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方法(1)从最简式入手 例题1、某化合物由C、H两种元素组成,其中 含C的质量分数为85.7％,在标准状况下,11.2L此 化合物气体的质量为14g,求此化合物的分子式。 解:该烃的摩尔质量为Mr=14 g÷(11.2L／22.4 L／mol)=28 g／mol且分子中C、H原子个数比为 85.7％／12:(1-85.7％)／1=1:2,所以该烃的最简式为 CH2,分子式应为(CH2)n,再结合其摩尔质量则有(12+ 2)n=28,得出n=2,故此化合物的分子式为C2H4。
Method (1) Starting from the simplest example, a certain compound consists of two elements: C and H. The mass fraction of C is 85.7%. Under standard conditions, the mass of 11.2 L of this compound gas is 14 g. The molecular formula of the compound. Solution: The molar mass of this hydrocarbon is Mr=14 g÷(11.2 L/22.4 L/mol)=28 g/mol and the number of C and H atoms in the molecule is 85.7%/12:(1-85.7%)/ 1=1:2, so the simplest formula of this hydrocarbon is CH2, the molecular formula should be (CH2)n, and combined with its molar mass, (12+2)n=28, which gives n=2, so the molecular formula of this compound is For C2H4.