Question

a certain anesthetic contains 64.9 percent c, 13.5 percent h, and 21.6 percent o by mass. at 121.0 °c and 781 mmhg, 3.50 l of the gaseous compound weighs 8.26 g. what is the molecular formula of the compound? note: reference the fundamental constants table for additional information.

Explanation:

Step1: Find empirical formula

Assume 100g sample: 64.9g C, 13.5g H, 21.6g O.
Moles: $n_C=\frac{64.9}{12.01}\approx5.40$, $n_H=\frac{13.5}{1.008}\approx13.4$, $n_O=\frac{21.6}{16.00}=1.35$.
Ratio: $5.40/1.35=4$, $13.4/1.35≈10$, $1.35/1.35=1$. Empirical formula: $C_4H_{10}O$.

Step2: Calculate molar mass

Use $PV=nRT$, $n=\frac{m}{M}$ → $M=\frac{mRT}{PV}$.
$P=\frac{781}{760}≈1.0276$ atm, $T=121+273.15=394.15$ K, $V=3.50$ L, $m=8.26$ g, $R=0.0821$ L·atm/(mol·K).
$M=\frac{8.26×0.0821×394.15}{1.0276×3.50}≈74$ g/mol.

Step3: Find molecular formula

Empirical mass: $4×12 + 10×1 + 16=74$ g/mol.
Molar mass = empirical mass → molecular formula = empirical formula.

Answer:

$C_4H_{10}O$