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% 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
% 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
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

molarity calculations part 1
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.
slide5

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?

dilution c 1 v 1 c 2 v 2
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?
dilution
Dilution
  • Describe how to make 250.0 mL of 1.50M HCl solution from a 12.0 M HCl stock solution.
acid base titrations
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?
molarity part ii solution stoichiometry
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?