An organic compound having $C, H$ and $O$ has $13.13 % H, 52.14 % C$ . Its molar mass is $46.068$ g. What are its empirical and molecular formulae?

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$C_{2} H_{6} O_{2}, C_{3} H_{9} O_{4}$

$C_{2} H_{6} O, C_{2} H_{6} O$

${C H}_{3} O, C_{2} H_{6} O_{2}$

$C_{2} H_{6} O, C_{4} H_{12} O_{2}$

answer is C.

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Detailed Solution

Percentage composition $= \frac{mass of element}{total mass of compound} \times 100$

mass of carbon $= \frac{52.14 \times 46.068}{100} \approx 24 g$

$= 2$ atoms of $C$ .

max of Hydrogen $= \frac{13.13 \times 46.068}{100} = 6 g$

$= 6$ atoms of ' $H$ "'

mass of oxygen $= \frac{[100 - (13.13 + 52.14)] \times 46.068}{100}$

$\approx 16 = 1$ atom of ' $O$ '.

So, molecular formula $= C_{2} H_{6} O$

Since it cannot be further simplified, the empirical Formula is also the same.

Hence, option (C) is correct.

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