A 1.638 Gram pure sample of a compound containing only carbon hydrogen and oxygen was burned in excess oxygen gas 3.117 g of carbon dioxide and 1.911 g of water were produced find the empirical formula of the compound

C2H6O

Explanation:

First, find the masses of C, H, and O. I used ratios.

3.117g CO2  x  1 mol CO2/44.01g CO2  x  1 mol C/1 mol CO2  x  12.01g C/1 mol C = 0.850606 g C

1.911g H2O  x  1 mol H2O/18.016g H2O  x  2 mol H2O/1 mol H2O  x  1.008g H/1 mol H = 0.213842 g H

To find the mass of oxygen, just add the masses of C and H together and subtract that from the total mass of the sample.

1.638 - (0.850606 + 0.213842) = 0.57355 g O

Second, find the moles of C, H, and O.

0.850606g C  x  1 mol C/12.01g C = 0.070825 mol C

0.213842g H  x  1 mol H/1.008g H = 0.212145 mol H

0.57355 g O  x  1 mol O/16g O = 0.035847 mol O

Finally, divide the number of moles of each element by the smallest value of moles calculated. In this case, the smallest value is 0.035847.

0.070825/0.035847 = 2

0.212145/0.035847 = 6

0.035847/0.035847 = 1

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"