4.6g有机物在氧气中完全燃烧,测得生成co2的质量为8.8g生成H2O为5.4g确定该有机物的化学式

解: ∵n(C)=n(CO2)=8.8g/(44g/mol)=0.2mol,n(H)=2n(H2O)=2*5.4g/(18g/mol)=0.6mol,∴m(C)=0.2mol*12g/mol=2.4g,m(H)=0.6mol*1g/mol=0.6g,根据质量守恒定律可得:此有机物中含有O元素,且m(O)=4.6g-2.4g-0.6g=1.6g,∴n(...