Question

 

Combustion of 0.2000 g sample of Vitamin C yields 0.2998 g CO₂ and 0.0819 g H₂O. What is the mass percent of each element? What is the empirical formula for Vitamin C?

Answer 1

moles of CO2 = 0.2998 / 44 = 6.81 x 10^-3

moles of C = 6.81 x 10^-3

mass of C = 6.81 x 10^-3 x 12

               = 0.0818

moles of H2O = 0.0819 / 18

                     = 4.55 x10^-3

moles of H = 2 x 4.55 x10^-3

                 = 9.1 x 10^-3

mass of H = 9.1 x 10^-3

mass of O = 0.2 - (0.0818 +9.1 x 10^-3 )

                = 0.1091 g

moles of O = 0.1091 / 16

                 = 6.82 x 10^-3

moles ratio C: H : O = 6.81 x 10^-3 : 9.1 x 10^-3 : 6.82 x 10^-3

                              = 1 : 1.33 : 1

                              = 1 : 4 / 3 : 1

                             = 3 : 4 : 3

empirical formula = C₃H₄O₃

mass percent of C= 0.0818 x 100 / 0.200

                              = 40.9%

mass percent of H = 9.1 x 10^-3 x 100 / 0.2

                                 = 4.55%

mass of pecent of O = 100 - (40.9 + 4.55)

                               = 54.6 %