Concentration Conversion The Amount of Solute in the solvent Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Expressing Concentration January 19 Components of Solution Mixtures: Variable components, retains properties of its component. Homogeneous systems: Solutions Solution - Homogeneous mixture of two or more substances Components of solution Solute - Substance being dissolve Solvent - Substance in which solute is dissolved in. If solvent is water, then solution is considered aqueous. 2 Expressing Concentration January 19 Expressing Concentration 5 ways of expressing concentration- Molarity (M) - moles solute / Liter solution Molality* (m) - moles solute / Kg solvent Conc. by parts (% m)- (solute [mass] / solution [mass]) * 100 w/v [mass solute (g) / volume solution (ml)] * 100 v/v [vol solute (mL) / vol solution (mL)] * 100 χ mole fraction ( A) - moles solute / Total moles solution Normality (N) - Number of equivalent / Liter solution 3 Expressing Concentration January 19 Concentration Relationship Molecular Weight moles mass } Solute Molc’ Wt moles mass } Solvent χ MassSolution m %m Density Solution M* VolSolution Equivalence/mol N * Volume of solution must be used and not just volume of solvent 4 Expressing Concentration January 19 Concentration Relationship * Volume of solution must be used and not just volume of solvent 5 Expressing Concentration January 19 Concentration by Parts Solute (mass or volume) x multiplier Solution (mass or volume) % Concentration w/w = Wt Solute• g •100 g % (pph) Wt Soln g w/v = Wt Solute• g • 100 g % (pph) Vol Soln ml v/v = Vol Solute• ml • 100 g % (pph) Vol Soln ml ppm & ppb (For dilute solution) m/m = mass Solute • g •106 g ppm (ppm) mass Soln g v/v = Vol Solute• ml • 109 g ppb (ppb) Vol Soln ml 6 Expressing Concentration January 19 Interconverting Concentration: A Calculation Example Example: A perchloric acid (HClO4 MWt = 100.5 g/mol) solution is 10.0 %m:m (by mass). The density of solution is 1.060 g/cc. What is the Molarity, molality, mole fraction. 10.00 g → 9.95•10-2 mole → 9.95•10-2 mole → → 9.95•10-2 mole 100 g solution → 94.34 cc 0.090 Kg H O 2 → 5.00 mol H2O Answer Molarity = 1.05 M molality = 1.11 m χA = .0195 7 Expressing Concentration January 19 Interconverting Concentration: A Calculation Example Example: A perchloric acid (HClO4 MWt = 100.5 g/mol) solution is 10.0 %m:m (by mass). The density of solution is 1.060 g/cc. What is the Molarity, molality, mole fraction. 10.00 g → 9.95•10-2 mole → 9.95•10-2 mole → → 9.95•10-2 mole 100 g solution → 94.34 cc 0.090 Kg H O 2 → 5.00 mol H2O Molarity = 1.05 M molality = 1.11 m χA = .0195 8 Expressing Concentration January 19 Calculating molality (m) and mole fraction (χ) from volume percent Example # 1: An alcoholic beverage is 80.00 proof (40.00% alcohol v:v). Calculate the molality and mole fraction of ethanol in the beverage. ρETOH = 0.789 g/cc 40.00% ethanol = 40.00mL EtOH in 100.0 mL solution, MWETOH = 46.07 g/mol Moles solute Moles solute Molality = , mole faction = Kg Solvent Moles solute + Moles solvent Amount solute (EtOH) = 40.00 mL, ρ = 0.7890 g/cc EtOH 1. Moles solute (EtOH): 1a. Convert volume EtOH to mass EtOH by density 0.7890 g 1b. Convert mass EtOH to moles EtOH using molar mass. 1a: 40.00ml EtOH ∗ = 31.56 g EtOH 2. Kg solvent: 100ml solution - 40.00ml EtOH = 60.00ml H O 1 cc 2 2a. Convert 60.0 mL of water to kg water ρ = 1.00 g/cc H O 2 1 mol 1b: 31.56 g EtOH * = 0.6850 mol EtOH 3. Moles of solvent (H O): 2 46.07g 3a. Convert mass of H O to moles of H O using molar mass. 2 2 1.000 g 2a: 60.00mL H O ∗ = 60.00 g H O → 0.06000Kg H O 2 1 cc 2 2 1 mol 3a: 60.00 g H O * = 0.03330 mol H O 2 18.02g 2 Moles solute Moles solute Molality = , mole faction = Kg Solvent Moles solute + Moles solvent Amount solute (EtOH) = 40.00 mL, ρ = 0.7890 g/cc EtOH 1. Moles solute (EtOH): 1a. Convert volume EtOH to mass EtOH by density 1b. Convert mass EtOH to moles EtOH using molar mass. 2. Kg solvent: 100ml solution - 40.00ml EtOH = 60.00ml H O 2 2a. Convert 60.0 mL of water to kg water ρ = 1.00 g/cc H O 2 3. Moles of solvent (H O): 2 3a. Convert mass of H O to moles of H O using molar mass. 2 2 0.7890 g 1a: 40.00ml EtOH ∗ = 31.56 g EtOH 1 cc 1 mol 1b: 31.56 g EtOH * = 0.6850 mol EtOH 46.07g Expressing Concentration 1.000 g 2a: 60.00mL H O ∗ = 60.00 g H O → 0.06000Kg H O 9 2 2 2 January 19 1 cc 1 mol 3a: 60.00 g H O * = 0.03330 mol H O 2 18.02g 2 Calculating Weight Percent & mole fraction (χ) from from molality (m) Example # 1: An solution has a concentration of 12.50 m EtOH. Calculate the weight percent and mole fraction of ethanol in the beverage. ρETOH = 0.789 g/cc 12.50 m ethanol = 12.50 moles EtOH in 1.000 kg H2O, MWETOH = 46.07 g/mol Mass Solute Moles solute Weight % = ∗100, mole faction = Mass solute + Mass solvent Moles solute + Moles solvent Amount solute (EtOH) = 12.50 moles EtOH, 1. Mass solute (EtOH): 1a. Convert moles EtOH to mass EtOH by molar mass 46.07 g 2. Mass water: There is 1.000 Kg water 1a: 12.50 mol EtOH ∗ = 575.88 g EtOH 2a. Convert 1.000 kg of water to g 1 mol 3. Moles of solvent (H O): 2 3a. Convert mass of H O to moles of H O using molar mass. 1000 g 2 2 2a: 1.000 kg H O ∗ = 1000. g H O 2 1 Kg 2 1 mol 3a: 1000. g H O * = 55.49 mol H O 2 18.02g 2 Mass Solute Moles solute Weight % = ∗100, mole faction = Mass solute + Mass solvent Moles solute + Moles solvent Amount solute (EtOH) = 12.50 moles EtOH, 1. Mass solute (EtOH): 1a. Convert moles EtOH to mass EtOH by molar mass 2. Mass water: There is 1.000 Kg water 2a. Convert 1.000 kg of water to g 3. Moles of solvent (H O): 2 3a. Convert mass of H O to moles of H O using molar mass. 2 2 46.07 g 1a: 12.50 mol EtOH ∗ = 575.88 g EtOH 1 mol 1000 g 2a: 1.000 kg H O ∗ = 1000. g H O Expressing Concentration 10 2 1 Kg 2 January 19 1 mol 3a: 1000. g H O * = 55.49 mol H O 2 18.02g 2 Calculating molality (m) and mole fraction (χ) from mass of solute and solvent Example # 1: 50.00 g EtOH is added to 500.0 g H2O. What is the molality and mole fraction of the solution? ρETOH = 0.789 g/cc , MWETOH = 46.07 g/mol Moles solute Moles solute Molality = , mole faction = Kg Solvent Moles solute + Moles solvent 1. Moles solute (EtOH): 1a. Convert mass EtOH to moles EtOH using molar mass. 2. Kg solvent: There is 500.0 g H O 2 2a. Convert 500.0 g of water to kg water 3. Moles of solvent (H O): 2 3a. Convert mass of H O to moles of H O using molar mass. 1.0 mol 2 2 1a: 50.00g EtOH ∗ = 1.085 mol EtOH 46.07 g 1.0 kg 2a: 500.0 g H O * = 0.5000 kg H O 2 1000 g 2 1 mol 3a: 500.0 g H O * = 27.75 mol H O 2 2 Moles solute Moles solute Molality = , mole faction = Kg Solvent Moles solute + Moles solvent 18.02g 1. Moles solute (EtOH): 1a. Convert mass EtOH to moles EtOH using molar mass. 2. Kg solvent: There is 500.0 g H O 2 2a. Convert 500.0 g of water to kg water 3. Moles of solvent (H O): 2 3a. Convert mass of H O to moles of H O using molar mass. 2 2 1.0 mol 1a: 50.00g EtOH ∗ = 1.085 mol EtOH 46.07 g 1.0 kg 2a: 500.0 g H O * = 0.5000 kg H O 2 1000 g 2 1 mol 3a: 500.0 g H O * = 27.75 mol H O 11 2 18.02g 2 Expressing Concentration January 19 Calculating molality (m) and mass solute from mole fraction (χ) and mass solvent Example # 1: Given a 0.250 mole faction (x) of ethanoic solution in 1.000 L, what is the molality and mass of solute in the solution? ρETOH = 0.789 g/cc , MWETOH = 46.07 g/mol Moles solute Molality = , mass solute = moles solute *MW Kg Solvent solute Moles solute Given: mole faction = Best to setup equation- 1000 mL 1.0 g Moles solute + Moles solvent 1a. 1.000L * * = 1000 g 1 L 1 mL 1. Mass of H O 2 1a. Convert volume H O to mass H O. 2 2 1 mol 2. Moles of H O (solvent) 2a. 1000 g * = 55.49 mol H O 2 2 2a. Convert mass H O to moles of H O. 18.02g 2 2 3. Moles of solute 3a. Solve for moles solute by plugging moles of H O into the mole fraction equation 2 moles solvent 55.49 moles H O and solve for moles solute 3a moles solute = = 2 Moles solute Moles solute 1 3 0.250 = → Moles solute + Moles solvent = -1 Moles solute + Moles solvent 0.250 moles solute 0.250 - moles solute = moles solvent → 4 moles solute - 1 mole solute = moles solvent 0.250 moles solvent moles solute = = 18.50 moles EtOH 1 -1 0.250 4.