A. 1ding the Simplest Formula 36.58 1. Mass of empty crucible + cover g Shiny, light gray, 2. Initial appearance of the magnesium ribbon 3. Mass of crucible + cover + magnesium 36.79 g 4. Mass of crucible + cover + oxide product 36.94 g 5. Mass of magnesium 0.21 g 6. Mass of magnesium compound 0.36 g 7. Mass of oxygen in the product 0.15 g 8. Moles of Mg Mg atomic weight = 24.305 g/mol 0.009 mole 0.009 mole Calculations mol of Mg = (0.21g ÷ 24.305 g/mol) = 0.009 moles 9. Moles of O (Show calculations.) O atomic weight = 16.00 g/mol mol of O = (0.15g ÷ 16.000 g/mol) = 0.009 moles Mg and O have similar 10. Which number of moles (Mg or O) is smaller 0 009 number of moles 1 moles of Mg (rounded to a whole #) 1 moles of O (rounded to a whole #) moles of Mg = 0 009 0 009 moles O = 0 009 Formula: Mg O subscripts Questions and Problems Q1 Using the rules for writing the formulas of ionic compounds, write the ions and the correct formula for magnesium oxide. Magnesium: Oxide ion: Magnesium oxide: Mg2+ O2MgO Q2 Write a balanced equation for the reaction of the magnesium and the oxygen (O 2 ), including their physical states. 2 Mg(s) + O2(g) 2 MgO(s) Q3 Calculate the simplest formula for each of the following compounds: a. 0.200 mole of Al and 0.600 mole of Cl Al = (0.200 mole ÷ 0.200 mole) = 1 Cl = (0.600 mole ÷ 0.200 mole) = 3 Empirical formula = AlCl3 b. 0.080 mole of Ba, 0.080 mole of S, 0.320 mole of O Ba = (0.080 mole ÷ 0.080 mole) = 1 S = (0.080 mole ÷ 0.080 mole) = 1 O = (0.320 mole ÷ 0.080 mole) = 4 Empirical formula = BaSO 4 Q4 When 2.50 g of copper reacts with oxygen, the copper oxide compound has a mass of 2.81 g. What is the simplest formula of the copper oxide? moles of Cu = (2.50g ÷ 63.55 g/mol) = 0.039 moles moles of O = ((2.81g – 2.50g) ÷ 16.00 g/mol) = 0.019 moles Cu = (0.039 moles ÷ 0.019 moles) = 2 O = (0.019 moles ÷ 0.019) = 1 Empirical Formula = Cu2O B. Formula of a Hydrate 1. Mass of crucible 32.16 g 2. Mass of crucible and hydrate 40.94 g 3. Mass of crucible and anhydrate 36.48 g 3a Mass of crucible and anhydrate (second heating) g Calculations 4. Mass of hydrate 8.78 g 5. Mass of anhydrate 4.32 g 6. Mass of water lost 4.46 g 7. Percent water (Show calculations.) 50.80 % 0.248 moles 0.036 moles % water = (4.46 g ÷ 8.78 g) x 100 = 50.80 % 8. Moles of water (Show calculations.) moles of water = (4.46 g ÷ 18.015 g/mol) = 9. 0.248 Moles of salt (anhydrate) (Show calculations.) moles of MgSO4 = (4.32 g ÷ 120.366 g/mol) = 0.036 10 7 Ratio of moles of water to moles of hydrate Ratio = (0.248 moles ÷ 0.036 moles) = 7 7 MgSO 4  Formula of hydrate H2O THIS IS THE ANSWER Questions and Problems Q5 Using the formula you obtained in B11, write a balanced equation for the dehydration of the MgSO 4 hydrate used in the experiment. MgSO4 • 7H2O (s) MgSO4 (s) + 7 H2O(g)

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