Question

cyanidin a pigment found in many red berries, such as grapes, raspberries, and cranberries, contains carbon, hydrogen, and oxygen. the complete combustion of a 0.80000g sample of this compound in excess O2 produces 1.83856 g CO2 and 0.27604 g H2O.

a) What is the empirical formula for cyanidin?

b) Given that the molar mass of cyanidin is 290 g/mol, what is its molecular formula?

4. Cyanidin, a pigment found in many red berries, such as grapes, raspberries, and cranberries, contains carbon, hydrogen, and oxygen. The complete combustion of a 0.80000 g sample of this compound in excess O2 produces 1.83856 g CO2 and 0.27604 g Но. Show all your work to receive a credit. (10 points) a) What is the empirical formula for cyanidin? b) Given that the molar mass of cyanidin is 290 g/mol, what is its molecular formula?

Answer 1

1,83856 g CO.*44 01g CO,*1 mol CO, 1.83856 Cox 1mol CO, Imol C 12.01g C a)%C= 1 mol C <100% = 62.7% C 0.80000 g 0.2760 1 mol H2O 2 mol H 1.008 g H 18.02g H, 1 mol H, 0 1 mol H %H = - -100% = 3.86% H 0.80000 g %O = 100-%C+%H) = (100-62.7-3.86)% = 33.44% HOX- - - - - - - - - - - - - - - 100 g of compound contains 62. g C, 33.44 g 0 and 3.86 g H Number of moles of C = 62.7 g x = 5.22 mol C 12.01 g 1 mol H - 3.83 mol H Number of moles of H = 3.86 g = 1.008 g Number of moles of O = 33.44 g = 2.09 mol H 16.00 g Among the values of relative no of atoms of 3 elements, choose the least one (209) and divide the other values by this least one to get the simplest ratio of atoms C = 5.22 = 2.5 = x6 = 15 2.09 H = 3.83 = 1.3 = x6 = 11 209 2 = 1 = x6 = 6 Place these simplest ratio values as the lower subscript to the concerned element to get the empirical formula CH6 b) Empirical formula mass of C5H11% = 287.2 g/mol MOlar mass of C5H. = 290 g/mol Molecular formula= nxempirical formula molecular mass -Xempirical formula empirical formula mass 290 g/mol 287.2 g/mol (C5H1.) = 1x(C54,0) = C134,