n CO2 =0,672/22,4 = 0,03(mol)
n H2O = 0,63/18 = 0,035(mol)
n H2O > n CO2 nên X no
Bảo toàn nguyên tố C,H :
n C = n CO2 = 0,03(mol)
n H = 2n H2O = 0,07(mol)
n O = (m X - m C - m H)/16 = (0,51 -0,03.12 - 0,07)/16 = 0,005(mol)
Ta có :
n C : n H : n O = 0,03 : 0,07 : 0,005 = 6 : 14 : 1
Vậy CTPT của X là C6H14O
CH3-CH2-CH2-CH2-CH2-CH2OH
CH3-CH(CH3)-CH2-CH2--CH2-OH
CH3-CH2-CH(CH3)-CH2-CH2-OH
CH3-CH2-CH2-CH(CH3)-CH2-OH
CH3-CH(CH3)-CH(CH3)-CH2-OH
CH3-CH2-C(CH3)2-CH2-OH
CH3-C(CH3)2-CH2-CH2-OH
CH3-CH2-CH(C2H5)-CH2-OH
CH3-CH2-CH2-CH2-CH(OH)-CH3
CH3-CH(CH3)-CH2-CH(OH)-CH3
CH3-CH2-CH(CH3)-CH(OH)-CH3
CH3-CH2-CH2-C(CH3)(OH)-CH3
CH3-CH(CH3)-C(CH3)(OH)-CH3
CH3-C(CH3)2-CH(OH)-CH3
CH3-CH2-CH2-CH(OH)-CH2-CH3
CH3-CH(CH3)-CH(OH)-CH2-CH3
CH3-CH2-C(CH3)(OH)-CH2-CH3