$M=5,286.28=148$
Giả sử có 100g anetol.
$m_C= 81,08g \Rightarrow n_C=\frac{81,08}{12}= 6,7 mol$
$m_H= 8,1g \Rightarrow n_H=8,1 mol$
$m_O= 100-81,08-8,1=10,82g \Rightarrow n_O=\frac{10,82}{16}= 0,67 mol$
$n_C : n_H : n_O= 6,7:8,1:0,67=10:12:1$
$\Rightarrow $ CTĐGN $(C_{10}H_{12}O)_n$
$M=148\Rightarrow n=1$
Vậy CTPT anetol là $C_{10}H_{12}O$