Question

determinations - (1) mass of covered crucible and Mg, g (2) mass of crucible and cover, g P.8 g 17.355 0.45g 18.095 0.749 (3) mass of Mg, g - - - (4) mass of covered crucible and magnesium oxide, g (5) mass of magnesium oxide, (6) mass of O that combined with Mg, g (7) number of moles of Mg (8) number of moles of O (9) ratio of number of moles of Mg to the smaller of the two numbers of moles of each (10) ratio of number of moles of O to the smaller of the two numbers of moles of each (11) empirical formula of the oxide of magnesium Calculations (Show all your work. Use additional paper if necessary)

Answer 1

ul penal produit 3) Mass of Mg 0.45g = 0.29g =0.018125ma 16 g mol' (9) Ratio of Mg = 0.01851 mol = 1.02 0.018125 mol 7 number of moles of Mg = mass atomic mass 0.45g = 1 24.305 gmat (10) Ratio of no = =0.01851 mol 11) Emporial formula - Mg 1.020, 6) Mass of 0 that combined = 0.29 g with Mg Moles of o (nog) = mass Thus, emporial formula of Magnesium oxide was found to be Mgo mass atomic mass 8)