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Conversion

The Amount of Solute in the

Dr. Fred Omega Garces 201 Miramar College

1 Expressing Concentration January 19 Components of : Variable components, retains properties of its component.

Homogeneous systems: Solution - Homogeneous 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 [] / solution [mass]) * 100

w/v [mass solute (g) / solution (ml)] * 100

v/v [vol solute (mL) / vol solution (mL)] * 100

χ 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 of solution is 1.060 g/cc. What is the Molarity, , .

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 . 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. mass of solute 40.07 g 4a. Convert moles solute to mass solute 4a: 18.50 moles EtOH * = 741.2 g EtOH 1 mole

Moles solute Molality = , mass solute = moles solute *MW Kg Solvent solute

Moles solute Given: mole faction = Best to setup equation- Moles solute + Moles solvent

1. Mass of H O 2 1a. Convert volume H O to mass H O. 2 2 2. Moles of H O (solvent) 2 2a. Convert mass H O to moles of H O. 2 2 3. Moles of solute 3a. Solve for moles solute by plugging moles of H O into the mole fraction equation 2 and solve for moles solute Moles solute Moles solute 0.250 = → Moles solute + Moles solvent = Moles solute + 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 (ETOH) 4a. Convert moles EtOH to mass EtOH

1000 mL 1.0 g 1a. 1.000L * * = 1000 g 1 L 1 mL

1 mol 2a. 1000 g * = 55.49 mol H O 18.02g 2

moles solvent 55.49 moles H O 3a moles solute = = 2 Expressing Concentration 1 3 -1 12 0.250 January 19

= 18.50 moles EtOH

40.07 g 4a: 18.50 moles EtOH * = 741.2 g EtOH 1 mole ...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 .

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 O = 1000 g vol = 1000 mL H O 55.70 g 2 2 % m = ⋅ 100 = 5.276 % vol = 50.0 mL ethylene Glycol 1055.7 g 1.114 g 50.00 mL ⋅ = 55.70 g mL mol glycol, 55.70 g ⇒ 0.89737 mol mol H O, 1000 g ⇒ 55.56 mol 2 mass H2O + mass ethylene glycol D = 2 .89737 mol vol H O +vol ethylene glycol molality = = 0.8974 m 1.00kg 1055.70 g g .89737 mol D = = 1.0054 = 1.005 Molarity = = 0.8546 M 1050.0 mL mL 1.050 L

13 Expressing Concentration January 19 Practice Problems Harris 7th ed p18

1. The density of 70.5 Wt% aqueous perchloric acid, HClO4, 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 HClO4 are in 1.000L? 1180 g

(c) How many moles of HClO4 in 1.000L? 11.7 mol

2. An 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, C6H12O6) 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•102 m and and average depth of 10.0 m. (V = π r2 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 H2SO4 are required to react with 4.35 g of solid containing 23.2 m:m% Ba(NO3)2 if the reaction produces BaSO4 precipitate. 133, 1.29 mL

14 Expressing Concentration January 19 Solution at a Glance Solutions can be describe by the following:

Solvent Solute The component of a solution present in The component of the greatest solution present in quantity the lesser quantity

Solution A homogeneous mixture of two or more Concentration of a substances in which Solution each substance retains The amount of solute its chemical identity in a specific amount of solution.

Molarity (M) moles of solute Liters of solution

15 Expressing Concentration January 19 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) st nd rd th th th 1 (low), 2 , 3 , 4 , 5 , 6 (high),

A HF 18.0 %

B CH3OH 1.50 m

C C6H12O6 15.0 %

D NaI 0.750 m

E CH3CO2H 5.00 %

F KNO3 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 Grams Molality Mole Fraction of Mole Fraction (Ranking) Compound Water Compound 1st (low), 2nd, 3rd, 4th, 5th, 6th(high),

A Na2CO3 40.5 155.0

B C3H7OH 250. 2.55

C NaNO3 555 0.0334

D Pb(NO3)2 800. 3.45

E Sr(OH)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? ______(Answer)

What is the molality (m) of the solution? ______(Answer)

What is the concentration in ppm? ______(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)

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