Calculating the empirical formula of organic compound 3/12

Organic can be found by:
- Burning
- Collecting, weighing product
- Mass, mole that can be calculated
Ex. 5g of CxHy burn, then produce 15g of CO2 + 8.18g H20, Let the Empirical foumula - CxHy
mol CO2 = 15.0g CO2 x ( 1 mol of CO2) / (44.0g CO2) = 0.341 mol

mol H20 = 8.18g H2O x (1 mol H2O) / (18.0 g H2O) = 0.454 mol

- 0.454 water , 0.341 Carbon dioxide
#2

mol C = 0.341 mol CO2 x ( 1 mole C) / ( 1 mol CO2) = 0.341 mol
mol H = 0.454 mol H20 x  ( 2 mole H) /  ( 1 mol H2O) = 0.908 mol

0.908 H, 0.341 C > CxHy > could be C 0.341 H 0.908

ratios of moles = (0.908 mol H) / (0.341 mol C) = (2.66) / (1) (as we can't have any decimal in empirical formula, we have to round it up to the closest whole number)(2.66) x 3 > 7.98 > 8
1        x 3 >     3   > 3

The empirical formula > C3H8
DONE !!

As we learned all those empirical formula and molecular formula from the past, the picture from below is a little summary that conver all we had learned since term 2 has started! It's the relationship between EF, MF and Organic compound!