Moles, Formulas, Reactions & Stoichiometry

1. Moles, Formulas, Reactions & Stoichiometry • Lecture 3 Topics Brown, chapter 3 • 1. Moles 3.4 • Molecular mass (aka molecular weight) 3.3 • 2. Molar converstions • Percent composition 3.3 • Empirical formulas 3.5 • 3. Stoichiometry: balancing chemical equations 3.1 • 4. Patterns of Chemical Reactivity 3.2 • Combination • Decomposition • Combustion • Exchange • 5. Stoichiometry & Conversions 3.6 • Limiting reactants 3.7 • Theoretical & percent yield

2. Conversions: moles to molecules to atoms Percent composition is molar. Empirical formulae are molar. Molecular formulae are molar.

3. Interconversion of moles, mass, atoms Chemistry students are often frustrated by the subject’s lack of cannedequations. Problem solving depends much more on strategy & process: logic.Use these ‘devices’ to develop problem-solving strategies. mass(grams) moles atoms /molecules molar mass (MW) g/mole Avogadro’s number molecules/mole The Conversion Flower can be used toconvert:moles <--> grams moles <--> molecules moles A <--> moles B moles <--> liter p.93 - 94

4. Conversions with moles How many grams is 0.55 mol Cu? STRATEGY: moles ------------> grams atomic mass 0.55 mol 63.55 g = 35 g 1 mol How many moles contain 6.96x1024 atoms of Na? STRATEGY: atoms -------------------> moles the mole 6.96x1024 Na atoms 1 mole = 11.6 mol 6.02x1023 atoms How many molecules in 25 g of NaOH? STATEGY: g --------------> moles ---------------> molecules molar mass the mole 25 g 1 mole 6.02x1023 molecules = 3.8x1023g 40.00 g 1 mole p.93 - 94

5. Percent composition • Percent composition: what percent of a molecule’s mass is contributed by each element? • Calculated using atomic and molar masses. Percentage composition of C12H22O11? FW/MW = 342 amu % C = (12)(12.0 amu)x 100% = 42.1% 342 amu Calculate mass of each element. Calculate MW (sum masses in 1.) Calculate each element’s mass as % of the MW.= (part/whole)(100) % should sum to 100! % H = (22)(1.0 amu)x 100% = 6.4% 342 amu % O = (11)(16.0 amu)x 100% = 51.5% 342 amu H2O (2.02/18.02)(100) = 11.2% H(16.00/18.02)(100) =88.8% O C6Cl5OH (72.07/266.32)(100) = 27.06% C(177.25/266.32)(100) = 66.56% Cl (16.00/266.32)(100) = 6.01% O (1.01/266.32)(100) = 0.38% H p.88

6. Empirical formula from chemical analysis Empirical formulae can be calculated from % composition data obtained via simple chemical analysis. Steps:1. Begin by assuming you have 100 g of the substance & calculate g each element.2. Convert g to moles. Using atomic mass.3. Calculate mole ratios: divide all by the smallest number of moles.4. Convert to formula ratios. You may need to multiply by small whole numbers to get to whole number ratios. A pure hydrocarbon (molecule made wholly of C & H) is 85.7% C.Calculate the empirical formula of the compound. C 85.7 g 1 mol = 7.14 mol /7.14 = 1.00 -> 1 so CH2 is empirical formula 12.01 g H 14.3 g 1 mol = 14.2 mol /7.14 = 1.99 -> 2 1.01 g An Fe/O compound is 69.9% Fe. Fe 69.9 g/55.85 = 1.25 mol/1.25 = 1 x 2 = 2 O 30.1 g/16.00 = 1.88 mol/1.25 = 1.50x2 = 3 So the formula is Fe2O3 p.95

7. Calculate molecular from empirical formula A molecular formula can be calculated from an empirical formula ONLYif the MW is known. 1. Calculate empirical formula (previous slide), & then empirical weight.2. Calculate the ratio of MW/EW. 3. Multiply empirical formula subscripts by that ratio. Vitamin C (ascorbic acid, MW 176 g/mol) has a percent composition of: 40.19% C 4.58% H 54.50% O 40.19 g/12.01 g/mol = 3.35 mol/3.35 = 1.00 x 3 = 3 4.58 g/1.01 g/mol = 4.58 mol/3.35 = 1.37 x 3 = 4 EF= C3H4O3 54.50 g/15.99 g/mol = 3.41 mol/3.35 = 1.02 x 3 = 3 EW = 88.1 g/mol MW= 176 g/mol = 2 So multiply the empirical subscripts by 2 to EW 88.1 g/mol get the molecular formula -> C6H8O6 Mesitylene, a minor component of crude oil, has an empirical formula ofC3H4. Its molar mass is 121 amu. What’s its molecular formula? empirical mass: (3)(12.011) + (4)(1.00794) = 40.065 amu 121 amu/40.065 amu = 3 multiple --> C9H12 p.95