Answer
$$C_{18}H_{20}O_2$$
Work Step by Step
1. Find the empirical formula of equilin.
$M(CO_2) = 44.0 \space g/mol$
$M(H_2O) = 18.0 \space g/mol$
$M(O_2) = 32.0 \space g/mol$
$$n(C) = n(CO_2) = \frac{39.61 \space g}{44.0 \space g/mol} = 0.900 \space mol$$ $$n(H) = 2 \times n(H_2O) = 2 \times \frac{9.01 \space g}{18.0 \space g/mol} = 1.00 \space mol$$
$$m(C) = 0.900 \space mol \times 12.0 \space g/mol = 10.8 \space g$$ $$m(H) = 1.00 \space mol \times 1.0 \space g/mol = 1.0 \space g$$
$$m(equilin) = m(C) + m(H) + m(O)$$ $$m(O) = m(equilin) - m(C) - m(H)$$ $$m(O) = 13.42 \space g - 10.8 \space g - 1.0 \space g = 1.6 \space g$$
$$n(O) = \frac{1.6 \space g}{16.0 \space g/mol} = 0.10 \space mol$$
$$equilin = C_{0.900}H_{1.00}O_{0.10}$$
$\div 0.10$
$$equilin = C_{9}H_{10}O_{1}$$
2. Calculate its molar mass and find the proportion.
$M(C_9H_{10}O) = 12.0 * 9 + 1.0 * 10 + 16.0 = 134.0 \space g/mol$
$$\frac{268.34 \space g/mol}{134.0 \space g/mol} \approx 2$$
Thus:
$$equilin = 2 * C_9H_{10}O = C_{18}H_{20}O_2$$
