Question

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.

• A. $C_5{H}_{10}{O}_3$
• B. $C_4{H}_8{O}_4$
• C. $C_6{H}_{14}{O}_2$
• D. $C_3{H}_2{O}_5$

Answer 1

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 = $0.00535\times 22400=120$

$\begin{array}{cccccc}Element& Mass\%& Atomicmass& AtomicRatio& SimplestAtomicRatio& SimplestwholeNo.Ratio\\ C& 40.00& 12& \frac{40}{12}=3.33& \frac{3.33}{3.33}=1& 1\times 2=2\\ H& 6.66& 1& \frac{6.66}{1}=6.66& \frac{6.66}{3.33}=2& 1.5\times 2=3\\ O& 53.34& 16& \frac{53.34}{16}=3.33& \frac{3.33}{3.33}=1.03\approx 1& 1\times 2=2\end{array}$

$\therefore$ Simplest whole no. ratio of C : H : O is 1 : 2 : 1.
$\therefore$ Empirical formula $=C{H}_{2}O$

Empirical Formula mass of $C{H}_{2}O$
$=12+2+16=30$
$n=\frac{\text{Molecular Mass}}{\text{Empirical Mass}}=\frac{120}{30}=4$
$\therefore$ Molecular Formula = n $\times$ Empirical formula
$=4\times C{H}_{2}O$
$=C_4{H}_8{O}_4$