91Ó°ÊÓ

To find the molar mass of the hydrocarbon, we need to find the mass of carbon and hydrogen in the sample. We can do this by calculating the mass of CO₂ and H₂O, since they are the products of the combustion reaction. The volume of the mixture after combustion is four times the volume of the hydrocarbon, so their molar ratios are the same. Because PV = nRT, we can relate the pressure and temperature of the mixture with the moles of hydrocarbon (CₓHᵧ) and CO₂ and H₂O. Let's call "n" the number of moles of the hydrocarbon. For the hydrocarbon: \(0.959 \mathrm{~atm} \cdot V = n \cdot R \cdot 298 \mathrm{~K}\) For the mixture of CO₂ and H₂O: \(1.51 \mathrm{~atm} \cdot 4V = (nCO₂ + nH₂O) \cdot R \cdot 375 \mathrm{~K}\) We can also relate the number of moles of CO₂ and H₂O with their densities, masses, and molar masses. Let mCO₂ and mH₂O be the masses of carbon dioxide and water vapor: \(mCO₂ = nCO₂ \times MCO₂\) \(mH₂O = nH₂O \times MH₂O\) The density of the mixture is 1.391 g/L, and since the volume of the mixture is 4 times the volume of the hydrocarbon, the total mass of CO₂ and H₂O can be expressed as: \(4V \times 1.391 \mathrm{~g} / \mathrm{L} = mCO₂ + mH₂O\)

We will eliminate the unknown masses of mCO₂ and mH₂O by substituting the equations for the masses with their corresponding equations of moles: \(4V \times 1.391 \mathrm{~g} / \mathrm{L} = nCO₂ \times MCO₂ + nH₂O \times MH₂O\) Now, we will relate the number of moles of CO₂ and H₂O with the number of moles of the hydrocarbon using stoichiometry of the combustion reaction: \(CₓHᵧ + aO₂ \rightarrow bCO₂ + cH₂O\) From the above combustion reaction, we find: \(nCO₂ = b \cdot n\) \(nH₂O = c \cdot n\) Now, substitute the values of nCO₂ and nH₂O into the density equation: \(4V \times 1.391 \mathrm{~g} / \mathrm{L} = (b \cdot n) \times MCO₂ + (c \cdot n) \times MH₂O\) Thus, we have an equation of the form: \(4V \times 1.391 (\frac{g}{L}) = n(\frac{b \times M_{CO_{2}} + c \times M_{H_{2}O})\)

Divide both sides of the equation by Vn: \(4 \times 1.391 (\frac{g}{L}) = \frac{b \times M_{CO_{2}} + c \times M_{H_{2}O}}{V}\) Now, we can simplify this equation between n, b, and c: \(M_{C_xH_y} = \frac{4 \times 1.391}{V} \times (b \times M_{CO_{2}} + c \times M_{H_{2}O})\)

From the equation in step 3, the molar mass of the hydrocarbon is a multiple of the sum of the molar masses of COâ‚‚ and Hâ‚‚O. We can now find the lowest whole number ratio of carbon and hydrogen atoms in the molecular formula. We must assume x moles of carbon and y moles of hydrogen in the given mass. By trial and error or using the method of continuous variation, we find that the molecular formula of the hydrocarbon that satisfies this condition is Câ‚‚Hâ‚„. So, the molecular formula of the hydrocarbon is Câ‚‚Hâ‚„, also known as ethene or ethylene.