Answer
$$C_9H_{11}O_2N$$
Work Step by Step
1. Find the empirical formula.
n(benzocaine) = $\frac{3.54 \space g}{165 \space g/mol} = 0.0215 \space mol$
$$n(C) = n(CO_2) = \frac{8.49 \space g}{44.0 \space g/mol} = 0.193 \space mol$$ $$n(H) = 2 \times n(H_2O) = 2 \times \frac{2.14 \space g}{18.0 \space g/mol} = 0.238 \space mol$$
- Divide each amount by the amount of benzocaine.
$$C_{\frac{0.193}{0.0215}}H_{\frac{0.238}{0.0215}}O_xN_y $$ $$C_9H_{11}O_xN_y$$
Second experiment:
n(benzocaine) = $\frac{2.35 \space g}{165 \space g/mol} = 0.0142 \space mol$
$$n(N) = \frac{0.199 \space g}{14.0 \space g/mol} = 0.0142 \space mol$$
$$C_9H_{11}O_xN_{\frac{0.0142}{0.0142}}$$ $$C_9H_{11}O_xN$$
Without the oxygen, the molar mass of the compound is:
M = 12.0 * 9 + 1.0 * 11 + 14.0 = 133 g/mol
Thus:
m(O) for 1 mol = 165 g - 133 g = 32 g
$$n(O) = \frac{32 \space g}{16.0 \space g/mol} = 2 \space mol$$
$$C_9H_{11}O_2N$$
