[ClarkKent]

When 1.24g of an organic compound with the formula CxHyOz is burned in excess oxygen, 1.76g of carbon dioxide and 1.08g of water vapor are obtained.  What is the empirical formula of the compound?

i found the mmol's of CO2 and H2O (39.99mmol and 59.95mmol) then tried to find the percent of C and H in the compound and I got like 34.335%C and 387.35%H which is obviously Wrong with a capitol W...  (59.99mmol * 1.008)/1.24g = 387.35%H is how i did it... I am lost now.

[Yggdrasil]

It should be 0.05995mol*1.008/1.24g since the atomic mass of hydrogen is in units of g/mol not g/mmol

[xiankai]

here's a simple way to figure it out: convert given data to moles.

1.76g of CO2                 = 0.04 mol of CO2
1.08g of H2O                 = 0.06 mol of H2O
(1.76+1.08-1.24)g of O2 = 0.05 mol of O2

based on molar ratios,

nCxHyOz + 5O2       --> 4CO2 + 6H2O

xC + yH + zO + 10O --> 4C + 12H + 14O

x=4, y=12, z=4

C4H12O4

empirical formula... u should know how to simplify it.

[sdekivit]

importnt to remember is that no mass will be lost or gained, like xiankai did.