichiometry of the combustion reaction, we can calculate the moles of C and H in the hydrocarbon: moles of C = moles of CO2 (because 1 mole of C in the hydrocarbon produces 1 mole of CO2) moles of H = moles of H2O × (2/1) (because 2 moles of H in the hydrocarbon produce 1 mole of H2O)

Divide the moles of C and H by their respective smallest whole number to get the ratio of C and H in the empirical formula. Finally, to determine the molecular formula, multiply the empirical formula by the smallest whole number that gives the molecular formula with the correct ratio of C and H.