The empirical formula of a compound is CH $_{2}$ O. If its vapour density is 90; find out the molecular formula of the compound.

C_{5} H_{10} O_{5}

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C_{4} H_{8} O_{4}

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

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

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Solution

The correct option is C $C_{6} H_{12} O_{6}$
Emperical mass of $C H_{2} O$ = 30g
Vapour density = 90
Molecular mass = $2 \times 90$ = 180g
Therefore n = $\frac{180}{30}$ = 6
Thus, the formula of the compound is ; $(C H_{2} O)_{6}$ = $C_{6} H_{12} O_{6}$

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The empirical formula of a compound is CH2O. If its vapour density is 90; find out the molecular formula of the compound.

The correct option is C C6H12O6Emperical mass of CH2O = 30gVapour density = 90Molecular mass = 2×90 = 180gTherefore n = 18030 = 6Thus, the formula of the compound is ; (CH2O)6 = C6H12O6

The empirical formula of a compound is CH $_{2}$ O. If its vapour density is 90; find out the molecular formula of the compound.

The correct option is C $C_{6} H_{12} O_{6}$
Emperical mass of $C H_{2} O$ = 30g
Vapour density = 90
Molecular mass = $2 \times 90$ = 180g
Therefore n = $\frac{180}{30}$ = 6
Thus, the formula of the compound is ; $(C H_{2} O)_{6}$ = $C_{6} H_{12} O_{6}$