An organic compound contains 40% of C, 6.66 % of H, 53.34 % of O. Predict it's molecular formula. The density of the compound at STP is 0.00535 g/mL.

C_{5} H_{10} O_{3}

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

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C_{6} H_{14} O_{2}

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C_{3} H_{2} O_{5}

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Solution

The correct option is B $C_{4} H_{8} O_{4}$
Given,
The density of the compound at STP is 0.00535 g/mL.
Mass of 1 mL = 0.00535 g
Molar mass = mass of 22400 mL = $0.00535 \times 22400 = 120$

$\begin{matrix} E l e m e n t & M a s s % & A t o m i c m a s s & A t o m i c R a t i o & S i m p l e s t A t o m i c R a t i o & S i m p l e s t w h o l e N o . R a t i o C & 40.00 & 12 & \frac{40}{12} = 3.33 & \frac{3.33}{3.33} = 1 & 1 \times 2 = 2 H & 6.66 & 1 & \frac{6.66}{1} = 6.66 & \frac{6.66}{3.33} = 2 & 1.5 \times 2 = 3 O & 53.34 & 16 & \frac{53.34}{16} = 3.33 & \frac{3.33}{3.33} = 1.03 \approx 1 & 1 \times 2 = 2 \end{matrix}$

$∴$ Simplest whole no. ratio of C : H : O is 1 : 2 : 1.
$∴$ Empirical formula $= C H_{2} O$

Empirical Formula mass of $C H_{2} O$
$= 12 + 2 + 16 = 30$
$n = \frac{Molecular Mass}{Empirical Mass} = \frac{120}{30} = 4$
$∴$ Molecular Formula = n $\times$ Empirical formula
$= 4 \times C H_{2} O$
$= C_{4} H_{8} O_{4}$

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