The empirical formula of an organic compound is $C H_{2} O$ . Its vapour density is $45$ . The molecular formula of the compound is:

C_{2} O

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

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C_{3} H_{6} O_{3}

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C_{6} H_{12} O_{6}

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Solution

The correct option is A $C_{3} H_{6} O_{3}$
Molecular weight $= 2 \times$ Vapour density

$= 2 \times 45 = 90$

Let the molecular formula be $C_{x} H_{2 x} O_{x}$ .

Therefore, $90 = 12 x + 2 x + 16 x = 30 x$

we get, $x = 3$ .

Hence, the molecular formula is $C_{x} H_{2 x} O_{x} = C_{3} H_{6} O_{3}$ .

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The empirical formula of an organic compound is CH2O. Its vapour density is 45. The molecular formula of the compound is:

The correct option is A C3H6O3Molecular weight =2× Vapour density =2×45=90Let the molecular formula be CxH2xOx.Therefore, 90=12x+2x+16x=30xwe get, x=3.Hence, the molecular formula is CxH2xOx=C3H6O3.

The empirical formula of an organic compound is $C H_{2} O$ . Its vapour density is $45$ . The molecular formula of the compound is: