1. What is the empirical formula of a compound made of 22.26 g of sodium with 7.74 g of oxygen? 22.26 g Na x 1 mole = 0.968 moles Na 23.0 g 7.74 g O x 1 mole = 0.484 moles O 16.0 g moles Na = 0.968 = 2 moles O 0.484 1 Na2O

A compound has an empirical formula of CH2 and a molecular mass of 84. What is the molecular formula? Empirical mass of CH2 = 14 84 = 6 14 molecular formula = C6H12

70.0 g x 1 mole = 1.25 mole Fe 55.8 g 30.0 g x 1 mole = 1.88 mole O 3. What is the empirical formula of a compound which is 70.0 g iron and 30.0 g oxygen? 70.0 g x 1 mole = 1.25 mole Fe 55.8 g 30.0 g x 1 mole = 1.88 mole O 16.0 g 1.88 mole O = 1.5 = 3 1.25 mole Fe 1 2 Fe2O3

21.6% oxygen. What is the empirical formula? A compound is 67.6% Hg, 10.8% S, and 21.6% oxygen. What is the empirical formula? 67.6% → 67.6 g 67.6 g x 1 mole = 0.337 mole Hg (1) 200.6 g 10.8% → 10.8 g 10.8 g x 1 mole = 0.336 mole S (1) 32.1g 21.6% → 21.6 g 21.6 g x 1 mole = 1.35 mole O (4) 16.0 g HgSO4

A compound is 30.4% nitrogen and 69.6% oxygen with a molecular mass of 92. What is the molecular formula? 30.4% → 30.4 g x 1 mol = 2.17 mol N 14.0 g 69.6% → 69.6 g x 1 mol = 4.35 mol O 16.0 g 4.35 = 2 = NO2 2.17 1 Empirical mass NO2 = 46 92 = 2 46 Molecular formula = N2O4

A student used electrolysis to break down a compound to find its formula. The following data was collected:   Mass container = 62.75 g Mass container + compound = 66.25 g After electrolysis: Mass container + iron = 63.20 g The other substance produced was iodine.

Use the data to determine the following:   1. The mass of iron and the mass of iodine in the compound. 2. The percentage (by mass) of iron and iodine. 3. The moles of iron and moles of iodine 4. The mole ratio of iodine to iron and the formula of the compound. 5. Name the resulting compound.

Mass container + iron - Mass container = Fe 63.20 g - 62.75 g = 0.45 g Fe Mass container + compound – Mass container 66.25 g - 62.75g = 3.50 g compound   3.50 g - 0.45 g Fe = 3.05 g I or Mass container + compound – Mass container + iron = I 66.25 g - 63.20 g = 3.05 g I  

  % Fe = 0.45 g x 100 = 13% 3.50 g  % I = 3.05 g x 100 = 87.1 = 87% 3.50 g

  0.45 g Fe x 1 mole = 0.0081 mole Fe 55.9 g 3.05 g I x 1 mole = = 0.0240 mole I 126.9 g   I 0.0240 mole = 3 FeI3 Fe 0.0081 mole 1 iron (III) iodide