Gọi $x$, $y$ là số mol $C_6H_{14}$, $C_8H_{18}$
$\Rightarrow 86x+114y=2,86$ $(1)$
$n_{CO_2}=\dfrac{4,48}{22,4}=0,2(mol)$
Bảo toàn $C$: $6n_{C_6H_{14}}+8n_{C_8H_{18}}=n_{CO_2}$
$\Rightarrow 6x+8y=0,2$ $(2)$
$(1)(2)\Rightarrow x=0,02; y=0,01$
$\%m_{C_6H_{14}}=\dfrac{0,02.86.100}{2,86}=60,14\%$
$\%m_{C_8H_{18}}=39,86\%$