When 0.5000 grams of an unknown hydrocarbon,C_xH_y,is completely combusted with excess oxygen,1.037 L CO₂ gas and is produced at 98.3 °C and 1.000 atm.What is the empirical formula of the hydrocarbon? (R = 0.08206 L⋅atm/mol⋅K) A) CH B) CH₂ C) C₂H₃ D) C₃H₅ E) C₃H₈

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Quizplus · Accepted Answer

From the given hydrocarbon, we can see that it contains carbon (C), hydrogen (H), and possibly sulfur (S). First, we need to find the number of moles of CO2 produced in the combustion: 0.5000 g hydrocarbon x (1 mol hydrocarbon / molar mass hydrocarbon) x (1 mol CO2 / 1 mol hydrocarbon) = 0.0213 mol CO2 Next, we can use the ideal gas law to find the volume of CO2 produced: PV = nRT (1.000 atm) (1.037 L) = (0.0213 mol) (0.08206 L⋅atm/mol⋅K) (371.3 K) Solving for V, we get: V = 0.871 L CO2 Now we can use the mole ratio between CO2 and the hydrocarbon to determine the empirical formula. Since the carbon in the hydrocarbon goes to CO2 and the hydrogen goes to H2O, we need to balance the equations: C + O2 → CO2 2H + O2 → H2O For every mole of CO2 produced, one mole of carbon is consumed, and for every 2 moles of water produced, one mole of hydrogen is consumed. Therefore: Empirical formula = CxHySz x = 1 (one carbon atom for every mole of CO2 produced) y = 2 (two hydrogen atoms for every 2 moles of water produced) z = 1 (no sulfur atoms since there is no sulfur in the empirical formula) Therefore, the empirical formula is CH2, and the only choice that matches is E, C₃H₈.

When 05000 Grams of an Unknown Hydrocarbon,CxHy,is Completely Combusted with Excess

When 05000 Grams of an Unknown Hydrocarbon,C_xH_y,is Completely Combusted with Excess