问题
多选题
完全燃烧2.00g有机物,可生成4.40g CO2和1.44g H2O,则该有机物的化学式为( )
A.C5H8
B.C5H8O
C.C5H8O2
D.C10H16O4
答案
n(CO2)=
=0.10mol,4.40g 44g/mol
则2.00g有机物中:n(C)=n(CO2)=0.10mol,m(C)=0.10mol×12g/mol=1.20g,
n(H2O)=
=0.08mol,1.44g 18g/mol
则2.00g有机物中:n(H)=2n(H2O)=0.16mol,m(H)=0.16mol×1g/mol=0.16g,
因(1.20g+0.16g)<2.00g,
所以有机物中还应还用O元素,且m(O)=2.00g-1.20g-0.16g=0.64g,n(O)=
=0.04mol,0.64g 16g/mol
则有机物中:n(C):n(H):n(O)=0.10mol:0.16mol:0.04mol=5:8:2,
所以该有机物的最简式为C5H8O2,
分子式为C5H8O2或C10H16O4,
故选CD.