Answer
$$C_{18}H_{22}O_2$$
Work Step by Step
1. Find the empirical formula of estrone.
$M(CO_2) = 44.0 \space g/mol$
$M(H_2O) = 18.0 \space g/mol$
$M(O_2) = 32.0 \space g/mol$
$$n(C) = n(CO_2) = \frac{5.545 \space g}{44.0 \space g/mol} = 0.126 \space mol$$ $$n(H) = 2 \times n(H_2O) = 2 \times \frac{1.388 \space g}{18.0 \space g/mol} = 0.154 \space mol$$
$$m(C) = 0.126 \space mol \times 12.0 \space g/mol = 1.51 \space g$$ $$m(H) = 0.154 \space mol \times 1.0 \space g/mol = 0.154 \space g$$
$$m(estrone) = m(C) + m(H) + m(O)$$ $$m(O) = m(estrone) - m(C) - m(H)$$ $$m(O) = 1.893 \space g - 1.51 \space g - 0.154 \space g = 0.229 \space g$$
$$n(O) = \frac{0.229 \space g}{16.0 \space g/mol} = 0.0143 \space mol$$
$$estrone = C_{0.126}H_{0.154}O_{0.0143}$$
$\div 0.0143$
$$estrone = C_{8.8}H_{10.8}O_{1} \approx C_9H_{11}O$$
2. Calculate its molar mass and find the proportion.
$M(C_9H_{11}O) = 12.0 * 9 + 1.0 * 11 + 16.0 = 135.0 \space g/mol$
$$\frac{270.36 \space g/mol}{135.0 \space g/mol} \approx 2$$
Thus:
$$estrone = 2 * C_9H_{11}O = C_{18}H_{22}O_2$$
