Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670 g of H2O. What is the empirical formula of the compound?

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1 Answer
Feb 10, 2018

#"Empirical formula"=C_3H_8O#

Explanation:

We assess the moles of carbon by measuring the moles of carbon dioxide...

#"Moles of carbon dioxide"=(1.043*g)/(44.00*g*mol^-1)=0.0237*mol#...and there were thus.............. #0.0237*molxx12.00*g*mol^-1=0.284*g# #"carbon"# in the initial sample...

And the moles of hydrogen by assessing the moles of water...

#"Moles of water"=(0.5670*g)/(18.01*g*mol^-1)=0.0315*mol#...and there were thus.............. #2xx0.0315*molxx1.00794*g*mol^-1=0.0635*g# #"hydrogen"# in the initial sample...why did I multiply by 2?

And the balance of the mass was oxygen....

#(0.600-0.284-0.0635)*g=0.2525*g#...

...a molar quantity of #(0.2525*g)/(15.999*g*mol^-1)=0.01578*mol#.

And so we gots molar quantities with respect to #C#, #H#, AND #O#, we can finally address the empirical formula by dividing thru by the lowest molar quantity to give an empirical formula of.....

#C_((0.0237*mol)/(0.01578*mol))H_((0.0635*mol)/(0.01578*mol))O_((0.01578*mol)/(0.01578*mol))=C_(1.50)H_4O#

But by definition, the #"empirical formula"# is the SIMPLEST WHOLE number ratio...so #"empirical formula"-=C_3H_8O_2#, and this represents a saturated organic molecule.

And finally, we would address the #"molecular formula"#, the which is a whole number multiple of the #"empirical formula"# had we a determination of molecular mass. The question is INCOMPLETE.

i.e. #"molecular formula"=nxx"empirical formula"#