Concentration Conversion
Transcript of Concentration Conversion
January 19 1 Expressing Concentration
Concentration Conversion
The Amount of Solute in the solvent
Dr. Fred Omega Garces Chemistry 201 Miramar College
January 19 2 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.
January 19 3 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 (χA) - moles solute / Total moles solution Normality (N) - Number of equivalent / Liter solution
January 19 4 Expressing Concentration
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
January 19 5 Expressing Concentration
Concentration Relationship
* Volume of solution must be used and not just volume of solvent
January 19 6 Expressing Concentration
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
January 19 7 Expressing Concentration
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.
January 19 8 Expressing Concentration
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
January 19 9 Expressing Concentration
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
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
January 19 10 Expressing Concentration
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):
3a. Convert mass of H2O to moles of H
2O using molar mass.
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):
3a. Convert mass of H2O to moles of H
2O using molar mass.
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
January 19 11 Expressing Concentration
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
Molality = Moles soluteKg Solvent
, mole faction = Moles soluteMoles 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
2O
2a. Convert 500.0 g of water to kg water3. Moles of solvent (H
2O):
3a. Convert mass of H2O to moles of H
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
Molality = Moles soluteKg Solvent
, mole faction = Moles soluteMoles 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
2O
2a. Convert 500.0 g of water to kg water3. Moles of solvent (H
2O):
3a. Convert mass of H2O to moles of H
2O using molar mass.
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
January 19 12 Expressing Concentration
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
Molality = Moles soluteKg Solvent
, 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
January 19 13 Expressing Concentration
...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 %
January 19 14 Expressing Concentration
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
January 19 15 Expressing Concentration
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)