Question

0.6908 g of an organic compound containing only C, H, and O was burned in excess of pure O2, producing 1.7604 g of CO2 and 0.4504 g of H2O.

6.a) Calculate the number of grams of C and H in the combustion products and therefore in the 0.6908 g of the compound. Also calculate the number of grams of O in the 0.6908 g. Show your work.

6.b) What is the empirical formula of this compound?

Answer 1

6.a)

weight of organic compound = 0.6908 g

44 grams of CO₂ contains 12 grams of 'C'

1.7604 g of CO₂ contains --? grams of 'C' = (12/44) x 1.7604 = 0.4801 g

18 grams of H₂O contains 2 grams of 'H'

0.4504 g of H₂Ocontains --? grams of 'H' = (2/18) x 0.4504 = 0.05 g

weight of 'O' in the compound = Total weight - (wight of 'C' + weight of 'H')

                                           = 0.6908 - (0.4801 +0.05) = 0.6908-0.5301 = 0.1607 g

6.b)                                                             C                            H                          O

weights in the compound (g)                       0.4801                      0.05                   0.1607

weight/ atomic weight                                0.4801/12                  0.05/1                 0.1607/16

                                                             0.04                         0.05                    0.01

Ratio of C,H and O in the compound = 4:5:1

Empirical formula = C₄H₅O