Calculate The Empirical Formula Of Organic Compound~

An organic compound : any substance that contain CARBON

When we do the calculation, we can write a balanced chemical equation for the burning of CxHy
(CxHy + zO2 -----> xCO2 + y/2H2O)

1. We should calculate the moles of CO2 and H2O produced
2. Find the mole of C and H in the CO2 and H2O
3. Find the ratio of C : H
4. Multiply the ratios to get a whole number

For example...
1.  What is the empirical formula of a compound that burns to produce 8.45g of CO2 and 1.73g of H2O?
mol of CO2 = 8.45/44 = 0.19        mol of C in CO2 = 0.19C
mol of H2O = 1.73/18 = 0.10       mol of H in H2O = 0.2H
Mole ratio = 1:1
Therefore,the empirical formula is CH

2. When 6.28g of an organic compound is burned,10.22g of CO2 and 5.18g of H2O is produced?What is the empirical formula?
mole of CO2 = 10.22/44 = 0.232   mol of C in CO2 = 0.232C
mole of H2O = 5.18/18 = 0.288     mol of H in H2O = 0.576H
Check Mass :
0.232C x 10.22g = 2.371g
0.576H x 5.18g = 2.98g
6.28g - 2.371g - 2.98g = 0.929g ----->   Oxygen
0.929/16 = 0.058mol Oxygen
Mole ratio = 0.232 : 0.576 : 0.058
                  = 4        : 10      : 1
Therefore,the empirical formula is C4H10O