Answer
Both the empirical and molecular formula of this compound are $C_{3}H_{6}O_{2}$.
Work Step by Step
Let's denote the molecular formula 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 $1.114g$ of carbon dioxide is:
$m(C) = \frac{Ar(C)}{Mr(CO_{2})}\times m(CO_{2})=\frac{12}{44}\times 1.114g = 0.3038g$
Therefore, carbon percentage in the compound is equal to $\omega (C) = \frac{0.3038g}{0.625g}=48.611\%$
The mass of hydrogen present in $0.455g$ of water is:
$m(H) = \frac{2\times Ar(H)}{Mr(H_{2}O)}\times m(H_{2}O)=\frac{2}{18}\times 0.455g = 0.0506g$
Therefore, hydrogen percentage in this compound is equal to $\omega (H) = \frac{0.0506g}{0.625g}=8.089\%$.
The rest of the mass is oxygen, so $\omega(O) = 100\% - \omega(C) - \omega(H) = 43.3\%$
Now we can obtain the following equations:
$\frac{\omega(C)}{\omega(H)}=\frac{48.611\%}{8.089\%}=\frac{12x}{y} \implies y=2x$
$\frac{\omega(C)}{\omega(O)}=\frac{48.611\%}{43.3\%}=\frac{12x}{16z} \implies 3z=2x$
a) Since $x$, $y$ and $z$ need to be integers, by taking $x=3$, we will obtain $y=6$ and $z=2$, so the empirical formula of this compound is $C_{3}H_{6}O_{2}$.
b) We can obtain the molecular formula of this compound by dividing its real molecular mass by molecular mass of its empirical formula:
$k = \frac{74.1\frac{g}{mol}}{74\frac{g}{mol}}=1$, so the number of each of the atoms in this compound equals the number of atoms in its empirical formula. Hence, the molecular formula of this compound is $C_{3}H_{6}O_{2}$.
