Alkanols have the formula CnH2n+1OH\mathrm{C}_{n}\mathrm{H}_{2n+1}\mathrm{OH}CnH2n+1OH.
⇒ 12n+(1×(2n+1))+16.0+1.0=6012n + (1 \times (2n+1)) + 16.0 + 1.0 = 6012n+(1×(2n+1))+16.0+1.0=60
= 14n +18 = 60
n= 60−1814= 4214=3n = \;\frac{60-18}{14} = \;\frac{42}{14} = 3n=1460−18=1442=3
Hence, the alkanol is gotten by putting n=3,
=C3H7OH\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{OH}C3H7OH
