What is the molecular formula of a compound with the empirical formula $C_{3} H_{6} O_{2}$ and a mass of 148 amu?

C_{6} H_{12} O_{4}

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C_{3} H_{6} O_{2}

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C_{9} H_{18} O_{6}

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None of the above

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Solution

The correct option is A $C_{6} H_{12} O_{4}$
$M o l e c u l a r f o r m u l a = E m p i r i c a l f o r m u l a \times n - f a c t o r$
Also, $M o l a r m a s s o f M F = M o l a r m a s s o f E F \times n - f a c t o r$
$M o l a r m a s s o f M o l e c u l a r f o r m u l a = 148 a m u$ (given)
$M o l a r m a s s o f E m p i r i c a l f o r m u l a (C_{3} H_{6} O_{2}) = 36 + 6 + 32 = 74 u$
So, $148 = 74 \times n - f a c t o r$
$2 = n - f a c t o r$
$∴ M o l e c u l a r f o r m u l a = 2 \times (C_{3} H_{6} O_{2}) = C_{6} H_{12} O_{4}$

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