1) The document describes the simplest formula calculation for combustion of hydrocarbons in excess oxygen. 2) It involves a 5 step process of calculating the grams of carbon, hydrogen, and oxygen based on the combustion reaction and determining the moles of each element. 3) The ratios of each element are then calculated by dividing the moles of each element by the smallest value, in this case oxygen, to determine the simplest formula of C5H10O1.

1. Simplest Formula Prepared by: Stephen BoylanOctober 20, 2009

2. Simplest Formula ...

3. Simplest Formula - Simple001A. Combustion in Excess oxygen 12.795 g. HC ----> 32.689 g. CO2 + 13.377 g. H2OB. 5 Step Calculation C. The Simplest Formula.CxHy Oz

4. Molar Mass CO2 Element Number Atomic Mass Formula n Units, amu Mass, C 1 12.01 12.01 O 2 16.00 32.00 44.01grams per mole CO2

5. Molar Mass H2O Element Number Atomic Mass Formula n Units, amu Mass, n(amu) H 2 1.008 2.016 O 1 16.00 16.00 18.02 grams per mole H2O

6. 2.0 Calculate the grams of carbon, hydrogen and oxygen. 12.795g. HC --> 32.689 g CO2 + 13.377 g. H2OC:32.689 grams CO2 12.01 grams C = 8.921 g. C 44.01 grams CO2H:13.377 grams H2O 2x1.008 grams H = 1.497 g H 18.02 grams H2OO:12.795 g HC - 8.921g C - 1.497 g H= 2.377 g O

7. Moles of carbon, hydrogen and oxygen. C: 8.921 g. C 1 mole C = 0.7429 moles C 12.01 grams C H: 1.497 g. H 1 mole H = 1.485 mole H 1.008 grams H O: 2.377 g. O 1 mole O = 0.1485 mole O 16.00 grams O

8. Ratios Divide by smallest C: 0.7429 moles C 1 = 5.003 ~ 5 moles C 0.1485H: 1.485 mole H 1 = 10 moles H 0.1485O: 0.1485 mole O 1 = 1 moles O 0.1485

9. C5 H10 O1

10. The End