2. The Mole and Molar Mass Moles are the currency of chemistry 1 mole = 6.02 x 10 23 particles = Avogadro’s number Molar mass of an element = mass of one mole of that element Molar mass = atomic mass of element in grams - Average mass of one atom of C = 12.01 amu - Average mass of one mole of C = 12.01 g Molar mass of a compound = sum of molar masses of each atom in the compound Example: What is the molar mass of MgCl 2 ? Molar mass of Mg = 24.31 g, Molar mass of Cl = 35.45 g Molar mass of MgCl 2 = 24.31 g + (2 x 35.45 g) = 95.21 g *Note: use as many sig. figs. as given in the atomic masses (don’t round molar masses to 1/10 of a gram)

5. Calculations using Molar Mass Molar mass is used to convert between moles and grams of a substance Chemical equations are balanced with atoms, and moles are an extension of that (they also balance) Quantities of reactants and products are measured by mass (grams) in the lab, but we use moles to balance equations Example 1: How many grams of H 2 O are in 0.100 mol H 2 O? Molar mass of H 2 O = (2 x 1.008 g) + 16.00 g = 18.02 g Grams H 2 O = 0.100 mol x (18.02 g/1 mol) = 1.80 g H 2 O Example 2: How many moles are in 0.250 g of HCl? Molar mass of HCl = 1.008 g + 35.45 g = 36.46 g Mol HCl = 0.250 g x (1 mol/36.46 g) = 6.86 x 10 -3 mol HCl

6. Percent Composition Molecular formula tells you how many moles of each type of atom are in one mole of a compound Using molar mass, you can calculate the % by mass of each element in a compound Example: Calculate the % composition by mass of CO 2 Molar mass of CO 2 = 12.01 g + (2 x 16.00 g) = 44.01 g %C = (12.01 g/44.01 g) x 100% = 27.29% %O = (2 x 16.00 g/44.01 g) x 100% = 72.71%

7. Empirical Formulas Empirical formulas give the lowest whole-number ratio of atoms in a compound Molecular formula = multiple of empirical formula Example: empirical formula for glucose = CH 2 O molecular formula = C 6 H 12 O 6 Can calculate empirical formula from masses or from % composition by mass Example: Calculate empirical formula for compound with 8.0 g C and 2.0 g H 8.0 g C x (1 mol/12.01 g) = 0.6661 mol C 2.0 g H x (1 mol/1.008 g) = 1.984 mol H 0.6661 mol C/0.6661 mol = 1.000 C = 1 C 1.984 mol H/0.6661 mol = 2.979 H = 3 H’s Empirical formula = CH 3

8. Mole Relationships and Mole Calculations Atoms, moles and mass are conserved in a chemical reaction (shown in balanced equation) Using chemical formulas, molar masses and mole ratios (from balanced chemical equation), amounts of products and reactants can be calculated Example: For C 3 H 8 + 5O 2  3CO 2 + 4H 2 O How many moles of O 2 are needed to produce 2.0 moles of H 2 O? 1.Have moles H 2 O, want moles O 2 2.5 mol O 2 = 4 mol H 2 O 3.2.0 mol H 2 O x (5 mol O 2 /4 mol H 2 O) = 2.5 mol O 2

11. Mole Relationships and Mass Calculations Example: C 3 H 8 + 5O 2  3CO 2 + 4H 2 O How many grams of CO 2 are produced when 10.0 g of C 3 H 8 are consumed? 1.Have grams C 3 H 8, want grams CO 2 (g C 3 H 8  mol C 3 H 8  mol CO 2  g CO 2 ) 2.Molar mass C 3 H 8 = (3 x 12.01 g) + (8 x 1.008 g) = 44.094 g Mol C 3 H 8 = 10.0g x (1 mol/44.094 g) = 0.2268 mol 3.Mol CO 2 = 0.2268 mol C 3 H 8 x (3 mol CO 2 /1 mol C 3 H 8 ) = 0.6804 mol 4.Molar mass CO 2 = 12.01 g + (2 x 16.00 g) = 44.01 g Grams CO 2 = 0.6804 mol CO 2 x (44.01 g/1 mol) = 29.9 g

12. Percent Yield Theoretical (or ideal) yield is calculated from balanced chemical equation and molar masses Actual yield = what you got (usually < ideal) % yield = (actual yield/theoretical yield) x 100% Example:2Fe + 3S  Fe 2 S 3 What is the % yield if 1.00 g Fe makes 1.50 g Fe 2 S 3 ? 1.Calculate theoretical yield: (g Fe  mol Fe  mol Fe 2 S 3  g Fe 2 S 3 ) 1.00 g Fe x (1 mol/55.85 g) = 0.0179 mol Fe 0.0179 mol Fe x (1 mol Fe 2 S 3 /2 mol Fe) = 0.00900 mol Fe 2 S 3 Molar mass Fe 2 S 3 = (2 x 55.85 g) + (3 x 32.06 g) = 207.88 g 0.00900 mol Fe 2 S 3 x (207.88 g/1 mol) = 1.87 g Fe 2 S 3 2.% yield = (1.50 g Fe 2 S 3 /1.871 g Fe 2 S 3 ) x 100% = 80.2%

13. Limiting Reactant Limiting Reactant = reactant that is completely used up Example: If 1.0 mol of Fe are reacted with 3.0 mol of S, which is the limiting reactant? - Since 2.0 mol Fe are required to react with 3.0 mol of S, then 1.0 mol of Fe can only react with 1.5 mol of S - So, all the Fe is use up and 1.5 mol of S is leftover - Fe is the limiting reactant

14. Equilibrium Constants For reversible reactions at equilibrium: - Rate of forward reaction = rate of reverse reaction - Amounts of reactants and products are constant (but not necessarily equal) Ratio of products to reactants = equilibrium constant (K eq ) For a reaction aA + bB  cC + dD K eq = [C] c [D] d /[A] a [B] b, where [A] = concentration of A Example: for 2H 2 + O 2  2H 2 O K eq = [H 2 O] 2 /[H 2 ] 2 [O 2 ] If K eq > 1, equilibrium favors products If K eq < 1, equilibrium favors reactants