Prentice Hall © 2003Chapter 3 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations CHEMISTRY The Central Science 9th Edition

Prentice Hall © 2003Chapter 3 Lavoisier: mass is conserved in a chemical reaction Chemical equations: descriptions of chemical reactions Two parts to an equation: reactants and products: 2H 2 + O 2  2H 2 O 3.1: Chemical Equations

Prentice Hall © 2003Chapter 3 The formation of water: two hydrogen molecules reacting with one oxygen molecule to form two water molecules: 2H 2 + O 2  2H 2 O

Prentice Hall © 2003Chapter 3 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g) “Sodium metal reacts with liquid water to produce aqueous sodium hydroxide and hydrogen gas” 2K (s) + 2H 2 O (l)  2KOH (aq) + H 2(g) Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products They are all in their simplest form (no multiples!)

Prentice Hall © 2003Chapter 3 Text, P. 77

Prentice Hall © 2003Chapter 3 Law of conservation of mass: matter cannot be lost in any chemical reactionsText, P. 78 O

Prentice Hall © 2003Chapter 3 WRITING EQUATIONS Translate everything from the sentence –States (s), (l), (g), (aq) –Catalysts (Heat = Δ, Current = e.c., Catalyst = MnO 2 ) are written above the arrow Write correct formulas Diatomic molecules! H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 Balance the equation –Balance elements that appear in one place in both the reactants and products first –Usually balance polyatomic molecules as 1 unit (don’t break apart)

Prentice Hall © 2003Chapter 3 Examples # 5 & 7 (WS)

Prentice Hall © 2003Chapter 3 Combination reactions have fewer products than reactants: 2Mg(s) + O 2 (g)  2MgO(s) Mg has combined with O 2 to form MgO Categories: 1.Metal + O 2  Metal oxide (Mg + O 2 ) 2.Nonmetal + Nonmetal  covalent compound (ammonia formation) 3.Metal + Nonmetal (other than O 2 )  salt (Mg + Cl 2 ) 4.Metal oxide + H 2 O  metal hydroxide (NaOH) 5.Nonmetal oxide + H 2 O  oxyacid (CO 2 + H 2 O) 6.Metal oxide + Nonmetal oxide  salt (Na 2 O + SO 2 ) 3.2: Some Simple Patterns of Chemical Reactivity no change in charge on central atom

Prentice Hall © 2003Chapter 3 Decomposition Reactions have fewer reactants than products: 2NaN 3 (s)  2Na(s) + 3N 2 (g) (the reaction that occurs in an air bag) NaN 3 has decomposed into Na and N 2 gas Categories: 1.Binary Compound  elements (NaCl  ) (don’t forget diatomics!) 2.Metal Carbonate  metal oxide + CO 2 (Na 2 CO 3  ) 3.Metal hydroxide  metal oxide + H 2 O (NaOH  ) 4.Metal chlorate  chloride salt + O 2 (NaClO 3  ) 5.Oxyacid  nonmetal oxide + H 2 O (H 2 CO 3  ) These are the opposite of the combination reactions

Prentice Hall © 2003Chapter 3 Combination and Decomposition Reactions Text, P. 80

Prentice Hall © 2003Chapter 3 Combination and Decomposition Reactions

Prentice Hall © 2003Chapter 3 Combustion in Air Combustion is the burning of a substance in oxygen from air: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l) Hydrocarbon + O 2  CO 2 + H 2 O (complete combustion)

Prentice Hall © 2003Chapter 3 Examples # 13 & 11 (WS) and classification/prediction practice (WS) NaCl

Prentice Hall © 2003Chapter 3 Formula and Molecular Weights Formula weights (FW): sum of AW for atoms in a formula FW (H 2 SO 4 ) = 2AW(H) + AW(S) + 4AW(O) = 2(1.01 amu) + (32.06 amu) + 4(16.00 amu) = amu Molecular weight (MW) is the weight of the molecular formula MW(C 6 H 12 O 6 ) = 6(12.01 amu) + 12(1.01 amu) + 6(16.00 amu) = amu 3.3: Formula Weights

Prentice Hall © 2003Chapter 3 Percentage Composition from Formulas Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100: Don’t forget hydrates!

Prentice Hall © 2003Chapter 3 Examples # 15, 17, 19

Prentice Hall © 2003Chapter 3 Mole: convenient measure chemical quantities. 1 mole of something =  of that thing Experimentally, 1 mole of 12 C has a mass of 12 g Molar Mass Molar mass: mass in grams of 1 mole of substance (units of g/mol, g*mol -1 ) Mass of 1 mole of 12 C = 12 g 3.4: The Mole

Text, P. 88

Prentice Hall © 2003Chapter 3 Examples # 25 &

Prentice Hall © 2003Chapter 3 Text, P. 88

Prentice Hall © 2003Chapter 3 This photograph shows one mole of solid (NaCl), liquid (H 2 O), and gas (N 2 ).

Prentice Hall © 2003Chapter 3 Interconverting Masses, Moles, and Number of Particles Molar mass: sum of the molar masses of the atoms: molar mass of N 2 = 2  (molar mass of N) Periodic table Formula weights are numerically equal to the molar mass Text, P. 90

Prentice Hall © 2003Chapter 3 Relating particles, Mass and Moles Molar mass (g) 6.02x10 23 particles 1 mole Molecules (molecular compounds) Formula Units (ionic compounds) Atoms (elements) ions are made of atoms are made of

Prentice Hall © 2003Chapter 3 Macroscale vs. Microscale Interpretations of the Mole 1 formula unit of NaCl 1 mole of Cl - ions 1 mole of Na + ions1 ion of Na + 1 ion of Cl - Microscale 1 mole of NaCl Macroscale

Prentice Hall © 2003Chapter 3 Examples # 27, 29, 33 & 35

Prentice Hall © 2003Chapter 3 Recall: the Empirical Formula is the simplest formula Start with mass % of elements (i.e. empirical data) and calculate a formula, or Start with the formula and calculate the mass % elements 3.5: Empirical Formulas from Analyses

Text, P. 93

Prentice Hall © 2003Chapter 3 Examples # 39, 41

Prentice Hall © 2003Chapter 3 Common DecimalEquivalent FractionMole Ratio Example /8 1 : converts to 1 : 9/8 multiply throughout by 8 to give 8 : /4 1 : 0.25 converts to 1 : 1/4 multiply throughout by 4 to give 4 : /3 1 : 1.33 converts to 1 : 4/3 multiple throughout by 3 to give 3 : /8 1 : converts to 1 : 11/8 multiply throughout by 8 to give 8 : /2 2 : 1.5 converts to 2 : 3/2 multiply throughout by 2 to give 4 : /8 1 : converts to 1 : 13/8 multiply throughout by 8 to give 8 : /3 2 : 1.66 converts to 2 : 5/3 multiply throughout by 3 to give 6 : /8 1 : converts to 1 : 7/8 multiply throughout by 8 to give 8 : 7

Prentice Hall © 2003Chapter 3 Molecular Formula from Empirical Formula Once the empirical formula is known, we need the MW to find the molecular formula Subscripts in the molecular formula are always whole- number multiples of subscripts in the empirical formula

Prentice Hall © 2003Chapter 3 Examples # 43, 45

Prentice Hall © 2003Chapter 3 Combustion Analysis Empirical formulas are determined by combustion analysis:

Prentice Hall © 2003Chapter 3 Examples # 47, 93 And hydrates # 49

Prentice Hall © 2003Chapter 3 Balanced chemical equation states the number of molecules that react to form products These ratios are called stoichiometric ratios Stoichiometric ratios are ideal proportions Real ratios of reactants and products in the laboratory need to be measured (in grams and converted to moles) 3.6: Quantitative Information from Balanced Equations

Text, P : Quantitative Information from Balanced Equations

Prentice Hall © 2003Chapter 3 Examples # 55, 57, 59 & 61

Prentice Hall © 2003Chapter 3 If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess) Limiting Reactant: The reactant that is used up first in a reaction It controls the amount(s) of the other reactant(s) that are used It controls the amount of product produced (“maximum amount of product”) It produces less product than the other reactants possibly could produce (use this for problem solving) 3.7: Limiting Reactants

Prentice Hall © 2003Chapter 3 Examples # 67, 69, 71, 73

Prentice Hall © 2003Chapter 3 Theoretical Yields The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:

Prentice Hall © 2003Chapter 3 Examples # 75, 77

Prentice Hall © 2003Chapter 3 End of Chapter 3: Stoichiometry: Calculations with Chemical Formulas and Equations