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Chemical Measurements
Daniel C. Harris
Chapter 1
Dr. Sa’ib Khouri
AUM- JORDAN
Introduction
Analytical Chemistry: The science of chemical measurements that studies
and uses chemical methods and instruments.
Types of Questions Asked in Analytical Chemistry:
a) What is in the sample? (qualitative analysis that identifies analytes)
b) How much is in the sample? (quantitative analysis that determines the numerical
amount or concentration)
How pure is it? quantitative analysis
What are the impurities? qualitative analysis
b) Instrumental Methods: spectrometry, chromatography, conductomertry,
electrophoresis, voltammetry,…etc.
Techniques used in Analytical Chemistry:
a) Wet Chemical Methods: Separation, extraction, distillation,titration, color-forming
reactions, precipitations, ….etc.
Identification may be based on differences in color, odor, melting point, boiling point,
solubility, radioactivity or reactivity.
Fundamental units (base units) from which all
others are derived
Quantity Unit (Symbol)
Length Meter (m)
Mass Kilogram (kg)
Time Second (s)
Electric current Ampere (A)
Temperature Kelvin (K)
Amount of
substance
Mole (mol)
SI Units (metric units)
From the French système international d'unités
used by scientists around the world
SI-derived units with special names
Prefixes to express large (multiples) or small quantities
(fractions)
Example
At an altitude of 1.7 x 104
meters above the Earth’s surface, the
pressure of ozone over Antarctica reaches a peak of 0.019 Pa.
Express these numbers with prefixes from the table
An ozone “hole” forms each year in the
stratosphere over the South Pole at the
beginning of spring in October.
Converting Between Units
H.W. 1.5 and 1.9
Express the rate of energy used by a person walking 2 miles per hour (consuming
91 kcal per hour per 100 pounds of body mass) in kilojoules (kJ) per hour per
kilogram (kg) of body mass.
Example
Solution
1 cal = 4.184 J, so 1 kcal = 4.184 kJ 1 lb = 0.454 kg; so 100 lbs = 45.4 kg
or in one long calculation:
The rate of energy consumption is therefore
Chemical Concentrations
Concentration states how much solute (minor species) is contained in a given
volume or mass of homogeneous solution or solvent (major species).
Molarity and Molality
Molarity (mol/L, or M) is the number of moles of a substance per liter of solution.
Chemical concentrations, denoted with square brackets, are usually expressed in
moles per liter (M). Thus “[H+
]” means “the concentration of H+
.”
Molarity (M) =
moles of solute
liters of solution
The molarity of a strong electrolyte is called sometimes the formal concentration (F)
e.g. Magnesium chloride is a strong electrolyte. In 0.44 M MgCl2 solution, 70% of
the magnesium is free Mg2+
and 30% is MgCl+
. The concentration of MgCl2
molecules is close to 0.
The “molecular mass” of a strong electrolyte is called the formula mass (FM),
and it is the sum of atomic masses of atoms in the formula.
Molarity of Salts in the Sea
Example
solution
Calculate the formula mass of CaSO4. What is the molarity of CaSO4 in a solution
containing 1.2 g of CaSO4 in a volume of 50 mL? How many grams of CaSO4 are
in 50 mL of 0.086 M CaSO4? (Answer: 136.14 g/mol, 0.18 M, 0.59 g)
H.W.
For a weak electrolyte such as acetic acid, CH3CO2H, some of the molecules dissociate into
ions in solution:
Molality (mol/kg, or m ) is concentration expressed as moles of substance per kilogram of
solvent (not total solution)
Molality is independent of temperature
Molality (m) =
moles of solute
kg of solvent
Percent Composition
Weight percent (wt%):
Volume percent (vol%):
e.g. A common form of ethanol (CH3CH2OH) is 95 wt%; this expression means
95 g of ethanol per 100 g of total solution.
Note: mass is usually implied when units are absent.
Example
Find the molarity and molality of 37.0 wt% HCl. Its density is 1.19 g/mL.
Solution
For molarity, we need to find the moles of HCl per liter of solution
The mass of a liter of solution is (1.19 g/mL )(1000 mL) = 1.19 x 10
3
g.
The molecular mass of HCl is 36.46 g/mol, so the molarity is
For molality, we need to find the moles of HCl per kilogram of solvent (H2O).
100 g of solution contains 37 g of HCl and 63 g of H2O (0.063 kg)
37.0 g of HCl = 37.0 /(36.46 g/mol) = 1.01 mol HCl
HW
Calculate the molarity and molality of 49.0 wt% HF, assuming
a density of 1.16 g/mL.
Parts per Million (ppm) and Parts per Billion (ppb)
mean grams of substance per million or billion grams of total solution or mixture.
The density of a dilute aqueous solution is close to 1.00 g/mL (i.e 1 mL ≡ 1 g)
1 ppm corresponds to 1 μg/mL (1 mg/L), and
1 ppb is 1 ng/mL (1 μg/L)
Question: What does one part per thousand and one part per trillion mean?
ppt: g/L (mg/mL)
ppm: mg/L (µg/mL)
ppb: µg/L (ng/mL)
pptr: ng/L (pg/mL)
34 ppb means there are 34 ng of C29H60 per gram of rainwater, which is
nearly the same as 34 ng/mL. Find the molarity of this concentration.
Example
Number of grams of C29H60 in a liter is 1000 Multiply by 34 ng/mL and gives
34000 ng/L (or 34μg/L) of C29H60 of rainwater.
solution
The molecular mass of C29H60 is 408.8 g/mol
HW. How many ppm of C29H60 are in 23 μM C29H60? (Answer: 9.4 ppm)
Preparing Solutions
To prepare a solution with a desired molarity from a pure solute
(solid or liquid), we weigh out the mass of solute using analytical
balance and dissolve it in a volumetric flask.
Example
solution
Using a volumetric flask: The procedure is to place 0.999 g of
the solid into a 500-mL volumetric flask, add about 400 mL of
distilled water, and swirl to dissolve the reagent. Then dilute
with distilled water up to the 500-mL mark and invert the flask
several times to ensure complete mixing.
HW. Find the formula mass of anhydrous
CuSO4. How many grams should be
dissolved in 250.0 mL to make a 16.0 mM
solution? (Answer:159.61 g/mol, 0.638 g)
Dilution
Dilute solutions can be prepared from concentrated solutions
The number of moles in a concentrated (conc) solution that
taken is equal to the number of moles placed in the dilute
(dil) solution, i.e. no changes in number of moles.
To make 0.1 M HCl, we would dilute 8.26 mL of conc.
HCl up to 1 L.
Note: Dilute HCl solution here needs standardization
to find its accurate concentration.
Stoichiometry is the calculation of quantities of substances
involved in a chemical reaction.
Gravimetric analysis: chemical analysis based on weighing a
final product .
e.g. Iron (Fe) from a dietary supplement tablet can be measured by
dissolving the tablet and then converting the iron into solid Fe2O3
(ferric oxide) and from the mass of this solid one can calculate the
mass of iron in the original tablet.
Stoichiometry Calculations for Gravimetric Analysis
Gravimetric and titrations analysis were practiced long before
electronic instruments to make chemical measurements and called as
“classical” or “wet chemical” methods, and still have a place in modern
analytical chemistry to be more accurate than instrumental methods
usually used to prepare standards for instrumental methods of
analysis.
The steps in the procedure:
FeO(OH) ·xH2O
Example
In a gravimetric analysis, we need enough product to weigh accurately. Each
tablet provides about 15 mg (0.015 g) of iron. How many tablets we should use
to analyze in order to produce 0.25 g of solid Fe2O3?
Introduction to Titrations
In a titration, increments of reagent solution (the titrant and it is usually
delivered from a buret) are added to analyte until their reaction is complete.
Requirements
• Reaction must be quick
• Reaction must proceed according to a defined
equation
• Availability of indicator
• Reactions must be complete or have a large
equilibrium
From the quantity of titrant required, the quantity of
analyte concentration can be calculated
Common titrations rely on acid-base, precipitation,
oxidation-reduction, or complex formation reactions.
e.g. 5 mol of oxalic acid react with 2 mol of permanganate in hot acidic solution.
At the equivalence point
M • V (oxalic acid ) = 5/2 M • V (permanganate)
# moles of oxalic acid = 5/2 # moles of permanganate
End point: is marked by a sudden change in a physical property of the
solution, e.g. color change of a reactant or an external indicator added.
The equivalence point: The point when the quantity of added titrant is the
exact amount necessary for stoichiometric reaction with the analyte.
Primary standard
• At least 99.9% pure
• Stable (low reactive) under ordinary storage
• Stable when dried by heat or vacuum
• High molar mass (to minimize weighing errors)
A reagent is pure enough to be weighed and used directly, it should be
• Non-toxic
• Ready and cheap availability
• Non-hygroscopic
Some examples of primary standards
• Potassium hydrogen phthalate (KHP)
• Sodium carbonate (Na2CO3)
• Potassium iodate (KIO3)
• oxalic acid (C2H2O4)
By a procedure, called standardization, the concentration of titrant
(e.g. KMnO4 solution) can be determined, and then the titrant became a
standard solution.
Example
If 10.0 mL of 0.0106 M oxalic acid (C2H2O4) require 48.4 mL of
potassium permanganate (KMnO4) solution for titration by reaction:
What is the molarity of the permanganate solution?
At the equivalence point:
Solution
# moles of oxalic acid = 5/2 # moles of permanganate
M • V (oxalic acid ) = 5/2 M • V (permanganate)
0.0106 mol/L x 0.0100 L = 5/2 M x 0.0484
M = 0.000876 mol/L
Example
Solution

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Analytical chemistry ch01 chemical measurements

  • 1. Chemical Measurements Daniel C. Harris Chapter 1 Dr. Sa’ib Khouri AUM- JORDAN
  • 2. Introduction Analytical Chemistry: The science of chemical measurements that studies and uses chemical methods and instruments. Types of Questions Asked in Analytical Chemistry: a) What is in the sample? (qualitative analysis that identifies analytes) b) How much is in the sample? (quantitative analysis that determines the numerical amount or concentration) How pure is it? quantitative analysis What are the impurities? qualitative analysis b) Instrumental Methods: spectrometry, chromatography, conductomertry, electrophoresis, voltammetry,…etc. Techniques used in Analytical Chemistry: a) Wet Chemical Methods: Separation, extraction, distillation,titration, color-forming reactions, precipitations, ….etc. Identification may be based on differences in color, odor, melting point, boiling point, solubility, radioactivity or reactivity.
  • 3. Fundamental units (base units) from which all others are derived Quantity Unit (Symbol) Length Meter (m) Mass Kilogram (kg) Time Second (s) Electric current Ampere (A) Temperature Kelvin (K) Amount of substance Mole (mol) SI Units (metric units) From the French système international d'unités used by scientists around the world
  • 4. SI-derived units with special names
  • 5. Prefixes to express large (multiples) or small quantities (fractions)
  • 6. Example At an altitude of 1.7 x 104 meters above the Earth’s surface, the pressure of ozone over Antarctica reaches a peak of 0.019 Pa. Express these numbers with prefixes from the table An ozone “hole” forms each year in the stratosphere over the South Pole at the beginning of spring in October.
  • 9.
  • 10. Express the rate of energy used by a person walking 2 miles per hour (consuming 91 kcal per hour per 100 pounds of body mass) in kilojoules (kJ) per hour per kilogram (kg) of body mass. Example Solution 1 cal = 4.184 J, so 1 kcal = 4.184 kJ 1 lb = 0.454 kg; so 100 lbs = 45.4 kg or in one long calculation: The rate of energy consumption is therefore
  • 11. Chemical Concentrations Concentration states how much solute (minor species) is contained in a given volume or mass of homogeneous solution or solvent (major species). Molarity and Molality Molarity (mol/L, or M) is the number of moles of a substance per liter of solution. Chemical concentrations, denoted with square brackets, are usually expressed in moles per liter (M). Thus “[H+ ]” means “the concentration of H+ .” Molarity (M) = moles of solute liters of solution The molarity of a strong electrolyte is called sometimes the formal concentration (F) e.g. Magnesium chloride is a strong electrolyte. In 0.44 M MgCl2 solution, 70% of the magnesium is free Mg2+ and 30% is MgCl+ . The concentration of MgCl2 molecules is close to 0. The “molecular mass” of a strong electrolyte is called the formula mass (FM), and it is the sum of atomic masses of atoms in the formula.
  • 12. Molarity of Salts in the Sea Example solution
  • 13. Calculate the formula mass of CaSO4. What is the molarity of CaSO4 in a solution containing 1.2 g of CaSO4 in a volume of 50 mL? How many grams of CaSO4 are in 50 mL of 0.086 M CaSO4? (Answer: 136.14 g/mol, 0.18 M, 0.59 g) H.W. For a weak electrolyte such as acetic acid, CH3CO2H, some of the molecules dissociate into ions in solution: Molality (mol/kg, or m ) is concentration expressed as moles of substance per kilogram of solvent (not total solution) Molality is independent of temperature Molality (m) = moles of solute kg of solvent
  • 14. Percent Composition Weight percent (wt%): Volume percent (vol%): e.g. A common form of ethanol (CH3CH2OH) is 95 wt%; this expression means 95 g of ethanol per 100 g of total solution. Note: mass is usually implied when units are absent.
  • 15. Example Find the molarity and molality of 37.0 wt% HCl. Its density is 1.19 g/mL. Solution For molarity, we need to find the moles of HCl per liter of solution The mass of a liter of solution is (1.19 g/mL )(1000 mL) = 1.19 x 10 3 g. The molecular mass of HCl is 36.46 g/mol, so the molarity is
  • 16. For molality, we need to find the moles of HCl per kilogram of solvent (H2O). 100 g of solution contains 37 g of HCl and 63 g of H2O (0.063 kg) 37.0 g of HCl = 37.0 /(36.46 g/mol) = 1.01 mol HCl HW Calculate the molarity and molality of 49.0 wt% HF, assuming a density of 1.16 g/mL.
  • 17. Parts per Million (ppm) and Parts per Billion (ppb) mean grams of substance per million or billion grams of total solution or mixture. The density of a dilute aqueous solution is close to 1.00 g/mL (i.e 1 mL ≡ 1 g) 1 ppm corresponds to 1 μg/mL (1 mg/L), and 1 ppb is 1 ng/mL (1 μg/L) Question: What does one part per thousand and one part per trillion mean?
  • 18. ppt: g/L (mg/mL) ppm: mg/L (µg/mL) ppb: µg/L (ng/mL) pptr: ng/L (pg/mL)
  • 19. 34 ppb means there are 34 ng of C29H60 per gram of rainwater, which is nearly the same as 34 ng/mL. Find the molarity of this concentration. Example Number of grams of C29H60 in a liter is 1000 Multiply by 34 ng/mL and gives 34000 ng/L (or 34μg/L) of C29H60 of rainwater. solution The molecular mass of C29H60 is 408.8 g/mol HW. How many ppm of C29H60 are in 23 μM C29H60? (Answer: 9.4 ppm)
  • 20. Preparing Solutions To prepare a solution with a desired molarity from a pure solute (solid or liquid), we weigh out the mass of solute using analytical balance and dissolve it in a volumetric flask. Example solution
  • 21. Using a volumetric flask: The procedure is to place 0.999 g of the solid into a 500-mL volumetric flask, add about 400 mL of distilled water, and swirl to dissolve the reagent. Then dilute with distilled water up to the 500-mL mark and invert the flask several times to ensure complete mixing. HW. Find the formula mass of anhydrous CuSO4. How many grams should be dissolved in 250.0 mL to make a 16.0 mM solution? (Answer:159.61 g/mol, 0.638 g)
  • 22. Dilution Dilute solutions can be prepared from concentrated solutions The number of moles in a concentrated (conc) solution that taken is equal to the number of moles placed in the dilute (dil) solution, i.e. no changes in number of moles.
  • 23. To make 0.1 M HCl, we would dilute 8.26 mL of conc. HCl up to 1 L. Note: Dilute HCl solution here needs standardization to find its accurate concentration.
  • 24. Stoichiometry is the calculation of quantities of substances involved in a chemical reaction. Gravimetric analysis: chemical analysis based on weighing a final product . e.g. Iron (Fe) from a dietary supplement tablet can be measured by dissolving the tablet and then converting the iron into solid Fe2O3 (ferric oxide) and from the mass of this solid one can calculate the mass of iron in the original tablet. Stoichiometry Calculations for Gravimetric Analysis Gravimetric and titrations analysis were practiced long before electronic instruments to make chemical measurements and called as “classical” or “wet chemical” methods, and still have a place in modern analytical chemistry to be more accurate than instrumental methods usually used to prepare standards for instrumental methods of analysis.
  • 25. The steps in the procedure: FeO(OH) ·xH2O
  • 26. Example In a gravimetric analysis, we need enough product to weigh accurately. Each tablet provides about 15 mg (0.015 g) of iron. How many tablets we should use to analyze in order to produce 0.25 g of solid Fe2O3?
  • 27.
  • 28. Introduction to Titrations In a titration, increments of reagent solution (the titrant and it is usually delivered from a buret) are added to analyte until their reaction is complete. Requirements • Reaction must be quick • Reaction must proceed according to a defined equation • Availability of indicator • Reactions must be complete or have a large equilibrium From the quantity of titrant required, the quantity of analyte concentration can be calculated Common titrations rely on acid-base, precipitation, oxidation-reduction, or complex formation reactions.
  • 29. e.g. 5 mol of oxalic acid react with 2 mol of permanganate in hot acidic solution. At the equivalence point M • V (oxalic acid ) = 5/2 M • V (permanganate) # moles of oxalic acid = 5/2 # moles of permanganate End point: is marked by a sudden change in a physical property of the solution, e.g. color change of a reactant or an external indicator added. The equivalence point: The point when the quantity of added titrant is the exact amount necessary for stoichiometric reaction with the analyte.
  • 30. Primary standard • At least 99.9% pure • Stable (low reactive) under ordinary storage • Stable when dried by heat or vacuum • High molar mass (to minimize weighing errors) A reagent is pure enough to be weighed and used directly, it should be • Non-toxic • Ready and cheap availability • Non-hygroscopic Some examples of primary standards • Potassium hydrogen phthalate (KHP) • Sodium carbonate (Na2CO3) • Potassium iodate (KIO3) • oxalic acid (C2H2O4)
  • 31. By a procedure, called standardization, the concentration of titrant (e.g. KMnO4 solution) can be determined, and then the titrant became a standard solution. Example If 10.0 mL of 0.0106 M oxalic acid (C2H2O4) require 48.4 mL of potassium permanganate (KMnO4) solution for titration by reaction: What is the molarity of the permanganate solution? At the equivalence point: Solution # moles of oxalic acid = 5/2 # moles of permanganate M • V (oxalic acid ) = 5/2 M • V (permanganate) 0.0106 mol/L x 0.0100 L = 5/2 M x 0.0484 M = 0.000876 mol/L