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Empirical & Molecular Formula

0.75 mol of s...

Question

$0.75$ mol of solid $X_{4}$ and $2$ mol of gaseous $O_{2}$ are heated to react completely in sealed vessel to produce only one gaseous compound $Y$ . After the compound is formed, the vessel is brought to the initial temperature, the pressure is found to be the half of the initial pressure. The molecular formula of compound is :

X_{3} O_{4}

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X_{2} O_{4}

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X_{4} O_{4}

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none of the above

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Solution

The correct option is C $X_{3} O_{4}$
Let the molecular formula of the compound is $X_{x} O_{y}$
The balanced chemical equation for combution is
$X_{4} + \frac{2 y}{x} O_{2} \to \frac{4}{x} X_{x} O_{y}$
1 mole of $X_{4}$ will completely react with $\frac{2 y}{x}$ mol of $O_{2}$ .
0.75 mole of $X_{4}$ will completely react with $\frac{1.5 y}{x}$ mol of $O_{2}$ .
But it is equal to 2 moles.
$\frac{1.5 y}{x} = 2$ .....(1)
$1.5 y = 2 x$
Initial pressure $0.75 + \frac{1.5 y}{x}$
Final pressure $\frac{3}{x}$
But, the pressure is found to half the initial pressure.
$0.75 + \frac{1.5 y}{x} = \frac{2 \times 3}{x}$
$0.75 = - \frac{1.5 y}{x} = \frac{6}{x}$
$0.75 x = 6 - 1.5 y$ ......(2)
Substitute (1) into (2)
$0.75 x = 6 - 2 x$
$2.75 x = 6$
$x = 3$
$y = 4$
Hence, the empirical formula is
$X_{3} O_{4}$

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