Answer
a) $C_{3}H_{5}O_{2}$.
b) $C_{6}H_{10}O_{4}$.
Work Step by Step
Let's denote the molecular formula adipic acid 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 $2.96g$ of carbon dioxide is:
$m(C) = \frac{Ar(C)}{Mr(CO_{2})}\times m(CO_{2})=\frac{12}{44}\times 2.96g = 0.8073g$
Therefore, carbon percentage in adipic acid is equal to $\omega (C) = \frac{0.8073g}{1.638g}=49.284\%$
The mass of hydrogen present in $1.01g$ of water is:
$m(H) = \frac{2\times Ar(H)}{Mr(H_{2}O)}\times m(H_{2}O)=\frac{2}{18}\times 1.01g = 0.1122g$
Therefore, hydrogen percentage in adipic acid is equal to $\omega (H) = \frac{0.1122g}{1.638g}=6.851\%$.
The rest of the mass is oxygen, so $\omega(O) = 100\% - \omega(C) - \omega(H) = 43.865\%$
Now we can obtain the following equations:
$\frac{\omega(C)}{\omega(H)}=\frac{49.284\%}{6.851\%}=\frac{12x}{y} \implies 7.194y=12x$
$\frac{\omega(C)}{\omega(O)}=\frac{49.284\%}{43.865\%}=\frac{12x}{16z} \implies 17.977z=12x$
a) Since $x$, $y$ and $z$ need to be integers, by taking $x=3$, we will obtain $y=5$ and $z=2$, so the empirical formula of adipic acid is $C_{3}H_{5}O_{2}$.
b) We can obtain the molecular formula of this acid by dividing its real molecular mass by molecular mass of its empirical formula:
$k = \frac{146.1\frac{g}{mol}}{73\frac{g}{mol}}=2$, so the number of each of the atoms in adipic acid is twice the number of atoms in its empirical formula. Hence, the molecular formula of adipic acid is $C_{6}H_{10}O_{4}$.
