※ 引述《walk0123 (誓言)》之铭言： : 在0.959atm和298K状态下，一个由C和H所组成的样本 : 在燃烧之後，可以得到一1.51atm和375K的CO2和H2O的混合气体， : 此混合气体的密度是1.39g/L : 且混合物的气体体积是原来纯碳氢化合物的4倍， : 问碳氢化合的分子式? : 答案是C2H6。 : 要如何计算~???? : 原始题目如下! : Consider a sample of a hydrocarbon(a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 K. : Upon combusting the entire sample in oxygen, : you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375K. : 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. P V = n R T n = (P V)/(R T) n前 : n後 = (0.959*V)/(0.082*298) : (1.51*4V)/(0.082*375) = 1 : 5 设sample分子式为CaHb,且初始有 1 mol CaHb + (a+b/4) O2 ---> a CO2 + b/2 H2O 得 a + b/2 = 5 -----(1) 由气体密度得===> (44a + 9b)/4V = 1.39 V = (0.082*298*1)/0.959 = 25.48 将V带回得===> 44a + 9b = 142 -----(2) 由(1),(2)式可得, a = 2 b = 6 -- 我是这样算的@@ 不知道对不对 -- --

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