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QUANTITATIVE CHEMISTRY UNIT HL1

QUANTITATIVE CHEMISTRY UNIT HL1. IB Chemistry Gr.11 IB Topic 1. Topics 1.1 The mole and Avogadro’s Constant 1.2 Formulas. IB STANDARDS. Upon completion of this unit the SWBAT: 1.1.1 Apply the mole concept to substances (5.2.12.B.3)

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QUANTITATIVE CHEMISTRY UNIT HL1

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  1. QUANTITATIVE CHEMISTRY UNIT HL1 Chem2_Dr.Dura IB Chemistry Gr.11 IB Topic 1

  2. Topics 1.1 The mole and Avogadro’s Constant 1.2 Formulas Chem2_Dr.Dura

  3. IB STANDARDS Upon completion of this unit the SWBAT: • 1.1.1 Apply the mole concept to substances (5.2.12.B.3) • 1.1.2 Determine the number of particles and the amount of substance (in moles) (5.2.12.B.3, 9.1.12.A.1) • 1.2.1 Define the terms relative atomic mass and relative molecular mass (5.2.12.B.3) • 1.2.2 Calculate the mass of one mole of a species from its formula (5.2.12.B.3) • 1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass, and molar mass Chem2_Dr.Dura

  4. 1.1The Mole and Avogadro’s Constant. The mole (mol) is the amount that contains the same number of chemical species as there are atoms in exactly 12.00 grams of 12C. • Avogadro’s constant (L) 1 mol = L = 6.022x 1023 Chem2_Dr.Dura • The relative atomic mass of an element is the average mass of an atom of the element taking into account all its isotopes and their relative abundance, compared to one atom of carbon–12 expressed in atomic mass units (amu)

  5. Molar mass is the mass of 1 mole of a species expressed in grams and has units of grams/mole 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li Chem2_Dr.Dura Number of particles (N) = number of moles (n) x Avogadro’s Constant (L) N = nL

  6. Keep these relationships in mind: use molar mass grams use Avogadro’s number molecules moles Remember – the critical link between moles and grams of a substance is the molar mass. IT’S SIMPLE – THINK IN TERMS OF PARTICLES! 3.7

  7. 1 mol C3H8O x 60.0 g C3H8O Example 1: How many H atoms are in 72.5 g of propanol, C3H8O ? Solution: 1 mol C3H8O = (3 x 12.0) + (8 x 1.0) + 16.0 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H Chem2_Dr.Dura Step 1: Convert mass to moles: = 1.21 moles C3H8O 72.5 g C3H8O

  8. 8 mol H atoms 6.022 x 1023 H atoms x x = 1 mol C3H8O 1 mol H atoms STEP 2: CONVERT MOLES C3H8O TO MOLES H ATOMS = 9.68 moles H atoms 1.21 moles C3H8O Step 3: Convert moles H atoms to atoms H: Chem2_Dr.Dura 9.68 moles H atoms 5.83 x 1024 atoms H

  9. 1 mol K 6.022 x 1023 atoms K x x = 1 mol K 39.10 g K Example 2: How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K Chem2_Dr.Dura 0.551 g K 8.49 x 1021 atoms K

  10. IB Standards • 1.2.4 Distinguish between the terms empirical formula and molecular formula • 1.2.5 Determine the empirical formula from the percentage composition or from other experimental data (9.1.12.A.1) • 1.2.6 Determine the molecular formula when given both the empirical formula and experimental data (5.3.B1, 5.6.A1, 9.1.12.A.1 ) Chem2_Dr.Dura

  11. 1.2 CHEMICAL FORMULA • Molecular Formula gives the actual number of atoms of each element present in a molecule of a compound. • Empirical formula gives the ratio of the atoms of different elements in a compound. It is the molecular formula expressed as its simplest ratio. • What is the empirical formula of glucose, C6 H12O6? • Answer: The simplest ratio: CH2O Chem2_Dr.Dura

  12. Formula Mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.5 amu 1 mole NaCl = 58.5 g NaCl Chem2_Dr.Dura

  13. 3 x 40.1 = 120.3 3 Ca 2 P 2 x 31.0= 62.0 8 O + 8 x 16.0=128.0 Do NOW: What is the formula mass of Ca3(PO4)2 ? 1 formula unit of Ca3(PO4)2 Chem2_Dr.Dura 310.3 amu What is the molar mass of Ca3(PO4)2 ? Answer: 310.3 grams

  14. TOPIC1.3 CHEMICAL EQUATIONS IB Standards • Deduce chemical equations when all reactants and products are given • Identify the mole ratio of any two species in a chemical equations • Apply the state symbols (s), (l), (g), and (aq) Chem2_Dr.Dura

  15. reactants products 1.3 Chemical Equations A chemical equation is a shorthand notation of a chemical reaction. Chem2_Dr.Dura

  16. Types of Reactions 1. Combination: A + B  AB CO2 + H2O → H2CO3 2. Decomposition: AB  A + B CaCO3 →CO2 + CaO 3. Combustion (Burning fuels) CH4 + 2 O2 → CO2 + 2 H2O4. Single Replacement: A + BC AC + B 2Al + Fe2O3→ Al2O3 + 2Fe 5. Double Replacement: AB+ CD  AD + CB HCl + NaOH → NaCl + H2O Chem2_Dr.Dura

  17. C2H6 + O2 CO2 + H2O NOT 2C2H6 C4H12 Balancing Chemical Equations • Write the correct formula(s) for the reactants and the products. Ethane reacts with oxygen to form carbon dioxide and water Chem2_Dr.Dura • Change the coefficients of the chemical formulas to make the number of atoms of each element equal on both sides of the equation. Do not change the subscripts.

  18. 1 carbon on right 6 hydrogen on left 2 hydrogen on right 2 carbon on left C2H6 + O2 C2H6 + O2 C2H6 + O2 CO2 + H2O 2CO2 + H2O 2CO2 + 3H2O • Start by balancing those elements that appear in only one reactant and one product. start with C or H but not O multiply CO2 by 2 Chem2_Dr.Dura multiply H2O by 3

  19. multiply O2 by 4 oxygen (2x2) + 3 oxygen (3x1) 2 oxygen on left C2H6 + O2 2CO2 + 3H2O C2H6 + O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O 7 7 2 2 • Balance those elements that appear in two or more reactants or products. = 7 oxygen on right Chem2_Dr.Dura remove fraction multiply both sides by 2

  20. 4 C (2 x 2) 4 C Reactants Products 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) 4 C 4 C 12 H 12 H 14 O 14 O 2C2H6 + 7O2 4CO2 + 6H2O • Check to make sure that you have the same number of each type of atom on both sides of the equation. Chem2_Dr.Dura

  21. TOPIC 1.4 Mass and Gas Volume Relationships in chemical reactions Chem2_Dr.Dura

  22. IB STANDARDS • Calculate theoretical yields from chemical equations • Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given • Solve problems involving theoretical, experimental, and percentage yield (9.1.12.A.1) Chem2_Dr.Dura

  23. 1.4 Mass and Gaseous Volume Relationships in Chemical Reactions Quantities of reactants and products in a balanced chemical equation has many applications. Example 3: How many moles of O2 are needed to react with 4.26 moles of H2? 2 H2 + O2 → 2 H2O 4.26 mol H2 X ( 1 mole O2 / 2 mole H2) = 2.13 mole O2 Chem2_Dr.Dura

  24. 2CH3OH + 3O2 2CO2 + 4H2O grams CH3OH moles CH3OH moles H2O grams H2O 4 mol H2O 18.0 g H2O 1 mol CH3OH x = x x 2 mol CH3OH 32.0 g CH3OH 1 mol H2O Example 4: Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? molar mass CH3OH molar mass H2O coefficients chemical equation Chem2_Dr.Dura 209 g CH3OH 235 g H2O

  25. Start: 8 moles NO and 7 moles O2 ? NO2 Limiting Reactants What is the limiting reagent? NO Why? Chem2_Dr.Dura Then, O2 is the excess reactant.

  26. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Fe2O3 160. g Fe2O3 1 mol Al = x x x 27.0 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 Example 5: In one process, 124 g of Al are reacted with 601 g of Fe2O3 , Calculate the mass of Al2O3 formed. Chem2_Dr.Dura 367 g Fe2O3 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent

  27. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Al2O3 g Al2O3 1 mol Al x 27.0 g Al 1 mol Al2O3 102. g Al2O3 = x x 2 mol Al 1 mol Al2O3 Use limiting reagent (Al) to calculate amount of product that can be formed. 234 g Al2O3 124 g Al Chem2_Dr.Dura

  28. % Yield = x 100 Actual Yield Theoretical Yield Reaction Yield Theoretical Yield is the maximum amount of product That can be produced. Actual Yield is the amount of product actually obtained from a reaction. Chem2_Dr.Dura

  29. Class Exercise If the theoretical yield of iron was 30.0 g and the actual yield was 26.8 grams, calculate the per cent yield. 2Al + Fe2O3→ Al2O3 + 2Fe Chem2_Dr.Dura

  30. IB STANDARDS • Apply Avogadro’s Law to calculate reacting volumes of gases (9.1.12.A.1) • Apply the concept of molar volume at standard temperature and pressure in calculations • Solve problems using the relationships between temperature, pressure, and volume for a fixed mass of an ideal gas (9.1.12.A.1) • Solve problems using the ideal gas equation • Analyze graphs relating to the ideal gas equation (9.1.12.A.1, 9.1.12.B.1) Chem2_Dr.Dura

  31. Molar Volume of a Gas STP: standard temperature and pressure: 0 0C and 100 kPa. Previous standard is 1 atm = 101.3 kPa. Molar Volume at STP = 22.4 dm3 1 dm3 = 1 L Chem2_Dr.Dura Molar Volume at RTP = 24 dm3

  32. The Gas Laws: Boyle’s Law Chem2_Dr.Dura Pa 1/V Constant temperature Constant amount of gas P x V = constant P1 x V1 = P2 x V2

  33. Relationship of Gas Volume (V) with Temperature at Constant Pressure Charles’s Law VaT Temperature must be in Kelvin T (K) = t (0C) + 273.15 Relationship Between Gas Temperature And Pressure (P) Gay-Lussac’s Law: PaT Temperature in Kelvins P = 0 at absolute zero (- 273 oC) Chem2_Dr.Dura

  34. Avogadro’s Law Chem2_Dr.Dura

  35. EXAMPLE 6 Calculate the volume (in L) occupied by 7.40 g of NH3 at STP. (1 L = 1 dm3 ) Chem2_Dr.Dura

  36. Boyle’s law: Pa (at constant n and T) Va nT nT nT P P P V = constant x = R 1 V Ideal Gas Equation Charles’s law: VaT(at constant n and P) Avogadro’s law: V a n(at constant P and T) Chem2_Dr.Dura R is the gas constant PV = nRT

  37. COMBINED GAS LAW: P1V1/ T1 = P2V2/T2 • REMINDER Use SI units when using the ideal gas equation. R= 8.31 J/Kmol P in Pa (pascal) V in m3 T in Kelvin Chem2_Dr.Dura

  38. p = m V M= PM = RT pRT P Density (p) Calculations m is the mass of the gas in g M is the molar mass of the gas P is the pressure of the gas Molar Mass (M) of a Gaseous Substance Chem2_Dr.Dura p is the density of the gas in g/L

  39. TOPIC1.5SOLUTIONS AND SOLUTION CONCENTRATION IB Standards • Distinguish between the terms solute, solvent, solution, and concentration • Solve problems involving concentration, amount of solute, and volume of solution Chem2_Dr.Dura

  40. Solution Solvent Solute aqueous solutions of KMnO4 A solution is a homogenous mixture of 2 or more substances. The solute is (are) the substance(s) present in the smaller amount(s). The solvent is the substance present in the larger amount. Chem2_Dr.Dura H2O Soft drink (l) Sugar, CO2 Air (g) N2 O2, Ar, CH4 Pb Sn Soft solder (s)

  41. moles of solute M = molarity = liters of solution Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Chem2_Dr.Dura

  42. How to prepare a 1.00 M NaCl solution: mol solute M = L of solution Note – you do NOT add 58.5 g NaCl to 1.00 L of water. The 58.5 g will take up some volume, resulting in slightly more than1.00 L of solution – and the molarity would be lower. 5.5

  43. Preparing a Solution of Known Concentration Chem2_Dr.Dura

  44. When ions (charged particles) are in aqueous solutions, the solutions are able to conduct electricity. • Pure distilled water (nonconducting) • Sugar dissolved in water (nonconducting): a nonelectrolyte • NaCl dissolved in water (conducting): an electrolyte 5.6

  45. CaCO3 (s) CaO (s) + CO2 (g) - CaO (s) + H2O (l) Ca2+ (aq) + 2OH (aq) - Mg2+ (aq) + 2OH (aq) Mg(OH)2(s) Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l) Mg2+ + 2e- Mg 2Cl- Cl2 + 2e- MgCl2(aq) Mg (l) + Cl2(g) Chemistry in Action: Metals from the Sea Chem2_Dr.Dura

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