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% by Mass  Empirical Formula

% by Mass  Empirical Formula. CH 3 # 88 26.7% P 12.1 % N 61.2 % Cl Molar mass = 580 g/ mol. Limiting and Excess Problem. CH 3 # 124 Abbreviations and moalr masses: - Chlorobenzene = CB, 1 mol = 112.5 g - Chloral = CL, 1 mol = 147.5 g - DDT, 1 mol = 354.5 g.

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% by Mass  Empirical Formula

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  1. % by Mass  Empirical Formula • CH 3 # 88 26.7% P 12.1 % N 61.2 % Cl Molar mass = 580 g/mol

  2. Limiting and Excess Problem • CH 3 # 124 Abbreviations and moalr masses: - Chlorobenzene = CB, 1 mol = 112.5 g - Chloral = CL, 1 mol = 147.5 g - DDT, 1 mol = 354.5 g

  3. CH 4: Molarity Calculations

  4. Molarity Calculations – Part 1 • Molarity of solution from the mass of solute and volume of solution. • How to make a specific volume of a given concentration (M). • Concentration of ions in a solution of strong electrolytes • Moles of solute present in a given volume of solution • How to dilute solutions to a specific concentration.

  5. What is the molarity of a solution made by dissolving 15.0 g of NaOH in enough water to make 350.0 mL of solution?

  6. Describe how to make 500.0 mL of a 0.150 M CaCl2 solution.

  7. What are the concentrations of ions present in a 0.25 M AlCl3 solution?

  8. How many moles of HCl are present in 25.0 mL of 0.60 M HCl?

  9. DilutionC1V1 = C2V2 • What is the concentration of a solution made by diluting 25.0 mL of 6.0 M HCl to a total volume of 500.0 mL?

  10. Dilution • Describe how to make 250.0 mL of 1.50M HCl solution from a 12.0 M HCl stock solution.

  11. Acid Base Titrations 2 NaOH + H2SO4 Na2SO4 + 2 H2O • The complete reaction of 7.00 mL of H2SO4 requires 23.50 mL of 0.250 M NaOH. • What is the concentration of the H2SO4 solution?

  12. Molarity Part II: Solution Stoichiometry Pb(NO3) 2 + 2 NaCl  PbCl2 (S) + 2NaNO3 • What volume of 1.50 M NaCl is needed to fully react 25.0 ml of 0.750 M Pb(NO3) 2 solution? • What mass of PbCl2 will be made?

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