theo sơ đồ

`=>` $\rm M=74(amu)$

$\rm \%O=100-64,86-13,51=21,63$

gọi CTTQ: $\rm C_xH_yO_z$

$\rm x:y:z=\dfrac{64,86}{12} :\dfrac{13,51}{1} : \dfrac{21,63}{16}$

$\rm =5,405:13,51:1,352=4:10:1$

`=>` $\rm CT$ đơn giản nhất $\rm (C_4H_{10}O)_n$

$\rm M=74 \to 74n=74 \to n=1$

`=>`$\rm CTHH: C_4H_{10}O$

`=>` CTCT:

$\rm 1.CH_3CH_2CH_2CH_2OH$

$\rm 2.CH_3CH(CH_3)CH_2OH$

$\rm 3.CH_3CH_2CH(OH)CH-3$

$\rm 4.CH_3C(CH_3)(OH)CH_3$

$\rm 5.CH_3CH(CH_3)OCH_3$

$\rm 6.CH_3OCH_2CH_2CH_3$

$\rm 7.CH_3CH_2OCH_2CH_3$

`=>` số chất thỏa mãn alcohol : 4 (1,2,3,4)