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一、由等分子量变换总结通式原子量上:3O=4C(48);1CH_4=1O(16);1C=12H(12)。解题时,在分子量相等的前提下,可以去掉分子中一些原子而加入另一些原子。例1(91高考题):A、B 两有机物分子式和最简式均不同,无论以何种比例混和 A、B,只要总质量不变,完全燃烧生成的水量也不变,试举出6组符合条件的A 和 B。解:符合题设的 A 和 B 所含氢的百分含量(H％)应相同。一是分子量和氢原子数相同的 A 和 B,二是氢
First, from the molecular weight transformation summary formula Atomic weight: 3O = 4C (48); 1CH_4 = 1O (16); 1C = 12H (12). When solving the problem, under the premise of equal molecular weight, some atoms in the molecule can be removed and other atoms can be added. Example 1 (91 college entrance examination question): The molecular formulae of A and B are different from the simplest formula. No matter what ratio is used to mix A and B, as long as the total mass is constant, the amount of water generated by complete combustion will not change. Group A and B that meet the conditions. Solution: The percentage content of hydrogen (H%) in A and B should be the same. One is A and B with the same molecular weight and number of hydrogen atoms, and the other is hydrogen