Question

A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields 15.48 mg CO2 and 7.39 mg H20. The molar mass of the compound is 182.2 g/mol What are the empirical and molecular formulas of the compound? (Enter the elements in the order: C, H, O) The empirical formula: The molecular formula

Answer 1

No of mol of co2 produced = 15.48/44 = 0.352 Mmol

No of mol of c in sample = 0.352 Mmol

No of mol of H2O produced = 7.39/18 = 0.410 mmol

No of mol of H in sample = 0.41*2 = 0.82 mmol

mass of O in sample = 10.68 - (0.352*12+0.82*1) = 5.636 mg

no of mol of O in sample = 5.636/16 = 0.352 mmol

simplest ratio

C = 0.352/0.352 = 1

H = 0.82/0.352 = 2.33

O = 0.352/0.352 = 1

simplest whole number ratio : 3:7:3

empirical formula = C3H7O3

empirical formula mass = 91

n = M.wt/E.wt = 182.2/91 = 2

molecular formula = 2*(C3H7O3)

                  = C6H14O6