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有一类有机混合物计算型选择题,表面上看已知条件不足,求解较困难。但如能对混合物的分子式进行恰当的处理,即进行合并或分拆,则问题立即就会变得简单、明朗。下面略举两例: 一、合并法 例1 由苯、苯乙烯、苯酚组成的混合物,经测定其中氢的质量百分含量为7％,则氧的质量百分含量为( )。 A.6％ B.9％ C.11％ D.15％ 解析:由三种物质的分子式:苯(C_6H_6)、苯乙烯(C_8H_8)、苯酚(C_6H_6O)可知,不论把混合物按何种比例混合,C与H的物质的量之比总为1:1,即可把三种混合物合并成C_nH_nO_m,此时,碳的质量百分含量为C％=12×7％=84％,则氧的质量百分含量为:O％=1-7％-84％=9％。答案应选B。 二、分拆法 例2 由环己烷、乙醇、乙醚组成的混合物,经测定其中碳的质量百分含量为72％,则氧的质量百分含量为( )。 A.14.2％ B.16％ C.17.8％ D.19.4％ 解析:此题和例一初看很相似,但实际上有差别。
There is a type of organic compound calculation multiple-choice questions. On the surface, the known conditions are insufficient and the solution is difficult. However, if the molecular formula of the mixture can be properly handled, that is, merged or split, the problem will become simple and clear immediately. Here are two examples: First, the combined method Example 1 is a mixture of benzene, styrene, and phenol. The mass percentage of hydrogen is determined to be 7%, then the mass percentage of oxygen is (). A.6% B.9% C.11% D.15% Analysis: It can be seen from the molecular formulae of three substances: benzene (C_6H_6), styrene (C_8H_8), and phenol (C_6H_6O), regardless of the mixture ratio The total amount of substances between C and H is 1:1, and the three mixtures can be combined into C_nH_nO_m. In this case, the mass percentage of carbon is C%=12×7%=84%, then the oxygen The mass percentage is: O%=1-7%-84%=9%. The answer should be B. Second, the separation method Example 2 from the cyclohexane, ethanol, ethyl ether mixture, the determination of which the mass percentage of carbon is 72%, then the mass percentage of oxygen is (). A.14.2% B.16% C.17.8% D.19.4% Analysis: This question is very similar to the first example, but it is actually different.