concentration conversion

January 19 1 Expressing Concentration

Concentration Conversion

The Amount of Solute in the solvent

Dr. Fred Omega Garces Chemistry 201 Miramar College


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.


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 (χA) - moles solute / Total moles solution Normality (N) - Number of equivalent / Liter solution


Concentration Relationship

moles mass

moles mass

χ m %m

M*

} Solute

} Solvent

Molecular Weight

Molc’ Wt

MassSolution

* Volume of solution must be used and not just volume of solvent

VolSolution

Density Solution

N Equivalence/mol


Concentration Relationship

* Volume of solution must be used and not just volume of solvent


Concentration by Parts

% 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

Solute (mass or volume) Solution (mass or volume)

x multiplier


Interconverting Concentration: A Calculation Example

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 H2O

molality = 1.11 m

→ 9.95•10-2 mole → 5.00 mol H2O

χA = .0195

Answer

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.


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 100 g solution → 94.34 cc

Molarity = 1.05 M

→ 9.95•10-2 mole → 0.090 Kg H2O

molality = 1.11 m

→ 9.95•10-2 mole → 5.00 mol H2O

χA = .0195


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

Molality = Moles soluteKg Solvent

, mole faction = Moles soluteMoles solute + Moles solvent

Amount solute (EtOH) = 40.00 mL, ρEtOH

= 0.7890 g/cc1. 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

2O

2a. Convert 60.0 mL of water to kg water ρH2O

= 1.00 g/cc

3. Moles of solvent (H2O):

3a. Convert mass of H2O to moles of H

2O using molar mass.

1a: 40.00ml EtOH ∗ 0.7890 g1 cc

= 31.56 g EtOH

1b: 31.56 g EtOH * 1 mol46.07g

= 0.6850 mol EtOH

2a: 60.00mL H2O∗

1.000 g1 cc

= 60.00 g H2O→ 0.06000Kg H

2O

3a: 60.00 g H2O * 1 mol

18.02g = 0.03330 mol H

2O



Amount solute (EtOH) = 40.00 mL, ρEtOH

= 0.7890 g/cc1. 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

2O

2a. Convert 60.0 mL of water to kg water ρH2O

= 1.00 g/cc

3. Moles of solvent (H2O):



1a: 40.00ml EtOH ∗ 0.7890 g1 cc

= 31.56 g EtOH

1b: 31.56 g EtOH * 1 mol46.07g

= 0.6850 mol EtOH

2a: 60.00mL H2O∗

1.000 g1 cc

= 60.00 g H2O→ 0.06000Kg H

2O

3a: 60.00 g H2O * 1 mol

18.02g = 0.03330 mol H

2O


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

Weight % = Mass SoluteMass solute + Mass solvent

∗100, mole faction = Moles soluteMoles 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 g3. Moles of solvent (H

2O):



1a: 12.50 mol EtOH ∗ 46.07 g1 mol

= 575.88 g EtOH

2a: 1.000 kg H2O∗

1000 g1 Kg

= 1000. g H2O

3a: 1000. g H2O * 1 mol

18.02g = 55.49 mol H

2O

Weight % = Mass SoluteMass solute + Mass solvent

∗100, mole faction = Moles soluteMoles 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 g3. Moles of solvent (H

2O):



1a: 12.50 mol EtOH ∗ 46.07 g1 mol

= 575.88 g EtOH

2a: 1.000 kg H2O∗

1000 g1 Kg

= 1000. g H2O

3a: 1000. g H2O * 1 mol

18.02g = 55.49 mol H

2O


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



1. Moles solute (EtOH): 1a. Convert mass EtOH to moles EtOH using molar mass.2. Kg solvent: There is 500.0 g H

2O

2a. Convert 500.0 g of water to kg water3. Moles of solvent (H

2O):


2O using molar mass. 1a: 50.00g EtOH ∗ 1.0 mol

46.07 g = 1.085 mol EtOH

2a: 500.0 g H2O * 1.0 kg

1000 g = 0.5000 kg H

2O

3a: 500.0 g H2O * 1 mol

18.02g = 27.75 mol H

2O



1. Moles solute (EtOH): 1a. Convert mass EtOH to moles EtOH using molar mass.2. Kg solvent: There is 500.0 g H

2O

2a. Convert 500.0 g of water to kg water3. Moles of solvent (H

2O):



1a: 50.00g EtOH ∗ 1.0 mol46.07 g

= 1.085 mol EtOH

2a: 500.0 g H2O * 1.0 kg

1000 g = 0.5000 kg H

2O

3a: 500.0 g H2O * 1 mol

18.02g = 27.75 mol H

2O


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


, mass solute = moles solute *MWsolute

Given: mole faction = Moles soluteMoles solute + Moles solvent

Best to setup equation-

1. Mass of H2O

1a. Convert volume H2O to mass H

2O.

2. Moles of H2O (solvent)

2a. Convert mass H2O to moles of H

2O.

3. Moles of solute 3a. Solve for moles solute by plugging moles of H

2O into the mole fraction equation

and solve for moles solute

0.250 = Moles soluteMoles solute + Moles solvent

→ Moles solute + Moles solvent = Moles solute0.250

moles solute0.250

- moles solute = moles solvent → 4 moles solute - 1 mole solute = moles solvent

moles solute = moles solvent1

0.250-1

4. mass of solute 4a. Convert moles solute to mass solute

1a. 1.000L * 1000 mL1 L

* 1.0 g1 mL

= 1000 g

2a. 1000 g * 1 mol18.02g

= 55.49 mol H2O

3a moles solute = moles solvent1

0.250-1

= 55.49 moles H

2O

3

= 18.50 moles EtOH

4a: 18.50 moles EtOH * 40.07 g1 mole

= 741.2 g EtOH Molality = Moles solute

Kg Solvent, mass solute = moles solute *MW

solute

Given: mole faction = Moles soluteMoles solute + Moles solvent

Best to setup equation-

1. Mass of H2O

1a. Convert volume H2O to mass H

2O.

2. Moles of H2O (solvent)

2a. Convert mass H2O to moles of H

2O.

3. Moles of solute 3a. Solve for moles solute by plugging moles of H

2O into the mole fraction equation

and solve for moles solute

0.250 = Moles soluteMoles solute + Moles solvent

→ Moles solute + Moles solvent = Moles solute0.250

moles solute0.250

- moles solute = moles solvent → 4 moles solute - 1 mole solute = moles solvent

moles solute = moles solvent1

0.250-1

4. mass of solute (ETOH) 4a. Convert moles EtOH to mass EtOH

1a. 1.000L * 1000 mL1 L

* 1.0 g1 mL

= 1000 g

2a. 1000 g * 1 mol18.02g

= 55.49 mol H2O

3a moles solute = moles solvent1

0.250-1

= 55.49 moles H

2O

3

= 18.50 moles EtOH

4a: 18.50 moles EtOH * 40.07 g1 mole

= 741.2 g EtOH


...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

2O = 1000 g vol = 1000 mL H

2O

vol = 50.0 mL ethylene Glycol

50.00 mL ⋅ 1.114 gmL

= 55.70 g

D = mass H2O + mass ethylene glycol vol H2O +vol ethylene glycol

D = 1055.70 g 1050.0 mL

= 1.0054 = 1.005 gmL

mol glycol, 55.70 g ⇒ 0.89737 mol mol H

2O, 1000 g ⇒ 55.56 mol

molality = .89737 mol 1.00kg

= 0.8974 m

Molarity = .89737 mol 1.050 L

= 0.8546 M

% m = 55.70 g1055.7 g

⋅ 100 = 5.276 %


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 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, 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


Solution at a Glance Solutions can be describe by the following:

Solvent The component of a solution present in

the greatest 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.

Solute The component of solution present in the lesser quantity

Molarity (M)

moles of solute Liters of solution

21

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) 1st (low), 2nd, 3rd, 4th, 5th, 6th(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 Compound

Grams Water

Molality Mole Fraction of Compound

Mole Fraction (Ranking) 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)

concentration conversion

Documents