Gọi số mol $C_{2}H_{4}$: x
$C_{n}H_{2n-2}$:y
$nCO_{2}=\frac{22,4}{22,4}=1$
Bảo toàn nguyên tố "C"
$2x+ny=nCO_{2}=1$
⇒$ny=1-2x$ (1)
$C_{2}H_{4}+Br_{2} \to C_{2}H_{4}Br_{2}$
$C_{n}H_{2n-2}+2Br_{2} \to C_{n}H_{2n-2}Br_{4}$
$nBr_{2}=\frac{500.16\%}{160}=0,5$
$x+2y=0,5$
⇒$2y=0,5-x$(2)
$28x+(14n-2)y=13,8$
⇔$28x+14ny-2y=13,8$ (3)
Thay (1)(2) vào (3) ta được
⇒$28x+14(1-2x)-(0,5-x)=13,8$
⇔$28x+14-28x-0,5+x=13,8$
⇒$x=0,3$
$\%mC_{2}H_{4}=\frac{0,3.28}{13,8}.100=60,87\%$
$\%mZ=100-60,87=39,13\%$