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QUANTITATIVE CHEMISTRY

QUANTITATIVE CHEMISTRY. IB Chemistry Gr 12 IB Topic 1. REVIEW (MOLE, AVOGADRO’S CONSTANT, FORMULAS). 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

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  1. QUANTITATIVE CHEMISTRY IB Chemistry Gr 12 IB Topic 1

  2. REVIEW (MOLE, AVOGADRO’S CONSTANT, FORMULAS) 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 • 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 )

  3. 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 • 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 C–12 expressed in atomic mass units (amu)

  4. 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 Number of particles (N) = number of moles (n) x Avogadro’s Constant (L) N = nL

  5. 1.2.3 SOLVE PROBLEMS INVOLVING SUBSTANCES IN MOLES, MASS AND MOLAR MASS • http://www.youtube.com/watch?v=KhK_OazhcMk • 1.2.4 Distinguish between Empirical and Molecular Formula • http://www.youtube.com/watch?v=4CLXCu3DffE • The molecular formula shows the number of each type of atom in a molecule eg C6H6, C6H12O6, NH3 • The empirical formula shows the simplest ratio of the atoms in a molecule. eg CH, CH2O, NH3

  6. 1.2.5 Determine empirical formula from the percent composition or from other data. • http://www.youtube.com/watch?v=mfbjOJhVnIQ • 1.2.6 Determine molecular formula given empirical formula & experimental data • http://www.youtube.com/watch?v=SIvLZ4xnLTM

  7. 1.3 CHEMICAL EQUATIONS Objectives • 1.3.1 Deduce chemical equations when all reactants and products are given. • 1.3.2 Identify the mole ratio of any two species in a chemical equations • 1.3.3 Apply the state symbols (s), (l), (g), and (aq) • In chemical equations it may be important to know the states, which are abbreviated as:(s) solid(l) liquid(g) gas(aq) aqueous (in water)Other states are possible but you don't need to know them for IB

  8. reactants products 1.3 Chemical Equations A chemical equation is a shorthand notation of a chemical reaction. In balancing equations, do not change subscripts . You can change coefficients. 2Al + Fe2O3→ Al2O3 + 2Fe Is this equation balanced?

  9. 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

  10. REVIEW (STOICHIOMETRY) • 1.4.1 Calculate theoretical yields from chemical equations • 1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given • 1.4.3 Solve problems involving theoretical, experimental, and percentage yield (9.1.12.A.1) • 1.4.4 Apply Avogadro’s Law to calculate reacting volumes of gases (9.1.12.A.1) • 1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations • 1.4.6 Solve problems using the relationships between temperature, pressure, and volume for a fixed mass of an ideal gas (9.1.12.A.1) • 1.4.7 Solve problems using the ideal gas equation • 1.4.8 Analyze graphs relating to the ideal gas equation (9.1.12.A.1, 9.1.12.B.1)

  11. 1.4 Mass and Gaseous Volume Relationships in Chemical Reactions Quantities of reactants and products in a balanced chemical equation has many applications. Example: 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

  12. 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: 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 209 g CH3OH 235 g H2O

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

  14. 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: In one process, 124 g of Al are reacted with 601 g of Fe2O3 , Calculate the mass of Al2O3 formed. 367 g Fe2O3 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent

  15. 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

  16. % 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.

  17. Class Exercises If the theoretical yield of iron was 30.0 g and the actual yield was 25.0 g, calculate the per cent yield. 2Al + Fe2O3→ Al2O3 + 2Fe Answer: 83.3% Calculate the % yield if 26.8 g. of iron was collected in the same reaction.

  18. 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 Molar Volume at RTP = 24 dm3

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

  20. 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)

  21. Avogadro’s Law

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

  23. 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) R is the gas constant PV = nRT

  24. 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

  25. 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 p is the density of the gas in g/L

  26. TOPIC:1.5SOLUTIONS AND SOLUTION CONCENTRATION IB Standards • 1.5.1 Distinguish between the terms solute, solvent, solution, and concentration • 1.5.2 Solve problems involving concentration, amount of solute, and volume of solution

  27. 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. H2O Soft drink (l) Sugar, CO2 Air (g) N2 O2, Ar, CH4 Pb Sn Soft solder (s)

  28. 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.

  29. 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

  30. Preparing a Solution of Known Concentration

  31. 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

  32. INTEGRATIVE PRACTICE PROBLEM How many grams of water can be obtained from burning 20.0 mL of ethanol (C2H5OH; density =0.789 g/mL) Hint: Start by writing the balanced equation for the combustion of ethanol.

  33. 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

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