A hydrocarbon contains $10.5 g$ C-per gram atom of hydrogen, 1 litre hydrocarbon at $127^{o} C$ and 1 atm pressure weight $2.8 g$ . Find out the empirical formula and the molecular formula.

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Solution

Given $P = 1 a t m V = 1 L, T = 127^{o} C = 127 + 273 = 4000 K$
$P V = n R T$ (Ideal gas equation)
Or $n = P V / R T = 1 \times \frac{1}{0.082} \times 400 = 0.0304$
$M o l a r m a s s = \frac{M a s s}{M o l e s} = \frac{2.8}{0.0304} = 92.10$

Carbon 10.5 $\frac{10.5}{12} = .875$ $\frac{.875}{.875} = 1 \times 7 = 7$
Hydrogen 1 $\frac{1}{1} = 1$ $\frac{1}{.875} = 1.14 \times 7 = 8$

Emperical formula, $w t = 12 \times 7 + 1 \times 8 = 92$
Emperical formula = $C_{7} H_{8}$
$n = \frac{M o l . w t}{E m p . w t} = \frac{92.10}{92} = 1$
molecular formula $= n \times$ empirical formula
$= 1 (C_{7} H_{8}) = C_{7} H_{8}$

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Carbon	10.5	$\frac{10.5}{12} = .875$	$\frac{.875}{.875} = 1 \times 7 = 7$
Hydrogen	1	$\frac{1}{1} = 1$	$\frac{1}{.875} = 1.14 \times 7 = 8$

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