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"
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