Transcript 3-A
Chapter 3 Formulas, Equations, and Moles Chemistry, 5th Edition McMurry/Fay The Structure of Atoms Atomic Mass Unit 1 amu = 1/12 of the mass of on atom of Carbon-12 1 amu = 1.6605 x 10-24 g 2 Atomic and Molecular Mass mass of carbon 12 atom 1 amu 12 Mass: proton = 1.00728 amu neutron = 1.0086 amu electron = 0.0005486 12C atom = 12.00000 amu 13C atom = 13.00335 amu 3 Atomic and Molecular Mass • The atomic masses as tabulated in the periodic table are the averages of the naturally occurring isotopes. • Mass of C = average of 12C and 13C = 0.9889 x 12 amu + 0.0111 x 13.0034 amu = 12.011 amu 4 Atomic and Molecular Mass The mass of a molecule is just the sum of the masses of the atoms making up the molecule. m(C2H4O2) = 2·mC + 4·mH + 2·mO = 2·(12.01) + 4·(1.01) + 2·(16.00) = 60.06 amu 5 Avogadro and the Mole • One mole of a substance is the gram mass value equal to the amu mass of the substance. • One mole of any substance contains 6.02 x 1023 units of that substance. • Avogadro’s Number (NA, 6.022 x 1023) is the numerical value assigned to the unit, 1 mole. 6 Avogadro and the Mole 7 Avogadro and the Mole • Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H) 8 Avogadro and the Mole • The Mole: Allows us to make comparisons between substances that have different masses. 9 Balancing Chemical Equations • A balanced chemical equation represents the conversion of the reactants to products such that the number of atoms of each element is conserved. reactants products limestone quicklime + gas Calcium carbonate calcium oxide + carbon dioxide CaCO3(s) CaO(s) + CO2(g) 10 Balancing Chemical Equations CaCO3(s) CaO(s) + CO2(g) The letters in parentheses following each substance are called State Symbols (g) → gas (l) → liquid (s) → solid (aq) → aqueous 11 Balancing Chemical Equations A balanced equation MUST have the same number of atoms of each element on both sides of the equation. H2 + O2 → H2O Not Balanced H2 + ½O2 → H2O Balanced 2H2 + O2 → 2H2O Balanced 12 Balancing Chemical Equations The numbers multiplying chemical formulas in a chemical equation are called: Stoichiometric Coefficients (S.C.) 2H2 + O2 → 2H2O Balanced Here 2, 1, and 2 are stoichiometric coefficients. 13 Balancing Chemical Equations Hints for Balancing Chemical Equations: 1) Save single element molecules for last. 2) Try not to change the S.C. of a molecule containing an element that is already balanced. 3) If possible, begin with the most complex molecule that has no elements balanced. 14 Balancing Chemical Equations Hints for Balancing Chemical Equations: 4) Otherwise, trial and error!! 15 Balancing Chemical Equations CH4 + O2 → CO2 + H2O Example 1: Balance O2 last C is already balanced Start by changing S.C. of H2O to balance H CH4 + O2 → CO2 + 2H2O 16 Balancing Chemical Equations Example 1: CH4 + O2 → CO2 + 2H2O Now C and H are balanced Balance O by changing the S.C. of O2 CH4 + 2O2 → CO2 + 2H2O BALANCED! 17 Balancing Chemical Equations Example 2: B2H6 + O2 → B2O3 + H2O Balance O last B is already balanced Start by changing S.C. of H2O: B2H6 + O2 → B2O3 + 3H2O 18 Balancing Chemical Equations Example 2: B2H6 + O2 → B2O3 + 3H2O B and H are balanced Balance O by changing S.C. of O2 B2H6 + 3O2 → B2O3 + 3H2O BALANCED! 19 Balancing Chemical Equations Example 3: MnO2 + KOH + O2 → K2MnO4 + H2O Balance O last Mn is already balanced Change S.C. of KOH to balance K MnO2 + 2KOH + O2 → K2MnO4 + H2O 20 Balancing Chemical Equations Example 3: MnO2 + 2KOH + O2 → K2MnO4 + H2O Mn, K, and H are balanced (H was balanced by chance) Balance O MnO2 + 2KOH + ½O2 → K2MnO4 + H2O or 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O 21 Balancing Chemical Equations Example 4: NaNO2 + H2SO4→ NO + HNO3 + H2O + Na2SO4 Hard one (no single element molecules) S is balanced Start with NaNO2 to balance Na 2NaNO2 + H2SO4→ NO + HNO3 + H2O + Na2SO4 22 Balancing Chemical Equations Example 4: 2NaNO2 + H2SO4→ NO + HNO3 + H2O + Na2SO4 S, Na, and N are balanced Cannot balance H without changing S.C. for H2SO4! Boo! Option 1: trial and error Option 2: Go on to next problem! 23 Balancing Chemical Equations • Balance the following equations: C6H12O6 → C2H6O + CO2 Fe + O2 → Fe2O3 NH3 + Cl2 → N2H4 + NH4Cl KClO3 + C12H22O11 → KCl + CO2 + H2O 24 25 Balancing Chemical Equations • Balance the following equations: C6H12O6 → 2C2H6O + 2CO2 Fe + O2 → Fe2O3 NH3 + Cl2 → N2H4 + NH4Cl KClO3 + C12H22O11 → KCl + CO2 + H2O 26 Balancing Chemical Equations • Balance the following equations: C6H12O6 → 2C2H6O + 2CO2 4Fe + 3O2 → NH3 + Cl2 → 2Fe2O3 (balance O first) N2H4 + NH4Cl KClO3 + C12H22O11 → KCl + CO2 + H2O 27 Balancing Chemical Equations • Balance the following equations: C6H12O6 → 2C2H6O + 2CO2 4Fe + 3O2 → NH3 + Cl2 → 2Fe2O3 (balance O first) N2H4 + NH4Cl N:H is 1:3 on left, must get 1:3 on right! 28 Balancing Chemical Equations NH3 + Cl2 → N2H4 + NH4Cl N:H is 1:3 on left, must get 1:3 on right! 4NH3 + Cl2 → N2H4 + 2NH4Cl 29 Balancing Chemical Equations • Balance the following equations: C6H12O6 → 2C2H6O + 2CO2 4Fe + 3O2 → 4NH3 + Cl2 → 2Fe2O3 N2H4 + 2NH4Cl KClO3 + C12H22O11 → KCl + CO2 + H2O (tough!) 30 Balancing Chemical Equations • Balance the following equations: KClO3 + C12H22O11 → KCl + CO2 + H2O balance C KClO3 + C12H22O11 → KCl + 12CO2 + H2O balance H KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O balance O 8KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O 31 Balancing Chemical Equations • Balance the following equations: 8KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O balance K (and hope Cl is balanced) 8KClO3 + C12H22O11 → 8KCl + 12CO2 + 11H2O Balanced! 32 Balancing Chemical Equations • Write a balanced equation for the reaction of element A (red spheres) with element B (green spheres) as represented below: 33 Avogadro and the Mole 34 Stoichiometry • Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities. 35 Stoichiometry Aqueous solutions of NaOCl (household bleach) are prepared by the reaction of NaOH with Cl2: 2 NaOH(aq) + Cl2(g) NaOCl(aq) + NaCl(aq) + H2O(l) How many grams of NaOH are needed to react with 25.0 g of Cl2? 36 Stoichiometry 2 NaOH + Cl2 → NaOCl + NaCl + H2O 25.0 g Cl2 reacts with ? g NaOH 25.0 g Cl2 1 mole Cl2 0.353 moles Cl2 70.90 g Cl2 1 mole Cl2 2 moles NaOH 40.0 g NaOH 25.0 g Cl2 28.2 g NaOH 70.90 g Cl2 1 mole Cl2 1 mole NaOH 37 Avogadro and the Mole • Calculate the molar mass of the following: Fe2O3 (Rust) C6H8O7 (Citric acid) C16H18N2O4 (Penicillin G) • Balance the following, and determine how many moles of CO will react with 0.500 moles of Fe2O3. Fe2O3(s) + CO(g) Fe(s) + CO2(g) 38 Avogadro and the Mole Fe2O3 + CO → Fe + CO2 Balance (not a simple one) Save Fe for last C is balanced, but can’t balance O In the products the ratio C:O is 1:2 and can’t change Make the ratio C:O in reactants 1:2 Fe2O3 + 3CO → 2Fe + 3CO2 39 Avogadro and the Mole Fe2O3 + 3CO → 2Fe + 3CO2 3 mole CO 0.500 moles Fe2O3 1.50 moles CO 1 mole Fe2O3 40 Stoichiometry 41 Stoichiometry • Aspirin is prepared by reaction of salicylic acid (C7H6O3) with acetic anhydride (C4H6O3) to form aspirin (C9H8O4) and acetic acid (CH3CO2H). Use this information to determine the mass of acetic anhydride required to react with 4.50 g of salicylic acid. How many grams of aspirin will result? How many grams of acetic acid will be produced as a by-product? 42 Stoichiometry Salicylic acid + Acetic anhydride → Aspirin + acetic acid C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2 Balanced! Equal # moles for all 43 Stoichiometry 4.50 g Salicylic acid (C7H6O3) = ? moles MW C7H6O3 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00 = 138.12 g/mole 1 mole S . A. 4.50 g S . A. 0.0326 moles S . A. 138.12 g S . A. 44 Stoichiometry Since all compounds have the same S.C., there must be 0.0326 moles of all 4 of them involved in the reaction. g Aspirin (C9H8O4) = 0.0326 moles x MW Aspirin = .0326 x [9x12.01 + 8x1.008 + 4x16.00] =.0326 mole x 180.15 g/mole 5.87 g Aspirin 45 Stoichiometry • Yields of Chemical Reactions: If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate a percentage yield. Actual product yield % yield 100% Theoretica l product yield 46 Stoichiometry • Dichloromethane (CH2Cl2) is prepared by reaction of methane (CH4) with chlorine (Cl2) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%? 47 Stoichiometry CH4 + Cl2 → CH2Cl2 + HCl Balance CH4 + 2Cl2 → CH2Cl2 + 2HCl 1.85 kg CH4 = ? moles CH4 48 Stoichiometry CH4 + 2Cl2 → CH2Cl2 + 2HCl 1.85 kg CH4 = ? moles CH4 MW CH4 = 1x12.01 + 4x1.008 = 16.04 g/mole 1000 g 1 mole CH 4 1.85 kg CH 4 115 moles CH 4 kg 16.4 g CH 4 49 Stoichiometry CH4 + 2Cl2 → CH2Cl2 + 2HCl 115 moles CH4 in theory we should produce: 115 moles of CH2Cl2 and 230 moles of HCl And use up 230 moles of Cl2 50 Stoichiometry CH4 + 2Cl2 → CH2Cl2 + 2HCl 115 moles of CH2Cl2 = ? g MW CH2Cl2 = 12.01 + 2x1.008 + 2x35.45 = 84.93 115 moles x (84.03 g/mole) = 9770 g 51 Stoichiometry CH4 + 2Cl2 → CH2Cl2 + 2HCl Expect 9770 g CH2Cl2 but the yield is 43.1% So we produced just 0.431 x 9770 g 4.21 kg CH2Cl2 52 Stoichiometry CH4 + 2Cl2 → CH2Cl2 + 2HCl Suppose the reaction went to completion (100% yield) Is mass conserved? 53 Stoichiometry CH4 + 2Cl2 → CH2Cl2 + 2HCl Start with 115 moles CH4 and 230 moles Cl2 total mass = 115x16.04 + 230x70.90 = 1850 + 16300 = 18150 only 3 sig. figs. → 18.2 kg 54 Stoichiometry CH4 + 2Cl2 → CH2Cl2 + 2HCl End with 115 moles CH2Cl2 and 230 moles HCl total mass = 115x84.93 + 230x36.46 = 9770 + 8390 = 18160 only 3 sig. figs → 18.2 kg 55