The empirical formula of a compound is $C H_{2} O$ . If the vapour density is 60, what would be the molecular formula of the compound?

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

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C H_{2} O

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

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

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Solution

The correct option is D $C_{4} H_{8} O_{4}$
As we know,
Molecular weight $=$ Vapour density $\times 2 = 2 \times 60 = 120$
Molecular formula $=$ Empirical formula $\times$ n
$n = \frac{M o l e c u l a r w e i g h t}{E m p i r i c a l f o r m u l a w e i g h t} = 120 / 30 = 4$
so molecular formula $= 4 \times (C H_{2} O) = C_{4} H_{8} O_{4}$

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The empirical formula of a compound is CH2O. If the vapour density is 60, what would be the molecular formula of the compound?

The correct option is D C4H8O4As we know,Molecular weight = Vapour density ×2=2×60=120Molecular formula = Empirical formula × nn=MolecularweightEmpiricalformulaweight=120/30=4so molecular formula =4×(CH2O)=C4H8O4

The empirical formula of a compound is $C H_{2} O$ . If the vapour density is 60, what would be the molecular formula of the compound?