The structure of ethanol

Determination of the molar mass

$2.3\; g$ of ethanol is vaporized and we note that its vapor occupies at $ 100^oC$ and $1\;atm$ a volume of $1.53\;L$. → Calculation

$pV=nRT$ $n=\frac{pV}{RT}$ = $\frac{1\cdot1.53}{0.082\cdot373.15}$ = $0.05 mol$ $M=\frac{m}{n}$ = $\frac{2.3}{0.05}$ = $46\frac{g}{mol}$

Determination of the molecular formula

Elemental analysis revealed that ethanol contains $52,17\%\;C$ and $13,04\%\;H$ and oxygen. → Calculation

1 mole =$46 \;g$ contain: 1) $\frac{52.17\cdot46}{100}= 24g\; C$ $n_C$ $=$ $\frac{24}{12}$ $=$ $ 2mol\;C$ 2) $\frac{13.04\cdot46}{100}\;g$ $=$ $6\;g\; H$ $n_H$ $=$ $\frac{6}{1}$ $= $ $6\;mol\;H$ 3) $\frac{34.79\cdot46}{100}\;g$ $=$ $16\;g\; O$ $n_O$ $=$ $\frac{16}{16}$ $=$ $1 \;mol\;O$ Molecular formula: $C_2H_6O$

Determination of the structural formula

The empirical formula allows only two possible structural formulas. → Which ?

By reacting with an excess of sodium, $2.3\;g$ of ethanol provide $0.56\;L$ hydrogen under normal conditions. This allows to decide which of the above formulas is correct. → Explain !

$2.3\;g $ ethanol are $\frac{2.3}{46}$ $ =$ $0.05\; mol$ $0.56\;L $ $H_2$ n.t.p. are $\frac{0.56}{22.4}$ $ =$ $0.025\; mol$ $1\;mol$ ethanol produces $\frac{1}{2}\;mol \;H_2$ $1$ molecule ethanol liberates therefore $1$ atom $ H$ So we find one particular $H$ atom in each molecule ethanol, formula (1) is correct !