一定做到取3.4g只含羟基,不含其他官能团的液态饱和多元醇于5L氧气中点燃,醇完全燃烧生成二氧化碳和水,反应后气体体积减

0.125 ,0.30 ,0.10 ,5:12:4
V (CO2) = 2.80L ,n (CO2) = 0.125mol n (C) = 0.125mol
V (CO2) + V (O2剩) = 5L－0.56L = 4.44L ,V (O2剩) = 4.44L－2.80L = 1.64L ,
V (O2耗) = 5.00L－1.64L = 3.36L ,n (O2耗) = 0.15mol ,m (O2耗) = 4.80g
又m (H2O) + m (CO2) = 3.40g +4.80g ,m (H2O) = 8.20g－5.50g = 2.70g ,n (H2O) =0.15mol ,n (H) = 0.30mol ,则n (O) = 0.10mol

这醇大概是季戊四醇C(CH2OH)4
3.4g燃烧产生了2.8/22.4=0.125mol
则27.2g该物质中含有1个C
中学范围内，C,H,O原子量都是按整数计算的，让27.2*5=136
设一个它的分子含有5个C原子，3.4g就是0.025mol
因为是饱和的，就可以设它的分子式为C5H12Ox
C5H12Ox+(8-x/2）O2-...