CH 115 Fall 2014 Worksheet 19 1. Phenyl magnesium bromide is used as a Grignard reagent in organic synthesis. Determine its empirical and molecular formula if its molar mass is 181.313 g/mol and it contains 39.7458% C, 2.77956% H, 13.4050% Mg, and a unknown percentage of Br. All percentages must add up to 100 – can find the missing percentage of Br. 100-55.93036 = 44.06964% Br Assume 100 g of sample 39.7458 g C, 2.77956 g H, 13.4050 g Mg, 44.06964 g Br Convert grams to moles. 39.7458 g C / 12 g/mol = 3.309392173 mol C 2.77956 g H / 1 g/mol = 2.77956 mol H 13.4050 g Mg / 24.31 g/mol = 0.5514191691 mol Mg 44.06964 g Br / 79.9 g/mol = 0.5515599499 mol Br Divide through by the lowest number of moles. C=6 H=5 Mg = 1 Br = 1 Empirical formula = C6H5MgBr To find the molecular formula, first calculate the molar mass of the empirical formula. 181.27 g/mol Divide the molar mass of the molecular formula by the empirical formula to calculate the factor you must multiply the empirical formula by. In this case, 181.313/181.27 = 1, so the molecular formula = C6H5MgBr 2. A 3.585-gram sample of a hydrocarbon contains 1.388 grams of C, 0.345 grams of H, and 1.850 grams of O. Its molar mass is 62 g/mol. Determine its empirical and molecular formula. No percentages given – first need to calculate the percentages of C, H, and O present in this sample. 1.388 g C/3.585 g = 38.71687587% C 0.345 g H/3.585 g = 9.623430962% H 1.850 g O/3.585 g = 51.60390516% O Now, assume 100 g and follow the rest of the steps. 38.71687587 g C / 12 g/mol = 3.226406323 mol C 9.623430962 g H / 1 g/mol = 9.623430962 mol H 51.60390516 g O / 16 g/mol = 3.225244073 mol O Divide through by least number. C=1 H=3 O=1 Empirical formula = CH3O Molar mass of empirical formula = 31 g/mol 62/31 = 2 molecular formula = C2H6O2 CH 115 Fall 2014 3. Balance the following chemical equations. a). Br2 (l) + 2 H2O (l) + SO2 (g) 2HBr (aq) + H2SO4 (aq) b). 2MnO2 + 2 H2SO4 2MnSO4 + O2 + 2H2O c). 2FeCl3 + Be3(PO4)2 3BeCl2 + 2FePO4 d). 2C2H3O2Br + 3O2 4CO2 + 2H2O + 2HBr Worksheet 19