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Empirical & Molecular Formula

0.30g of an o...

Question

$0.30 g$ of an organic compound containing $C, H$ and $O$ on combustion yields $0.44 g$ ${C O}_{2}$ and $0.18 g$ $H_{2} O$ . If one mol of compound weighs $60$ , then molecular formula of the compound is:

{C H}_{2} O

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C_{3} H_{8} O

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C_{4} H_{6} O

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

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Solution

The correct option is D $C_{2} H_{4} O_{2}$
Moles of $C =$ moles of $C O_{2} = \frac{0.44}{44} = 0.01 m o l e s$

Moles of $H = 2 \times$ moles of $H_{2} O = 2 \times \frac{0.18}{18} = 0.02 m o l e s$

Weight of oxygen = ( 0.30 - weight of carbon and hydrogen ) = $(0.3 - 0.01 \times 12 - 0.02 \times 1) = 0.16 g m$

Moles of $O = \frac{0.16}{16} = 0.01 m o l e s$

Empirical formula $= C H_{2} O$
Empirical formula mass $= 30 g m$

$n = \frac{60}{30} = 2$

Formula $= (C H_{2} O)_{n} = C_{2} H_{4} O_{2}$

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0.30g of an organic compound containing C,H and O on combustion yields 0.44g CO2 and 0.18g H2O. If one mol of compound weighs 60, then molecular formula of the compound is:

$0.30 g$ of an organic compound containing $C, H$ and $O$ on combustion yields $0.44 g$ ${C O}_{2}$ and $0.18 g$ $H_{2} O$ . If one mol of compound weighs $60$ , then molecular formula of the compound is: