Question

2.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 136. g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass 5.18 g carbon dioxide 1.06 g water Use this information to find the molecular formula of X. ? X

Answer 1

1 mol Co, 1 mol C 12.01 g C 44.01g Co, 1mol CO, 1 mol C %C = — 4x100% = 70.7% C 2.00 g 1 mol H2O2 mol H 1.008 g H 1.06 H, 0X18020 HO 1 mol H, 1 mol H X100% = 5.9% H %H = - 2.00 g %O= 100-(%C+%H) = (100–70.7-5.9)% = 23.8% 100 g of compound contains 70.7 g C, 23.8 g 0 and 5.9 g H Number of moles of C = 70.7 g + 12.01 g 1 mol C - 5.89 mol C 1 mol H = 5.85 mol H Number of moles of H = 5.9g x- Number of moles of O = 23.8 g x = 1.5 mol H 16.00 g Among the values of relative no of atoms of 3 elements, choose the least one (1.5) and divide the other values by this least one to get the simplest ratio of atoms C = 5.894 1.5 H = 5.85 = 4 1.5 O = 1.3 = 1 -- -1 Place these simplest ratio values as the lower subscript to the concerned element to get the empirical formula CH,0 ----------- ------------------------------ Empirical formula mass of C,H,O=68g/mol MOlar mass of C,H,0 = 136 g/mol Molecular formula= nxempirical formula molecular mass - Xempirical formula empirical formula mass _136 g/mol cu 68 g/mol = 2x(CH.9) = CH.