Question

A 5.389 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 7.900 grams of CO2 and 3.235 grams of H2O are produced.

In a separate experiment, the molar mass is found to be 60.05 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

Let the organic compound sample be $C_xH_yO_z$

Moles of the sample$=\frac{5.389}{60.05}\approx0.09mol$

Moles of $CO_2$ formed$=\frac{7.900}{44}\approx0.18mol$

Moles of $H_2O$ formed$=\frac{3.235}{18}\approx0.18mol$

So from $1 mol$ $C_xH_yO_z$, $2 mol$ $CO_2$ and $H_2O$ are formed.

General combustion equation will be:

$C_xH_yO_z+O_2\to 2CO_2+2H_2O$

Hence $x=2$ and $y=4$

As molar mass of $C_xH_yO_z=60.05\frac{g}{mol}$, so $z=2$

Therefore, the molecular formula of the given sample of organic compound$=C_2H_4O_2$

And the empirical formula$=CH_2O$