Answer
$C_{29}H_{50}O_{2}$
Work Step by Step
Let's denote the molecular formula of vitamin E by $C_{x}H_{y}O_{z}$. The reaction equation of combustion of this compound is as follows:
$C_{x}H_{y}O_{z} + (x+\frac{y}{4}-\frac{z}{2})O_{2} \rightarrow xCO_{2}+\frac{y}{2}H_{2}O$
The mass of carbon present in $1.47g$ of carbon dioxide is:
$m(C) = \frac{Ar(C)}{Mr(CO_{2})}\times m(CO_{2})=\frac{12}{44}\times 1.47g = 0.401g$
Therefore, carbon percentage in the vitamin E is equal to $\omega (C) = \frac{0.401g}{0.497g}=80.67\%$
The mass of hydrogen present in $0.518g$ of water is:
$m(H) = \frac{2\times Ar(H)}{Mr(H_{2}O)}\times m(H_{2}O)=\frac{2}{18}\times 0.518g = 0.058g$
Therefore, hydrogen percentage in the vitamin E is equal to $\omega (H) = \frac{0.058g}{0.497g}=11.58\%$.
The rest of the mass is oxygen, so $\omega(O) = 100\% - \omega(C) - \omega(H) = 7.75\%$
Now we can obtain the following equations:
$\frac{\omega(C)}{\omega(H)}=\frac{80.67\%}{11.58\%}=\frac{12x}{y} \implies 7y=12x$
$\frac{\omega(C)}{\omega(O)}=\frac{80.67\%}{7.75\%}=\frac{12x}{16z} \implies 166.5445z=12x$
Since $x$, $y$ and $z$ need to be integers, by taking $x=29$, we will obtain $y=50$ and $z=2$, so the molecular formula of vitamin E is $C_{29}H_{50}O_{2}$.
