Concentration Conversion The Amount of Solute in the solvent Dr. Fred mega Garces Chemistry 201 Miramar College 1 Expressing Concentration
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
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 mole fraction ( χ A) - moles solute / Total moles solution Normality (N) - Number of equivalent / Liter solution 3 Expressing Concentration
Concentration Relationship Molecular Weight moles mass } Solute moles Molc Wt mass } Solvent χ Mass Solution m %m Density Solution M* Vol Solution Equivalence/mol N * Volume of solution must be used and not just volume of solvent 4 Expressing Concentration
Concentration Relationship * Volume of solution must be used and not just volume of solvent 5 Expressing Concentration
Concentration by Parts % Concentration Solute (mass or volume) Solution (mass or volume) x multiplier 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 10 6 g ppm (ppm) mass Soln g v/v = Vol Solute ml 10 9 g ppb (ppb) Vol Soln ml 6 Expressing Concentration
Interconverting Concentration: A Calculation Example Example: A perchloric acid (HCl 4 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 100 g solution 94.34 cc Molarity = 1.05 M 9.95 10-2 mole 0.090 Kg H 2 Answer molality = 1.11 m 9.95 10-2 mole 5.00 mol H 2 χa =.0195 7 Expressing Concentration
Interconverting Concentration: A Calculation Example Example: A perchloric acid (HCl 4 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 100 g solution 94.34 cc 9.95 10-2 mole 0.090 Kg H 2 9.95 10-2 mole 5.00 mol H 2 Molarity = 1.05 M molality = 1.11 m χa =.0195 8 Expressing Concentration
Molality =, mole faction = Kg Solvent Amount solute (EtH) = 40.00 ml, ρ EtH = 0.7890 g/cc 1. (EtH): 1a. Convert volume EtH to mass EtH by density + Moles solvent 1b. Convert mass EtH to moles EtH using molar mass. 2. Kg solvent: 100ml solution - 40.00ml EtH = 60.00ml H 2 2a. Convert 60.0 ml of water to kg water ρ H2 = 1.00 g/cc 3. Moles of solvent (H 2 ): 3a. Convert mass of H to moles of H using molar mass. 2 2 1a: 40.00ml EtH 0.7890 g = 31.56 g EtH 1 cc 1 mol 1b: 31.56 g EtH * = 0.6850 mol EtH 46.07g 2a: 60.00mL H 2 1.000 g 1 cc = 60.00 g H 0.06000Kg H 2 2 3a: 60.00 g H 2 * 1 mol 18.02g = 0.03330 mol H 2 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. ρ ETH = 0.789 g/cc 40.00% ethanol = 40.00mL EtH in 100.0 ml solution, MW ETH = 46.07 g/mol Molality = Moles, mole faction = Kg Solvent solute + Moles solvent Amount solute (EtH) = 40.00 ml, ρ EtH = 0.7890 g/cc 1. (EtH): 1a. Convert volume EtH to mass EtH by density 1b. Convert mass EtH to moles EtH using molar mass. 2. Kg solvent: 100ml solution - 40.00ml EtH = 60.00ml H 2 2a. Convert 60.0 ml of water to kg water ρ H2 = 1.00 g/cc 3. Moles of solvent (H 2 ): 3a. Convert mass of H 2 to moles of H 2 using molar mass. 1a: 40.00ml EtH 0.7890 g = 31.56 g EtH 1 cc 1b: 31.56 g EtH * 1 mol 46.07g = 0.6850 mol EtH 2a: 60.00mL H 2 1.000 g 1 cc = 60.00 g H 2 0.06000Kg H 2 3a: 60.00 g H 2 * 1 mol 18.02g = 0.03330 mol H 2 9 Expressing Concentration
Mass Solute Weight % = 100, mole faction = Mass solute + Mass solvent + Moles solvent Amount solute (EtH) = 12.50 moles EtH, 1. Mass solute (EtH): 1a. Convert moles EtH to mass EtH 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 2 ): 3a. Convert mass of H 2 to moles of H 2 using molar mass. 1a: 12.50 mol EtH 46.07 g 2a: 1.000 kg H 2 1000 g 1 mol = 575.88 g EtH 1 Kg = 1000. g H 2 3a: 1000. g H 2 * 1 mol 18.02g = 55.49 mol H 2 Calculating Weight Percent & mole fraction (χ) from from molality (m) Example # 1: An solution has a concentration of 12.50 m EtH. Calculate the weight percent and mole fraction of ethanol in the beverage. ρ ETH = 0.789 g/cc 12.50 m ethanol = 12.50 moles EtH in 1.000 kg H 2, MW ETH = 46.07 g/mol Weight % = Mass Solute Moles 100, mole faction = Mass solute + Mass solvent solute + Moles solvent Amount solute (EtH) = 12.50 moles EtH, 1. Mass solute (EtH): 1a. Convert moles EtH to mass EtH 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 2 ): 3a. Convert mass of H 2 to moles of H 2 using molar mass. 1a: 12.50 mol EtH 46.07 g 1 mol = 575.88 g EtH 2a: 1.000 kg H 2 1000 g 1 Kg = 1000. g H 2 3a: 1000. g H 2 * 1 mol 18.02g = 55.49 mol H 2 10 Expressing Concentration
Molality =, mole faction = Kg Solvent + Moles solvent 1. (EtH): 1a. Convert mass EtH to moles EtH using molar mass. 2. Kg solvent: There is 500.0 g H 2 2a. Convert 500.0 g of water to kg water 3. Moles of solvent (H 2 ): 3a. Convert mass of H 2 to moles of H 2 using molar mass. 1.0 mol 1a: 50.00g EtH = 1.085 mol EtH 46.07 g 1.0 kg 2a: 500.0 g H 2 * 1000 g = 0.5000 kg H 2 3a: 500.0 g H 2 * 1 mol 18.02g = 27.75 mol H 2 Calculating molality (m) and mole fraction (χ) from mass of solute and solvent Example # 1: 50.00 g EtH is added to 500.0 g H 2. What is the molality and mole fraction of the solution? ρ ETH = 0.789 g/cc, MW ETH = 46.07 g/mol Molality = Moles, mole faction = Kg Solvent solute + Moles solvent 1. (EtH): 1a. Convert mass EtH to moles EtH using molar mass. 2. Kg solvent: There is 500.0 g H 2 2a. Convert 500.0 g of water to kg water 3. Moles of solvent (H 2 ): 3a. Convert mass of H 2 to moles of H 2 using molar mass. 1.0 mol 1a: 50.00g EtH 46.07 g = 1.085 mol EtH 2a: 500.0 g H 2 * 1.0 kg 1000 g = 0.5000 kg H 2 3a: 500.0 g H 2 * 1 mol 18.02g = 27.75 mol H 2 11 Expressing Concentration
Molality = Kg Solvent, mass solute = moles solute *MW solute Given: mole faction = Best to setup equation- + Moles solvent 0.250 = + Moles solvent = + Moles solvent 0.250 moles solute - moles solute = moles solvent 4 moles solute - 1 mole solute = moles solvent 0.250 moles solvent moles solute = 1-1 0.250 4. mass of solute (ETH) 4a. Convert moles EtH to mass EtH 1000 ml 1a. 1.000L * 1 L * 1.0 g = 1000 g 1 ml 2a. 1000 g * 1 mol 18.02g = 55.49 mol H 2 moles solvent 3a moles solute = 1 = 55.49 moles H 2-1 3 0.250 = 18.50 moles EtH 4a: 18.50 moles EtH * 40.07 g = 741.2 g EtH 1 mole 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? ρ ETH = 0.789 g/cc, MW ETH = 46.07 g/mol Molality = Kg Solvent, mass solute = moles solute *MW solute Given: mole faction = + Moles solvent 1. Mass of H 2 1a. Convert volume H 2 to mass H 2. 2. Moles of H 2 (solvent) 2a. Convert mass H 2 to moles of H 2. 3. Moles of solute Best to setup equation- 3a. Solve for moles solute by plugging moles of H 2 into the mole fraction equation and solve for moles solute 0.250 = + Moles solvent = + Moles solvent 0.250 moles solute - moles solute = moles solvent 4 moles solute - 1 mole solute = moles solvent 0.250 moles solvent moles solute = 1 0.250-1 4. mass of solute 4a. Convert moles solute to mass solute 1. Mass of H 2 1a. Convert volume H 2 to mass H 2. 2. Moles of H 2 (solvent) 2a. Convert mass H 2 to moles of H 2. 3. Moles of solute 3a. Solve for moles solute by plugging moles of H 2 into the mole fraction equation and solve for moles solute 12 Expressing Concentration 1a. 1.000L * 1000 ml 1 L * 1.0 g 1 ml = 1000 g 2a. 1000 g * 1 mol 18.02g = 55.49 mol H 2 3a moles solute = moles solvent 1 0.250-1 = 18.50 moles EtH 4a: 18.50 moles EtH * 40.07 g 1 mole = 55.49 moles H 2 3 = 741.2 g EtH
...and ever more Examples Extra examples 50.00ml of ethylene glycol (ρ = 1.114 g/ml; MW = 62.07 g/mol) is added to 1.000-L water (ρ = 1.00 g/ml) at 20 C. Answer the following questions and assume additive volumes. i) What is the density of the mixture ii) Calculate the % mass of the ethylene glycol in the solution. iii) Calculate the molarity and molality of ethylene glycol in the solution. Mass H 2 = 1000 g vol = 50.0 ml ethylene Glycol 50.00 ml 1.114 g ml D = vol = 1000 ml H 2 = 55.70 g mass H2 + mass ethylene glycol vol H2 +vol ethylene glycol D = 1055.70 g 1050.0 ml = 1.0054 = 1.005 g ml % m = 55.70 g 1055.7 g mol glycol, 55.70 g 0.89737 mol mol H 2, 1000 g 55.56 mol molality = Molarity = 100 = 5.276 %.89737 mol 1.00kg.89737 mol 1.050 L = 0.8974 m = 0.8546 M 13 Expressing Concentration
Practice Problems Harris 7 th ed p18 1. The density of 70.5 Wt% aqueous perchloric acid, HCl 4, is 1.67 g/ml. 1.20 (a) How many grams of solution are in 1.000 L 1670 g (b) How many grams of HCl 4 are in 1.000L? 1180 g (c) How many moles of HCl 4 in 1.000L? 11.7 mol 2. An aqueous solution containing 20.0% wt% KI had a density of 1.168 g/ml. Find the molality, mole fraction, and molarity of the KI solution. 1.21 1.51 m 3. The concentration of sugar (glucose, C 6 H 12 6 ) in human blood ranges from about 80mg/100mL before meal to 120mg/100mL after eating. Find the molarity before and after eating. 1.22 4.4e-3M, 6.7e-3M 4. It is recommended that drinking water contain 1.6 ppm fluoride (F-) for preventing of tooth decay. Consider a reservoir with a diameter of 4.50 10 2 m and and average depth of 10.0 m. (V = π r 2 h) How many grams of fluoride should be added to give 1.6 ppm? How many grams of sodium fluoride, NaF contains this much fluoride? 1.25, 2.5e6 g F -, 5.6e6 g NaF 5. How many ml of 3.00 M H 2 S 4 are required to react with 4.35 g of solid containing 23.2 m:m% Ba(N 3 ) 2 if the reaction produces BaS 4 precipitate. 133, 1.29 ml 14 Expressing Concentration
Solution at a Glance Solutions can be describe by the following: Solvent The component of a solution present in the greatest quantity Solute The component of solution present in the lesser quantity Solution A homogeneous mixture of two or more substances in which each substance retains its chemical identity Concentration of a Solution The amount of solute in a specific amount of solution. Molarity (M) moles of solute Liters of solution 15 Expressing Concentration
Activity 1: Concentration Conversion / Score Name (last) (first) Lab Section: Day Time i Show your work in another sheet of paper and then fill in the blanks in the table. The solvent is water for these solutions. Your answer should contain the right number of significant figures with the correct units. If you do not know how to determine the number of significant figures an answer should contain, please review your chem 200 fundamentals. Compound Molality Weight Percent Mole Fraction Mole Fraction (Ranking) 1 st (low), 2 nd, 3 rd, 4 th, 5 th, 6 th (high), A HF 18.0 % B CH3H 1.50 m C C6H126 15.0 % D NaI 0.750 m E CH3C2H 5.00 % F KN3 0.0143 m ii. Fill in the blanks in the table. Your answer should contain the right number of significant figures with the correct units. Compound Grams Compound Grams Water Molality Mole Fraction of Compound Mole Fraction (Ranking) 1 st (low), 2 nd, 3 rd, 4 th, 5 th, 6 th (high), A Na2C3 40.5 155.0 B C3H7H 250. 2.55 C NaN3 555 0.0334 D Pb(N 3 ) 2 800. 3.45 E Sr(H)2 255. 0.0545 F Pt(NH3)2Cl2 75.4 205. iii. You wish to prepare an IV solution, NaCl, with a mole fraction of 2.90 10-4. Assume that the density of water = 1.000 g/cc How many grams (g) of NaCl must you combine with 1.000 L of water to make this solution? What is the molality (m) of the solution? What is the concentration in ppm? (Answer) (Answer) (Answer) iv What is the mass % (m:m) of physiologically correct saline solution also known as normal saline solution? (Use 3 significant figures) (Use the Internet and keyword physiologically normal saline concentration ) What are the molarity and the mole fraction of this solution? (Answers) Note: If you rip this page from the lab manual, be sure to trim the edge. (Reminder from your lab instructor) 21