Answer to Question #95880 in General Chemistry for Brittany Wallace

Question #95880
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.
1
Expert's answer
2019-10-07T06:42:33-0400

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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