Dry heating of $6.2 g$ fo $C_{x} (H_{2} O) y$ . produces $2.48 g$ carbon. The empirical formula of compound is:

C H_{2} O

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C H_{4} O_{2}

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C_{2} H_{2} O

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C_{3} H_{4} O_{2}

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Solution

The correct option is D $C_{3} H_{4} O_{2}$

Law of equivalent on $C$ atom
$molecule mass 12 x + 18 y = \frac{6.2}{12 x + 18 y} \times x = \frac{2.48}{12}$
$\frac{62 x}{2 x + 3 y} = \frac{2.48}{2} = \frac{x}{4} = \frac{2}{3}$
$\frac{x}{y} = \frac{2}{3}$
$x = y$
$\begin{matrix} molecular mass & = 12 x + 18 \times \frac{3}{2} x = 31 x \end{matrix}$
Hence $x = 3 & y = 2$
$C_{3} {(H_{2} θ)}_{2} = C_{3} H_{4} O_{2} - ------- -$
Correct answer $- D - -$

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Dry heating of 6.2g fo Cx(H2O)y. produces 2.48g carbon. The empirical formula of compound is:

The correct option is D C3H4O2Law of equivalent on C atom molecule mass 12x+18y=6.212x+18y×x=2.481262x2x+3y=2.482=x4=23xy=23x=y molecular mass =12x+18×32x=31xHence x=3&y=2C3(H2θ)2=C3H4O2–––––––––Correct answer −D––

Dry heating of $6.2 g$ fo $C_{x} (H_{2} O) y$ . produces $2.48 g$ carbon. The empirical formula of compound is: